STPM Biology - chapter 4(old syllabus) chapter 6 (new syllabus): Revision essay question 9

9. Describe the mechanism of action in synthesis of lipid by photosynthesis.

Answer

1. Lipid Synthesis
1. Fatty acid synthesis is catalyzed by fatty acid synthetase using the substrates acetyl-CoA and malonyl-CoA, the reductant NADPH, and a small protein called acyl carrier protein, which carries the growing fatty acid chain; the fatty acid is lengthened by adding two carbons at a time to its carboxyl end
2. Triacylglycerols are formed from the reduction of dihydroxyacetone phosphate (a glycolytic pathway intermediate) to glycerol 3-phosphate, which then undergoes esterification with two fatty acids to form phosphatidic acid; this can then be used to produce triacylglycerol
3. Phospholipids are also produced from phosphatidic acid using a cytidine diphosphate (CDP) carrier

•Lipid Anabolism - lipid synthesis is known as lipogenesis

•

Øglycerol is synthesized from dihydroxyacetone phosphate (an intermediate product of glycolysis)

Ø

Ømost lipids, including nonessential fatty acid chains and steroids, begin with acetyl-CoA

Ø

Ølipogenesis can use almost any organic substrate because lipids, amino acids, and carbohydrates can be converted to acetyl-CoA

Ø

Ønot every fatty acid chain that can be broken down can be built; thus, fatty acids that cannot be built must be obtained in the diet and are called essential fatty acids (e.g. linoleic acid and linolenic acid are essential fatty acids)

animation

http://ull.chemistry.uakron.edu/pathways/FA_synthesis/index.html

STPM Biology - chapter 4(old syllabus) chapter 6 (new syllabus): Revision essay question 10

10. Briefly describe the structure of chloroplast

Answer

STPM Biology - chapter 18: Revision essay question 4

4. Given a population of 35% of white mice. Knowing that the white colour skin is caused by the double recessive genotype, “aa”. Calculate the allele and genotype frequency for this population.

Answer


         White mice = 35
         Frequency of aa genotype, q² = 0.35
         Frequency of allele, q = 0.59
        
         Frequency of allele, p = 1 – q
                                             = 1 – 0.59
                                             = 0.41

         Frequency of genotype, AA = p²
                                                       = (0.41)²
                                                       = 0.17

         Frequency of genotype, Aa = 2pq
                                                      = 2 (0.41) (0.59)
                                                      = 0.48

         CHECKING:
            p² + 2pq + q²
         = 0.17 + 0.48 + 0.35
         = 1


Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 18: Revision essay question 3

3. In a given population, only the "A" and "B" alleles are present in the ABO system.  There are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)?


Answer


         Type A blood = homozygous AA = 200
         Type B blood = homozygous BB = 25
         Type AB blood = heterozygous AB = 75

       The frequency of A = 2 (200) + 75 
                                              2 (300)
                                        = 0.792
                                        = p
        Since p + q = 1,
        then q = 1 – p
                    = 1 – 0.792
                    = 0.208



Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 18: Revision essay question 2

2, Define the Hardy-Weinberg Law                         

Answer

A principle that in an infinitely large, randomly mating population in which selection, migration and mutation do not occur, the frequencies of alleles and genotypes do no change from generation to generation.

It can be represented by the Hardy-Weinberg equation.
                             p² + 2pq +q² = 1
           where, p = dominant allele frequency
                       q = recessive allele frequency

                       p² = frequency of homozygous dominant in a population
                       q² = frequency of homozygous recessive in a population
                       2pq = frequency of heterozygous in a population

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 18: Revision essay question 1

1. What is meant by:

Answer

                    i.            Gene pool

The sum total of all the genes and alleles that are present in a sexually reproducing population of a particular species pf organism at a given time.

                  ii.            Allele frequency

The relative number of gene compared to the total number of all alleles of the gene in a population.

                iii.            Genotype frequency in a population

                  The percentage of individuals with a specific genotype in a population

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 17: Revision essay question 6

6. Describe the chromosomal events which may have led to a person with the genetic disorder Down syndrome.

Answer

Down syndrome is an example of aneuploidy. There is an abnormal number of autosomes.it is a results of non-disjuction during meiosis. The two chromosomes number 21 fails to separateduring anaphase I or anaphase II of meiosis. The gametes produced contain 24 chromosomes (two copies of chromosome 21) and 22 chromosomes ( no chromosomes 21). When a sperm cell containing 23 chromosomes fuses with an ovum contaning 24 chromosomes, the zygote formed contains trisomy 21 (three chromosomes 21).

The induviduals with trisomy 21 may be males or females. They have flat, broad faces, slanted eyes with skin folds in the inner corner, protuding tongue, short palms, hanging mouth, wide nose and are mentally retarded. They usually have congenital heart defects, reduced resistance to diseases and a shortened lifespan.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 17: Revision essay question 5

5. Explain the meaning of the following term:

(i) non-disjunction of chromosomes

(ii) polyploidy

Answer

(i) non-disjunction is the failure of homologous chromosome pairs or sister chromatids to segregate into different daughter cells during mitosis or meiosis. Both homologous chromosomes and sister chromatids end up in the same daughter cell during cell division.

(ii) poluploidy is an example of euploidy. It describe the condition of a nucleus, cell or organism that contains three or more complete sets of chromosomes. Polyploidy can be divided into autopolyploidy in which the three or more sets of chromosomes are derive from the same species and allopolyploidy when they are derive from different species.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 17: Revision essay question 4

4. Using a named example, describe how a gene mutation may affect a phenotype of an organism.

Answer

An example of gene mutation is sickle-cell anaemia, where there is substitution of a base thymine by adenine.in the gene that codes for the beta polupeptide chain. Sickle-cell anaemia patients have abnormal haemoglobin molecules with valine instead of glutamic acid as the sixth amino acid for both the beta-polypeptide chains.

A sickle-cell haemoglobin easily crystallize at low concentrations of oxygen and the red blood cells are pilled into a sickle shape. These sickle red blood cell can easily break and clump together to clog the blood capillaries, impeding the flow of blood. This results in less oxygen is transported to the muscles and the internal organs. Sickle-cell patients usually suffer from heart problem, kidney failure, abdominal pains, paralysis and premature death.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 17: Revision essay question 3

3. Define a gene mutation.

Answer

A gene mutation is a change in the sequence of nucleotide bases of the DNA that corresponds to a particular gene. The different type of gene mutations are deletion, substitution, insertion, inversion, translocation and duplication.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 17: Revision essay question 2

2. Give five examples of mutagen and describe briefly the effects of each mutagen.

Answer

The five examples of mutagens are X-rays, ultraviolet (UV) rays, bromouracil (BrU), nitrous acid, colchicine.

X-rays cause extensive aberrations in chromosomes and genes, including breaks in DNA strand, destruction or modifications of nucleotide bases or sugars. It may also results in failure of organelles to function, prevent cell division or cause death of cells, for examples, death of bone marrow cells.

UV radiation causes breakage of A and T bases between the two complementary strands. Cross covalent bonding then occurs between two adjacent thymine  bases on the same DNA strand to form thymine dimer. This will cause distortion of the DNA helix, prevent normal base pairing and impede replication or transcription.

Bromouracil resembles thymine (has Br atom instead of methyl group). It can be incorporated into DNA and the keto form of BrU can pair with adenine.

Nitrous acid causes deamination of cytosine to produce uracil, and adenine to a guanine analogue would form H-bonds with cytosine instead of base thymine.

Colchicine inhibits spindle formation during cell division. The chromosomes do not separate to opposite poles, causing the doubling of chromosomes.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 17: Revision essay question 1

1.           Explain what is meant by term mutagen.

Answer

A mutagen is a natural environmental agent or human-made artificial agent, physiacal or chemical, which induces a higher rate of mutation by altering the structure or sequence of nucleotide bases in DNA.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W.

STPM Biology - chapter 16: Revision essay question 3

3. Do the results of test cross follow the Mendel’s Second Law? Explain your answer with a suitable example.

Answer

 No. An example is the test cross of Drosophilia flies which gives a ratio of 6:6:1:1 for the body colour and the size of wings. If the test cross follow Mendel’s Second Law, it will give a ratio of 1:1:1:1. The gene of the body colour and the size of wing are linked. When test cross occur, this result in the majority progeny of having the parental genotype and phenotype. Four types of progeny are produced. A small number of crossing over has occur resulting in new genetics combinations of linked gene in some gametes.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W. Tan and C.C.,Teh 2011/2012 (IBM)

STPM Biology - chapter 16: Revision essay question 2

2. Define Mendel’s second law.

Answer

Mendel’s second law state that each member of a pair of alleles may combine randomly with either of another pair during gametogenesis.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W. Tan and C.C.,Teh 2011/2012 (IBM)

STPM Biology - chapter 16: Revision essay question 1

1. A woman and a man who are both normal for blood clotting have a haemophiliac son. Haemophilia is a sex-linked recessive disorder.

Answer

A)  What is meant by sex-linked gene?

Sex-linked gene are genes located on the sex chromosomes but are not involved in determining of sex of an individual. Most of this gene are located on the X chromosomes. The allele located on the same sex chromosomes are not separated during meiosis and is inherited.

B)  What is the probability that their second child will be 1) a haemophiliac boy, 2) a daughter who is a carrier for haemophilia?

Both of the probability is 25%.

C)  Draw a simple pedigree diagram to show that the genotypes of the three individual ( the parents and the haemophiliac son). Explain how the pedigree is obtained.

The son is a haemophiliac. He inherit the X chromosome from the mother and the Y chromosomes from the father. The X chromosome carries the recessive alleles for haemophilia. The mother is phenotypically normal but is a heterozygous carrier. The son inherits the haemophlia recessive allele.

D) Explain why a human population will contain more colour-blind individuals than haemophiliacs?

 Under natural conditions, the selection pressure against haemophilia is greater than that against colour-blindness. Haemophiliacs are weaker than colour-blindness and maynot reach sexual maturiry. Some of the haemophiliac gene are therefore lost from population. However, haemophilia can now be treated by gene therapy and many of them live a normal life.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W. Tan and C.C.,Teh 2011/2012 (IBM)

STPM Biology - chapter 15: Revision essay question 2

2. Roles of hormones in ecdysis and metamorphosis

Answer

Metamorphosis and growth in arthropods are controlled by two hormones, which is juvenile hormone (JH) and ecdysone.

Neurosecretory cells in the pars intercerebralis of the brain are used to detect external stimuli such as food,light and temperature.

Neurosecretory cells secrete a hormone called prothoracicotrophic.

The hormone prothoracicotrophic transported by axoplasmic streaming along the axons and stored in a pair of glands called corpora cardiac.

When stimulated, impulses are transmitted down the nervicorpora cardiac and stimulate the release of prothoracicotrophic from the corpora cardiac into the haemolymph.

The prothoracicotrophic is then transported to the corpora allata to produce juvenile hormone.

Juvenile hormone actives the genes that control the retention of larval characteristic during development. It also suppresses the genes which control the production of adult characteristics.

This causes the prothoracicotrophic be transported to the prothracic glands. The prothoracic glands produce ecdysone.

The low level of juvenile hormone causes the level of ecdysone to increase and thus moulting occurred.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W. Tan and C.C.,Teh 2011/2012 (IBM)

STPM Biology - chapter 15: Revision essay question 1

1. Describe the pattern of each of the growth curve .

Answer

Lag phase, growth is slow in this phase. There is only a no cell division nor ell enlargement.

Log phase, growth is the fastest in this phase. The cells divide and enlarge rapidly as conditions for growth are at optimum condition.

Decelerating phase ,Growth begins to decrease due to the limiting by external or internal factors or both. For examples are depleting nutrients,limited space,competition, accumulation of excretory wastes and toxins.

Stationary phase, Growth are no longer occur.

Negative phase, the organism is ageing and the growth rate start to decrease.

Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W. Tan and C.C.,Teh 2011/2012 (IBM)

STPM Biology - chapter 11: Revision essay question 4

4. What are the hormones produced by the pancreas that regulate blood glucose?  How do they work together to regulate blood glucose levels?  How would consuming large quantities of starchy foods, such as pasta, benefit a person prior to running a marathon?

Answer 

The pancreas makes insulin, which is produced in higher concentrations as blood glucose concentration increases, such as after consuming a meal rich in carbohydrates.  This causes cells of the liver, muscle cells, and fat cells to absorb glucose.  As blood glucose levels fall, the pancreas produces less insulin.  The pancreas also makes glucagon, which is produced in higher concentrations when blood glucose levels drop too low.  Glucagon causes the conversion of glycogen stored in the liver into glucose, which enters the circulatory system and raises blood glucose concentrations back to normal range.   As blood glucose levels return to a normal range, the pancreas reduces its production of glucagon.

A common practice prior to running a marathon or participating in other long endurance sports, is to consume a large quantity of foods rich in starch that increase blood glucose levels.  In response, the pancreas produces more insulin, which causes liver cells to absorb glucose and convert it to glycogen, which is stored in the liver.  During a high endurance activity, such as a marathon, glycogen is converted back to glucose, which raises blood glucose levels back to a normal level and allows the person to continue their activity.  

STPM Biology - chapter 11: Revision essay question 3

3. Target cells of a hormone will often reduce the number of receptors for a hormone when the levels for that hormone become chronically elevated.  Conversely, the number of receptors for a hormone may increase if the concentration of a hormone is chronically reduced.  Explain why this occurs during these cases. 
 
Answer

Hormones are chemical messengers that are required in small quantities to produce their effects.  If hormone levels are elevated for long periods of time, cells with the correct receptors will only need a smaller number of receptors for maximum response.  Therefore, the cell has more receptors for this hormone than it needs and will respond by reducing the number of receptors.  Conversely, if tissue fluid levels, including blood, of hormones are chronically low, cells will produce more receptors to maximize their response to existing levels of hormones.   

Assignment submitted by Dhevan, 2011/2012(IBM)

STPM Biology - chapter 11: Revision essay question 2

2. What are the hormones produced by the pancreas that regulate blood glucose?  How do they work together to regulate blood glucose levels?  How would consuming large quantities of starchy foods, such as pasta, benefit a person prior to running a marathon?

Answer

The pancreas makes insulin, which is produced in higher concentrations as blood glucose concentration increases, such as after consuming a meal rich in carbohydrates.  This causes cells of the liver, muscle cells, and fat cells to absorb glucose.  As blood glucose levels fall, the pancreas produces less insulin.  The pancreas also makes glucagon, which is produced in higher concentrations when blood glucose levels drop too low.  Glucagon causes the conversion of glycogen stored in the liver into glucose, which enters the circulatory system and raises blood glucose concentrations back to normal range.   As blood glucose levels return to a normal range, the pancreas reduces its production of glucagon.

A common practice prior to running a marathon or participating in other long endurance sports, is to consume a large quantity of foods rich in starch that increase blood glucose levels.  In response, the pancreas produces more insulin, which causes liver cells to absorb glucose and convert it to glycogen, which is stored in the liver.  During a high endurance activity, such as a marathon, glycogen is converted back to glucose, which raises blood glucose levels back to a normal level and allows the person to continue their activity.  

Assignment submitted by Dhevan, 2011/2012(IBM)

STPM Biology - chapter 9: Revision essay question 5

5. Explain the function of the liver.

Answer


-Regulates blood and body carbohydrate level at 80-100mg/100ml of blood:
    ~Converts excess glucose into glycogen in the presence of insulin in a process called
      glycogenesis.
    ~Converts glycogen into glucose in the presence of glycagon in a process called glycogenolysis.
    ~Releases glucose into the blood when the blood glucose level drops below the normal level.
    ~Metabolise lactate(lactic acid) transported from muscle cells through the Cori cycle or converts  
      it into pyruvate and glucose.
    ~Forms glucose from non-carbohydrate sources through gluconeogenesis.
-Regulates body and blood lipid level.
    ~Synthesis and breakdown of fat.
    ~Synthesis of cholesterol and phospholipids.
-Regulates blood and body proteins level.
    ~Deamination of excess amino acid into keto acid and ammonia.
    ~Ammonia produced is then converts into urea through ornithine cycle.
    ~Carries out transamination in which, the amino group from an amino acid is transferred to a
      keto acid.Transamination is used to synthesis non-essential amino acids.
-Stores vitamins and minerals such as ferum ions and vvitamin A,B12,D,and K.
-Produces plasma protein such as prothrombin and glubulins.
-Kupffer cell of the liver breaks down the old/worn-out erythrocytes.
-Produces bile,which is stored in the gall bladder.
-Performs detoxification of toxic substances such as ammonia and alcohol.
-In the foetal stage, liver produces erythrocytes.
-Stores blood.
-An active organ that produces a high amount of heat that can be used in the regulation of body temperature.

STPM Biology - chapter 9: Revision essay question 4

4. By using a specific organisms, explain how an endothermic organisms              
                         maintains its body temperature when the external temperature drops.

Answer


Endotherm such as humans use metabolic energy(internally generated energy) to mantain a
 constant body temperature.
     -Decrease in external temperature is detected by thermoreceptors(bulb of Krausse) in the skin.
     -Impulse is sent to hypothalamus, the temperature regulatory centre.
     -Hypothalamus then sends action impulse to the following effectors:
         ~Vasocontriction(diameter reduce) of subcutaneous capillaries occurs to reduce blood flow to
            the skin.
         ~Since blood transports heat, less heat is lost through the skin.
         ~Erector muscles on the skin contracts, hair on the skin becomes erect to trap a thick layer of
            air.
         ~The layer of air acts as an insulating layer to reduce heat loss through the skin.
         ~The medulla oblongata release impulse to skeletal muscles, which causes the skeletel
            muscles
            to contract and relax alternatively.
         ~Shivering occurs, contraction of the skeletel muscle muscle release heat to increase the body
            temperature.
         ~Thyroid gland secretes thyroxin, which increase the basal metabolic rate of the body to
            increase heat production.
         ~Adrenal gland secretes adrenaline to increase the conversation rate of glycogen to glucose in
            the liver.
         ~The glusoce produced is used for heat production in the body.
    -All of these physiological changes maintaince the body temperature at the normal range(37ºc).

STPM Biology - chapter 9: Revision essay question 3

3. State the importance of homeostasis.

Answer


-Life of the organisms becomes less dependent on the extenal enviroment because the organisms
 is able to control the changes in its internal enviroment to compensate for the changes in the
 external condition.
    -Homeostasis ensures an optimum internal enviroment for the cell to function, the organisms to
 survive and reproduce efficiently.
    -Homeostasis enable biological systems to function efficiently and smoothly with minimum
 wastage of energy.
    -Organisms can live in a wider range of habitats because the organisms can live in areas with
 variable condition.
    -Organisms can increase or decrease its metabolic rate to suit its requirements.

STPM Biology - chapter 8: Revision essay question 4

4. Write down the control of heart beat by Sinoatrial node

 Answer

Heartbeat is myogenic,that is the beatings of the heart are started by cardiac muscles and not by external stimulation.The excitation that starts the heart beat originates  from the sinoatrial node(SAN).Sinoatrial node(SAN) is a patch of specialized cardiac muscle tissues located in the  wall of the right atrium near the entry of the vena cava.SAN is responsible for starting the heartbeat and the control of instrinsic rate of heart beat.The popential difference across the plasma membrane of SAN is -90mV.SAN cells have a permanent high permeability towards natrium ions.Natrium ions diffuse into SAN tissue cells continuously and depolarize these cells.This depolarization will generate wavw of excitation that are transmitted out of SAN to produce contraction of the heart.The wave of excitation is conducted from SAN to the walls of both atria.This causes both atria to contract simultaneously;atrial systole occurs.Atrioventricular node (AVN) is a specialized myocardial tissue that is similar to SAN.AVN is located in the wall between the atrium and the ventricle.AVN is connected to the Fine branches from the bundle of His are called Purkinje fibers. Purkinje fibres permeates the muscles of the entire ventricular wall.As the atrium begins to contract,the wave of excitation from SAN reaches AVN.AVN then transmits this excitation via the bundle of His to the apex of the ventricle.From this apex,the wave of excitation are then transmitted to the ventricular muscles via the Purkinje fibers.Such mode of transmission causes both ventricles to contract from the apex,forcing the blood out of the heart into the pulmonary artery and aorta.Contracting cardiac muscles cannot be stimulated by any other stimulation until it relaxes again.This period where the muscles are insensitive is called absolute refractory period.Cardiac muscles have a longer absolute refractory period compared to other muscles.Because of his nature,the heart can contract quickly and strongly without experiencing gatigue or oxygen debts.Ventricular systole is followedby ventricular diastole.

STPM Biology - chapter 8: Revision essay question 3

3. Write down the structure of the mammalian heart      

Answer

The heart of mammal has 4 chambers:left atrium,right atrium,left ventricle and right ventricle.Both atria walls are thin,but the left and right ventricles have thick muscular wall that can contract strongly to pump blood out of the heart.Walls of the left ventricle are 3-4 times thicker than the left ventricle has to pump blood through the whole body,while the right ventricle pumps blood into the lungs only.The left side of the heart is completely separated from the right side by the intraventricular septum.Atrium receives and collects blood from the body and lungs,which is then channeled into the ventricle.The right atrium receives deoxygenated blood from the body via the superior vena cava.The left atrium receives oxygenated blood from the lungs via the pulmonary vena.From the right atrium,blood enters the right ventricle through the tricuspid valve.Blood is then pumped from the right ventricle to the lungs,the left pulmonary artery transport blood to the left lung and right pulmonary artery transport blood to the right lung.Pulmonary artery is the only artery carrying deoxygenatedblood.All the other arteries transport blood rich in oxygen.When the right ventricle contracts,blood is prevented from re-entering the right atrium by the tricuspid valve.Blood enriched with oxygen in the lungs is channeled by left and right pulmonary veins into the left atrium.Pulmonary veins are the only veins in the body transporting oxygenated blood.Blood is pumped from the left atrium into the left ventricle.Contraction of the left ventricle wall forces the blood out to every part of our body via the aorta.Blood is prevented from flowing back into the left atrium by the bicuspid valve.Semi-lunar valves prevent blood from flowing back into the ventricles from the pulmonary artery or aorta.

STPM Biology - chapter 7: Revision essay question 4

4. Adaptations of haemoglobin to transport oxygen

Answer

Each haemoglobin molecule consists of 4 polypeptide chains.Each polypeptide in haemoglobin has a haem group.In the middle of the haem group is ferrum ion that could bind with oxygen molecules.So each haemoglobin molecule can bind with four oxygen molecules.In the lung,haemoglobin will only bind oxygen molecules,this is because haemoglobin has a high affinity for oxygen in the lungs.In tissues,haemoglobin will release oxygen.This is because haemoglobin has low affinity for oxygen in tissue.

STPM Biology - chapter 7: Revision essay question 3

3. Process involved in gaseous exchange in mammals

Answer

Gaseous exchange mammals occurs in the alveoli.Alveoli are the respiratory surface for human and mammals.Each alveolus is a tiny sac with just one cell thick.The inner surface of the alveolus is always moist.Right on the outside of alveoli is a network of blood capillaries.Blood in the pulmonary capillary has a low partial pressure of oxygen compared to the air in the alveolus.Hence,oxygen dissolves in the moisture on the alveolar surface and diffuses into blood capillary.Oxygen subsequently enters the red blood cell and bind to haemoglobin molecule,forming oxyhaemoglobin.The partial pressure of carbon dioxide in the blood of capillary is higher compared to the air in the alveolus.So,carbon dioxide diffuses from blood capillary into the alveolus to be exhaled out.Partial pressure of carbon dioxde in the interstitial fluid is higher compared to the blood in the capillary.Carbon dioxide diffuses from the interstitial fluid into the blood.The partial oxygen pressure in the blood capillary is higher compared to interstitial fluid.Oxygen diffuses from blood into the tissue cells.

STPM Biology - chapter 6: Revision essay question 6

6. Give the meaning and examples of the different types of symbiotic relationship.

Answer

·       Commensalism is an interaction between two animal or plant species that habitually live together in which one species (the commensal) benefits from the association while the other is not significantly affected. For example, the burrows of many marine worms contain commensals that take advantage of the shelter provided but do not affect the worm.

·       Mutualism is an interaction between two species in which both species benefit. A well-known example of mutualism is the association between termites and the specialized protozoans that inhibit their guts. The protozoans, unlike the termites, are able to digest the cellulose of the wood that the termites eat and release sugars that the termites absorb. The termites benefit by being able to use the wood as a foodstuff, while the protozoans are supplied with food and a suitable environment.

Parasitism is an association in which one organism (the parasite) lives on (ectoparasitism) or in (endoparasitism) the body of another (the host), from which it obtains its nutrients. Some parasites inflict comparatively little damage on their host, but many cause characteristic diseases (these are, however, never immediately fatal, as killing the host would destroy the parasite’s source of food). Parasites are usually highly specialized for their way of life, which may involve one host or several. The typically produce vast numbers of eggs, very few of which survive to find their way to another suitable host. Obligate parasites can only survive and reproduce as parasites; facultative parasites can also live as saprotrophs. The parasites of human include fleas and lice (which are ectoparasites), various bacteria, protozoans, and fungi (endoparasites causing characteristic diseases), and tapeworms, which lives in the gut.      

Assignment submitted by S.M., Kang, E.S., Ong, T.W., Tan, K.Y., Ho. X.H., Wong 2011/2012 (IBM)

STPM Biology - chapter 6: Revision essay question 5

5. What is the meaning of autotrophic and heterotrophic nutrition?

Answer

Autotrophic nutrition is a type of nutrition in which organisms synthesize the organic materials they require from inorganic sources. Chief sources of carbon and nitrogen are carbon dioxide and nitrates, respectively. All green plants are autotrophic and use light as a source of energy for the synthesis, i.e. they are photoautotrophic. Some bacteria are also photoautotrophic; others are chemoautotrophic, using energy derived from chemical processes.

Heterotrophic nutrition is a type of nutrition in which energy is derived from the intake and digestion of organic substances, normally plant or animal tissues. The breakdown products of digestion are used to synthesize the organic materials required by the organism. All animals obtain their food this way and they are heterotrophs.

Assignment submitted by S.M., Kang, E.S., Ong, T.W., Tan, K.Y., Ho. X.H., Wong 2011/2012 (IBM)

STPM Biology - chapter 5 (old syllabus) chapter 5 (new syllabus): Revision essay question 9

9. State the uses of fermentation in the industry.

Answer

There is wide range of applications in industry, especially in the food industry. For example, baking breads, manufacture beer, wine and alcoholic beverages, dairy product and ethanol. In the baking breads, yeast is needed to mix with flour. The carbon dioxide gas produced from the fermentation process enable the dough to rise and soften the breads when it is baked in the oven.

During the making of the dairy products such as yoghurt, the enzyme from bacteria acts on milk sugar (lactose) to produce lactic acid. The lactic acid can give the milk a sour taste and make milk into curd called yoghurt. In order to produce a variety of yoghurt which is different tastes and flavour can be added with various flavours like strawberry and blueberry.

The enzyme diastase in the malt derived from cereals or fruit juices can be manufactured beer, wine and alcoholic

beverages. The enzyme diastase converts starch and other sugars into maltose. Yeast is then added to allow fermentation to take place. During this process, the maltase converts the maltose into glucose. Glucose is converted into ethanol and carbon dioxide by the enzyme zymase. The fermented mixture is then distilled to produce beer or wine depends on ethanol concentrations.

Ethanol can be obtained from the alcoholic fermentation. Ethanol acts as a solvent which used in the pharmaceutical industry. Ethanol can also manufacture various organic compounds such as acetic acid in vinegar.

Assignment submitted by S.M., Kang, E.S., Ong, T.W., Tan, K.Y., Ho. X.H., Wong 2011/2012 (IBM)

STPM Biology - chapter 5 (old syllabus) chapter 5 (new syllabus): Revision essay question 8

8. Describe anaerobiosis in yeast and in animal muscle cell.
 Answer

Yeast and animal muscle tissue are one type of the organism which can live in oxygen deficient condition even they need oxygen for a certain length of time. In the absence of the oxygen, yeast undergoes the anaerobic respiration. Anaerobic respiration is defined as a type of respiration is partially oxidized, with the release of chemical energy, in a process not involving atmospheric oxygen. The process is

called fermentation. Yeast is a one type of facultative anaerobes. There are two types of fermentation processes which are alcoholic fermentation and lactate fermentation. Alcoholic fermentation occurs in plant cell whereas lactate fermentation occurs in animal cells. The alcoholic fermentation is carried out by yeast in the absence of oxygen. Yeast is added to allow fermentation to take place. The yeast for example saccharomyces cerevisiae breaks down glucose during fermentation, producing ethanol and carbon dioxide. First, glucose is converted to pyruvate, generates a net gain of 2ATP and NAD⁺ reduced to NADH. Pyruvate is then decarboxylated to become ethanol (acetaldehyde) and carbon dioxide is released. The decarboxylation is catalysed by the enzyme decarboxylase. Ethanal act as a hydrogen acceptor, accepts hydrogen atoms from NADH+H⁺ to be reduced to ethanol. This reaction required the enzyme NADH dehydrogenates to catalyse.

Overall equation for alcoholic fermentation:

 glucose→ 2ethanol + 2CO₂ + 2ATP

Lactate fermentation occurs in muscles during vigorous exercise. When the oxygen level in the muscle is limited, anaerobic respiration occurs. This is due to the rate of consumption exceeding the rate of supply. During glycolysis, glucose is converted to pyruvate. This will generates a net gain of 2ATP and NAD⁺ reduced to NADH. Pyruvate is then reduced to lactate by NADH and NAD⁺ is regenerated.

However, lactate accumulation in the muscle causes fatigue, lowers blood pH and cramping. To break down the lactate, oxygen is required and is known as the oxygen debt. After vigorous exercise, the lactate is oxidized back to pyruvate in the liver. The pyruvate is then oxidized to water and carbon dioxide.

Overall equation for lactate fermentation:

     glucose→2lactate + 2CO₂ + 2ATP

Assignment submitted by S.M., Kang, E.S., Ong, T.W., Tan, K.Y., Ho. X.H., Wong 2011/2012 (IBM)

STPM Biology - chapter 6: Revision essay question 4

4. Define the process of photosynthesis in Bacteria.

Answer

Light reaction occurs in photosynthetic membrane of bacteria.

                    -In present of light, bacteriochorophyll absorbs light energy.

                    -Energy absorbed is then channelled to reaction centre.

                    -Reaction centre becomes photoactivated and photosystem I becomes photoactivated.

                    -Electron of reaction centre becomes excited and escape.

                     Escaped electrons are then transferred through a series of electron carriers and finally

                     back to reaction centre.

                    -As electron carriers transfer electron down the energy level, energy released. Released  

                     energy is used for photophosphorylation of ADP to ATP.

                    -In purple bacteria, hydrogen sulphide is splitted into electrons, sulphur and hydrogen

                      ion. Reverse electron flow to oxidised NADP produces NADPH.

                    -Dark reaction is similar to plants except the process occurs in the cytoplasm of

                      bacteria.

STPM Biology - chapter 6: Revision essay question 3

3. How does the nutrition of the named endoparasite differ from the nutrition of Mucor?

Answer

                    -Taenia is an endoparasitic tapeworm living in the human gut and may harm the host.       

                     Mucor is a saprotroph living on dead decaying matter.

                    -The parasitic tapeworm absorbs soluble predigested food from the living host.

                      Exracellular digestion is not required. The fungal hyphae of Mucor penetrate th food

                      source. Enzymes are secreted externally (extracellular digestion) to digest the

                      complex food into soluble substances for absorption.

STPM Biology - chapter 6: Revision essay question 2

2. With reference to a named endoparasite,state the main adaptive features that enable it to live in its host.

Answer

An example is Taenia, an endoparasitic tapeworm in human intestine.

                      Main adaptive features are:

                    -Posses organs such as hooks and suckers for attachment to gut wall of host.

                    -Body covered with outer thick cuticle resistance to hydrolysis by host enzymes.

                    -Reduction of sense, digestive and locomotory organs as the parasite is less dependent 

                     on them. 

                    -Able to respire anaerobically in the gut.

                    -The tapeworm has a long flat ribbon-shaped body (up to 12m long) to increase surface  

                      area for absorption of pre-digested food.

                    -The body consists of numerous proglottids. Each proglottid contains male and female 

                      reproductive organs (hermaphrodile) allowing cross fertilisation or self-fertilisation if                                   

                      only tapeworm is present and production of many eggs.

                    -A secondary host is used as a vector for wider dispersal.

STPM Biology - chapter 5 (old syllabus) chapter 5 (new syllabus): Revision essay question 7

7. Describe the structure of mitochondrion.

Answer

-         Mitochondria are rod or spherical shaped.

-         Has length of about 1-10µ and width 0.25-10µm

-         It is covered by two membrane.

-         The outer membrane controls the movement of chemical substances in and out from the mitochondria.

-         The inner membrane is folded to form cristae to increase the surface area for electron transport system and oxidative phosphorylation to take place.

-         There are many stalked particles on the cristae surface.

-         The matrix of the mitochondrion is rich with proteins,enzymes,DNA, and ribosomes.