CHAPTER 10 143 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth (d) Three of the cells die and the one that is left is called primary embryo sac cell. (e) The nucleus then divides by three consecutive mitotic divisions to form eight nuclei. (f) They are arranged to form a matured embryo sac, as shown in Figure 10.18. 5. Insects pollinate the flower and fertilisation occurs later. The flowers have insect-pollinated characteristics. (a) The flower is brightly colour with red petals and with markings to lead insects to find food in the nectary. (b) Each flower is not very big, about 5 cm in diameter but many of them clustered in inflorescences (one stalk with many flowers), big enough for insects to see from far away. (c) The flower is scented, producing fragrance to attract insects. (d) At the base of the ovary, there is a nectary producing nectar. i.e. a sugary fluid as food for insects. Double Fertilisation 1. Double fertilisation is a process in the flowering plant life cycle in which there are two fertilisations; one fertilisation results in the formation of zygote, whereas the second results in the formation of endosperm. 2. Double fertilisation occurs after two male nuclei are produced at the tip of pollen tube after pollination when a pollen grain has landed on the stigma. After the pollen grain is formed, it has one haploid nucleus. Then, the one nucleus divides by mitosis to form two nuclei i.e. the tube nucleus and the generative nucleus. The tube nucleus controls germination of the pollen at the stigma and controls the pollen tube growing in the style to reach the ovule. The generative nucleus divides again by mitosis to form two male (sperm) nuclei (gametes). 3. Double fertilisation occurs in the embryo sac after the pollen tube is penetrated into it. The tip of the pollen tube is digested away by hydrolytic enzymes together with the tube nucleus. The two male nuclei that follow behind, are free to move into the embryo sac. 4. Double fertilisation occurs when there are two fusions of two pairs of nuclei. Each pair is from the opposite male and female parts. (a) The first fusion occurs when the first male nucleus fuses with the egg nucleus after it enters the egg cell to form a zygote. The zygote formed is a cell with diploid nucleus i.e. with two sets of chromosomes. One set from the female and one set from the male. (b) The second fusion occurs when the second male nucleus fuses with the two polar nuclei to form a primary endosperm nucleus. The endosperm nucleus is triploid with three sets of chromosomes. Two sets from the female and one set from the male.
CHAPTER 10 144 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 5. Double fertilisation has the advantage to ensure that the endosperm tissue, the food store, is formed only in fertilised ovules. The embryo formed from the zygote is supplied with nutrients from the endosperm during the development of the seed. 6. Double fertilisation is a process occurs only in flowering plants i.e. phylum Angiospermatophyta. However, double fertilisation has been reported in non-flowering plants Ephedra nevadensis and Gnetum gnemon. The process differs from double fertilisation in flowering plants in that an additional zygote, rather than endosperm, is produced. The second zygote later disintegrates. 7. Double fertilisation like any fertilisation process is to restore the diploid number of chromosomes. This will bring variation in the next generation of flowering plant in the form of seed. The variation is brought about by meiosis through independent assortment of chromosomes from male and female parents and crossing over of chromosome parts of the two parents too. 8. Double fertilisation thus brings about alternation of generations from haploid gametophyte generation to diploid sporophyte generation. Double fertilisation brings about a rapid end to the free pollen grain (microgametophyte) stage and the embryo sac (megagametophyte) stage within the ovule. The seed formed enables dispersal of the plant to far away places and to pass through unfavourable conditions of cold winter season or drought. The Embryonic Development in Seed and Formation of Fruit The embryonic development in seed 2011, 2015 1. Seed development starts immediately after double fertilisation. The triploid primary endosperm nucleus formed starts dividing fast by mitosis. This results in the formation of an embryo sac which is full of nuclei. The embryo sac also expands when the cells within divide and grow. 2. This is accompanied by the partitioning of the nuclei with the thin cell wall, forming endosperm cells or tissue. The endospermic cells provide nutrients for the development of the embryo. 3. The diploid zygote then divides by mitosis, forming a row of cells. This is the result of the plane of division of the cells which orientates only in one direction. 4. Further divisions of cells occur at the tip of the row, depending on whether it is a dicotyledonous or monocotyledonous plant. If it is a dicotyledonous plant, it will form a heart-shaped structure, as shown in Figure 10.19.
CHAPTER 10 145 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Integument Zygote Primary endosperm cell Zygote divides Endosperm cells formed Suspensor basal cell Endosperm cells formed Cotyledon Cotyledon Plumule Radicle Testa Figure 10.19 The development of embryo and seed 5. This causes further expansion of the embryo sac. The endosperm cells may completely disappear, being absorbed by the developing embryo. Some endospermic seeds still retain the endosperm, like cereal grains where most of the food is stored. 6. In most cases, food is stored in the cotyledon. The food stored is in the form of starch in the amyloplasts. Certain amount of oil is also stored, especially in peanut and oil palm seed. In beans, protein is also stored. In cereals, the protein is stored in the aleurone layer just beneath the husk which is the fruit wall. 7. As the seed matures, there is loss of water. Finally, only about 15% of water is left in the seed. The loss of water lowers the respiration rate and seed becomes dormant. 8. The integuments of the ovule form the testa, which has the same genetic content as the parent plant. Usually a pore called micropyle is also formed. Some seeds become stone-like so that they are not destroyed by the animals that eat the fruits or able to withstand digestion even if swallowed. 9. The matured seed has low water content and impervious testa so that it is dormant. Respiration is minimum. This would ensure the seed to be kept alive for a long time even to the extent of several hundred years. 10. Some seeds store up abscisic acid and other inhibitors to remain dormant for a longer period of time before germination can occur. This ensures the seed is dispersed over longer distance.
CHAPTER 10 146 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 11. The significance of the seed formation is to provide a structure that is resistant to unfavourable condition, be dormant until the condition is favourable. The seed is adapted for dispersal and ensures the survival of the next generation with food stored until photosynthesis of the seedling can occur. 12. The wall of the ovary forms the fruit. In some fruits such as coconut and maize, the ovary wall fuses with the integument to form pericarp. The seeds of these plants are intact the fruits. Formation of the fruit 1. Fruit is formed from the ovary aided by hormones. Therefore, fruits are matured ovaries or carpels. Some fruits are formed from the receptacle, the enlarged part of the flower stalk such as cashew nut and apple. 2. The hormones involved are auxin, gibberellin and cytokinin. One or more of these hormones are produced by the developing embryo. Pollination can cause hormonal production in the ovary to form fruit even without fertilisation called parthenocarpy e.g. banana and pineapple. 3. The epidermis of a developing ovary wall has green photosynthetic cells and food is stored within. The ovary slowly thickens to form fruit and the other parts of the flower wither and are shed. Organic acids, alkaloids and starch are also stored. This gives the typical bitter or acidic tastes of most unripen fruits. 4. The type of fruit formed is dependent on the genetic factors of the plants. Some are simple fruits, formed from one ovary of one flower, such as beans, cereal grains and oranges. Some are aggregate fruits, formed from many ovaries of one flower such as raspberry. Others are multiple fruits, formed from many combined ovaries of many flowers such as pineapple and Jackfruit. 5. As the fruit matures, alkaloids produce aroma, starch is changed to sugar, organic acid reduced, producing a typical flavour of ripen fruit. Thus, the fruit tastes sweet and becomes softened due to break down of starch granules and the cell walls. 6. This is aided by ethylene. The fruit changes its colour to bright orange, yellow or other bright colour to attract animals to disperse the seeds. This is due to the break down of chlorophyll into carotenoids in the epidermal cells or the formation of other pigments. 7. When the fruits are ripened, the seeds within are matured for dispersal. Dry fruits mature when they undergo senescence. The fruit of soy bean pod becomes dry and opens to disperse the seeds. 2011 STPM
CHAPTER 10 147 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 8. Certain seeds do not have a fleshy fruit. They are fibrous like that of coconuts. Cereal grains are fruits. The ovary wall forms the husk called pericarp, protecting the seed. 9. Different fruits have different forms adapted for dispersal. Some fruits such as those of rubber tree and Balsam plant are self-exploding to disperse the seeds when matured or dried. 10. Other fruits have wings, are feathery or light so that the wind can blow them to far away places for dispersal. Other fruits have sticky fine bristles that easily attached to the animals’ fur for the same purpose. 10.3 Seed Germination 10.3 Seed Germination The Mobilisation of Nutrients after Imbibition in Seed Germination 1. The first process of germination is the uptake of water into the seed. There are two ways in which water is taken up. One way is by imbibition where water is absorbed into the cell walls, including those of testa cells and through the micropyle, a tiny hole at one side of the seed near the scar left after detached from the ovary wall. This is due to absorption of water by substances within the seed. These include protein, starch and cell wall materials such as cellulose, hemicelluloses and pectic substances. The second way is by osmosis, where water enters the cell through the plasma membrane because of low water potential within. Water also moves from cell to cell by osmosis down water potential gradient. 2. Then, the protoplasm of the cells becomes hydrated. Hydration starts the metabolic process within the cells. Water activates and provides the medium for enzymes to act on soluble substrates, especially those of respiration in which energy in the form of ATP is produced. 3. Auxin, gibberellin and cytokinin are activated or produced by the embryo cells, especially in cereal grains where the hydrated embryo releases gibberellic acid. The hormone diffuses into the aleurone cells to stimulate the synthesis and release hydrolases. The hydrolases are released into the endosperm (scutellum) to hydrolyse the food reserve such as starch as shown in Figure 10.20. 4. Hydrolysis of stored food occurs. Starch forms the major food reserve of most seeds but some seeds have a substantial amount of oil e.g. peanut. In pea and bean seeds, protein is an additional reserve food. This is also the result of activation or synthesis of hydrolases within the storage cotyledon cells of non-cereal seeds. Hydrolysis of starch by _-amylase occurs in the amyloplasts. Fats and oils are hydrolysed by lipase. Proteins are hydrolysed by proteases. Learning Outcomes Students should be able to: (a) explain the mobilisation of nutrients after imbibition in seed germination; (b) state the external MHJ[VYZHɈLJ[PUN germination. Pericarp (Husk) Endosperm Aleurone Embryo Water Starch _-amylase Sugars Gibberellic acid Figure 10.20 Changes within a cereal grain during germination 2014/P2/Q11
CHAPTER 10 148 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 5. This is followed by mobilisation of the food to the growing embryo. The products of hydrolysis such as glucose, amino acids, glycerol and fatty acids are soluble; diffuse and are transported into the cells of the embryo. These are used as substrates for respiration or synthesis of new molecules in the seedling. 6. Expansion and division of cells occur. There is a lot of water absorbed into the seed. The cells inside the seed become turgid and break the soften seed coat. 7. Root emerges from the seed. The meristem at the tip of the radicle divides very fast, resulting the root coming out from the seed. The root system develops and anchors the young seedling in the soil. 8. Shoot then emerges. The emergence of shoot follows two typical patterns: the epigeal and hypogeal germinations. 9. The epigeal germination as in soy bean is shown in Figure 10.21. The hypocotyl, the part of the shoot just below the cotyledons, grows very fast and pulls the cotyledons out of the seed coat and out of the soil. The cotyledons and the shoot become green in colour when exposed to sunlight to begin photosynthesis. 2010 Radicle Testa Hypocotyl Tap root Figure 10.21 Germination of soy bean 11. The hypogeal germination such as maize is as shown in Figure 10.22. Coleoptile Coleorhiza Radicle Leaf Figure 10.22 Germination of maize grain INFO Seed Germination
CHAPTER 10 149 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth The epicotyl, the part of the shoot just above the cotyledon, grows very fast and emerges out from the soil. In maize, it is the coleoptile that emerges from the soil. The first leaf unrolls within and carries out photosynthesis. 2010 External Factors Affecting Germination 1. Water is the fundamental substance required to start the germination. The amount determines the rate and extent of germination. Water plays the following roles. (a) Water is required to provide a medium for metabolic reactions, also to activate enzymes and hormones. (b) Water is required as a substrate for hydrolysis of stored food such as starch and lipid. (c) Depending on the size of the seed, all the cells within the seed must have enough water for the seed to germinate successfully. Subsequently, water is required for the seedling to grow. Example, desert seeds will only germinate after a heavy rain, enough to dilute the abscisic acid in the seeds. So, the seeds are not only germinated but complete the life cycle within a few weeks. 2. Temperature. There is a minimum to optimum temperature range for any species of seed. The range normally is from 1°C to 45°C. (a) Temperature influences the kinetics of chemical molecules within the cell. It provides kinetic energy to increase the rate of reactions. (b) Temperature is of particular importance to the activities of enzymes that determine the metabolic processes and food mobilisation during germination. Maize tends to need a higher optimum temperature compared to wheat. (c) Some seeds such as those of apples, require prolonged exposure to low temperatures before their seeds break dormancy and germinate. This requirement ensures that seeds adapted to temperate climates germinate in the spring rather than in autumn. 3. Oxygen. Oxygen is required for aerobic respiration that break down food molecules, especially glucose to produce ATP as a source of energy. A temporary replacement of anaerobic respiration without oxygen is possible especially before the testa is ruptured. Germinating seeds respire rapidly and they will not germinate in total absence of oxygen. However, rice grains germinate in flooded soil requires hardly any oxygen and can carry out alcohol fermentation to obtain enough ATP for the germination and the seedlings to grow. 4. Time. Some seeds simply would not germinate even given the above three essential factors. They need to have a dormant period before they can germinate. (a) Orchid seeds have not fully developed yet after their dispersal. The seeds need some time to develop and complete formation of the embryo before the seed embryo can germinate. External factors such as temperature affect this period of maturation.
CHAPTER 10 150 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth (b) Beans and apricots that have impervious seed and testa will not germinate as water cannot be absorbed. The seeds need to wait for decay or passage through the animal’s gut for the testa to become pervious to water before they can germinate. Sometimes, physical damage (scarification) to the seed coat can make the seed germinate. (c) The fruits or seeds have inhibitors which prevent germination. Abscisic acid is the usual cause as in ash seeds. Soaking the seeds in water will helps to remove the inhibitor. Another way to break a chemical inhibited dormancy is to soak the seeds in a solution of gibberellic acid, a growth hormone. Tomato seeds contain high levels of abscisic acid which prevent germination of the seeds inside the tomato fruit. (d) Some seeds require pre-chilling or stratification treatment before they germinate. An example is the Rose family (Rosaceae), the seeds must be kept in the fridge overnight or a natural cold spell that mimics winter before they germinate. This is related to lowering the abscisic acid or increasing the gibberellic acid after the cold treatment. 5. Light (a) Some varieties of lettuce seeds require light especially red light for germination. This is related to phytochrome control (discussed in Chapter 9.2). Enough of phytochrome PFR formed after exposure to white or red light stimulates the synthesis of gibberellic acid. Gibberellic acid promotes germination if the seeds are soaked in it. Exposure to far-red light will not cause germination. Some wild temperate seeds like fox glove (Digitalis purpurea) will not germinate when buried too deep in soil. The seeds only germinate in spring when the photoperiod is increasing. (b) Some species of seeds are inhibited by light for germination. This is true with blue light and many hours are required to inhibit germination. The effect of inhibition with light is magnified if seeds are under water stress. In tomato, this inhibition is due to increase in synthesis of abscisic acid and also increase in abscisic acid under blue light. 6. Smoke. There are many wild species of plants in arid areas (heathland) in South Africa, Australia and North America that smoke has stimulatory effect on germination of their seeds. This is due to the presence of an organic compound (Butenolide) in the smoke even at 1 ppm (part per million). Some of these seeds, including pine cones, germinate readily after being burnt by fire. 7. Chemicals (a) Some chemicals especially plant hormones like auxin, gibberellic acid, cytokinin and even ethylene promote seed germination in some species when added. That plant growth regulators are activated when seeds absorb water. Some are synthesised by the embryos after the cells absorbed water.
CHAPTER 10 151 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth (b) Some organic compounds also promote germination in some species of seeds. The organic chemical include chloroform, acetone and alcohol in very small concentrations. (c) Inorganic compounds that promote seed germination in some species include hydrogen peroxide, potassium nitrate and even cyanide. (d) A lot of chemicals including the promoter chemicals inhibit seed germination in higher concentration. These inhibitors are hydrogen sulfide, ammonia, chlorine, sulfur dioxide, aldehydes, organic acids, alkaloids, tannins and short chain fatty acids. 8. Soil conditions. Soil conditions that promote seed germination are related to promoting the supply of water and oxygen like soil compactness, texture and salinity. However, soil conditions that promote seed germination for some species may inhibit seed germination in other species. For example, some variety of rice germinate better in soil with lower oxygen concentration, lower water potential and the presence of microbes including fungi. Other varieties of rice are inhibited for germination under such conditions. 10.4 Growth Curves and Patterns of Growth There are three different types of curves, namely absolute growth curve, absolute growth rate curve and relative growth rate curve. Absolute growth curve 2009 1. Absolute growth curve is a graph of absolute parameter plotted against time. The parameters can be length, height, mass, surface area, volume and number. Absolute growth is actual growth increase in size or mass with time. For example, the mass of mouse is plotted against time (or age) in weeks, as shown in Figure 10.23. Mass/ g Time/ weeks a b c d Figure 10.23 An absolute growth Learning Outcomes Students should be able to: (a) Explain the types of growth curves (absolute growth curve, absolute growth rate curve and relative growth rate curve); (b) explain with examples the patterns of growth (limited growth in humans, unlimited growth in perennial plant, allometric growth in humans, isometric NYV^[OPUÄZOHUK intermittent growth in insect); (c) explain the processes of ecdysis and metamorphosis in insects, and relate the role of hormones (neurosecretory hormone, juvenile hormone and ecdysone) in these processes.
CHAPTER 10 152 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 2. As shown in the graph, the curve is described as sigmoid, meaning S-shaped, and is typical of growth in unicellular organisms, plants and animals. It can be divided into four phases. a: Lag phase. This is the phase with little growth because the organism is young and just introduced to a new environment. It has not adapted to the new condition yet. Cell division and cell enlargement occur at a slow rate. b: Log or exponential phase. During this phase the growth proceeds exponentially. The rate of growth accelerates and at any point is proportional to the amount of materials or number of cells already present. The internal as well as external environments are optimum. The organism involved is well-adapted. There is an unlimited supply of food. In all cases of growth, the exponential increase will eventually decrease. The point at which this occurs is known as the inflexion point. Rate of growth is at its maximum at this point as the gradient is steepest. c: Decelerating or linear growth phase. The fast rate of growth is slowing down because of internal or external factor, on the interaction of both environmental inhibitors or limiting factors are present. One of the reasons is that the condition may not be so optimal. Overcrowding may be a reason. Competition for space and oxygen may occur. There might be accumulation of waste materials as another factor. d: Stationary phase or plateau phase. This is when the maximum growth has been achieved. The maximum size is determined by genetic factor. The overall growth has stop and the parameter under consideration remains constant. There may be some variation that follow depending on species. Rate of cell division is equal to the rate of cell death. There may be a slight increase called positive growth as in fish and some reptiles. It may be followed by negative growth e.g. decrease in mass during which aging occurs before the organism dies as in mammals including humans. At this time, rate of cell division is lower than rate of cell death. 3. The importance of this curve is that it shows the pattern, such as weight at different ages and the extent of growth i.e. the maximum reached. Absolute growth rate curve 1. Absolute growth rate curve is a graph obtained with rate of growth plotted against time. The rate of growth is calculated as change in unit parameter per unit time i.e. dm——dt changes with time (m is the mass of organism). 2. An example is the change in fresh mass of mouse in grams per week of mouse plotted against time i.e. weeks, as shown in Figure 10.24.
CHAPTER 10 153 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Time/ weeks Rate of growth (increase in mass / g per week) Inflexion point a b c d Figure 10.24 Absolute growth rate curve 3. The curve can also be divided into four phases (a, b, c and d) as in the absolute growth curve. The gradient at b shows the rate of increase whereas that of c shows the rate of decrease. The peak indicates the maximum rate of growth. This is the inflexion point that marks the beginning of decrease in rate before the adult size is reached. 4. The curve is usually a bell shape indicating the change of increase in rate is almost the same as the change in decrease in rate. However, the change in increase in rate may be steeper than the decrease in rate. The last phase may be just no rate of growth. It can go up as in positive growth or go beneath the x axis as in negative growth. Relative growth rate curve 1. Relative growth rate curve is a graph of relative growth rate i.e. increase over the previous parameter ( dm——dt . 1——m ) plotted against time, as shown in Figure 10.25. 2. Relative growth rate is the absolute growth rate divided by original measurement or growth in a given time period divided by measurement at start of time period. Relative growth rate = dm——dt = 1 —– m = Final measurement – initial measurement ——————————————–———————— Initial measurement = 1 ————— unit time There is no unit for relative growth rate, it is just expressed as decimal but can be more than 1. 3. Relative growth rate can be expressed in percentage of increase of the previous measurement. For example, relative growth rate can be expressed as gain in percentage of previous mass. It may be more than 100% initially and tapering off when the organism grows older. Relative growth rate = Final measurement – initial measurement ——————————————–———————— Initial measurement = 100%
CHAPTER 10 154 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 4. Relative growth rate takes into consideration of the existing size or mass of an organism. It is more meaningful than absolute growth rate due to the amount of growth depends on size or mass already present. 5. If both a baby mouse of 10 g and an older mouse of 100 g grow 15 g in one week, the baby mouse would grow in a much faster (150%) relative growth rate than the older one (15%). Even though both have grown 15 g, the difference in relative growth rate due to the origin size is taken into account. 6. The curve is a measurement of the efficiency of growth. If the relative growth rate is high, its efficiency is proportionately high at that time. When the drop is steep, the drop in efficiency is big. 7. Four phases (a, b, c and d) are observed as in the graph. Phase a is a slow gain or gentle drop, b is a drastic drop, c is a gentle drop whereas d is stationary. Usually, the highest point is at the beginning of life. Time/ weeks Percentage increase in mass 100 0 a b c d Figure 10.25 Relative growth rate curve 8. A comparison of relative growth rate curves for the same species with another species will show which is more efficient. Similarly, a comparison of organisms grown in different conditions will show which condition is most efficient. Exam Tips Remember the three types of growth curves and the shapes of each of them with reference to a named animal, e.g. human. Quick Check 1 1. Write out the formulae for absolute growth rate and relative growth rate. Patterns of Growth There are five patterns of growth, namely limited growth, unlimited growth, isometric growth, allometric growth and intermittent growth. Limited growth in humans 1. Limited growth, also known as definite or determinate growth, refers to the growth of an organism for a relative period of time before a maximum growth. Thereafter, the organism matures, reproduces, then ages and dies.
CHAPTER 10 155 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 2. During aging, the organism may undergo negative growth and the sigmoid curve is modified. In this graph, there is a dry mass decrease at the begining and at the end. Growth does not continue throughout life. The dry mass at the end is when the organism dies. Mass / kg Adolescent growth spurt Childhood growth Instant growth spurt 80 60 40 20 0 10 15 5 Time / years Conception Birth 20 25 70 50 30 10 Figure 10.26 Limited growth curve 3. A typical exmple of limited growth is human whose growth curve is as shown in Figure 10.26. In humans there are two growth spurts (phase of rapid growth); during infancy and the other during puberty. This is using mass as the growth parameter. (a) During infancy from birth to four years of age, growth rate is rapid and same for boys and girls. (b) During childhood, the rate of growth for boys is faster than girls. (c) At puberty, there is another growth spurt. The rate of growth for girls is faster and start at the age of 12 but ends at 16 years old. Boys start at 14 years old with a faster rate near the end of 18 years old. (d) After adulthood, the growth rate is almost zero. (e) When old, the growth rate is negative starting from about 30 years old. However, the changes of mass can be easily affected by over eating. Unlimited growth in perennial plant 1. Unlimited growth, also known as infinite or indeterminate growth, refers to non-stop growth over a relatively long life-span with unlimited growth, some slight growth continuous until death. 2. A typical example is a perennial plant. Its growth curve consists of a series of smaller sigmoid curves as shown in Figure 10.27.
CHAPTER 10 156 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Time/ years Dried weight/ kg Figure 10.27 Unlimited growth curve 3. Each sigmoid curve represents one year’s growth. Each curve is divided into four parts, each corresponding to a season. 4. In spring, temperature and light intensity are low, there is very little photosynthesis. This results in small increase in height of the plants. 5. In summer, temperature and light intensity are high; the rate of photosynthesis is high. This results in large increase in the height of the plants. 6. In autumn, temperature and light intensity are lower than those in summer. Less photosynthesis occurs and therefore, the increases in the height of plants is small. 7. In winter, there is no growth as there is no or very little photosynthesis. Isometric growth in fish 1. Isomertric growth refers to growth of an organ that is proportional to that of the whole body. So, isometric growth occurs when an organ grows at the same mean rate as the rest of the body. 2. It has the following characteristics: (a) The increase in size of an organism is not accompanied with a change in the shape of the organism. (b) The relative sizes of organs of the body remains constant. (c) The increase in surface area of the body is proportional to the square of its increase in length (A | l 2 ). (d) The increase in volume or mass of the body is proportional to the cube of its increase in length (V | l 3 or M | l 3 ). An animal showing a small increase in overall dimensions with time will show a significant increase in mass. For example, an increase in length of 10% is accompanied by a 33% increase in mass. 3. Examples of isometric growth are fish, as shown in Figure 10.28 and grasshopper, excluding the growth of wing and sex organs, as shown in Figure 10.29. 2013 STPM 2016 STPM
CHAPTER 10 157 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Fish Grasshopper Figure 10.28 0ZVTL[YPJNYV^[OPUÄZO Figure 10.29 Isometric growth in grasshopper Allometric growth in human 2016 1. Allomertric growth refers to growth of an organ that is not proportional to that of the whole body. The organ of the body grows at a different rate from the rest of the body. 2. It has the following characteristics: (a) The increase in size of an organism is accompanied with a change in the shape of the organism. (b) The increase in surface area of the body is not proportional to the square of its increase in length. (c) The increase in volume or mass of the body is not proportional to the cube of its increase in length. 3. A typical example of allometric growth is human. In humans, the growth of the organs such as the brain, thymus gland and sex organ are not proportional to that of the body, as shown in Figure 10.30. Relative growth/ % 200 100 Brain Sex organ Age / years Whole body Thymus gland 0 10 20 Figure 10.30 Isometric growth in human
CHAPTER 10 158 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 4. Human of different ages, especially in the foetus and baby, have the head made up of a large proportion of the body length. This is because of the different growth rate at different ages, as shown in Figure 10.31. There is a change in shape of the body when the foetus develops finally to a matured human. 2 months (foetus) 5 months (foetus) Newborn 2 years 6 years 12 years 21 years Figure 10.31 Development of foetus with the change in shape Intermittent growth in insect 1. It is a discontinuous growth that occurs at certain short periods when the insect is young. 2. This is a variation of the sigmoid growth, producing a step-like curve example of grasshopper, as shown in Figure 10.32. 3. Insect has exoskeleton that limits the growth. Growth only occurs just after ecdysis when the new exoskeleton is still soft. The body can expand and increase in length for the short period of time especially when the insect takes in air or water. Increase in dry weight can still continue after the exoskeleton is hardened but only to a limited extend. Length/ g Time/ days 1st instar 2nd instar 3rd instar 4th instar 5th instar Adult Figure 10.32 Step-like curve Quick Check 2 1. Both arthropods like crab and molluscs like snail have shells. Why do arthropods have intermittent and limited growth whereas some snails have unlimited growth? 2. Is there any advantage of allometric growth compared to isometric growth? Exam Tips 9LTLTILY[OLÄ]LKPɈLYLU[ types of growth patterns with the help of examples and graphs.
CHAPTER 10 159 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Ecdysis and Metamorphosis Ecdysis 1. Ecdysis is a moulting process, in which the old exoskeleton of arthropods is replaced with a new soft layer which later becomes harden again. 2. Ecdysis usually occurs when the arthropods are young and still growing. Examples in the larval stage or nymphal stage of insects. 3. The process is stimulated by environmental factors, such as after eating, temperature changes and day-length or photoperiod. 4. It is controlled by hormones. They are three hormones involved, as shown in Figure 10.33. (a) Thoracic gland stimulating hormone. It is a neurosecretion produced by cerebral ganglion after it receives suitable stimulation. It is released by a pair small bodies called corpus cardiaca found on either side of the thorax near to the cerebral ganglion. Its function is to stimulate the thoracic gland to produce ecdyson. (b) Ecdyson. It is the moulting hormone produced by the thoracic glands after they are stimulated by the thoracic gland stimulating hormone. Its function is to start the moulting process. (c) Juvenile hormone. It is the hormone produced by a pair of bodies called corpus allata (singular corpus allatum). Its function is to ensure a young exoskeleton or characteristic is retained in it after moulting. Cerebral ganglion Neurosecretory cells (neurosecretion) Corpus cardiacum (stores & releases neurosecretion) Corpus allatum (juvenile hormone) Thoracic gland (ecdysone) Figure 10.33 Endocrine system of insects
CHAPTER 10 160 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 5. The process in a bed bug Rhodnius well studied by Wigglesworth, proceeds as follows: (a) The initial stimulation is after a meal of blood, which distends the stomach of the bug. An impulse is sent to the cerebral ganglion to start the production of the neurosecretion called thoracic gland stimulating hormone. (b) It takes four days for the neurosecretion to accumulate to trigger the thoracic gland to produce ecdyson. Ecdyson is a steroid hormone that activates genes to produce protein. (c) It takes another ten days for the ecdyson to accumulate enough in the body to start the moulting process. (d) The ecdyson mainly affects the epidermis. It causes the cells to produce digestive enzymes, which are secreted into lower part of the epidermis. The enzymes digest the lower half of the exoskeleton. (e) The exoskeleton becomes thin and easily cracks or breaks apart. This enables the insect to literally crawl out of its own skin. (f) Other enzymes are also produced by the epidermis, which produces and secretes a new layer of exoskeleton. The colour of the arthropod is yellowish and the exoskeleton is very soft. It takes a day or two for the exoskeleton to harden and regains its original colour. (g) The process is affected by another hormone, the juvenile hormone. Juvenile hormone acts together with ecdyson to produce young skin or exoskeleton, as shown in Figure 10.34. During the young nymphal stage (or larval stage for other insects), more juvenile hormone is produced and the amount gets lesser and lesser when the young becomes older. Finally, when no juvenile hormone is produced, a matured insect is produced. 6. Ecdysis is important to overcome the problem of growth, as the body is encrusted within a rigid exoskeleton that prevents body expansion and increase in mass. When the new exoskeleton is still soft, the body can increase in size by taking air or water, followed by an increase in mass. Metamorphorsis in insects 1. It is a series of progressive steps in the development of the young insects until they become adults. 2. The process is controlled by hormones; a neuro-secretion called thoracic gland stimulating hormone, ecdyson and juvenile hormone. 3. There are two types of metamorphosis in insects i.e. the complete metamorphosis (holometaboly) and incomplete metamorphosis (hemimetaboly). Temperature Cerebral ganglion Neurosecretion Thoracic gland Released in corpus cardiacum Corpus allatum Ecdyson Juvenile hormone Epidermis Epidermis Adult skin Juvenile skin Food Light Figure 10.34 Control of ecdysis
CHAPTER 10 161 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth (a) Complete metamorphosis 2016 t $PNQMFUFNFUBNPSQIPTJTJOWPMWFTUIFGPSNBUJPOPGlarva and pupa. t &YBNQMFTPGJOTFDUTUIBUVOEFSHPDPNQMFUFNFUBNPSQIPTJTBSF butterflies and mosquitoes. t ćFSFBSFfive stages in the development of larva. The period for each stage is called an instar, which is between two ecdysises, as shown in Figure 10.34. Adult Eggs Larva Pupa Figure 10.34 Control of hormones in complete metamorphosis (Example: mosquito) Ecdyson + juvenile hormone Ecdyson Egg First Second Third Fourth Fifth Pupa Imago Instar larval stages (b) Incomplete metamorphosis t *ODPNQMFUFNFUBNPSQIPTJTJOWPMWFTUIFGPSNBUJPOPGnymphal stages without larva and pupa. t &YBNQMFT PGJOTFDUT UIBU VOEFSHPJODPNQMFUFNFUBNPSQIPTJT are cockraches and grasshoppers. t ćFSF BSF BMTP ĕWF JOTUBS TUBHFT CFGPSF UIF OZNQI CFDPNFT adult, as shown in Figure 10.35. Ecdyson + juvenile hormone Ecdyson Egg First Second Third Fourth Fifth Adult Instar nymphal stages
CHAPTER 10 162 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Adult Eggs Hatching 1st moult 3rd moult 4th moult 5th moult 2nd moult Instar 1st stage nymph Instar 2nd Instar 3rd stage nymph stage nymph Instar 4th stage nymph Instar 5th stage nymph Figure 10.35 Control of hormones in incomplete metamorphosis 4. External factors such as temperature, photoperiod and food influence the production of ecdyson that causes the ecdysis. A high concentration of juvenile hormone is produced when the larva or nymph is young. As the larval is getting older, the amount of juvenile hormone produced becomes lesser. At the last larva or nymphal stage, no juvenile hormone is produced. The larva becomes a pupa and reorganisation of tissues occurs inside the pupa to form adult tissues. Similarly, the last nymphal stage becomes an adult. Objective Questions 1. Which is true of gastrula? I The presence of dorsal neural tube to the notochord II The blastocoel is replaced by archenteron III A hollow fluid-filled ball of cells IV The presence of three germ layers A I and III C II and III B I and IV D II and IV 2. Which forms the umbilical blood vessels in human being? A amnion B allantois C yolk sac D chorion 3. Which membrane forms the outermost layer of the human placenta? A Allantois B Amnion C Chorion D Yolk sac 4. Which period has the lowest level of FSH, leutenising hormone, oestrogen and progesterone in the menstrual cycle? A Days 1 to 5 B Days 6 to 14 C Days 15 to 20 D Days 21 to 28 STPM PRACTICE 10
CHAPTER 10 163 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 5. Three hormones and their target tissues are shown in the table below. Hormone Target tissue Aldesteron I Kidney tubule Epinephrine II Mammary gland Prolactin III Sympathetic receptor Which of the following is correct? Aldesteron Epinephrine Prolactin A I II III B I III II C III I II D III II I 6. At birth, which role of hormone is not correct? A Prolactin is secreted by anterior lobe of pituitary gland. B Prostaglandin release is increased in the placenta. C Oestrogen level is lower than that of progesterone. D Oxytocin is secreted by the posterior lobe of pituitary gland. 7. What is the function of oxytocin during pregnancy? A For the dilation of cervix B For the contraction of uterus C For the production of prolactin D For the growth of the mammary glands 8. The flow chart below shows the flow of chorionic gonadotrophin hormone (HCG) through the major blood vessels of a pregnant woman. Uterine vein X Pulmonary artery Y Z Renal artery XYZ A Aorta Vena cava Pulmonary vein B Vena cava Pulmonary vein Aorta C Pulmonary vein Vena cava Aorta D Aorta Pulmonary vein Vena cava 9. The diagram below shows the structure of a cereal seed. 3 6 4 5 Which structure is responsible for producing hydrolytic enzymes after imbibition of water? A P B Q C R D S 10. Which combination correctly describes germination? Type of germination Description A Epigeal The cotyledon is pushed above the ground B Hypogeal The cotyledon remains in the ground C Epigeal The cotyledon remains in the ground D Hypogeal The cotyledon is pushed above the ground
CHAPTER 10 164 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth 11. Which hormone is involved in seed germination? A Auxin C Ethene B Cytokinin D Gibberellin 12. Which combination correctly describes the functions of gibberellin, aleurone and water in seed germination? Component Function I Gibberellin P Synthesises digestive enzymes II Aleurone Q Trigger the synthesis of |-amylase III Water R Releases gibberellin I II III A QPR B QRP C RPQ D RQP 13. The structure of a fertilised carpel of a flower is shown in the diagram below. X Z Y What are X, Y and Z after fertilisation has occured? XYZ A Testa Seed Pericarp B Testa Pericarp Seed C Pericarp Testa Seed D Pericarp Seed Testa 14. Which organism displays an intermittent growth pattern? A Dragonfly C Snake B Human D Fish 15. Which about the cells found in the anther and carpel of lily flower is correct? Diploid Haploid A Pollen grains, embryo sac and fused polar nuclei Pollen mother cells, embryo sac mother cells, male and female gametes B Pollen mother cells, embryo sac mother cells, male and female gametes Pollen grains, embryo sac and fused polar-nuclei C Pollen grains, embryo sac, male and female gametes Pollen mother cells, embryo sac mother cells and fused polar nuclei D Pollen mother cells, embryo sac mother cells and fused polar-nuclei Pollen grains, embryo sac, male and female gametes 16. Which of the following organisms shows allometric growth? A Centipede C Human B Catfish D Grasshopper 17. What does the curve below represent? Daily gain in mass/ g 0 10 20 30 40 50 60 70 Age / weeks A Absolute growth B Unlimited growth C Relative growth rate D Absolute growth rate 18. Which statement about isometric growth is correct? A An organ grows at the same rate as the rest of its body parts. B The change in size of an organism is accompanied by a change in shape of the organism. C The body parts’ relative proportion changes as the organism grows. D An example of an organism displaying this type of growth pattern is human.
CHAPTER 10 165 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Structured Questions 1. The following diagrams show the vertical sections of maize grain and soya bean. Maize grain Soya bean (a) State four structures that are common to both of them. [2] (b) Explain why a maize grain is considered as a fruit. [1] (c) Other than the feature mentioned in (b), state four differences between the maize grain and soya bean. [4] (d) Explain how seed and fruit are formed after fertilisation. [3] 2. The diagram below shows the change in shape of the foetus from the time it is formed to become an adult. 2 months (foetus) 5 months (foetus) Newborn 2 years 6 years 12 years 21 years (a) What type of pattern of growth is shown? Explain. [3] (b) Why is human development also considered to be limited growth? [2] (c) Why are invertebrates considered to have unlimited growth? [2] (d) Give an example of an organism with intermittent growth. Explain why it has such growth. [3] Essay Questions 1. (a) Describe the pattern of growth for (i) humans, [4] (ii) fish [4] (b) Describe how cacti can adapt to their habitat to overcome water loss. [7] 2. (a) The seed completes the process of reproduction in a flower. Describe the embryonic development in the dicotyledonous seed. [13] (b) State two importance of fruits for plants survival. [5]
CHAPTER 10 166 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth Quick Check 1 1. Absolute growth rate = Final measurement – initial measurement —————————————————————— initial measurement = unit time Relative growth rate = Final measurement – initial measurement —————————————————————— initial measurement = unit time or = Final measurement – initial measurement —————————————————————— initial measurement = 100% Quick Check 2 1. This is because of the body of arthropods such as crab is completely covered by the shell and growth can only occur after ecdysis when the shell is soft. The body then increases in size, followed by gain in weight. After some time, the process is repeated, especially during metamorphosis. This will produce intermittent growth. After the arthropod becomes adult, ecdysis stops and it finally dies, producing limited growth. On the other hand, the shell of molluscs such as snail does not completely cover the whole body. The mantle that produces the shell adds in new substances to the shell at the growing periphery. It produces a continuous and unlimited growth. 2. Allometric growth produces a change in the form of the body compared to isometric growth. This change in form allows flexibility in adaptation to the environment as the young develops or metamorphosises. The larva and adult may feed on different food, thus lessen competition between the young and adult. STPM Practice 10 Objective Questions 1. D 2. B 3. C 4. A 5. A 6. C 7. B 8. B 9. A 10. B 11. D 12. A 13. D 14. A 15. D 16. C 17. D 18. A Structured Questions 1. (a) Cotyledon, plumule, radicle and embryo. (b) The outer coat of maize grain is the pericarp formed from fusion of the ovary wall and integuments. (c) Maize grain has endosperm whereas soya bean has no endosperm. Maize grain has one cotyledon whereas soya bean has two cotyledons. Maize grain has coleoptile and coleorhiza whereas soya bean has no such structures. Maize grain has aleurone layer whereas soya bean has no such layer. (d) Seed is formed after the triploid endosperm primary nucleus divides repeatedly to form many nuclei and then cells to supply food for the embryo. The zygote divides first, forming a row of cells, then forms the cotyledon. The embryo is protected by the integument, forming the testa and the seed. Fruit is formed when the ovary wall thickened and later becomes fleshy or fibrous to protect the seed. 2. (a) It is allometric growth. The growth of the head is not proportional to that of the body. In the foetal state, the growth of the head is faster than that of the body. However, the other parts of the body grow faster after birth. (b) It is because human growth occurs before adulthood and stops thereafter. After a period of maximum growth during adolescence, human reproduces and ages and no further growth is observed until death at about 75 years old. (c) Most of them such as corals and molluscs do not stop growing. They do not have a period of maximum growth and continue to grow and reproduce at the same time. (d) Insects. During their development or metamorphosis, they cannot grow continuously as the exoskeleton limits it. Only after ecdysis, when the exoskeleton is still soft can the body grow i.e. expand and increase in dry mass. Essay Questions 1. B J t )VNBOHSPXUIJTBMMPNFUSJDffiUIFHSPXUI of organs is not proportional to that of the body t "OFYBNQMFJTUIFHSPXUIPGUIFIFBEJT very fast during embryonic and infant stages then almost stops after four years old t 4P UIF JODSFBTF JO TJ[F PG UIF CPEZ JT accompanied with the change of shape t ćF JODSFBTF JO TVSGBDF BSFB JT OPU proportional to the square of the increase in length JJ t 'JTIHSPXUIJTJTPNFUSJDffi UIF HSPXUI PG organs is proportional to that of the body t 4P UIFJODSFBTFJOTJ[FPGUIFCPEZJTOPU accompanied with the change of shape t ćF JODSFBTF JO TVSGBDF BSFB JT proportional to the square of the increase in length t ćFJODSFBTFJOWPMVNFJTQSPQPSUJPOBMUP the cube of the increase in length ANSWERS
CHAPTER 10 167 Biology Term 2 STPM Chapter 10 Reproduction, Development and Growth (b) t $BDUJ BEBQU UP EFTFSU IBCJUBU CZ IBWJOH succulent leaves to store water t 4JNJMBSMZ TPNFDBDUJIBWFTVDDVMFOUTUFNTGPS water storage t ćFZIBWFXBYZDVUJDMFUPQSFWFOUXBUFSMPTT t 4PNFIBWFIBJSZTVSGBDFTUPQSFWFOUXBUFSMPTT from the surfaces t 4PNF IBWF MJUUMF BOE OP MFBWFT UP QSFWFOU excessive loss of water t 4PNF IBWF OP TUPNBUB UP QSFWFOU XBUFS MPTT through transpiration t 4PNF IBWF UIFJS TUPNBUB PQFO EVSJOH UIF night for intake of carbon dioxide to minimise transpiration 2. B t &NCSZPOJD EFWFMPQNFOU TUBSUT BęFS EPVCMF fertilisation when pollen tube has grown into the ovule t 'JSTU UIFNBMF OVDMFVTGSPN UIF QPMMFO UVCF fuses with the egg nucleus forming zygote t ćFO BOPUIFS NBMF OVDMFVT GVTFT XJUI UXP polar nuclei forming the primary endosperm nucleus t 1SJNBSZ FOEPTQFSNJD OVDMFVT EJWJEFT immediately forming a mass of nuclei in the embryo sac t ćFTF DFMMT BDRVJSF UIJO DFMM XBMM FBDI UP form endosperm cells to supply nutrients for developing embryo t ;ZHPUFEJWJEFTMBUFSBęFSFOEPTQFSNDFMMTBSF formed t ćF[ZHPUFEJWJEFTJOPOFQMBOFGPSNJOHBSPX of cells t ćF DFMM BU UIF UJQ UIFO EJWJEFT JO EJČFSFOU planes forming a heart-shaped structure t ćFDFMMTBU UIF UXPFOETEJWJEFGPSNJOH UXP cotyledons in a dicotyledonous seed t ćFTUSVDUVSFCFUXFFO UIFTFDPUZMFEPOTGPSN the plumule and the radicle t 'JOBMMZ UIF FNCSZPJTGVMMZGPSNFEXJUI UXP cotyledons, the plumule and the radicle t ćF FNCSZP NBZ VTF VQ BMM UIF FOEPTQFSN cells before forming the seed t ćFFNCSZPMPTFTXBUFSCFGPSFNBUVSFT (b) t 5P QSPUFDU UIF TFFET XIJMF UIF TFFET BSF developing so guaranteeing the survival of the next generation t 5PIFMQJOTFFEEJTQFSTBMGBSGSPNQBSFOUQMBOUT so as to have a better chance for survival
CHAPTER HOMEOSTASIS Concept Map 11 Bilingual Keywords Homeostasis – Homeostasis Liver – Hati <S[YHÄS[YH[PVU¶Ultraturasan 9LHIZVYW[PVU¶Penyerapan semula :LJYL[PVU¶Perembesan 9LN\SH[PVU¶Pengawalaturan (KHW[H[PVU¶Penyesuaian diri Homeostasis Liver Osmoregulation The importance of homeostasis The role of stoma In mammals In plants The structure of liver Carbohydrate metabolism Protein metabolism Types of plant adaptations The homeostatic control system in mammals The physiological and behavioural control in thermoregulation of endotherms The process of ultrafiltration, reabsorption and secretion The role of ADH and aldosterone The regulation of pH of tissue fluid O
Learning Outcomes CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis :[\KLU[ZZOV\SKILHISL[V! H L_WSHPU[OLPTWVY[HUJL VMOVTLVZ[HZPZ" I KLZJYPIL[OL OVTLVZ[H[PJJVU[YVS Z`Z[LTPUTHTTHSZ" J L_WSHPU[OLWO`ZPVSVNPJHS HUKILOH]PV\YHSJVU[YVS PU[OLYTVYLN\SH[PVUVM LUKV[OLYTZ 11.1 Importance of Homeostasis 11.1 Importance of Homeostasis 1. Homeostasis is the maintenance of constant internal environment of an organism even though there are changes in the external environment. The internal environment refers to the tissue fluid and blood in the mammals. 2. Homeostasis is the basic function of an open biological system like cell or organism in which input has to be balanced with output in order for the system to work. The cell or the organism as a whole is exposed to different environmental factors. It is equipped with a ‘modulator’ that sets the different physiological parameters like temperature, pH, water potential and specific ionic concentrations. Then, any variation that affects the predetermined setting would be adjusted for the cell or organism to survive. 3. Homeostasis ensures an optimum internal environment for cells to function, the organism to survive and reproduces efficiently. For example, our body temperature at 37ºC is optimum for all physiological processes catalysed by enzymes. Endotherms therefore respond more rapidly than ectoterms to both external and internal stimuli. They are active even when temperature drops to very low during winter. 4. Homeostasis maintains almost constant condition in the body even though the external environmental conditions vary. For example, even though the external temperature varies, our body temperature is constant. The activities inside the body remain active all the time. 5. Homeostasis enables certain organisms to adapt to extreme environmental conditions such as living in the desert or tundra. For example, kangaroo rat can live in the desert while the polar bear is well adapted in the tundra region. Both animals face harsh environment but they survive and reproduce well. 6. Homeostasis enables the biological system to function efficiently and smoothly with minimum wastage of energy. In the course of evolution, organisms that could not survive efficiently would become extinct like the dinosaurs when the environment changed. Improvements in homeostatic control will determine the long term survival and also evolution of a new species. Homeostatic Control System in Mammals 1. The control system in mammals involves the following organs. (a) Receptors. These are special cells that can detect changes in certain factor and an impulse is sent to the brain. Examples are chemoreceptors that detect changes like water potential (osmoreceptors) located in the hypothalamus and anti-diuretic hormone (ADH) is stimulated to be released in the posterior part Summary Importance of homeostasis 1. Basic function for organism to survive 2. Ensures an optimum internal environment 3. Maintains constant internal despite changes outside 4. Enables certain organisms to adapt to extreme environment 5. Enables the biological Z`Z[LT[VM\UJ[PVULMÄJPLU[S` INFO Homeostasis
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis of pituitary gland. Other receptors that are involved in homeostatic control are located in strategic positions related to their functions like thermoreceptros located in the skin and hypothalamus. (b) Brain. A part of the brain acts as a control centre that receives impulses from receptors, coordinates them and sends impulses to effectors. The hypothalamus is the main control centre for most homeostatic process in our body. (c) Effectors. These are the muscles or glands that correct the changes in the internal environment after receiving impulse from the control centre. Examples are the constrictor muscles and thyroid glands. The effectors restore the changes to the norm i.e. the normal pre-set control reference point. 2. The system controls the following parameters: (a) Body temperature. (b) Water potential of the blood and inside the cells. (c) Glucose and chemical concentration in the blood. (d) pH of the blood and inside the cells. (e) Gaseous concentration in the blood. (f) Rate of heartbeat and breathing. 3. The internal environment of mammals is not static but dynamic, in which certain factors or physiological parameters vary within a certain narrow range. Incoming variation Outcome as original Control reference point (predetermined) Detector Effector Modulator 4. Homeostasis involves two feedback controls: negative feedback and positive feedback. (a) Negative feedback. The negative feedback mechanism has the following features: (i) It is the regulatory mechanism that always being used to restore the body to its original constant state after it is affected by any disturbance. (ii) If the physiological parameters were raised, certain process takes place to reduce it. For example, any increase in our body temperature triggers the negative feedback mechanisms to lower it. (iii) The reverse is true when the physiological parameter is lowered. When our body temperature is lowered in cold environment, thyroxine is produced to raise the metabolic rate and so raises the body temperature. Exam Tips 9LTLTILY[OLPTWVY[HUJL VMOVTLVZ[HZPZLZWLJPHSS`[V HUPTHSZ
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis (iv) The negative feedback mechanism is used more compared to the positive feedback mechanism as it brings about restoration to its constant steady state. (v) The action of the negative and positive feedback mechanisms is as shown in Figure 11.1. Further increase Increased Decreased Further decrease Disturbance Constant condition Disturbance Restoration mechanism Restoration mechanism Constant condition Positive feedback Positive feedback Negative feedback Negative feedback Figure 11.1 Feedback mechanisms (b) Positive feedback. The positive feedback mechanism has the following feedback: (i) It is the mechanism seldom used as it creates further chaos or extremes as shown in Figure 11.1. (ii) If a physiological parameter is increased, positive feedback further increases the factor. For example, when the temperature increases, positive feedback further increases the temperature. This happens as in too high a temperature like 50 ºC, the negative mechanism fails to work and the positive mechanism brings the temperature higher and ultimately, a certain death. (iii) This is also true when the physiological parameter decreases. For example, when the environmental temperature is lowered than –20 ºC, positive feedback also brings the temperature lower and a certain death. (iv) Certain positive feedback helps in homeostasis. The cascade effect of non-steroid hormones is a positive feedback in which it brings about a quick response. For example, adrenaline brings about a quick response to the production of a lot of glucose. (v) However, positive feedback works temporarily. For example, when the blood glucose level increases, more insulin is produced. However, it is only true for a short while. Then, no insulin is produced when excess glucose is converted into glycogen.
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis (vi) Positive feedback can bring about a synergistic effect especially for two hormones to work. For example, under extreme starvation, the low blood sugar level will not only trigger glucagon but also cortisol and adrenaline to enhance each other’s effect. This will maintain the blood glucose level. Thermoregulation of Endotherms 1. Endotherms are animals that can produce heat within the body to regulate their temperature, usually from respiration or basic metabolic rate, such as in mammals and birds. During the conversion of energy from one form to another, energy is lost such as when the energy in glucose is converted to produce ATP. Energy is lost as heat as the conversion is not 100% efficient. This keeps the body temperature of mammals and birds within a narrow range of variations. The old term of homoitherm indicates that their body temperature is more or less constant. 2. In endotherms, the organs that produced most heat are the liver and muscles. The liver carries out a wide range of activities in the metabolism of carbohydrates, proteins, fats and other food substances. The muscles including the heart produce mechanical movement that requires conversion of ATP to produce movement and release heat. 3. Factors that affect body temperature include environmental temperature, hormones such as thyroxine and adrenaline; and muscular contraction particularly striated muscles. 4. When the environment is cold, cold receptors in the skin detect the lowering of the body temperature and impulse is sent to the temperature regulatory centre in hypothalamus of the brain. The hypothalamus also has cold receptors directly detecting the lowering of blood temperature. The brain then sends impulse to the effectors to bring about changes to restore the temperature back to the norm. 5. The physiological controls involved when cold are: (a) Thyroxine is released into the blood from thyroid gland and thyroxine increases the basal metabolic rate of the body. This brings about heat production to increase the body temperature. (b) The subcutaneous blood capillaries constrict (vasoconstriction), resulting in less blood being brought to the skin and lost through radiation, convection and conduction. More blood is sent to the internal organs such as liver, spleen and pancreas where vasodilation occurs in the blood vessels. (c) Hair or feather erector muscles contract. The hair especially the fur or feather layer is raised, trapping air which forms a thicker insulating layer to prevent heat loss. (d) Sweating and panting are inhibited. Less heat is lost as a result of evaporation of water from the skin and the lungs. Exam Tips 9LTLTILY[OLKPɈLYLUJLZ IL[^LLUULNH[P]LHUK positive feedback TLJOHUPZTZ 2014/P2/Q9 STPM
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis (e) Adrenaline is released to produce uncontrollable shievering to generate heat from contracting muscles in extreme cold. This can be observed in small monkeys in the snow. 6. Behavioural control may also be involved when in cold: (a) A simple response is just move away to a place of higher temperature. This is especially for small mammals or birds which bathe in the morning sun. (b) Some birds and mammals sense the coming of winter. They will do the annual migration from the North Pole to the south. Geese and ducks will fly south from northern Canada. Reindeer will also move south from northern Russia. (c) Small birds and mammals will lower their metabolic rate or do the night hibernation. Examples are humming birds up the Andes mountain in South America. Some bats and small rodents also do this diurnal hibernation. (d) Bigger mammals like brown bears will hibernate during long cold winter. Rodents like ground squirrels and hamsters will also hibernate but they do wake up periodically to maintain their blood circulation. They might feed, urinate or defaecate. (e) Birds and some mammals huddle in cold conditions to decrease surface area to decrease heat loss. Penguins and snow monkeys are commonly seen to huddle together when facing extreme cold condition. 7. Endotherms can survive in extreme cold environment through adaptive features such as having a thicker layer of fat, thicker layer of fur and larger body. By having larger bodies, they have a smaller surface area to volume ratio to prevent heat loss. However, they are governed by Allen’s rule that states that species living in colder climates have smaller extremities than related species in warmer climates. For examples, artic foxes has smaller ears than those of tropical foxes. 8. When the temperature is hot, the heat receptor of the skin and thermoreceptors in the hypothalamus detect the rise in temperature and impulse is sent to the appropriate effectors to bring about the physiological control: (a) Thyroxine is not released and the metabolic rate is kept at minimum. Less energy and heat are generated in the body. That is why we feel sleepy when the environment is hot. (b) The subcutaneous capillaries dilate (vasodilation), bringing more heat to the skin to be dissipitated. This is also caused by impulses sent to internal organs including liver, pancreas and spleen to contract, so more blood flows to the skin. (c) Sweating and panting take place so that heat is lost through evaporation. These occur after exercise. (d) The erector muscles of the hair and feather relax. The fur or feather layer is lowered to the minimum so less heat is trapped.
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis 9. The behavioural control when hot is as follow: (a) The mammals and birds just move away from hot places. Birds and monkeys gather under the shade in hot tropical sun. (b) Birds and mammals migrate to the North Pole during summer. Birds from southern hemisphere fly to the north too. Reindeer move to the north in Russia. (c) Bathing in hot afternoon is common both for mammals and birds. Elephants spray themselves with water. Hippopotamus and ducks swim to cool themselves. Some birds do sand bathing to get rid of heat. (d) Dogs and chickens purposely pant to get rid of heat in the hot afternoon. (e) The Cape ground squirrel positions its tail to shade the body from direct rays of the sun. 10. Endotherms living in the tropics have adaptive characteristics of lesser hair and more skinny body. This is governed by Bergman’s rule, which states that animals living in the poles have larger body while those living in the tropics have smaller bodies. Smaller bodies like the tropical squirrels and mouse deer have larger surface area to volume ratio to get rid of excessive heat. 11. Thermoregulation in endotherms is summarised in Figure 11.2. Exam Tips 9LTLTILYOV^ LUKV[OLYTPJHUPTHSZ WYVK\JLOLH[ Hot thermoreceptors in skin and hypothalamus Thermoregulatory centre in hypothalamus Cerebral cortex Cold thermoreceptors in skin and hypothalamus Thermoregulatory centre in hypothalamus Cerebral cortex Skin arterioles vasoconstriction Skeletal muscles shivering Hair erector muscles contract and raise hair Adrenal glands release adrenaline and thyroid gland releases thyroxine causing metabolic rate increase Hypothalamus heat loss centre Hypothalamus heat gain centre Behavioural control: bathing, panting, move to the shade Behavioural control: huddling and expose to sun Skin temperature Blood temperature decreases Blood temperature increase RECEPTORS RECEPTORS EFFECTORS EFFECTORS CONTROL CENTRE CONTROL CENTRE Sweat glands increase sweating Skin arterioles vasodilation No shievering Metabolic rate Increase in skin decrease temperature Decrease in skin temperature nerve impulses nerve impulses nerve nerve impulses impulses hormonal control hormonal control negative feedback Figure 11.2 Summary of thermoregulation in endotherms
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis 11.2 Liver 11.2 Liver The Structure of Liver and the Roles of Components 1. Liver is the largest visceral organ in the body. It weighs about 3-5% of the body weight. It is located below the diaphragm in the right upper quadrant of the abdominal cavity. 2. The liver is separated into a right and left lobe, by the falciform ligament. The right lobe is much larger than the left lobe as shown in Figure 11.3. Left lobe Falciform ligament Right lobe Common bile - duct Portal vein - brings blood from the bowel Hepatic vein - takes blood back to the heart Gallbladder Hepatic artery - brings blood from the heart Figure 11.3 The structure of liver 3. Two blood vessels supply blood to the liver and one drains blood away from it as shown in Figure 11.3. (a) Hepatic artery supplies oxygenated blood to it and hepatic portal vein supplies blood from the gut bringing along most of the digested food to it. The two blood vessels fuse to form many capillaries called sinusoids that connect to the central vein as shown in Figure 11.4. (b) Hepatic vein drains blood away from the liver, bringing deoxygenated blood back to the heart. Secretions including plasma proteins and hormones; including glucose, amino acids and vitamins are sent away from the liver. Wastes like urea and excess mineral ions are also sent away. Learning Outcomes :[\KLU[ZZOV\SKILHISL[V! H KLZJYPIL[OLZ[Y\J[\YL VMSP]LYHUKL_WSHPU[OL YVSLZVMP[ZJVTWVULU[Z" I KLZJYPILJHYIVO`KYH[L TL[HIVSPZTPU[OL SP]LYNS`JVNLULZPZ NS`JVNLUVS`ZPZ NS\JVULVNLULZPZ" J KLZJYPILWYV[LPU metabolism [YHUZHTPUH[PVU KLHTPUH[PVUHUK\YLH MVYTH[PVUPU[OLSP]LY Quick Check 1 1. Define homeostasis. 2. State the role of the components involved in homeostatic control. 3. State the meanings of physiological control and behavioural control. INFO Liver
CHAPTER 11 fl Biology Term 2 STPM Chapter 11 Homeostasis 4. The outer protective layer of the liver consists of two layers. (a) Peritoneum is the outer smooth and moist layer. It prevents frictions against the surrounding structures. (b) Glisson capsule is the inner fibrous layer. It directly protects the inner cells of the liver. 5. The basic internal units consists of lobules and portal regions that are located at the edges as shown in Figure 11.4. (a) Portal regions. These triangular structures where branches of hepatic artery and hepatic portal vein enter into the lobules. Bile ducts are also found inside these regions. (b) Lobules. The lobules are basic functional units which are hexagonal structures. Each lobule is separated from the others by their fibrous membrane. 6. The working cells of the liver are known as hepatocytes. The hepatocytes are arranged in strands radiating from a central vein towards the periphery. The strands of hepatocytes are in close contact with special expandable capillaries called sinusoids. The sinusoids are actually fusion of hepatic arteriole and hepatic portal venule. The hepatocytes are large cuboidal cells, each with big nucleus and many organelles. They have a unique capacity to reproduce in response to liver injury. Liver regeneration can occur after surgical removal of a portion of the liver or after injuries that destory parts of the liver. Hepatocytes are very active cells producing many enzymes to carry out reactions and secrete out many blood proteins. Their secretions and products enter the sinusoids except bile. Central veins collect from sinusoid and join to the hepatic vein. Branch of hepatic portal vein Branch of hepatic artery Bile canaliculus Branch of bile duct Bile ductule Hepatic portal vein Portal region Bile duct Hepatic artery Bile canaliculus Sinusoid Kupffer cells Hepatocyte Liver lobules Central vein leading to hepatic vein Hepatic lacuna Figure 11.4 The structure of liver
CHAPTER 11 flfl Biology Term 2 STPM Chapter 11 Homeostasis 7. Special macrophages called Kupffer cells are attached to the inner wall of the sinusoid for phagocytosising old red blood cells, bacteria and other pathogens. 8. Bile is produced in the hepatocytes and secreted into the canaliculi, which are interconnecting fine network of tubules. The bile moves in opposite direction and emptied into the bile ducts at the portal regions. The bile is channelled into the gallbladder. 9. The gallbladder is located beneath the right lobe of the liver. The primary function of the gallbladder is to store and concentrate bile. Bile is comprised of bile salts, bilirubin, phospholipids, cholesterol, bicarbonate and water. Bile salts mix with ingested fats to promote absorption of fats from the small intestine. Bilirubin, cholesterol, and phospholipids are end products of metabolism. Bicarbonate and water are needed in the small intestine to help neutralise the stomach acid, as digestion and absorption require an alkaline environment. The bile collected in the gallbaldder is released through the bile duct to an opening in the duodenum. Central vein Canaliculus Bile duct Portal region Hepatic arteriole Hepatic portal venule Sinusoid Strand of hepatocytes Figure 11.5 The basic functional tissue within the lobule Carbohydrate Metabolism in The Liver Glycogenesis 2011 1. Glycogenesis is the formation of glycogen from glucose. The process is under the control of insulin. Insulin is released from the β cells of the Islets of Langerhans of the pancreas in response to high blood glucose level after a meal. The hepatic portal vein collects the glucose from the small intestine and sends it to both pancreas and liver. 2. Liver responds to insulin as liver cells have receptors for the insulin. Insulin binds to its receptor (1), which in turns starts many protein activation cascades (2). These include translocation of Glut-4 transporter to the plasma membrane and influx of glucose (3), glycogen synthesis (4), glycolysis (5) and fatty acid synthesis (6) as shown in Figure 11.6. 2014/P2/Q17
CHAPTER 11 flffi Biology Term 2 STPM Chapter 11 Homeostasis Insulin 1 Insulin receptor Glucose transporter-4 Fatty acid 6 Pyruvate 5 4 2 Glycogen Glucose 3 Figure 11.6 Effect of insulin on glucose uptake and metabolism 3. The process of glycogenesis from glucose is as follows: Glucose Glucose-6-phosphate Glucose-1-phosphate UDP-glucose UDP (Glucose)n (Glucose)n+1 Glycogen synthase Phosphoglucomutase Hexokinase ATP ADP UTP PPi Glucose-1-phosphate uridyltransferase Figure 11.7 Process of glycogenesis (a) Glucose is first converted into glucose-6-phosphate by the action of glucokinase or hexokinase. (b) Glucose-6-phosphate is then converted into glucose-1-phosphate by the action of phosphoglucomutase, passing through an obligatory intermediate step of glucose-1,6-bisphosphate. (c) Glucose-1-phosphate is converted into UDP-glucose by the action of uridyltransferase (also called UDP-glucose pyrophosphorylase) and pyrophosphate is formed, which is hydrolysed by pyrophosphatase into 2 molecules of Pi . (d) Glucose molecules are assembled in a chain by glycogen synthetase, which must act on a pre-existing glycogen primer or glycogenin (small protein that forms the primer). The mechanism for joining glucose units is that glycogen synthase binds to UDPG, causing it to break down into an oxonium ion, which is also formed in glycogenolysis. This oxonium ion can be readily added to the 4-hydroxyl group of a glucosyl residue on the free end of the glycogen chain.
CHAPTER 11 flffl Biology Term 2 STPM Chapter 11 Homeostasis (e) Branches are made by branching enzyme (also known as amyloα(1:4)→α(1:6) transglycosylase), which transfers the end of the chain onto an earlier part via α-1:6 glycosidic bond, forming branches, which further grow by addition of more α-1:4 glycosidic units Figure 11.8 1,4 bonds and 1,6 bonds in glycogen molecule 4. Glycogenesis is part of the homeostasis process to lower the blood glucose level. If the blood glucose level increases to more than 100 mg per 100 cm3 of blood, the condition is known as hyperglycaemia. This can lead to fainting as water is drawn out from the brain cells to the blood by osmosis. Hyperglycaemia can lead to glucosuria where glucose is found in the urine. These conditions usually occur as a result of diabetes in which effective insulin is not produced (type I diabetes) or has no effect on the liver (type II diabetes). Glycogenolysis 2011, 2017 1. Glycogenolysis is the breakdown of glycogen, the primary carbohydrate stored in the liver and muscle cells of animals. It is broken down into glucose to provide immediate energy and to maintain blood glucose levels. This occurs when the blood glucose level drops to lower than 90 mg per 100 cm3 of blood. 2. In glycogenolysis, glycogen stored in the liver and muscles is converted first to glucose-1-phosphate and then into glucose-6-phosphate. 3. Two hormones which control glycogenolysis are peptides, glucagon from the pancreas and epinephrine from the adrenal glands. 4. Glucagon is released from the pancreas in response to low blood glucose and epinephrine from adrenal medulla is released in response to a threat or stress. Both hormones act upon enzyme to stimulate glycogen phosphorylase to begin glycogenolysis and inhibit glycogen synthetase (to stop glycogenesis). 5. Another hormone, glucocorticoid or cortisol from adrenal cortex binds to receptor inside the liver cell. It stimulates the synthesis of enzymes of glycogenolysis and enzyme of synthesis of glucose. These occur in extreme starvation to maintain blood glucose level. 2014/P2/Q17
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis 6. Glycogen is highly branched polymeric structure containing glucose as the basic monomer. First individual glucose molecules are hydrolysed from the chain, followed by the addition of a phosphate group at C-1. Then, the phosphate is moved to the C-6 position to give glucose 6-phosphate, a cross road compound. 7. The process of glycogenolysis can be summarised in Figure 11.9. Cleavage of bond linking in glycogen by glycogen phosphorylase yields glucose-1-phosphate, which is converted into glucose-6- phosphate by phosphoglucomutase. Glycogen Glucose-1-phosphate Glucose-6-phosphate Glucose Phosphoglucomutase Glycogen phosphorylase Pi Glucose-6-phosphatase Figure 11.9 The steps of glycogen breakdown in the liver 8. Glucose-6-phosphate cannot exit the cell, and so, in the liver, glucose6-phosphate is transported into the endoplasmic reticulum, where it is converted into glucose by glucose-6-phosphatase. Glucose then exits the cell through the Glut-2 transporter. Muscles have no phosphatase, so glucose is not released from them. The glycogen is used for respiration within. 9. Glycogenolysis is part of the homeostatic process to raise the blood glucose level when it comes down. The blood glucose level has to be raised as not enough glucose sent to the brain cell will cause fainting. Under extreme starvation, glucocorticoid will cause the production of enzymes to break down proteins of the muscles to convert amino acids into glucose. Gluconeogenesis 2011 1. Gluconeogenesis is the synthesis of glucose from noncarbohydrate sources such as lactic acid (lactate), glycerol, fatty acids and some amino acids (that are glucogenic or can be converted into glucose). This occurs when the demand for glucose has exhausted the glycogen store in the liver. 2. Gluconeogenesis is one of the two main mechanisms humans and many other animals used to keep blood glucose levels from dropping too low (hypoglycaemia). The other means of maintaining blood glucose levels is through the degradation of glycogen (glycogenolysis).
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis 3. Gluconeogenesis is controlled by the hormones epinephrine and cortisol. As mentioned earlier, epinephrine is produced during threat and stress, controls both glycogenolysis and gluconeogenesis. Epinephrine stimulates lipolysis in the adipose tissue resulting in the rapid release of both fatty acids and glycerol. These fatty acids and glycerol serve as fuel in the muscles and also activate gluconeogenesis in the liver. 4. Cortisol is produced and released by adrenal cortex. Cortisol is a steroid hormone that stimulates the synthesis of enzymes which break down fat to release fatty acids and glycerol from fatty tissues. If the body is lack of fat, cortisol can cause the synthesis of protein hydrolytic enzymes and the release of amino acids from the muscles. 5. In the liver, cortisol stimulates the synthesis of the enzymes in the conversion of lactic acid, glycerol from fat and some amino acids from protein into glucose as shown in Figure 11.10. Gluconeogenesis 6 ADP 6 ATP Glycerol from lipids Dihydroxyacetone Some amino acids Oxaloacetic acid Glycogenesis Glycogen Store in liver, muscles Glucose-6-phosphate Lactic Acid Pyruvic Acid Glucose Blood ;OLZ[HY[PUNWVPU[VMNS\JVULVNLULZPZPZW`Y\]PJHJPKHS[OV\NOV_HSVHJL[PJHJPKHUK KPO`KYV_`HJL[VULWOVZWOH[LHSZVWYV]PKLLU[Y`WVPU[ Figure 11.10 Steps in gluconeogenesis (a) Conversion of fatty acid and glycerol from lipids to glucose 1. From the Figure 11.11, triglyceride in fat is hydrolysed into fatty acid and glycerol. Then, fatty acid can be broken down into acetylCoA by β oxidation. Then, acetyl-CoA is converted into pyruvate, phosphoenolpyruvate and reversal of glycolysis before changing to glucose. Glycerol is easier to be converted into glucose as glycerol can be converted into triose. .S\JVULVNLULZPZVJJ\YZ THPUS`PU[OLSP]LY^P[OHZTHSS HTV\[HSZVVJJ\YPUNPU[OL JVY[L_VM[OLRPKUL`=LY`SP[[SL NS\JVULVNLULZPZVJJ\YZPU[OL IYHPUZRLSL[HST\ZJSLZOLHY[ T\ZJSLZVYV[OLYIVK`[PZZ\L Info Bio
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis HO HO HO O CO2OPO OH OH HO OH HO OH OH O P O O O O O O O O HO O O POH2C OH Glucose-6-phosphate Fructose-6-phosphate Glucose-6-phosphatase Fructose-1,6-bisphosphatase Fructose-1,6-bisphosphate Dihydroxyacetone-phosphate Glyceraldehyde-3-phosphate 1.3-Bisphosphoglycerate Phosphoenolpyruvate PEP carboxykinase Oxaloacetate Pyruvate carboxylase Pyruvate Glycerol 3-Phosphoglycerate Triglyceride Hydrolysis Fatty acid Aceyl-CoA 2-Phosphoglycerate OH Glucose 2– 3 2– 3 Figure 11.11 Triglyceride is broken down into glycerol and acetyl CoA then to glucose (b) Conversion of amino acids to glucose 1. The conversion of the twenty amino acids to glucose in the liver is complicated as shown in Figure 11.12. 2. Some amino acids are directly converted into pyruvate, others are indirectly converted into acetyl-CoA while other can be converted into the intermediates of citric acid cycle (Krebs’ cycle) before they can be converted to form glucose.
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis Pyruvate Acetyl-CoA Acetoacetate Leucine Lysine Phenylalanine Tyrosine Tryptophan Isoleucine Alanine Cysteine Glycine Serine Arginine Isocitrate Histidine Citrate Citric Acid Cycle Oxaloacetate Proline Glutamine Isoleucine Methionine Succinate Succinyl-CoA Valine Threonine α-ketoglutarate Glutamate Malate Fumarate Tyrosine Phenylalanine Threonine Phosphoenolpyruvate Glucose Asparagine Aspartate Figure 11.12 The conversion of amino acids to glucose (c) Conversion of lactic acid (lactate) to glucose 1. Gluconeogenesis also includes the conversion of lactid acid (lactate) to glucose which is formed during anaerobic respiration when we exercise. This is part of the Cori cycle as shown in Figure 11.13 below: (i) In the muscles, epinephrine at (1) stimulates the enzymes to break down glycogen during exercise. Glycogenolysis at (2) is stimulated to make more glucose-6-phosphate. When the cells become anaerobic, glycolysis (3) continues if pyruvic acid is converted to lactic acid (4). The synthesis of lactic acid requires NADH from glycolysis and produces NAD+ so that glycolysis can continue even in anaerobic condition. The formation of lactic acid buys time and shifts part of the metabolic burden to the liver. Even though not as much ATP can be furnished by glycolysis alone, it is a significant source of ATP when muscular activity continues for any length of time. The final limiting factor in continued muscular activity is the build up of lactic acid. The lactic acid eventually produces muscular pain and cramps which force discontinuation of activity. Usually before this happens and after the activity has ceased, lactic acid diffuses out of the muscle cells and into the blood where it enters the liver. (ii) In the liver, after the lactic acid is sent in through the blood (5), the liver converts it back to pyruvic acid (6) and then to glucose through gluconeogenesis (8). The glucose can enter the blood (9) and be carried to muscles and immediately used. If by this time the muscles have ceased activity, the glucose can be used to rebuild supplies of glycogen through glycogenesis (10).
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis Epinephrine Muscle Cell Glycogen Glycose-6-phosphate Pyruvic acid Liver cell Glycogen Glycose-6-phosphate Gluconeogenesis ATP 1/6 Acetyl CoA Citric acid cycle ATP ATP 5/6 Glycogenesis Glycogenesis Glycogenolysis Glycolysis Pyruvic acid Lactic acid Blood Glucose Lactic acid Lactic acid 10 Figure 11.13 Cori cycle (iii) The Cori cycle also operates more efficiently when the muscular activity has stopped. At this time the oxygen debt can be made up so that the citric cycle and electron transport chain also begin to function again. In order for most of the lactic acid to be converted to glucose, some must be converted to pyruvic acid and then to acetyl CoA (7). The citric acid (Krebs’) cycle and electron transport chain must provide ATP to “fuel” the gluconeogenesis of the remainder of the lactic acid to glucose. Glucose Lactic acid Pyruvic acid To blood and brain Glycogenesis Stored in liver and muscle cells Glycogenolysis Gluconeogenesis Glycolysis Carbohydrates glucose, fructose, galactose Glycogen Glucose-6-phosphate Figure 11.14 The summary of carbohydrate metabolism
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis Protein Metabolism in the Liver Transamination 2011, 2013, 2017 1. Transamination as the name implies, refers to the transfer of an amine group from one molecule of amino acid to another. This reaction is catalysed by a family of enzymes called transaminases. Actually, the transamination reaction results in the exchange of an amine group on an amino acid with a ketone group on a keto acid. It is analogous to a double replacement reaction. Amino acid A R1 C NH2 – COOH + + Amino acid B R2 H NH2 – C – COOH Keto acid B R2 O Transaminase C – COOH Keto acid A R1 O C – COOH Alanine CH3 H NH2 – C – COOH + + Glutamate CH2 H NH2 – C – COOH CH2 COOH a-ketoglutarate C – COOH CH2 O CH2 COOH Pyruvate CH3 O C – COOH Figure 11.15 .LULYHSHUKZWLJPÄJ[YHUZHTPUH[PVU 2. The enzymes involved are transaminases or aminotransferases. These enzymes require pyridoxal-5’-phosphate (PLP) vitamin B6 derivative as co-enzyme. 3. The most usual and major keto acid involved with transamination reactions is α-ketoglutaric acid, an intermediate in the Krebs’ cycle. A specific example is the transamination of alanine to make pyruvic acid and glutamic acid. Other amino acids which can be converted after several steps through transamination into pyruvic acid include serine, cysteine, and glycine. 4. Aspartic acid can be converted into oxaloacetic acid, another intermediate of the Krebs’ cycle. Other amino acids such as glutamine, histidine, arginine, and proline are first converted into glutamic acid. Glutamine and asparagine are converted into glutamic acid and aspartic acid by a simple hydrolysis of the amide group. 5. All of the amino acids can be converted through a variety of reactions and transamination into a keto acid which is a part of or feeds into the Krebs’ cycle. The interrelationships of amino acids with the Krebs’ (citric acid) cycle are illustrated in Figure 11.16.
CHAPTER 11 ffi Biology Term 2 STPM Chapter 11 Homeostasis alanine glycine cystenine serine isoleucine leucine tryptophan asparic acid asparagine tyrosine phenylalanine aspartic acid Oxaloacetic acid Fumaric acid Succinyl CoA Ketoglutaric acid isoleucine methionine threonine valine leucine lysine phenylalanine tyrosine glutamic acid glutamine histidine proline arginine Pyruvic Acid Acetyl CoA Acetoacetyl CoA Citric Acid Cycle Figure 11.16 Amino acids in overall metabolism 6. Once the keto acids have been formed from the appropriate amino acids by transamination, they may be used for several purposes. The most obvious is the complete metabolism into carbon dioxide and water by the Krebs’ cycle. 7. However, if there are excess proteins in the diet, those amino acids converted into pyruvic acid and acetyl CoA can be converted into lipids by the lipogenesis process. If carbohydrates are lacking in the diet or if glucose cannot get into the cells (as in diabetes), then those amino acids converted into pyruvic acid and oxaloacetic acids can be converted into glucose or glycogen. The hormone cortisol from adrenal cortex stimulates the synthesis of glucose from amino acids in the liver and also functions as antagonist to insulin. 8. In addition to the catabolic function of transamination reactions, these reactions can also be used to synthesise amino acids needed or not present in the diet. An amino acid may be synthesised if there is an available “root” keto acid with a synthetic connection to the final amino acid. Since an appropriate “root” keto acid does not exist for eight amino acids, (lys, leu, ile, met, thr, try, val, phe), they are essential and must be included in the diet because they cannot be synthesised. In herbivores, the liver has the enzymes to form these essential amino acids. 9. Glutamic acid usually serves as the source of the amine group in the transamination synthesis of new amino acids as in Figure 11.17. The reverse of the reactions mentioned earlier are the most obvious methods for producing the amino acids alanine and aspartic acid. Several non-essential amino acids are made by processes other than transamination. Cysteine is made from methionine, and serine and glycine are synthesised from phosphoglyceric acid – an intermediate of glycolysis.
CHAPTER 11 ffifl Biology Term 2 STPM Chapter 11 Homeostasis Transamination HO – C – CH2CH2 – C – C – OH NH2 O O R – CH – C – OH O O α-ketoglutaric acid + HO – C – CH2CH2 – CH – C – OH NH2 O R – C – C – OH synthesis of any amino acid Nitrogen bases Keto acid O O O Glutamic acid + Nitrogen bases Nitrogen bases Transamination Figure 11.17 Transamination Deamination 1. Deamination is also an oxidative reaction that occurs under aerobic conditions in all tissues especially the liver. During oxidative deamination, an amino acid is converted into the corresponding keto acid by the removal of the amine functional group and the amine functional group is replaced by the ketone group. Deamination of amino acids results in the production of ammonia (NH3 ). The ammonia eventually goes into the urea cycle. Deamination also refers to the removal of amine group from any organic compounds including bases like adenine. 2. Deamination is known as oxidative deamination where oxygen is required. It occurs in all amino acid but primarily on glutamic acid because glutamic acid was the end product of many transamination reactions as shown below. Water Glutamic acid α-ketoglutaric acid + Ammonia (converted into urea via ornithine cycle) NAD+ NADH 3. The enzyme involved can be amino acid oxidase or glutamate dehydrogenase. The glutamate dehydrogenase is allosterically controlled by ATP and ADP. ATP acts as an inhibitor whereas ADP is an activator. 4. During the reaction, a coenzyme is usually reduced when the amino acid is oxidised to an imino acid by dehydrogenation. The coenzyme may be FAD or NAD depending on the enzyme. The reduced coenzyme can be used to generate ATP.
CHAPTER 11 ffiffi Biology Term 2 STPM Chapter 11 Homeostasis 5. The α-ketoglutaric acid formed is recycled to form glutamic acid and in turns used for the formation of new amino acids. The interconversion of α-ketoglutaric acid and glutamic acid is the central role of nitrogen metabolism. These molecules serve as the “collection and receiving agent” for nitrogen. The subsequent fate of the amino group is in new amino acids, any nitrogen bases, or any other nitrogen containing compounds. Oxidative deamination HO – C – CH2CH2 – C – C – OH NH2 O O R – CH – C – OH O O _-ketoglutaric acid + HO – C – CH2CH2 – CH – C – OH NH2 O R – C – C – OH Any amino acid Keto acid O O O Glutamic acid + NAD+ NADH + H+ NH3 H2O Transamination Figure 11.18 Deamination can be linked to synthesis of new amino acids 6. The keto acids may act as respiratory substrate to produce energy in the form of ATP. They are free to enter other metabolic pathways; several tricarboxylic (citric) acid cycle intermediates are produced in this way. Pyruvate is formed from deaminated amino acid alanine. Thus, oxidative deamination of excess amino acids taken in can be converted into carbohydrates or lipids. 7. The ammonia eventually is converted into the urea through the ornithine (urea) cycle. The process requires ATP and is described in detail later. 8. Ammonia may also be incorporated into other amino acids, nucleic acids and many other nitrogenous compounds. It plays a central role in nitrogen metabolism, as it is a major by-product of protein and nucleic acid catabolism. Ammonia is present in body fluid as both NH3 and NH4 +, and these are in equilibrium according to the equation: NH3 + H+ NH4 + 9. NH3 can diffuse freely across membranes via aquaporins and NH4 + is transported to the liver by an active transport system, the RhB glycoprotein. The blood ammonia concentration is normally below 35 μmol/L as ammonia is neuro-toxic at higher concentrations.
CHAPTER 11 ffiffl Biology Term 2 STPM Chapter 11 Homeostasis 10. The liver is the most important site of ammonia metabolism; it removes much of the toxic ammonia presented to it by urea and glutamine synthesis. By doing so, the liver also plays a major role in the metabolic regulation of systemic pH, because hydrogen ions released from NH4 + during the synthesis of urea neutralise the excess bicarbonate produced by the breakdown of amino acids. Urea is excreted by the kidney, and is normally present in plasma and body fluid at a concentration of 3.0–6.5 mmol/L. 11. Oxidative deamination of amino acids is part of the catabolic process of protein metabolism. This occurs when excess amino acids are taken into the liver. This also can occur during extreme starvation when the muscle proteins are broken down to form glucose or for respiration. Urea formation 1. Urea is formed in the liver and then transported to the kidneys where it is excreted. A small amount of urea is also excreted in the skin as sweat. Urea is the major end product of nitrogen metabolism in humans and mammals. 2. Urea is formed from ammonia. Ammonia, the product of oxidative deamination reactions, is toxic even in small amounts and must be removed from the body. 3. The ammonia goes through a cycle called ornithine cycle to form urea. The cycle is also known as urea cycle. The cycle is shown in Figure 11.19. L-Arginine Citrulline Argininosuccinate Fumarate L-Aspartate ATP AMP + PPi L-Ornithine Urea 2ATP Pi + ADP Pi H2O Carbamoyl phosphate Carbamoyl-phosphate synthetase (ammonia) Ornithine carbamoyltransferase Argininosuccinate synthetase Argininosuccinate lyase Arginase CO2 + NH3 1 1 2 3 4 5 KEY TO ENZYMES 2 3 5 4 Figure 11.19 The ornithine cycle
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis 4. The first step in the reaction is ammonia ion, carbon dioxide, water and ATP react to form a carbamoyl phosphate and ADP. This reaction is catalysed by carbomyl phosphate synthetase. 5. Then, the carbamoyl phosphate reacts with ornithine, an amino acid to form citrulline, another amino acid and giving out an inorganic acid (phosphoric acid). The reaction is catalysed by ornithine transcarbamylase. Both ornithine and citruline are not among the twenty amino acids that form protein. 6. The citrulline formed reacts with ATP and aspartate to form argininosuccinate, AMP and pyrophosphate. This reaction is catalysed by argininosuccinate synthetase. 7. The argininosuccinate formed is broken down by argininosuccinate lyase to form arginine and fumarate. Arginine is a normal amino acid. 8. The arginine formed is hydrolysed by arginase to form urea and ornithine again. The cycle can then be repeated and the ornithine is reused. 9. The overall urea formation is: 2 Ammonia + carbon dioxide + 3ATP ±A urea + water + 3 ADP 10. A complete block of any step in the urea cycle is fatal since there is no alternative pathway for the synthesis of urea. Inherited disorders from defective enzymes may cause a partial block in some of the reactions and results in hyperammonemia which can lead to mental retardation. Extensive ammonia accumulation leads to extensive liver damage and death. Liver cirrhosis caused by alcoholism creates an interference in the enzymes which produce carbamoyl phosphate in the first step on the cycle. 11. Urea is routinely measured in the blood as blood urea nitrogen (BUN). BUN levels may be elevated (a condition called uremia) in both acute and chronic renal (kidney) failure. Various diseases damage the kidney and cause faulty urine formation and excretion. Congestive heart failure leads to low blood pressure and consequently reduce filtration rates through the kidneys, therefore, BUN may be elevated. Urinary tract obstructions can also lead to an increased BUN. In severe cases, hemodialysis is used to remove the soluble urea and other waste products from the blood. Waste products diffuse through the dialysing membrane because their concentration is lower in the dialysing solution. Ions, such as Na+ and Cl– which are to remain in the blood, are maintained at the same concentration in the dialysing solution - no net diffusion occurs.
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis Aspartatearginino succinate shunt of citric acid cycle Urea cycle Citric acid cycle Carbamoyl phosphate α-keto acids NH3 Biosynthesis of amino acids, nucleotides, and biological amines Carbon skeletons Dietary protein Amino acids Intracellular protein Urea (nitrogen excretion product) Oxaloacetate Glucose (synthesised in gluconeogenesis) CO2 + H2O + ATP Figure 11.20 Summary of protein metabolism Other Metabolisms 1. Lipid metabolism. Liver does not store fat but is involved in the processing of lipids. (a) Liver synthesises fats in a process also known as lipogenesis. Liver converts excess carbohydrates into fatty acids and glycerol. Then, they are used to form fat and transported elsewhere where the fat is stored in adipose tissue. (b) Liver synthesises other lipids. Liver can form cholesterol and phospholipids when they are in short supply and the body requires them. (c) Liver breaks down fats. This occurs during exercise and starvation. The fatty acid and glycerol formed are respired with the fatty acid oxidised to acetyl coenzyme A before entering the Krebs cycle. Some fatty acids and glycerol are converted into glucose to maintain the blood glucose level. (d) Liver breaks down other lipids including excessive amounts of cholesterol and phospholipids. However, it may not be able to break down all of them. Cholesterol is converted into bile salts and it may be converted to gall stone. Phospholipids can be respired or converted to other substances.
CHAPTER 11 ffl Biology Term 2 STPM Chapter 11 Homeostasis (e) Liver aids in transport of lipids. Liver produces globulin to transport lipids away from it. Excessive consumption of alcohol results in too much fat being produced from it. The fat is not transported away and results in a condition called fatty liver. 2. Vitamin metabolism regulates body and blood vitamin concentration. Vitamins are not metabolised much in the liver except riboflavin and nicotinic acid for the production of FAD and NAD, respectively. (a) Most of the fat-soluble vitamins like A, D, E and K are stored. Vitamin A is stored to such an excessive amount in the liver that it can be toxic to the liver if an overdosage is taken. (b) Some water-soluble vitamins are also stored but to a lesser degree, such as vitamin B including biotin, folic acid, niacin and vitamin C. 3. Mineral metabolism. Mineral turnover that is the highest in the liver is iron. The iron ions mainly come from the breakdown of old red blood cells and they are temporarily stored in the liver as ferritin. These ions are transported to the bone marrow to be recycled. Other mineral ions stored in the liver in small amount include potassium, copper, zinc, cobalt and molybdenum. 4. Porphyrin metabolism. Liver breaks down haemoglobin obtained from old red blood cells, which are phagocytosised by Kupffer cells. Haemoglobin becomes haem and protein. The protein is hydrolysed to form amino acids that are recycled. The haem group has its ferrous ion separated to form porphyrin. The porphyrin part is first converted to biliverdin, a green bile pigment and later becomes bilirubin, a yellow bile pigment. The bile pigments are excreted into the canaliculi to form bile. If the bile pigments escape into the blood, this results in yellowing of the skin and a condition called jaundice develops. 5. Bile metabolism. Bile is a greenish yellow concentrated fluid produced by liver cells and stored in the gall bladder, before released into the duodenum. The release of the bile is stimulated by the presence of food from the stomach mediated by hormone pancreozymin or cholecystokinin. (a) Bile contains more than 98% water with dissolved bile salts, bile pigments, mineral salts, phospholipids and cholesterol. (b) The dissolved substances are mainly excretory products like bile pigments and excessive mineral ions and cholesterol. (c) Part of the cholesterol is converted into bile salts i.e. sodium glycocholate and sodium taurocholate. The bile salts help in the emulsification of fat i.e. breaking down of fat into fine droplets in the small intestine.