CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis 6. Hormone metabolism. One of the hormones synthesised in the liver is somatomedin that induces growth. Liver plays more important role in the breakdown of hormones. Testosterone and aldosterone are broken down very fast. Other hormones like insulin, glucagon, thyroxine, ADH, female sex hormones, intestinal hormones and adrenal hormones are broken down more slowly. 7. Detoxification metabolism. Toxins may include any chemical that brings about ill effects to the body if allowed to accumulate. As mentioned in earlier metabolisms, these chemicals include lactic acid from anaerobic respiration in muscles, ammonia from deamination of amino acids, heavy metallic ions and even vitamin A and cholesterol from excessive intake. Liver keeps these toxins to the minimum. Other toxins include pathogens and their harmful secretions, drugs, harmful gases and organic chemicals in the food. There are several ways the liver detoxifies the harmful chemicals through enzymes. (a) Oxidation. Certain chemicals when added with oxygen, removal of hydrogen ions or electrons become non-toxic or broken down. (b) Reduction. Similarly, certain chemicals when added with hydrogen ions or electrons or removal of oxygen become nontoxic or broken down. (c) Methylation. Certain chemicals when added with methyl groups become non-toxic too. (d) Addition. Addition of other organic or inorganic groups may also neutralise the toxins. 8. Liver metabolism for temperature regulation. Liver indirectly regulates the body temperature. It is the most active organ. The amount of heat produced contributes significantly in raising the body temperature, especially when the environment temperature is low. Thyroxine and adrenaline raises the metabolic rate of the liver too. When the subcutaneous blood vessels contract, the liver can dilate its own blood vessels to store part of the blood. When the environmental temperature is high and the subcutaneous blood vessels dilate, the liver can adjust the blood volume by contracting its blood vessels. Exam Tips 9LTLTILYOV^SP]LYJHU HJ[HZHOVTLVZ[H[PJVYNHU HUKP[ZYVSLHUKHJ[PVUVU ]HYPV\ZJSHZZLZVMMVVKLN WYV[LPU¶ :;74LZZH` X\LZ[PVU9LTLTILY*VYP HUKVYUP[OPULJ`JSLZ :;74LZZH`X\LZ[PVU Quick Check 2 1. If we can survive with no liver, what are the changes in the blood compositions? Explain.
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis Proteins amino acids Nitrogen pool Glycogen Glucose-6-phosphate Fats and Lipids Fatty acid, glycerol Pyruvic Acid Acetyl CoA Citric Acid Cycle 2H+ 2e– CO2 ADP ATP ADP Electron Transport Chain ATP ADP ATP O2 H2O NH3 Urea cycle Urea CO2 Fatty Acid Spiral Lipogenesis Lactic Acid Gluconeogenesis Glycolysis Glycogenolysis Carbohydrates glucose, fructose, galactose Glycogenesis Tissue protein Figure 11.21 Link between protein, carbohydrate and lipid metabolism 11.3 Osmoregulation in Mammals 11.3 Osmoregulation in Mammals 1. Osmoregulation and formation of urine for homeostatic control of blood water potential, ion concentration and pH are the roles of kidney coordinated by both the nervous and endocrine system. The internal structure of kidney together with nephron is as shown in Figure 11.22. Ureter Pelvis Medulla Cortex Renal vein Renal artery Nephron Renal corpuscle Aorta Renal artery Left kidney Ureter Urinary bladder Vena cava Right renal vein Right kidney Urethra Figure 11.22 Vertical section of kidney
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis 2. Renal artery supplies both oxygenated and ‘dirty’ blood to the kidney. Once inside the kidney, the artery branches to form arterioles and finally become fine afferent arterioles supplying blood to each glomerulus. In turn, the efferent arterioles branch to form networks called peritubular network and vasa recta around each nephron before connected to venules and renal vein draining ‘clean’ blood from the kidney. 3. The outer skin of the kidney is called capsule. This is a fibrous layer of membrane covering the kidney and protecting it. The outer layer of the kidney is called cortex. This region is where the Malphigian (renal) corpuscles are located. Beside the Malphigian corpuscles, it contains other parts of the nephron. The inner part of the kidney is called medulla. Other than the blood vessels, it contains mainly the loops of Henle and the collecting ducts of the nephron. The concave empty centre of the kidney is called pelvis. This is where the urine is flowed into before it is channelled to the urinary bladder through the ureter. 4. The basic unit of function in the kidney is the nephron. Its number is about a million in each of the kidney. The sum total of its functions is equal to the functions of the kidney. 5. The structure of the nephron is as shown in Figure 11.23. Distal convoluted tubule Bowman’s capsule Glomerulus Proximal convoluted tubule Peritubular capillaries Collecting duct Cortex Medulla Afferent arteriole Efferent arteriole Loop of Henle (descending limb) Loop of Henle (ascending limb) Vasa recta (capillaries) Venule Figure 11.23 The structure of a nephron Learning Outcomes :[\KLU[ZZOV\SKILHISL[V! H L_WSHPU[OLWYVJLZZ VM\S[YHÄS[YH[PVU YLHIZVYW[PVUHUK ZLJYL[PVUPU[OL MVYTH[PVUVM\YPUL" I L_WSHPU[OLYVSLVM(+/ HUKHSKVZ[LYVULHUK [OLYLSH[LKOVYTVULZPU YLN\SH[PUN^H[LYZVKP\T HUKWV[HZZP\TPVUZVM \YPUL" J L_WSHPU[OLYLN\SH[PVUVM W/VM[PZZ\LÅ\PK VIDEO INFO ;OL4HTTHSPHU 2PKUL` 2PKUL`
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis 6. The major function of the nephron or kidney is to produce urine in which waste (excretory) products of metabolism and chemicals taken in excess are excreted out of the body. However, before urine is formed, the kidney performs a series of processes that include ultrafiltration, selective reabsorption of substances, secretion of wastes, osmoregulation, regulation of mineral ions and pH before urine is produced with unwanted substances dissolved within. Formation of Urine Ultrafiltration 1. Ultrafiltration is a process in which smaller molecules are filtered from the blood in the glomerulus into the Bowman capsule. This is the first step in the formation of urine. The glomerulus is a knot of capillaries with the Bowman capsule in part of the nephron called Malphigian corpuscle. 2. Substances that cannot be filtered out include large proteins (bigger than molecular mass of 68, 000), platelets and blood cells that are too big to go out from the capillary wall. The capillary wall is made up of endothelium of simple squamous epithelium which has pores and gaps. The wall is also surrounded by a thin layer of protein fibres called basement membrane that prevents large protein molecules from going through. 3. Smaller molecules including water, mineral ions, glucose, amino acids, small polypeptides, vitamins, uric acid, creatinine and drugs can be filtered. This forms the ultrafiltrate that flows into the proximal convoluted tubule of the nephron. 4. The structure of the Malphigian corpuscle and the process of ultrafitration are as shown in Figure 11.24. Podocytes Bowman’s capsule Afferent arteriole Efferent arteriole Mesangial cells Glomerular capillaries Bowman’s space Proximal tubule Figure 11.24 4HSWOPNPHUYLUHSJVYW\ZJSLHUK\S[YHÄS[YH[PVU 5. There are several adaptations for the process of ultrafiltration. (a) The high hydrostatic pressure in the capillaries of the glomerulus helps in the process. This high pressure is the result of high blood pressure of the renal artery that is directly branched from the dorsal aorta.
CHAPTER 11 fflfl Biology Term 2 STPM Chapter 11 Homeostasis (b) Structurally the afferent arteriole is bigger than the efferent arteriole. Blood is forced into smaller capillaries magnifying the hydrostatic pressure to force small molecules out of the blood. (c) The fine pores (10 nm diameter) in the endothelium and the basement membrane consisting mesh work of fibres allow small molecules to go through. Larger molecules like proteins, platelets and blood cells are retained in the capillaries. (d) Low hydrostatic pressure outside the capillaries and in the Bowman’s capsule causes a continual flow of filtrate from glomerulus into the tubule of the nephron. (e) Similarly, the inner wall of the Bowman’s capsule that is made of special podocyte cells does not hinder ultrafiltration. The podocytes form a very porous slit-like network supporting the capillaries of each glomerulus. 6. The filtration pressure that is the net pressure that forces the liquid out from the blood can be calculated using the following formula. Filtration pressure = hydrostatic pressure of the blood – (colloidal solute potential + hydrostatic pressure of the glomerular filtrate) Selective reabsorption 1. This is the process in which certain substances are selectively reabsorbed from the glomerular filtrate into the cells of the proximal convoluted tubule and then into the capillary network surrounding each nephron to be carried away from the kidney. Almost 80% of the filtrate is reabsorbed this way. 2. Substances absorbed include the following: (a) 100% glucose, amino acids, vitamins and hormones. (b) 80% sodium chloride. (c) 40-50% urea. (d) At least 80% of water. 3. The mechanisms involved are as follows: (a) Sodium ions are actively transported out by sodium-potassium pumps at the outer peripheral membrane of the tubule. ATP is required to pump out three sodium ions for two potassium ion pumped in. (b) Then, glucose and sodium ions are co-transported by specific transport proteins into the tubular cells at the inner membrane. Both sodium ion and glucose diffuse down concentration gradient. (c) Facilitated diffusion is used to transport the glucose out of the tubular cells and into the capillaries around the nephron. Channel proteins are required.
CHAPTER 11 fflffi Biology Term 2 STPM Chapter 11 Homeostasis (d) Pinocytosis is used to transport small protein molecules into tubular cells. (e) Osmosis is used for water reabsorption in the proximal tubule. 4. Adaptations of the cells in proximal convoluted tubule include the following: (a) Microvilli greatly enlarge the surface to volume ratio for the absorptive cells of the proximal convoluted tubule. (b) Pinocytotic grooves are very small inlets at the base of the microvilli that enable pinocytosis of smaller protein molecules. (c) Spaces between the tubule cells enable easy passage of substances from the inside to the outside of the tubules. (d) Basal channels are spaces between the cells and the basement membrane. These channels increase the surface to volume ratio for substances to go out from the cells of the tubules. (e) Abundant mitochondria are present inside the cells for supplying energy needed for active transport as shown in Figure 11.25. Microvillus Pinocytic vesicle Nuclear envelope Nucleoplasma Intercellular space Mitochondrion Infolding of plasma membrane Basal channel Basement membrane Endothelial cell Blood Lumen of blood capillary Reabsorption of material by diffusion, active transport and pinocytosis Lumen of proximal convoluted tubule Figure 11.25 ;OLÅV^VMZ\IZ[HUJLZMYVTÄS[YH[L[VJHWPSSHY` 5. The transport pathway is as shown in Figure 11.25. (a) Small protein molecules move in by pinocytosis and other substances entered by active or passive transport into the cytoplasm of the cell. (b) Proteins are digested into amino acids when lysosomes fuse with the pinocytotic vesicles. Other simple substances just move to the peripheral membrane on the opposite side. (c) The substances then move across the membrane into the basement channels and cross the basal membrane into the blood capillaries by active or passive transport. (d) Once inside the capillary, the substances are carried away generating a lower concentration gradient between the inside and outside the tubule. So, there is a continual diffusion of substances into the peritubular capillary.
CHAPTER 11 fflffl Biology Term 2 STPM Chapter 11 Homeostasis Reabsorption of water 1. Reabsorption of water back into the blood capillaries of the kidney is governed by need and also by adaptations. (a) Animals which are adapted to live in the desert such as camel and kangaroo rat have a large medulla with long loop of Henle for this purpose. Water can be reabsorbed in a very large proportion from the glomerular filtrate and thus, very little urine is produced. (b) Humans, on the other hand, only reabsorb about 30 percent of water through this mechanism. (c) This mechanism involved is known as the counter-current multiplier mechanism. This process occurs in the loop of Henle and is shown in Figure 11.26. Water, glucose, amino acids, salt and urea Descending arm Collecting duct Glomerular filtrate Loop of Henle Urine Ascending arm K+ H+ H+ NaCl NaCl NaCl NaCl H2O H2O H2O H2O H2O H2O H2O K+ Na+ Cl– Bowman’s capsul Figure 11.26 The counter-current multiplying system (d) When the glomerular filtrate flows down the descending arm of the loop of Henle, water can move out of its wall, as it is permeable to water. There are transport proteins for this purpose. (e) Sodium chloride cannot move out from the descending arm as its wall is not permeable to sodium and chloride ions. (f) However, sodium and chloride ions are actively transported out from the ascending arm. (g) This results in the accumulation of the sodium chloride ions in the loop and their concentration decreases away from the tip of the loop.
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis (h) The longer the loop of Henle or correspondently the larger the medulla, more salt is accumulated at the loop or in the medulla, as in desert animals. (i) This enables more water to be reabsorbed from the urine through osmosis at the collecting duct as the urine passes through the medulla and very little urine is produced. (j) This reabsorption of water is called obligate reabsorption and always occurs irrespective of our body water potential. (k) Another reabsorption is called facultative with the help of hormone ADH depending on our body water potential. Both reabsorptions are involved in osmoregulation in our body. Active secretion 1. Active secretion into the peritubular capillary is the process in which substances are actively transferred first from the blood capillaries around the nephron, then through their wall and wall of nephron into the glomerular filtrate. Energy in the form of ATP is required. 2. This happens especially in the proximal convoluted tubule, but occur in other parts of the nephron as well. Substances secreted are as follows: (a) Nitrogenous wastes, including urea, ammonia, uric acid and creatinine, are secreted mainly at the later part of the nephron. (b) Ions, including hydrogen and potassium ions, especially when they are in excess in the blood are secreted. (c) Drug is secreted. This is especially true for antibiotic like penicillin that is eliminated from the blood once it enters the kidney. This also explains why penicillin is needed in large dosage to be effective for curing diseases. 3. Active secretion serves as an additional way in which excretory wastes and foreign substances are effectively removed from the blood. This prevents the accumulation of these harmful substances in the blood. 4. This is also a part of the homeostatic function of the kidney to maintain a constant internal environment. This tubular secretion where most organic substances are removed from blood into the nephron cells then into the filtrate and disposed in the urine. This process also serves to dispose of hydrogen and hydrogen carbonate ions for controlling blood pH.
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis Pituitary gland secretes less ADH Pituitary gland secretes more ADH Osmoreceptor in hypothalamus Osmoreceptor in hypothalamus Normal WP level Normal WP level Permeability of DCT decreases Permeability of DCT increases Increases Decreases Concentrated urine formed Negative feedback Dilute urin formed Negative feedback Figure 11.27 The regulation of the blood water potential (Key: WP= water potential, DCT=distal convoluted tubule) 2. When the water potential of the blood increases as in the case of too much water intake, the following events occur: (a) Osmoreceptor in the hypothalamus can detect the changes, impulse is sent to the pituitary gland. (b) Little or no ADH is secreted from the pituitary gland into the blood. The release of ADH is inhibited. (c) Little water is reabsorbed back into the blood from the glomerular filtrate in the distal convoluted tubule and collecting duct, as their membranes are impermeable to water. (d) Diuresis occurs, a condition in which large amount of dilute urine is produced. (e) This results in the decrease in water potential of the blood and the water potential becomes normal again. 3. When the water potential of the blood decreases as in the case of too much sweat is lost or much salt is taken, the following events occur: (a) Osmoreceptor in the hypothalamus can detect a fall in blood water potential, impulse is sent to the pituitary gland. (b) ADH is secreted into the blood from the posterior pituitary gland. It travels in the blood to the kidney. (c) It increases the permeability of water in the distal convoluted tubule and collecting duct. This is the result of ADH binds with a receptor protein. A cascade of reactions and vesicles with aquaporins (water channel proteins) fuse with the inner membrane of the tubular cells as shown in Figure 11.28. Role of ADH and Aldosterone Role of ADH 1. ADH plays an important role in the regulation of water potential (or osmoregulation) as shown in Figure 11.27.
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis Cross-section of collecting duct Filtrate 300 mOsM Exocytosis of vesicles 600 mOsM Vesicles Second messenger signal cAMP Aquaporin water pores 600 mOsM Medullary fluid Collecting duct lumen Collecting duct cell Vasa recta 700 mOsM H2O H2O H2O H2O ADH receptor ADH ADH binds to membrane receptor Receptor activates cAMP second messenger system Cell inserts AQP water pores into apical membrane Water is absorbed by osmosis into the blood Figure 11.28 The mechanism of action of ADH (d) When ADH secretion is stopped, the process goes into reverse by endocytosis. Vesicles with aquaporins are taken back into the cells for recycling. (e) A lot of water is reabsorbed back into the blood from the glomerular filtrate. Little and concentrated urine is produced. (f) ADH also causes thirst sensation in the throat. After drinking water, the water potential is increased to normal.
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis Role of Aldosterone 1. Aldosterone plays an important role in the homeostatic control of sodium ion in the blood. The mechanisme is as shown in Figure 11.29. Blood volume and pressure rises Blood volume and pressure falls Less renin produced More renin produced Less angiostensin II formed More angiostensin II formed Adrenal cortex produces less aldosterone Adrenal cortex produces more aldosterone Normal blood sodium concentration Normal blood sodium concentration Na+ rises Na+ Na rises + falls Na+ falls Figure 11.29 The regulation of the blood pH 2. When the blood sodium ion decreases leading to the blood volume decreases, aldosterone is produced from the adrenal cortex. The decrease in blood volume resulting in decrease in blood pressure stimulates a special group of cells called juxta glomerular complex found between the efferent arteriole and distal convoluted tubule to produce renin. (a) Renin is an enzyme that converts angiostensinogen to angiostensin I. Another enzyme converts angiostensin I to angiostensin II. (b) Angiostensin II stimulates the release of aldosterone from adrenal cortex. (c) Aldosterone binds to receptors in the nucleus of the distal convoluted tubule and the collecting duct. It increases the synthesis of carrier protein for the sodium potassium pumps. (d) Aldosterone increases the absorption of sodium ions from the glomerular filtrate by the distal convoluted tubule and the collecting duct. This is due to aldosterone increases the activity of the sodium-potassium pump at the basolacteral membrane resulting more sodium ions actively transported into peritubular capillaries. Water then follows into the capillary and so are chloride ions that maintain the electrochemical balance. (e) Aldosterone also increases the uptake of sodium in the small intestine, thus increasing sodium level in the blood. (f) Aldosterone also increases the secretion of potassium ions into the glomerular filtrate, thereby maintaining sodium-potassium ion balance in the blood. Negative feedback mechanism adjusts the sodium ion back to its normal level.
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 3. When the sodium ion increases leading to the blood volume and pressure increases, little or no aldosterone is produced. Little or no sodium ion is absorbed back into the blood from the glomerular filtrate in the nephron. The sodium ion in the blood decreases to the normal level with the help of negative feedback mechanism. Potassium ions are absorbed from the filtrate. Regulation of pH of Tissue Fluid 1. Kidney regulates blood pH. This is the control of both tissue and blood pH level so that the blood pH is constant at around 7.4. Proteins in the blood buffer a slight change in pH due to an increase in carbon dioxide concentration from respiration. Other buffers in the blood are in the forms of hydrogencarbonate and phosphate. More pronounced change of pH requires the intervention of the nephron. 2. When the pH falls, the proximal and distal convoluted tubules can actively secrete hydrogen ions into the glomerular filtrate. The tubules withhold and actively absorb back the hydrogencarbonate ions from the glomerular filtrate. 3. In more extreme case, the hydrogen ion combines with ammonia produced by the cells of the distal convoluted tubule. The process is catalysed by an enzyme that removes amino group from glutamine. The ammonium ion formed is excreted in the urine. 4. When the pH rises, the proximal and distal convoluted tubules can actively secrete hydrogencarbonate ions and retain all hydrogen ions. 5. The blood pH is kept at 7.4 for enzymes and other proteins to function optimally. Any deviation away from this may result in fatality as the enzymes and proteins are denatured. 6. Kidney forms urine. The formation of urine is as in Figure 11.30. Filtrate Useful substances Water Mineral ions pH Drugs Wastes in urine Nephron Figure 11.30 The formation of urine in the nephron
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 11.4 Osmoregulation in Plants 11.4 Osmoregulation in Plants Role Of Stomata In The Regulation Of Water Loss and Importance of Transpiration Role of stomata in the regulation of water loss 1. The stomata in the leaves open all the time during the day, when the supply of water in the soil is enough for the roots. When there is light, the stomatal opening mechanism ensures the opening of the stomata. The opening of the stomata is for gas exchange for photosynthesis with loss of water through transpiration. There is no need to control the loss of water. 2. The stomata of some species open even at night, when the soil has extremely high water content. This opening presumably has the advantage of depleting soil water and thus preventing root anaerobiosis. This may due to overall turgidity of all the cells including the guard cells. This may also due to the production of auxin in the roots in response to high water content. The auxin is transported in the xylem of some species to cause the stomata to open. 3. The stomata close when the soil water content drops. This is due to the supply of water is not enough for the plant, all the cells in the leaves including those of guard cells become flaccid. When the guard cells become flaccid, the guard cells collapse upon each other closing the stoma. In addition, the leaves with flaccid cells would droop or roll up. This would prevent direct sunlight shinning on the leaves. This lowers the temperature of the leaves; less water is lost through evaporation. In the evening when the supply of water is enough, the cells in the leaves would be turgid again. 4. The stomata close when the guard cells response to abscisic acid (ABA). ABA is produced in the leaves in response to high temperature and low water potential in the leaves. It appears that as the soil dries, the roots “sense” the lower soil water potential, and synthesise ABA. This ABA is transported to the leaves in the transpiration stream via the xylem, and produces the same effect on stomata as ABA that is produced from the leaf. This mechanism may allow plants to anticipate drought conditions and respond accordingly even before the leaf water potential has been affected. 5. The stomata close at high environmental temperatures not only due to the production of ABA but due to direct heat. High temperatures increase transpiration due to high water molecular movement and greater vapour diffusion gradient between inside the leaves and the atmosphere. The resulted greater loss of water causes the guard cells to lose water and the stomata to close. Learning Outcomes :[\KLU[ZZOV\SKILHISL[V! H KLZJYPIL[OLYVSLVM Z[VTH[HPUYLN\SH[PVUVM ^H[LYSVZZHUKL_WSHPU [OLPTWVY[HUJLVM [YHUZWPYH[PVU" I KLZJYPIL[OL]HYPV\Z [`WLZVMWSHU[ HKHW[H[PVUZ[VWYL]LU[ ^H[LYSVZZOHSVWO`[LZ HUK_LYVWO`[LZ 2011
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 6. Stomata may open in deleteriously high leaf temperatures in some species of plant. The effect occurs even in darkness. This appears to be a strategy designed to decrease leaf temperatures through evaporative cooling. This opening occurs irrespective of the supply of water in the soil but eventually the stomata will close as the guard cells can no longer be turgid in continual short water supply. 7. The stomata usually open due to short supply of carbon dioxide if the guard cells are not facing short supply of water. Carbon dioxide deficit results in increase of pH that activates the amylase to digest starch to form maltose in the guard cell. This takes a minute or so for the maltose to decrease the water potential in the guard cells to absorb water due to osmosis and becoming turgid. 8. Stomata of Crussulaceae, pineapple family and some cacti open their stomata at night. This is a physiological adaptation to conserve water for xerophytes living in dry areas with hot day temperatures and cold night. The opening of stomata at night enables the minimal loss of water. The plants carry out fixation of carbon dioxide into malate throughout the night and release the carbon dioxide for normal photosynthesis during the day. 9. Certain plants especially xerophytes have stomata adapted to live in dry areas. There are a number of ways the stomata are adapted. (a) In most plants especially dicotyledonous plants, stomata are found only in the lower epidermis. This would prevent excessive loss of water as the lower epidermis is on the shaded side of the leaves with lower temperature. (b) In dessert plants especially cacti, the number of stoma is greatly reduced. In extreme cases, there is no leaf at all as the leaves are modified to become spines. (c) Certain plants have very fine stomata. This would cut down the loss of water from transpiration. (d) In others, the structure of the stomata is modified and becomes sunken. They are found among fine hairs called trichomes or in special grooves as in the case of Nerium or Ammophila. The hairs or grooves cause the accumulation of water vapour around the stomata decreasing the vapour potential gradient for diffusion between the inside of leaves and the atmosphere. Importance of Transpiration 1. Transpiration produces a force called transpiration pull on the xylem of the stems and roots to uplift water to the leaves. This is due to transpiration creates a lower water potential in the cells of the leaves. Water will move down water potential gradient from xylem to the mesophyll cells where evaporation of water occurs from the wet cell walls. Since there are so many leaves in a plant, the force created on
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis a hot day is very high. Therefore, transpiration can help to transport water up from the soil even to more than 100 meters such as in the giant Sequoia tree of California. 2. Since mineral ions dissolved in water, transpiration helps in the transport of the ions from the roots or soil to the leaves where they are most needed. This is especially important for ions such as nitrates, sulphates, phosphates, calcium, magnesium and iron which are required for the synthesis of amino acids, nucleic acids, phospholipids and chlorophyll in the leaves. Other micronutrients like copper and cobalt ions are also important as activator for enzymes. 3. Besides, it also helps in the transport of water-soluble organic products such as the plant hormones especially ethylene and cytokinin. Ethylene is a gas produced in most cells and easily soluble in water to be carried by xylem. It helps the auxin in suppressing growth of lateral buds. On the other hand, cytokinin is produced at the root tip in higher amount in some plants and gets transported up to the shoot tip by xylem. Here, it antagonises auxin and promotes the growth of the lateral buds. 4. Transpiration helps to cool the leaves. This is especially so in a hot afternoon under the tropical sun when a steady stream of water from the soil is needed to prevent over-heating of leaves. Evaporation of water helps to absorb heat and lower temperature from the leaves so that they can keep on photosynthesising. 5. Greater amount of transpiration stimulates the development of mechanical tissues in plants. The plants become healthier and the cell walls become more compact, thicker and cutinised. Therefore, the plants are able to resist the attack of bacteria and not so easily sway in the wind. 6. Transpiration maintains an optimum degree of turgor in cells. Under favourable conditions plants absorb excess amount of water, which is given off by transpiration to maintain the optimum turgor for better metabolism and growth. 7. Transpiration can improve the taste of fruits. This is due to the solutes inside the cell become more concentrated when transpiration is rapid. Then, the concentration of sugar solution in the cells of fruits increases and fruits taste sweeter. Adaptations of Halophytes and Xerophytes to Reduce Water Loss Halophytes 1. They are plants that can live in soil with more than 0.5% of salt. They live in deserts, estuaries, and seashores. Summary Importance of Transpiration 1. Causes transpiration pull to uplift water into leaves from soil through stem 2. Transport mineral ions into leaves from soil 3. Transport plant hormones produced from root to shoot 4. Cool the leaves 5. Stimulates development of mechanical tissues 6. Maintain optimal turgor in leaves 7. Improve sugar content of fruits Exam Tips 9LTLTILY[OLYVSLVM Z[VTH[HPUYLN\SH[PVU VM^H[LYSVZZHUKYLSH[L P[[V[OLZ[Y\J[\YLHUK HKHW[H[PVUZMVY_LYVWO`[LZ [VZ\Y]P]LPUKY` HYLHZ
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis Figure 11.31 Examples of halophytes 2. Salt accumulators, such as saltbush (Atriplex), smooth cordgrass (Spartina alterniflora), saltgrass (Distichlis spicata), and tamarisk (Tamarix pentandra), have specialized cells called salt glands located on the surfaces of their leaves, used for storing excess sodium chloride and burst releasing the salt. 3. Epidermal cell of leaves of mangrove (Avicennia germinans) can excrete salt resulting the surface become whitish. 4. Succulent halophytes, such as pickleweed (Salicornia virginica) transporting sodium and chloride ions into the central vacuole. 5. Salt excluders Some salt excluders avoid salt stress by defoliation or abscission (leaf release with toxic level of salt). The root epidermis (outer layer of cells) of some halophytes may not allow the passage of sodium and chloride ions through the cell membrane. In some plants, root cells are capable of actively pumping excess sodium and chloride ions out into the surrounding soil. Xerophytes 1. They are plants that are adapted to survive in an arid environment, such as a desert or sandy shore. Therefore, some of them are also halophytes. Figure 11.32 Examples of xerophytes 2016 STPM
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis 2. They are adapted to conserve water, and commonly also to store large quantities of water, during dry periods e.g. succulents like cacti. Other species may be adapted to survive long periods of desiccation of their tissues, during which their metabolic activity may effectively shut down e.g. Joshua tree. 3. They have long deep roots like pine tree to obtain water at low water table or profuse surface roots like Bromeliads to absorb surface water rapidly. Adaptation of Halophytes and Xerophytes 1. Most dicotyledonous halophytes like mangroves and xerophytes have their stomata located in the shaded underside of leaves to reduce water loss. The shaded underside has lower temperature. Thus, loss of water is minimal. 2. Many halophytes and xerophytes have adaptations to trap vapour outside their stomata to produce high localised humidity to reduce diffusion gradient so as to reduce transpiration loss. These are by having: (a) Rolled leaves. Ammophila that live in sandy shore roll their leaves. This will create moist air of high humidity within the rolled leaves. Their stomata are facing the humid air. Thus, the moist air reduces the diffusion of vapour out of the stomata. (b) Hairs on epidermis. Many xerophytes like cacti and semi-desert plants have trichomes or long hairs on their leaf surface to trap vapour to reduce transpiration loss. (c) Sunken and hidden stomata. Many cacti and halophytes have sunken stomata. Their stomata have pits that can trap vapour. Some xerophytes like Allamanda have grooves or folds together with hairs to trap vapour as well. 3. Many halophytes and xerophytes have adaptations to reduce surface for transpiration. (a) Small leaves. Many cacti have small leaves but big succulent stem. Thus, they have less leaf surface for transpiration. Some salt marsh halophytes similarly have small and few leaves. (b) Leaves modified into spines. Many cacti have done away with leaves. Their leaves are modified into spines. Photosynthesis is carried out by leaf-like stems with no stomata to reduce transpiration. Pickleweeds have their leaves modified into scales or drop their leaves. 4. Many halophytes and xerophytes have leaves adapted to reduce surface : volume ratio for transpiration. These leaves are thick or succulent as in many cacti and desert plants including Aloe vera. Halophytes have thin slender leaves. Mangroves have thick leaves. The main function of the succulent leaves is for storage of water.
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 5. Almost all halophytes and xerophytes have thick waxy cuticle to decrease permeability for evaporation loss. The cuticle is made up of wax and other lipids. 6. Many halophytes and xerophytes are adapted to reduce rate of diffusion of water from inner cells. These are made possible by having: (a) Reflective cuticle. Most waxy cuticle is reflective. Thus, it can help to reflect light as well as heat away so that the leaves are not easily heated up and have a high rate of diffusion. Halophytes have salt bladders or cells that burst to release salt that also acts to cover the leaf surface to cut down heat. (b) Epidermis with thick-walled cells. Almost all xerophytes and some halophytes have thick-walled epidermis. The wall is impregnated with suberin or lignin that are impervious to water. Some epidermal cells are full of oily resin. Thus, the epidermis can prevent diffusion and evaporation of water from its surface. (c) Several layers of hypodermis. Many xerophytes including cacti, Allamanda and pine leaves have several layers of thickwalled cells beneath the epidermis. These cells are sclerenchyma with lignified wall that further reduce rate of diffusion of water across them. Some cacti and Ammophila are covered with layers of sclerenchyma and have so few living cells within that they are almost non-living when there is no rain. (d) Few stomata. Many halophytes and xerophytes cut down the number of stomata as in cacti. This would reduce the loss of water in the form of vapour through the stomata. 7. Most xerophytes and some halophytes produce abscisic acid under stress of lack of water and high temperature to close stomata to reduce transpiration. The abscisic acid produce in the leaves will diffuse to the guard cells and bind with receptors at the membrane. Then, it will stimulate the active transport of potassium ions out of guard cells causing the cells to become flaccid and close the stomata. 8. Crassulaceae family of plants and many cacti including halophytes close their stomata during the day and open at night to reduce transpiration. This is due to day temperature is high and night temperature is low. By opening the stomata at night, the loss of water from transpiration is reduced. These plants have C4 cycle for carbon fixation and store up malic acid throughout the night. This malic acid will release carbon dioxide during the day when there is light for photosynthesis. 9. Many halophytes and xerophytes are adapted physiologically to reduce water loss. They have C4 cycle of carbon fixation and Kranz anatomy in the leaves. They reduce the minimal loss of water through the almost close stomata during the hot afternoon like sorghum and photosynthesise efficiently. Summary Adaptation of Xerophytes and Halophytes to Reduce Water Loss 1. Stomata located in the shaded underside of the leaves 2. Rolled leaves, hairs on epidermis, sunken & hidden stomata trap vapour to produce high humidity to reduce diffusion gradient 3. Small leaves & leaves TVKPÄLKPU[VZWPULZ to reduce surface for transpiration 4. Thick or succulent leaves to reduce surface : volume ratio for transpiration 5. Thick waxy cuticle decreases permeability for evaporation loss 6. 9LÅLJ[P]LJ\[PJSLZL]LYHS layers of hypodermis, epidermis with thick walled cells & few stomata to reduce rate of diffusion of water 7. Abscisic acid produced to close stomata when hot to reduce transpiration 8. Stomata are closed during the day and open at night to reduce transpiration 9. ,MÄJPLU[WO`ZPVSVNPJHS control of stomata and C4 J`JSLVMJHYIVUÄ_H[PVU
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis Quick Check 3 1. Can plants survive without stoma? Explain. 2. An adaptation of desert plants is that the plants can complete their life cycle in around two weeks. What characteristics do these plants possess? Objective Questions 1. What happens if the body temperature of Homo sapiens decreases? A The metabolic rate decreases. B The hair erector muscle relaxes. C The peripheral blood vessel dilates. D The skeletal muscle contracts rapidly. 2. Which of the physiological control in thermoregulation of endotherms is correct? A The body temperature is influenced by the environment. B The body heat is higher than that of the environment. C The body generates small amount of heat. D The body metabolic rate is low. 3. The graph below shows the rates of production of glucagons and insulin plot against that of blood glucose concentration. 40 Glucagon Insulin 0 80 120 Blood glucose concentration (mg/ 100 ml) 160 200 240 Rate of glucagons Secretion units (Arbitrary units) Rate of insulin secretion (Arbitrary) Which statement is false of the graph? I The secretion of glucagon increases with the blood glucose concentration increases till 80 mg/100 ml II The insulin secretion decreases with the increase in blood glucose concentration from 80 mg/100 ml till 200 mg/100 ml III Glucagon stimulates the use of glucose whereas insulin stimulates the release of glucose into the blood A I only C I and II B II only D I, II and III 4. Which condition involves homeostatic regulation? A The varying rate of heartbeat according to activity B A nerve impulse that propagates through an axon C A person who goes into a state of hyperthermia D The contraction of uterus due to the secretion of oxytocin during childbirth 5. What happens to the rate of glycogenesis, glycogenolysis and gluconeogenesis when the glucose level in the blood falls below normal level? Glycogenesis Glycogenolysis Gluconeogenesis A Increase Decrease Decrease B Decrease Increase Increase C Increase Decrease Increase D Decrease Decrease Decrease STPM PRACTICE 11
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 10. Which terms correspond correctly with their respective biological processes? Term Biological process I Transamination II Glycogenesis III Deamination IV Gluconeogenesis w Removal of amino group from amino acid to form ammonia x Synthesis of glucose from noncarbohydrate substance y Synthesis of new amino acid by the enzymatic transfer of amino group z Conversion of glucose to glycogen I II III IV A wyxz B xwz y C ywx z D y zwx 11. What is the function of the microvilli on the proximal convoluted tubule of nephron? A To provide a large surface area for the movement of materials into cells B To act as a filter to prevent the movement of large molecule into nephron C To generate a high concentration gradient between tissue fluid and blood plasma in renal medulla D To supply ATP for the absorption of glucose via active transport 6. Which is not true of diabetes mellitus? I Causes excessive urination II β-cells fail to secrete insulin III Insulin receptor is not functional IV The secretion of ADH is low A I and II B I and IV C II and III D III and IV 7. Which statement about the hormone secreted by `-cells of the islets of Langerhans is correct? A It activates enzymes which speed up gluconeogenesis in liver cells B It causes greater glucose uptake by muscle cells C It increases blood glucose level D It is a steroid hormone 8. What is the function of the liver? A Synthesis of glucagon B Synthesis of cholesterol C Regulation by pH D Production of corticosteroid 9. The diagram below shows the ornithine cycle in liver cells. Fumarate Aspartate T S R H2O Urea CO2 What do R, S and T represent? RST A Arginine Citrulline Ornithine B Citrulline Arginine Ornithine C Ornithine Arginine Citrulline D Ornithine Citrulline Arginine
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 12. What causes the higher water potential in the two arms of the loop of Henle as shown in the diagram? –1200 –300 –100 Descending arm Ascending arm A Na+ and Cl– are actively moved out of the ascending arm B Water diffuses out from the ascending arm by osmosis C The descending arm is permeable to dissolved solutes D The ascending arm is permeable to water 13. When the osmoreceptor in the hypothalamus detects an increase in water potential of blood, A more dilute urine is produced. B more urea is diffused out of the filtrate. C the posterior pituitary gland releases more ADH. D the collecting ducts reabsorb more water from the filtrate. 14. Excessive sweating might A stimulate the secretion of ACTH B stimulate the secretion of ADH C decrease the secretion of aldosterone D decrease the secretion of adrenaline 15. Which of the followings is true when there is a decrease in blood sodium level? A Aldosterone will inhibit the sodiumpotassium pumps in the distal convoluted tubule B Renin will catalyse the formation of angiotensin C Juxtaglomerular complex will not be stimulated D Blood volume will increase 16. Which buffer is the most important in the blood? A Amino acids C Carbonic acid B Haemoglobin D Plasma proteins Structured Questions 1. The diagram below shows the ornithine cycle. (a) What is deamination? [1] (b) Label W, X, Y and Z. [2] (c) What is the purpose of the cycle? [1] (d) Write a formula to summarise the cycle [2] (e) State two similarities among ornithine cycle and Cori cycle. [2] (f) State two differences between ornithine cycle and Cori cycle. [2] NH3 NH3 H2O H2O CO2 H2O Z X Y Deamination W
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis 2. Dolly drank a glass of sweetened fruit juice in morning after fasted overnight. The concentration of insulin and glucose in her blood is sown in the diagram below. 0 15 45 Drink sweetened fruit juice 60 75 Time/min Insulin Glucose Concentration of substance in blood 90 105 120 135 150 (a) Explain the relationship between the two curves (i) in the first 30 minute time interval after drinking the sweetened fruit juice, [3] (ii) one hour after drinking the sweetened fruit juice. [3] (b) Explain how a person suffers from hyperglycemia even with normal functioning pancreas. [4] Essay Questions 1. (a) Explain the concept of homeostasis and state its significance for organisms. [7] (b) Explain positive and negative feedback mechanisms with reference to endothermic and ectothermic animals. [8] 2. (a) Blood glucose level will drop to below the normal level when fosting. Describe the role of the liver in order to return the blood glucose level to the normal level. [8] (b) Describe how extra of amino acids are excreted from the human body. [7] 3. (a) Describe the structure of kidney. [6] (b) Explain how kidney performs its functions in homeostasis. [9]
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis Quick Check 1 1. Homeostasis is the maintenance of constant internal environment despite change in the external environment. 2. The components are receptor, brain and gland or muscle. Receptor detects changes in both internal and external environment. Impulse is then sent to the brain. The brain acts as coordinating centre. It receives impulse of the changes and send impulse to gland or muscle to correct the changes. Gland or muscle acts as effector. Gland secrete substance such as hormone to correct the change. Muscle contract to bring about the restoration. 3. Physiological control means functional control especially it involves hormones or impulses. Behavioural control involves changes of behaviours such as moving or bathing. Quick Check 2 1. Our blood glucose, amino acids or other nutrients will be very high after a meal and become very low later, as they are stored after taken in. Urea concentration will be nil, as there is no liver to produce it. Our old red blood cells will still be accumulated in the blood as little is destroyed elsewhere. Lactic acid will be very high in the blood after exercise. Quick Check 3 1. Yes. They will survive only in places with cool temperature, as too high a temperature will kill the plants. There is no stoma to allow transpiration to cool the plants. The plants will be small and live in area with good water supply. There is no stoma to generate transpirational pull so water and mineral ions will not be transported in the plants. 2. The seed and later on, the seedling must be able to absorb water while the soil is still wet. The plants must have water storage characteristics. Loss of water by transpiration will not take place and they will not have any stoma. Germination and growth are very fast, so that flowering, pollination, fertilisation and production of seeds must be completed within such short period. STPM Practice 11 Objective Questions 1. D 2. B 3. D 4. A 5. B 6. C 7. B 8. B 9. D 10. D 11. A 12. B 13. A 14. A 15. C 16. B Structured Questions 1. (a) It is a reaction in which amino acid is removed of its amino group with the help of enzyme deaminase. (b) W is citrulline, X is arginine, Y is ornithine and Z is urea. (c) The purpose of the cycle is to produce urea from ammonia, and carbon dioxide. (d) 2 NH3 + CO2 ±A (NH2 )2 CO + H2 O (e) Both take place in the liver. Both require an enzyme in each step of its cycle. (f) Ornithine cycle is to produce urea whereas Cori cycle is to recycle lactic acid. Ornithine cycle requires one organ to complete the whole cycle whereas Cori cycle requires two organs, i.e. the liver and muscle to complete the cycle. 2. (a) (i) This is due to more blood flows to the ` cells of pancreas. The ` cells have receptors for glucose and response by releasing more insulin. Insulin is released directly to the blood with increases blood glucose concentration. (ii) Insulin concentration still increases when blood concentration becomes maximum after one hour. This is due to the continual increase in blood glucose level which keeps stimulating the ` cells. More insulin is released into the blood and peaks slightly later. (b) This is due to the excessive intake of carbohydrateds or sugars causing the insulin production not being able to cope with such high level of blood glucose. The person is suffering from type II diabetes, the liver is not responding to the insulin secretion. Essay Questions 1. B t )PNFPTUBTJT JT NBJOUFOBODF PG B DPOTUBOU internal environment of an organism even though there are changes in the external environment. t ćFJOUFSOBM FOWJSPONFOU SFGFST UP UIF UJTTVF fluid, blood and protoplasm within the cell, especially in plants. t *UJT OPU B TUBUJD FOWJSPONFOU CVU B EZOBNJD one in which certain factors or physiological parameters vary within a certain narrow range. t ćF GBDUPST JOWPMWFE BSF CPEZ UFNQFSBUVSF water potential, glucose and mineral concentration in the blood. ANSWERS
CHAPTER 11 Biology Term 2 STPM Chapter 11 Homeostasis t ćF PSHBOT JOWPMWFE BSF SFDFQUPST UIBU DBO detect changes in certain factors and send impulses to the brain. The brain receives impulses from receptors, coordinates them and sends the impulses to the effectors to correct the deviation. t ćF TJHOJĕDBODF PGIPNFPTUBTJTJT UP CBMBODF input with output in order for the open system within a cell or organism to work. t )PNFPTUBTJT FOTVSFT BO PQUJNVN JOUFSOBM environment, for example a temperature of 37 °C for cells to function, and the organism to survive and reproduces efficiently. t )PNFPTUBTJT NBJOUBJOT BMNPTU DPOTUBOU condition in the body even though the external environmental conditions vary. Homeostasis enables certain organisms to adapt to extreme environmental conditions such as living in the desert or tundra. t )PNFPTUBTJT FOBCMFT UIF CJPMPHJDBM TZTUFN to function efficiently and smoothly with minimum wastage of energy. (b) t ćF OFHBUJWF GFFECBDL NFDIBOJTN JT B regulatory mechanism that is always being used to restore the body to its original constant state after it is affected by any disturbance. t 8IFOUIFUFNQFSBUVSFPGFOEPUIFSNJDBOJNBM is raised, process like sweating takes place to reduce it. t 8IFOUIFUFNQFSBUVSFPGFOEPUIFSNJDBOJNBM is lowered in cold environmental, thyroxine is produced to raise the metabolic rate and raise the body temperature. t ćF QPTJUJWF GFFECBDL NFDIBOJTN JT B mechanism seldom used as it creates further chaos or extremes. t 8IFOUIFUFNQFSBUVSFPGFDUPUIFSNJDBOJNBM is raised during basking, positive feedback occurs and its temperature would increase. t *GUIJTIBQQFOTJOUPPIJHIBUFNQFSBUVSFMJLF 50°C and the negative feedback mechanism fails to work in the endothermic animals, the positive feedback mechanism brings the temperature higher and results in death for both endothermic and ectothermic animals. t ćJT JT BMTP USVF XIFO UIF FOWJSPONFOUBM temperature is lower than -20°C, where further decrease in temperature triggers positive feedback in endothermic animals. The temperature is lower and results in death. t 1PTJUJWF GFFECBDL POMZ XPSLT UFNQPSBSJMZ JO endothermic animals. For example, when more thyroxine is produced, the body temperature becomes higher. However, it is only true for a short while before the set point temperature is reached. 2. B t ćFSPMFPGUIFMJWFSEVSJOHGBTUJOHJTUPSFMFBTF glucose into the blood t .PSFHMVDPTFJTSFMFBTFEJOUPUIFCMPPEXIFO the blood glucose level drops further to raise the blood glucose level to normal t ćF MJWFS NBJOUBJOT UIF CMPPE HMVDPTF JO response to the hormone glucagon t .PSF HMVDPTF JT SFMFBTFE CZ UIF MJWFS JO response to higher blood level of glucagon released by the pancreas alpha cells stimulated by the low level of blood glucose t ćFMJWFSDFMMTIBWFSFDFQUPSTPOUIFDFMMTVSGBDF membrane for the glucagon to bind t "GBTU BDUJWBUJPO PG FO[ZNFTXJUIJO UIFMJWFS cells is stimulated after the glucagon molecules bound to each of the receptors t ćJT CSJOHT BCPVU B DBTDBEF PG SFBDUJPOT UIBU break down of glycogen stored in the liver t ćFHMZDPHFOJOUIFMJWFSDFMMTBSFCSPLFOJOUP glucose in the liver cells t ćF MJWFS SFMFBTFT UIF HMVDPTF QSPEVDFE JOUP the sinusoid blood stream in the liver to raise the blood glucose level (b) t &YUSBBNJOPBDJETBSFĕSTUEFBNJOBUFEJOUIF liver i.e. with the amino group removed from each amino acid molecule t ćFEFBNJOBUJPOSFBDUJPODPOWFSUTUIFBNJOP acids into keto acids after the removal of amino groups t ćFBNJOPHSPVQTSFNPWFEBSFDPOWFSUFEJOUP urea by ornithine cycle in the liver cells t ćFVSFBQSPEVDFEJOUIFMJWFSJTUSBOTQPSUFE by the blood into the kidney to be excreted in the urine t ,FUPBDJETDBOCFCSPLFOEPXOCZSFTQJSBUJPO into carbon dioxide and water t ćF DBSCPO EJPYJEF JT FYDSFUFE JO UIF MVOHT during breathing t ćFXBUFSJTVTVBMMZFYDSFUFEJOUIFLJEOFZBT urine 3. B t ćFIVNBOLJEOFZJTBCFBOTIBQFPSHBO EBSL red in colour and the size of our fist. t *U IBT B SFOBM BSUFSZ TVQQMZJOH CMPPE UP JU B renal vein to drain the blood away and a ureter to drain away the urine formed in it. t *UIBTBOPVUFSDBQTVMFUIBUDPOTJTUTPGĕCSPVT protein. t ćFPVUFSQPSUJPOPGJUJTDBMMFEDPSUFYXIFSF Malphighian corpuscles are located. t ćFJOOFSQPSUJPOJTDBMMFENFEVMMBXIFSFUIF loops of Henle and collecting ducts are located. t *O UIF DFOUSF UIFSF JT B TQBDF DBMMFE QFMWJT where the urine is collected before it is drained to the ureter.
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis t ćF CBTJD VOJU PG GVODUJPO JT UIF OFQISPO which numbered about a million in each kidney. t *U JT TFSWFE XJUI B CSBODI PG UIF BSUFSJPMF which enters the glomerulus surrounded by Bowman capsule. A fine tubule drains the filtrate and loops in and out from the medulla and finally joins with the collecting duct in which urine is drained out. (b) t ćFLJEOFZQFSGPSNTUISFFNBKPSIPNFPTUBTUJD functions, namely osmoregulation, regulation of mineral ions and pH of the blood. t ćFLJEOFZDBSSJFTPVUPTNPSFHVMBUJPOJOUXP ways i.e. by adaptation and with the help of hormone. t "OJNBMT MJWJOH JO UIF EFTFSU MJLF DBNFM IBT large medulla and long loop of Henle where urine is concentrated to the maximum and little water is excreted. t "OUJEJVSFUJD IPSNPOF JT QSPEVDFE XIFO the body lacks water. It causes the kidney to produce little urine. No such hormone is produced when the body has a lot of water and a lot of water is excreted. t ,JEOFZSFHVMBUFTNJOFSBMJPOTXJUIUIFIFMQPG specific hormones. An example is aldosterone that helps to regulated blood sodium ion concentration. t 8IFO UIF CPEZMBDLT TPEJVN BMEPTUFSPOFJT produced and causes the nephron to reabsorb back all the sodium ions and does not excrete any sodium ions. t 8IFO UIF CPEZ IBT IJHI TPEJVN JPO concentration, little aldosterone is produced. The kidney reabsorbs little sodium ions; most are excreted. t 4JNJMBSMZ XIFO UIF Q)JO UIF CMPPEJT IJHI meaning there is a high hydrogencarbonate ion concentration; the kidney excretes most of it and retains hydrogen ions to neutralise it. t 8IFO UIF Q) JT MPX NFBOJOH UIFSF JT a high hydrogen ion concentration; the kidney excretes most of it and retains hydrogencarbonate ions to neutralise it. t ,JEOFZ DBO QSPEVDF BNNPOJB UP SFBDU XJUI excessive high hydrogen ions and excrete them in the form of ammonium ions.
CHAPTER IMMUNITY Concept Map 12 Bilingual Keywords Immunity – Keimunan Lymphatic system – Sistem limfa Antibody – Antibodi Lymphocyte – Limfosit Organ transplant – Pemindahan organ Immune disorder – Imun tidak tertib Tissue rejection – Penolakan tisu Immunity B cells and T cells Cell-mediated and humoral immune responses Tissue rejection in organ transplant Antigenantibody reactions Cell-mediated and humoral immune Immune disorder Antibody Antigen Epitope Lymphatic system Development of immunity related to Concept of self and non-self Autoimmune disorder Systemic Lupus Erythematosus (SLE)
Learning Outcomes CHAPTER 12 219 Biology Term 2 STPM Chapter 12 Immunity Students should be able to: (a) describe human lymphatic system, and explain its function in relation to immunity; (b) describe antibody (structure and function), antigen, epitope, and the development of B and T cells; (c) describe the roles of macrophages, B cells and T cells 12.1 Immunity 12.1 Immunity 1. Immunity was defined by Sir Macfaclane Burnet as ‘the capacity to recognise the intrusion of foreign material to the body and to mobilise cells and cell products to help remove that particular sort of foreign materials with greater speed and effectiveness’. 2. Immunity is the protection against disease provided by the body’s defence or immune system. Human Lymphatic System and its Function in Relation to Immunity 1. The structures and functions of the components of the lymphatic system are already explained in Chapter 8.1. The primary function of the lymphatic system is to provide an accessory route for excess blood plasma to get returned to the blood. Lymph is essentially recycled blood plasma. The excess blood plasma gets into the tissue fluid after coming out from blood capilaries. 2. The function of the lymphatic system in relation to immunity are: (a) Allows the foreign materials (antigens) directly to go from the points of entry into the lymphoid tissues or organs. (b) Transports the foreign materials (antigens) directly from the tissue fluid into the lymphatic capillaries and lymphatic system, then to lymphoid tissue or organ. (i) From the air breathed in, the antigens may get into the adenoids, tonsil and the lymphoid tissues in the trachea. (ii) From the food, the antigens may get into the tonsil and lymphoid tissues in the mucosa of the oesophagus, stomach, small intestine (Peyer patches), appendix and large intestine. (c) Bone marrow is also a key component of the lymphatic system, producing the lymphocytes that support the body’s immune system (Figure 12.1). Marrow Bone graft Macrophage Erythrocytes Eosinophil Basophil Mast cell Megakaryocyte Platelets Dendritic cell B lymphocyte Natural killer cell T lymphocyte Lymphoid progenitor cell Monocyte Bone Hematopoietic stem cell Multipotential stem cell Myeloid progenitor cell Neutrophil Figure 12.1 Bone marrow gives rise to cells of immune system Sir Macfaclane Burnet was an Australian virologist and has been awarded the 1960 Nobel Prize together with another virologist for physiology and medicine for the discovery of acquired immunological tolerance, the concept on which tissue transplantation is found. Info Bio INFO Immunity
CHAPTER 12 220 Biology Term 2 STPM Chapter 12 Immunity (d) Transports stem cells especially the immature T cells (lymphocytes) from the bone marrow into the thymus gland where the T cells become matured. (e) Provides thymus gland with thymocytes that secrete thymosine for maturity of T cells. (f) Transports matured T cells and B cells to the lymphoid tissues and lymph nodes where they wait for antigens. (g) Transports antibodies and other complement proteins produced by white blood cells including T cells and B cells. (h) Provides the sites i.e. the lymphoid tissues, lymph nodes and spleen where the antigens are trapped and got rid of. Antigen, Epitope and Antibody Antigen 1. An antigen is a protein in which the body recognises as foreign or non-self and stimulates a specific immune response. 2. Antigen can be carbohydrate or some other macromolecules. Some are glycoproteins i.e. conjugated proteins. 3. The antigen is found on the surface of microorganisms especially pathogens. In viruses, the individual globular protein unit (capsomere) can act as antigen. In bacteria and infectious yeast, the protein or part of the peptide structure on the wall is also an antigen. 4. The protein on the plasma membrane of infectious eukaryotic cells like Plasmodium also acts an antigen. The free protein molecules produced by infectious pathogen are considered as antigen too. 5. The proteins found on the plasma membrane of cancerous cells can also become antigens. 6. Proteins produced in one’s body cannot act as antigens in that individual’s body but when transferred into another individual’s body, they act as antigens. Therefore, antigen can be just any ordinary protein that is determined by a specific gene. It is synthesised in the cell. 7. Antigens are very specific in stimulating immune response. Specific antibody molecules are produced to bind them. Specific T and B lymphocytes are produced to get rid of them. This is due to the specific sequence of amino acids in the antigen. So, this allows specific response that are customised to specific pathogens having such antigen. 2013/P2/Q17 STPM
CHAPTER 12 221 Biology Term 2 STPM Chapter 12 Immunity Epitope 1. Epitope is a small specific part of a larger antigen which is capable of bringing a specific antibody to bind with it (Figure 12.2). Protein (antigen) Antibody 1 Antibody 3 Epitope 2 Epitope 3 Epitope 1 Antibody 2 Figure 12.2 (WYV[LPUHU[PNLU^P[O[OYLLLWP[VWLZMVY[OYLLKPɈLYLU[HU[PIVKPLZ[VIPUK[V 2. Epitopes is an antigenic determinant. A larger antigen might have two or more such determinants. 3. Each epitope has a specific three dimensional shape composed of certain side chains of amino acids in the polypeptide or combined with sugar units. The side chains of the amino acids are not necessarily that of adjacent amino acids but they form a part of the antigen after the polypeptide is folded or coiled. The specific shape is complementary to the shape of the antibody binding site. 4. A bacterial cell may have many different epitopes. Each epitope is capable of causing a specific antibody to bind with it. The epitope may be part of the cell wall structure. 5. Viruses have fewer epitopes. Therefore, fewer type of antibodies are involved in removing them. Epitopes as short polypeptides or with specific shape can bind with specific surface membrane protein for “presentation”. 6. When macrophage or B cell has endocytosised bacteria or viruses, their antigens cut into epitopes of short polypeptides. These epitopes are presented on the surface of the cell for helper T cell to recognise and bind to start the immune response. Our body cells can also present the epitope if they are attacked by viruses. 7. Some epitopes are more effective than others at stimulating an immune response. This may be due to the shape of the polypeptide does not protrude well from the surface of antigen or when combined with membrane protein when presented by antigen presenting cells like macrophage or B cells. Antibody 1. Antibody is a specific protein released by a mature lymphocyte called plasma cell that acts against a particular antigen. The antibody binds and destroys the antigen. Antibody molecules found on the surface of ‘virgin’ B cells act as receptors to bind with antigens to trigger immune response leading to the destruction of the antigens.
CHAPTER 12 222 Biology Term 2 STPM Chapter 12 Immunity 2. Antibodies are all globular glycoproteins and form the group of plasma proteins called immunoglobulins. 3. The structure of each type of antibody is Y-shaped as shown in Figure 12.3. Attachment sites for binding antigen Disulfide bond Light chains Heavy chains Constant part of polypeptide where the amino acids sequence is the same for all the same type Variable part of polypeptide with sequence of amino acids differing from other types of antibodies Figure 12.3 Structure of antibody molecule 4. As shown in the Figure 12.3, each molecule of antibody composes of two identical heavy chains (H chains) and two identical light chains (L chains) that are linked together by disulfide bonds. 5. The lower portion of the molecule is constant part of polypeptide where the amino acids sequence is the same for a specific category of antibody. There are two similar variable parts at the upper portion of the molecule. 6. Each of the variable parts forms a binding site and can bind specifically to an antigen like ‘lock and key’. So, the shape of the binding site is complementary to the shape of antigen. Both binding sites are identical, binding to the same type of antigen. So, each antibody molecule can bind to two identical antigens. 7. Human body is capable of producing more than 100 million types of antibodies, each binding to a specific antigen that might enter our body. Such large number is possible due to the large size of the genes whose mRNA can be cut and recombined in almost unlimited combination. 8. The immunoglobulins are divided into five classes as follows: (a) IgM. This is the first type of antibodies produced in an immunity response. IgM starts the cascade effect of complement proteins and it can agglutinate antigen. Antigens, including microbes, that are bound with IgM or other complement proteins are easily phagocytosised. One molecule of IgM consists of five Y-shaped antibody molecules joined at the base. It has 10 antigen binding sites. It cannot cross placenta to react with antigens of the foetus. (b) IgG. It can activate the production of complement proteins and neutralise various toxins. It can exist long in the blood and able to cross placenta and secrete into the milk so that the baby can obtain passive natural immunity from the mother. 2014, 2016 STPM
CHAPTER 12 223 Biology Term 2 STPM Chapter 12 Immunity (c) IgA. It is usually found on the surface with mucus such as in the respiratory tract, digestive reproductive canals and initial breast milk called colostrum. Breast-fed babies have IgA in their digestive canal. IgA can destroy various pathogens. (d) IgE. It is produced as a result of invasion by parasitic worms or other pathogens. It binds with basophil and mast cell, causing the production of complement proteins. The complement proteins then causes inflammation. (e) IgD. It is found on the B cell surfaces and serves to act as receptor for antigen. 9. The differences between the five classes of immunoglobulins are shown in Table 12.1 and Figure 12.4. Table 12.1 +PɈLYLUJLZIL[^LLUJSHZZLZVMHU[PIVKPLZ Antibody class Relative molecular mass Number of antigen binding sites Sites of action Functions (a) Immunoglobulin M (IgM) 970 000 10 Blood tissue fluid and cannot cross placenta t $BVTFTBHHMVUJOBUJPO t "DUJWBUFTDPNQMFNFOU proteins (b) Immunoglobulin G (IgG) 150 000 2 Blood tissue fluid and can cross placenta t $BVTFTBHHMVUJOBUJPO t "DUTBTBOUJUPYJOT t "DUJWBUFTDPNQMFNFOU proteins t "JETUIFBDUJWJUZPG macrophages (c) Immunoglobulin A (IgA) 160 000 or 320 000 2 or 4 Tears, saliva, respiratory secretions, mucous secretions of gut, sex organs and colostrum/ breast milk t 1SFWFOUTCBDUFSJBHSPXUI on mucous membrances t *OIJCJUTCBDUFSJBGSPN sticking to host cells (d) Immunoglobulin E (IgE) 180 000 2 Tissue t 3FTQPOEUPJOGFDUJPOTCZ worms t "DUJWBUFNBTUDFMMTUP release histamine t "MMFSHFOJDSFTQPOTFTUP dust and pollen (hay fever) (e) Immunoglobulin D (IgD) 150 000 2 Virgin B cells t 4FSWFTBTSFDFQUPSGPS antigen
CHAPTER 12 224 Biology Term 2 STPM Chapter 12 Immunity IgG IgM IgA IgE IgD Figure 12.4 Structure of various immunoglobulins Lymphocytes 1. Lymphocytes are white blood cells that belong to the group called agranulocytes. They are smaller than phagocytes. 2. There are two types of lymphocyte, both of which are produced before birth in bone marrow. (a) B lymphocytes (B cells) (b) T lymphocytes (T cells) The development of T cells 1. The T cells are originated from stem cells that are formed in the bone marrow. The T cells leave the bone marrow and collect in the thymus where they mature. The thymus is a gland that lies in the chest just beneath the sternum. 2. The T lymphocytes then mature and differentiate into helper T cell, cytotoxic T cell, inducer T cell and suppressor T cell. Each has different T cell receptor (TCR) on its surface. The thymus gland, has cells that develop into thymocytes. The thymocytes produces hormone called thymosine for differention and maturity of T cells. 3. Helper T cells have T4 molecule as TCR. Suppressor T cells and cytotoxic T (killer) cells have T8 molecules as TCR. Helper T secretes interleukin-2 (IL-2) to stimulate division of immune cells in a process called clonal expansion. Cytotoxic T secretes perforin that can make holes in infected body cells. The holes cause the cells to be hydrolysed together with the pathogens within. 4. There are about 10 million types of each T cell with different types of receptors. The different types of TCR are formed using TCR loci located in chromosome 7 and 14 during the process of maturing or differentiating. Only one type of T cell can bind with any possible antigen that enters our body. Only mature lymphocytes can carry out immune responses. Info Bio
CHAPTER 12 225 Biology Term 2 STPM Chapter 12 Immunity Stem cell Mitosis Matured T cells each with a different TCR Maturation in thymus gland Figure 12.5 Maturation of T lymphocytes 5. These differentiated matured T cells after passing through the thymus gland, are known as ‘virgin’ T cells. They are found in the blood circulatory system and the tissue fluid. They take up positions in the lymph nodes, spleen and lymphoid tissues of the tonsil and intestine. They are ready to meet any antigen that may enter the body. The development of B cells 1. B lymphocyte cells originate from stem cells that are formed in the bone marrow. They remain in the bone marrow until they are mature and then spread throughout the body concentrating in lymph nodes and the spleen. 2. When they become matured, they form antibodies but only of one type in each cell. These Y-shaped antibody molecules attached to the plasma membrane of the cells with two tips acting as receptors. As explained earlier, during the maturing process, loci from antibody genes in chromosome 6 are used in combination to form the heavy chain of the antibody molecule. These are the ‘virgin’ B lymphocytes. Mitosis Matured B cells each with a different antibody receptor Maturation Figure 12.6 Maturation of B lymphocytes 3. When mature, all these B cells circulate between the blood and the lymph. This is to ensure that they are distributed throughout the body so that they come into contact with any pathogens. Immune responses depend on B and T cells interacting with each other to give an effective defence. Exam Tips Remember the T cells and B cells are produced in the bone marrow but they TH[\YLPUKPɈLYLU[WSHJLZ; cells mature in the thymus whereas B cells mature in bone marrow.
CHAPTER 12 226 Biology Term 2 STPM Chapter 12 Immunity Roles of Macrophages, B cells and T cells 2010, 2011 Macrophages 1. Macrophages are large phagocytes with a diameter of 21 μm compared with neutrophils of 12 μm. 2. Macrophages originated from stem cells that are formed in the bone marrow. They leave the bone marrow and travel in the blood as monocytes, which develop into macrophages once they leave the blood and settle in tissues and organs. They remove any foreign matters that are found there. 3. Macrophages have oval nucleus without lobes unlike granulocytes. The cytoplasm has less lysosomes than granulocytes. Macrophages have many mitochondria, endoplasmic reticulum and Golgi apparatus. Some monocytes change into dendritic cells almost having the same functions as macrophages. 4. Large numbers of mature macrophages take up positions in connective tissue, in the mucosa of digestive tract, in the lungs, spleen, lymph nodes and the sinusoids of liver. They are given name as Kupffer cells in the liver. 5. Macrophages carry out two major functions i.e. phagocytosis and antigen presenting. 6. In the lungs, macrophages can come out into the air sacs to engulf foreign matter including microbes entering with the inhaled air. The macrophages can secrete elastase to digest their way through the wall of alveoli. 7. In the liver, Kupffer cells will engulf and destroy old and worn out red blood cells together with microbes. The destroyed haemoglobin molecules are passed to the liver cells where bile pigments are secreted and iron ions are recycled. 8. In the spleen, macrophages can also engulf and destroy old red blood cells, white red blood cells and plateles. Macrophages will also filter microbes and antigens in the blood. Macrophages can engulf other worn out cells in the body. 9. An important role in non-specific immunity is to engulf microbes and antigens. A macrophage can engulf about one hundred bacteria during its lifespan. The macrophages stay in one place to catch the bacteria that pass by. 10. Macrophages can release interleukin-1 (IL-1) that enhance the inflammatory response. This occurs when gram-negative bacteria come in contact with the macrophage. 11. Macrophages can identify and engulf microbes or antigens marked with antibodies. The macrophages have receptors on their plasma membrane that match the constant region of the antibody. The process is called opsonisation (Figure 12.7).
CHAPTER 12 227 Biology Term 2 STPM Chapter 12 Immunity Nucleus Phagocytosis IgG Bacterium Complement activation CR1 Fc receptor C3b Phagocyte Figure 12.7 Marking the bacteria (opsonisation) before phagocytosis 12. Another important role is to act as antigen-presenting cells (APC) in specific immunity. (a) In this capacity, the protein in the virus or any antigen after phagocytosis is ‘presented’ i.e. exhibited on the macrophage’s surface membrane. The protein is cut into polypeptides or epitope and combined with the MHC class II protein (major histocompatibility complex II) and shown like bristles all over the surface. (b) ‘Virgin’ helper T cell with the complementary shape TCR can bind with the presented epitope. This binding will stimulate the macrophage to release a cytokine called interleukin-1 (IL-1). This protein IL-1 will cause the helper to release another cytokine called interleukin-2 (IL-2). The final outcome of this series of reaction is to destroy viral DNA in the host cell and also leave behind immunological memory cells that react quickly to kill the same type of virus if they ever infect again. T Cells 2008 1. T cells are lymphocytes that mature in the thymus gland. They are the smallest agranulocytes with spherical shape of 10 μm diameter containing little cytoplasm and a large spherical nucleus. 2. The T cells are divided functionally into four types. They are helper T cells, cytotoxic T cells, inducer T cells and suppressor T cells but physically they are indistinguishable. All have T-cell receptor (TCR) glycoproteins on their plasma membrane and the TCR can be complementary bound with any of the varieties of antigens that may enter the body. Antigen fragment Macrophage Class II MHC molecule T cell receptor Helper T cell (TH) Figure 12.8 Macrophage presents antigen with MHC protein
CHAPTER 12 228 Biology Term 2 STPM Chapter 12 Immunity (a) Helper T cells (i) Helper T cells also known as CD4 cells, have glycoprotein designated CD4. The TCR cannot bind directly with antigen but bind to epitope or antigenic polypeptide that is cut and presented together with MCH class II protein on the macrophage surface as shown in Figure 12.8. (ii) Helper T cells are activated and release cytokine known as interleukin-2 (IL-2) after their TCR bound to presented antigenic polypeptide. This is in response to the macrophage IL-1 causing the helper T cell to release interleukin-2 (IL-2). IL-2 stimulates the bound helper T cells to divide by mitosis called clonal expansion to form million of affector helper T cells and some memory helper T cells. (iii) The effector helper T cells will die after the infection is over but memory cells will not die. Memory helper T cells will response very fast and release IL-2 if the same antigen is presented again in the future called secondary response. So, memory cells have immunological memory. (b) Cytotoxic T cells (i) Cytotoxic T cells are known as killer T cells. They are stimulated by IL-2 when bound to presented antigen. They also respond by dividing forming effector cytotoxic T cells and memory cytotoxic T cells. (ii) Effector cytoxic T cells respond to IL-2 by releasing perforins. These protein molecules can punch holes in the body cells that are infected by the virus. Effector cytotoxic T only release the perforin when it binds to the presented antigen. The infected body cell together with the viral DNA will be destroyed by hydrolases released by the infected body cell when water rushes in to break the lysosome. This is also called apoptosis, a programmed death caused by perforin. (iii) Effector cytotoxic T cells will recognise cancer cells and foreign cells with foreign antigens. Cancer cells have their membrane surface protein changed resulting being attacked by perforins. (c) Inducer T cells Inducer T cells seem to release cytokines to activate all the T cells and even the B cells when antigens are present in the body. They are not well studied and they may be just the helper T cells during the initial response to antigens. (d) Suppressor T cells Suppressor T cells also release cytokines but at the end of infection. The cytokines seem to suppress the activities of B cells as well. Summary Helper T – CD4 cells Cytotoxic T – killer T cells – also known as TC Inducer T Suppressor T T cells
CHAPTER 12 229 Biology Term 2 STPM Chapter 12 Immunity Infected cell CD8 Class I MHC molecule Antigen fragment T-cell receptor Perforin TC cell Pore Ions and water Figure 12.9 Cytotoxic T cell releases perforin to kill infected cell B Cells 1. B cells are lymphocytes that mature in the bone marrow and then spread throughout the body. They can be distinguished from T cells by the presence of receptor protein on the B cell’s outer surface. This protein knowns as B cell receptor (BCR) is a type of immunoglobulin. This specialised receptor protein allows a B cell to bind to a specific antigen. 2. B cells have different functions compared to T cells. B cells are located at different locations in the spleen, lymph nodes and lymphoid tissues. 3. There are two identical binding sites on the receptor that can bind directly to any free antigen example bacterial protein toxin, or any surface protein of microbe or detached cell. There are million types of these ‘virgin’ B cells each with two unique binding sites for any free antigen that may come into the body. 4. After the ‘virgin’ B cells bond to a free antigen at any binding site, the antigen is endocytosised and cut into antigenic polypeptide or epitope. This antigenic polypeptide or epitope then presented on B cell’s surface by attaching to membrane protein. So, B cell can present the antigen on its own. 5. This presented antigenic polypeptide will be detected and bound with helper T cell. The helper T cell will release IL-2 for the B cell to divide by clonal expansion. Free antigens of the same type also can stimulate the B cell to divide. 6. After clonal expansion, two types of cells, the plasma cells and few memory B cells are formed. The plasma cells become much bigger with rough endoplasmic reticulum and Golgi apparatus to release millions of free antibody molecules. The molecules only bind to the specific protein that is detected earlier. 7. The function of B cells is to produce antibody molecules specifically to get rid of a particular antigen. Memory B cells left behind in the body will quickly produce the type of antibody to get rid of the same antigen in the future.
CHAPTER 12 230 Biology Term 2 STPM Chapter 12 Immunity CD4 T cell receptor (TCR) Helper T cell Peptide Class II histocompatibility molecule B cell B cell receptor (BCR) Lysosome Antigen Figure 12.10 B cell can engulf and present the antigen 8. Though there is only one type of B cells but five different types of immuloglobulins molecules are produced. These types are produced for specific purpose as mentioned in earlier section under antibody. Learning Outcomes Students should be able to: (a) explain cell-mediated and humoral immune responses; (b) outline the antigenantibody reactions (precipitation, aggulutination, neutralisation, JVTWSLTLU[Ä_H[PVU Quick Check 1 1. Explain why so many different classes of immunoglobulins are produced. 2. Explain what phagocytes are. Why phagocytes are important in the defence system of the body? 12.2 Development of Immunity 12.2 Development of Immunity Cell-mediated Response 1. ‘Virgin’ T cells cannot bind directly with any unprocessed antigen. When antigen like virus enters the body, macrophage engulfs them. After the virus is endocytosised, it is cut into short polypeptides or epitopes. These epitopes are displaced on the surface bound to major histocompatibility complex (MHC) class II proteins found in the plasma membrane of the macrophage. Therefore, macrophage acts as an antigen-presenting cell (APC). Such cell processes the antigen, forming antigen-MHC complex all over its surface so that T cells can detect it. 2013/P2/Q13 STPM
CHAPTER 12 231 Biology Term 2 STPM Chapter 12 Immunity 2. Virgin T lymphocytes with the right TCR are attracted to the APC and bind with it. However, only macrophage bound to helper T cell releases cytokine or interleukin–1. Cytokine is a complement protein that triggers a cascade of reactions. 3. Cytokine released by the macrophage stimulates the helper T cell to release other type of cytokine or interleukin–2. 4. Cytokine released by helper T cells stimulates cytotoxic T cells and themselves to divide. Each forms a clone of thousands of cytotoxic effector T cells and smaller number of cytotoxic memory cells. Similarly, helper T cells too divide to form both clones of effector cells and memory cells. Inducer T cells also release cytokines to help in the rapid formation of clones. 5. The action of cytotoxic T cells is as follows: (a) Effector cytotoxic T cells recognise and bind with the body cells infected with the virus. Such cells also display the same epitope combined with MHC class I protein to form on the cell surface antigen-MHC complex. (b) As soon as the effector cytotoxic T cell binds with the epitope, the effector cytotoxic T releases another complement protein called perforin. (c) Perforin perforates or punches holes on the infected body cells or tumour cells. Such protein molecules line themselves on specific spots forming large holes on the plasma membrane. (d) As soon as holes are formed in the membrane water rushes into the infected cells. The cell responds by autolysis in which hydrolases are released fom lysosome to digest its ownself, including the viral DNA within. (e) The cytotoxic T cell detaches itself before the infected cell lyses and it starts to attack other infected cells. 6. This is called primary response when a particular pathogen attacks the body for the first time. The sucessful defence of the body usually take about two weeks. The effector T cells then die and left behind the memory T cells which will respond for the second invasion of the same type of pathogen. The cell division process is dampened by suppressor T cells when all the pathogens are killed. 7. It is called a secondary response when a particular pathogen attacks for a second time. In this second attack, as soon as the memory T cells recognise the presence of that pathogen, a very rapid sequence of reactions, as in the primary invasion, takes place and the pathogens are killed without showing any symptom of that particular disease. Each subsequent attack increases the number of memory cells and antibody concentration in the blood. If not, the memory cells might dwindle in number and finally disappear in a few years. 2011
CHAPTER 12 232 Biology Term 2 STPM Chapter 12 Immunity 8. The process of immune response mediated by cells is as shown in Figure 12.11. Exam Tips Remember the development or mechanism of cellmediated immunity. Virus Engulfed by macrophage Antigen - MHC complex TCR Cytotoxic T cell cells divides, forming clone and memory T cells Detached and attack other infected cells Perforins Interleukin-2 Infected cell lysed together with virus within Helper T cell divides forming clone and memory T helper MHC protein at macrophage surface Nucleus Cytotoxic T cells bind to infected body cell Infected body cell Interleukin-1 Figure 12.11 Formation of cell-mediated immune response Exam Tips Remember the development or mechanism of humoral immune response and Figure 12.12. You should be able to list the similarities and KPɈLYLUJLZIL[^LLU[OL[^V immunities. Humoral Immune (plasma cell) Response 1. When the ‘virgin’ B cells meet with free antigens such as protein toxin released by bacteria, the toxin is engulfed by phagocytosis. The antigen is cut into epitopes and they are presented on the plasma membrane as antigen-MHC complex. B cells also posses MHC class II protein that bind with epitope for display. So, B cells can do their own presenting. 2. Helper T cells with their correct TCR can recognise the type of antigen presented and bind with the antigen-MHC complex. The process of binding activates the helper T cell and the B cell prepare to divide. Then, they separate. 3. More free unprocessed antigens bind with the B cell. At the same time, helper T cells secrete cytokine or interleukin–2. The B cell is stimulated by the free antigens and the cytokine to divide quickly to form a large clone of effector B cells and some memory B cells. These effector B cells mature to become plasma cells to produce free antibodies or immunoglobulins (Ig).
CHAPTER 12 233 Biology Term 2 STPM Chapter 12 Immunity 4. Figure 12.12 shows how humoral immune system works. Unprocessed antigen molecules MHC protein Antigen receptor (IgD) on 'virgin' B cell Antigen-MHC complex Antigen endocytosised and processed TCR attached with presented antigen Helper T cell with TCR Interleukin-2 Detached Unprocessed antigen Rapid cell division and differentiation Effector B cell Antibodies in circulation Memory B cell Bacterium destroyed Bacterial toxin Figure 12.12 Formation of humoral immune response Antigen–antibody Reactions There are four ways in which antibody can react with antigen. 1. Neutralisation is the simplest way in which the antibody binds and blocks the activity of the antigen. The following are examples of neutralisation. (a) When the antibody binds to a toxin that is an enzyme, the toxin will be neutralised. The enzyme cannot work after binding with an antibody. Other protein toxins are detoxified when bound likewise. (b) The antibody can neutralise a virus by attaching to the surface protein that the virus must use to gain entry into the host cell. So the virus cannot enter the host cell and later, the virus is removed by phagocytes. Four ways antigen-antibody reactions: (a) Neutralisation (b) Agglutination (c) Precipitation K *VTWSLTLU[Ä_H[PVU F i ib Summary
CHAPTER 12 234 Biology Term 2 STPM Chapter 12 Immunity (c) The antibody molecules can similarly neutralise a bigger microbe such as a pathogenic bacterium. A bacterium coated with antibodies is readily being engulfed by macrophage. This process is called opsonisation where the receptors on the surface membrane of macrophage bind to the constant ends of the antibody and wrap around the bacterium in phagocytosis. 2. Agglutination is the clumping of viruses or bacteria together to form a bigger aggregate. Each IgG antibody has two antigen-binding sites to link viruses or bacteria together. IgA has four antigen-binding sites and IgM has ten antigen-binding sites to link together forming an even bigger aggregate of viruses and bacteria. Such bigger units are easier for phagocytes to engulf. 3. Precipitation is the cross-linking of soluble antigen molecules. After cross-linking, the soluble antigens are precipitated into an immobile insoluble mass. The process is made easier with antibodies that have four or ten binding sites. Precipitates are easily recognised and engulfed by phagocytes. 4. Antigen can be disposed of by complement fixation, for example the activation of complement system by antigen-antibody complexes. The complement proteins have more than 20 serum proteins that are inactive without infection. During infection, a cascade of activation steps will activate the complement proteins. This will results in lysis of attacking viruses and pathogenic cells. There are two ways to achieve this disposal of antigens. (a) Classical pathway IgM or IgG antibodies bind to antigens on pathogen’s plasma membrane. Then, complement proteins are attached to link any two of such attached antibody molecules. A cascade of reactions is activated to form a membrane attack complex (MAC) that penetrates the pathogen’s membrane in step-by-step sequence forming a pore of 7-10 nm in diameter. Ions and water rush into the cell, causing it to swell and lyse. (b) Alternative pathway This is caused by substances that are naturally present on many bacteria, yeast, viruses and protozoan parasites after they have entered into our body. These original chemicals activate our complement proteins to cause inflammation. Histamine is released by basophils and mast cells and histamine increases the permeability of blood vessels. More complement proteins are activated to attract phagocytes to the site. Some complement proteins coat the pathogen cells like bacteria or antigens to facilitate opsonisation. This result in a phenomenon called immune adherence, for example a nonspecific immunity with complement proteins working with antibodies to attract phagocytes to get rid of pathogen cells or antigens.
CHAPTER 12 235 Biology Term 2 STPM Chapter 12 Immunity Antigen Neutralisation Agglutination Precipitation Complement fixation Bacterial cell Complement Antibodies block active sites on viruses and bacterial toxins which means then can no longer bind to receptor sites on tissue cells to cause injury. Particles such as bacteria, viruses or foreign blood cells clump together. Soluble antigens are made insoluble and then settle out of the solution. Foreign cells are tagged for distruction by phagocytes and complement proteins. More B cells are recruited. Leads to rupture of cells. Antibody Antibody-antigen complex Enchances phagocytosis Figure 12.13 Antigen-antibody reactions 12.3 Concept of Self and Non-self 12.3 Concept of Self and Non-self Concept of Self and Non-self and Relation to Tissue Rejection in Organ Transplant 1. This concept states that lymphocytes in the body can distinguish own proteins from foreign ones. Own proteins or cells are not attacked. This is known as immunological tolerance. Foreign proteins or antigens will be attacked and destroyed. 2. When tissue such as skin is grafted from one part of the body on another part, called autograft, there is no rejection. When it is grafted from an identical twin to another, called isograft, there is no rejection too. However, when it is grafted on another person, called allograft, the tissue may be rejected. This applies to organ transplant such as kidney transplant. Learning Outcomes Students should be able to: (a) explain the concept of self and non-self and relate this to tissue rejection in organ transplant; (b) explain the mechanism of immune suppression (HIV infection).
CHAPTER 12 236 Biology Term 2 STPM Chapter 12 Immunity Cell membrane Vital infection MHC class I Antigenic peptide Antigenic peptide MHC class I Antigenic peptide MHC class II Infection cell Antigen-presenting cell uses MHC Class I or II Figure 12.14 Marker of self: MHC class I and class II 3. A healthy immune system is able to distinguish between the body’s own cells, recognised as ‘self ’, and foreign cells, or ‘non-self ’. The body’s immune system normally coexists peacefully with cells that carry distinctive ‘self ’ marker molecule. When the immune system encounters foreign cells or organisms carrying ‘non-self ’ marker, it will quickly launches an attack. 4. The markers are the MHC proteins which are divided into three classes I, II and III. They are coded in the short arm of chromosome 6. MHC class I protein is found in body cell whereas class II protein is found in surface membrane of phagocytes and lymphocytes. Class III proteins are complement proteins. 5. MHC class I protein determines the tissue type or marker of a person. There are many variations of the proteins so are the tissue types. Every one of us has tissue type determined by the MHC class I protein as ‘self marker’. 6. MHC class II protein is found in phagocytes and lymphocytes. During embryonic development, our phagocytes or lymphocytes that react with the ‘self ’ marker or bind with our own MHC class I protein would die or being destroyed. So, our body has no immune cells that destroy our own protein or cells. 7. Foreign proteins (antigens) are recognised by our phagocytes and lymphocytes. These proteins or cells are ‘non-self’, the marker protein are different. The antigens or cells bearing the antigens will be attacked. 8. The phagocytes such as neutrophils and macrophages will engulf the antigens or microbes with such antigens. The antigen or microbes are destroyed by lysosomes within the phagocytes. 2014/P2/Q13 STPM
CHAPTER 12 237 Biology Term 2 STPM Chapter 12 Immunity 9. Antigen-presenting cells such as macrophages or dendritic cells will endocytosise the antigens and cut them into short polypeptide or epitope to combine with their MHC class II protein and present them on the surface. 10. A ‘ virgin’ B cell can bind to the antigen, microbe or transplanted cell with such antigens. The B cells can endocytosise the antigen and cut it into epitope to present them with its MHC class II protein on the surface of the cell. 11. Helper T and cytotoxic T cells will bind with the presented epitope by the macrophage or ‘virgin’ B cells. This will result in cell-mediated response or humoral response to get rid of non-self antigens, pathogen cells or transplanted organs. 12. The following steps can be taken to avoid rejection of transplanted organs: (a) Tissue matching. The tissues of donor and recipient are checked to see whether they matched. Usually tissues type of close relatives are matching or closest to matching. Tissues of donor that are identical to recipient will not result in any rejection. However, the closest possible match may also be used. (b) After transplantation, the bone marrow is irradiated with X-ray to slow down the production of white blood cells. This slows down the rejection, as the number of phagocytes is smaller. (c) Immuno-suppression drug such as cyclosporin is administered to slow down the immune response. The recipient might have to take these drugs for life. (d) More advanced or specific drug such as cylosporin A and FK 506 may be taken to act only on activation of helper T cell. The nonspecific immunity and the humoral response are not affected by the drug. This will cut down the risk of secondary attack by other opportunistic pathogens as the body is weakened. (e) In the near future, organs of transgenic animals that have human markers can be engineered or cultured for human transplant. This will produce a perfect match. Exam Tips Remember the concept of self and non-self with regard to organ transplant or skin grafting, its application in preventing rejection. Quick Check 2 1. What are MHC? How do they control antigens? 2. Why is there usually a need to take immunosuppresing drugs for life after organ transplant?
CHAPTER 12 238 Biology Term 2 STPM Chapter 12 Immunity Mechanism of Immune Suppression (HIV Infection) 1. AIDS is a disease with many symptoms caused by the attack of HIV on helper T (with CD4 molecules on plasma membrane) lymphocytes, B lymphoytes and brain cell macrophages, resulting in loss of immunity. When the patient loses his immunity, many opportunistic diseases will attack him and finally he dies from such diseases. In the first 10 years, after the individual is attacked by HIV, he shows no sign of the sickness except that he is HIV-positive; i.e. having HIV particles and antibodies in his blood and viral genes in the lymphocytes. 2. AIDS originated from a region in central Africa, purportedly arisen from a virus that attacked chimpanzees there. It is unknown whether the virus is exactly identical or a mutated version of the original virus. It is also unclear how it crossed species to infect human beings. 3. The virus was brought to Haiti, then to US and from there to all over the world. Initially, the virus is only affecting the gay communities but now, it is affecting both sexes almost equally. 4. The causative agent was first discovered in US in 1981 and Europe in 1983. The virus was given different names but finally settled as HIV or human immunodeficiency virus. Now, there are two major strains of the virus, HIV-1 and HIV-2. HIV-1 is more widely distributed and more virulent strain. 5. HIV is a retrovirus, a complex type of virus with two molecules of RNA as carrier of genetic materials. It is surrounded by a layer of globular proteins (capsid) and an outermost lipoprotein layer as shown in Figure 12.15. HIV - an enveloped virus with RNA ( x 27 000 000) Glycoprotien (gp120) molecules projecting from the envelope Capsomere Capsid The lipid bilayer is derived from host cell. Viral proteins contribute to the envelope RNA Reverse transcriptase Figure 12.15 Structure of HIV
CHAPTER 12 239 Biology Term 2 STPM Chapter 12 Immunity 6. HIV brings along an enzyme called reverse transcriptase that uses the RNA as a template to form DNA once inside the host cell. 7. The virus is very sensitive to temperature changes and die easily when the temperature dropped below 37ºC. 8. The mechanism of transmission is limited as HIV survives only in body fluid. Therefore, the only way of transmission is through blood and seminal fluid. The usual ways are as follows: (a) Sexual intercourse. Infected seminal fluid ejaculated into the vagina, mouth or rectum can transmit HIV to the partner. At the beginning, most of the transmission is among the gay communities but later, transmission from male to female is most common, though female to male transmission is not 100%. (b) Blood. HIV-infected blood or infected lymphocytes can be easily transmitted to another person. This can be done through the following ways. (i) Sharing of contaminated needle is the main way of spreading the disease among drug addicts. (ii) Contaminated blood received by haemophiliac or recipient was common in the past. (iii) Wound-to-wound contact, even from patient to doctor, was reported. (c) Mother to foetus. This is through the placenta, at around 25-50% chances and also through mother’s milk. 9. Replication of HIV in helper T cell produces the ultimate immune suppression. It occurs as follows: (a) Initially, HIV binds to the receptor at the surface of helper T lymphocyte with the help of coreceptors (CXCR4). (b) Endocytosis occurs as the lipoprotein of membranes of HIV and lymphocyte fuse. Then, viral RNA and reverse transcriptase enzyme enter the lymphocyte. (c) The RNA is used as a template to form DNA by reverse transcriptase. (d) The DNA formed enters the nucleus and incorporated into the DNA of the host cell. This is called the provirus form. (e) Each time the host’s DNA replicates, the viral DNA also replicates. Gradually, most of the newly formed helper T cells have the viral DNA. Viral DNA when replicating produces changes that cause the viral proteins difficult to be detected by the immune cells. (f) The viral DNA remains dormant for at least 6 years, i.e. without any visible symptoms of the disease exhibited. During the first year, HIV particle concentration seems to decrease due to the increase in antibody production. After the second year, the HIV particle concentration gradually increases to full-bloomed AIDS at around the tenth years (Figure 12.16).
CHAPTER 12 240 Biology Term 2 STPM Chapter 12 Immunity (g) HIV particles are produced by the transcription of the HIV gene to form RNA and some portions are translated into viral proteins. (h) Then, viral particles are formed in the lymphocytes and the viruses are released through budding. (i) Finally, the virus is spread to all the helper T cells and killing them, resulting in the destruction of the immune system or immune suppression. HIV gp 120 (a) The receptor of helper T cell binds to viral surface protein. The contents of the virus enter the cell by endocytosis CD4+ cell CD4 receptor Viral RNA CCR5 or CXCR4 coreceptor Reverse transcriptase Doublestranded DNA DNA Viral RNA (b) Reverse transcriptase catalyses the synthesis of a DNA strand. The host cell then forms a complementary strand of DNA, thus forming a complete DNA molecule. Host DNA Viral DNA (c) The viral DNA incorporated into the host DNA and becomes part of the host DNA (d) When the host cell divides, the daughter cells also have the viral DNA. This carries on for six years. The viral DNA is said to be dormant. DNA Viral protein Viral DNA (e) For no reason, the double-stranded DNA directs the synthesis of both HIV RNA and HIV proteins. Viral exit by exocytosis in macrophages Viral exit by cell lysis in T cells (f) HIV particles are assembled and buds out of the host cell by exocytosis. Figure 12.16 Replication of HIV
CHAPTER 12 241 Biology Term 2 STPM Chapter 12 Immunity 10. Immune suppession will results in symptoms as follows: (a) The first sign of infection is the same as an attack of cold and after a few days, the individual recovers. Nothing happens until more than 6 years later. (b) Then, there is rapid loss of weight. This is due to much energy of the body is used to get rid of the HIV. (c) Fever and sweating occur at night. The body temperature increases as the immune response is combating the HIV. (d) The lymph nodes enlarge. Both B and T cells dividing in the lymph nodes. (e) Diarrhoea, tiredness and dementia follow. Simple bacteria taken into the gut can cause diarrhoea. The body has used up all food reserves and some brain cells are infected. (f) Opportunistic diseases like cough, yeast infection in the mouth and sexual organs, pneumonia and Kaposi sarcoma infect the patient. (g) Finally, the patient dies because of such diseases that normally would not kill a normal person with healthy immune system. 11. Preventions of AIDS can be achieved as follows: (a) Use of condom during sexual intercourse prevents the mixing of body fluid. (b) Limiting to one sexual partner prevents infection through sexual contact. (c) Avoid sharing of injection needle cuts down transmission among drug users. (d) Hospitals ensure all bloods received are HIV-free. No infected blood is passed from the infected donor to the recipient of blood or its product. (e) Educate the public on the mode of transmission through mass media. Once the mode of transmission is understood, the chances of infection is cut down to minimum. (f) More screening of people to detect carrier. This ensures the carrier do not pass the infection, without his knowledge. Quick Check 3 1. How do HIV replicate themselves? 2. How can a HIV-positive person be cured? Exam Tips Remember how HIV can cause AIDS symptoms. Remember also the causes, mechanism of HIV infection, symptoms and prevention of AIDS. VIDEO HIV Real Life Stories
CHAPTER 12 242 Biology Term 2 STPM Chapter 12 Immunity 12.4 Immune Disorder 12.4 Immune Disorder Systemic Lupus Erythematosus (SLE) Systemic lupus erythematosus (SLE) is a systemic autoimmune disease (or autoimmune connective tissue disease) that can affect any part of the body, including the heart, joints, skin, lungs, blood vessels, liver, kidneys, and nervous system. The immune system attacks the body’s cells and tissue, resulting in inflammation and tissue damages. Symptoms of SLE 1. Symptoms of SLE vary widely and come and go unpredictably. SLE is always mistaken for other illnesses and it often mimics other diseases. Some people suffering unexplained symptoms of untreated SLE for years as doctors always give the wrong diagnosis. 2. The most common symptoms of SLE is as shown in the Figure 12.17. Learning Outcomes Students should be able to: (a) describe autoimmune disorder (Systemic Lupus Erythematosus (SLE)). %XWWHUÀ\UDVKRQIDFH Muscles –Aches Mouth and nose –Ulcers Psychological –Fatigue –Loss of appetite Face –Butterfly rash Pleura –Inflammation Pericardium –Inflammation Fingers and toes –Poor circulation Systemic: –Low-grade fever –Photosensitivity Joints –Arthritis Figure 12.17 The most common symptoms of SLE INFO Immune Disorder