Excel biology 3.pdf

Excel & Succeed Senior Secondary Biology Form 3 Davie G. Nserebo Jacinta Akatsa Harun Mwaura Distributed throughout Malawi by: Grey Matter Ltd P.O. Box 2608 Lilongwe, Malawi Tel: 01755411/01920788 Fax: 01755430 Email: [email protected] Under agreement with the Publishers: Longhorn Publishers Funzi Road, Industrial Area, P.O. Box 18033, Nairobi, Kenya © D. G. Nserebo, J. Akatsa, H. Mwaura, 2012 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the Publisher. First published 2012 ISBN 978 9996 014 401 Printed by English Press, Enterprise Road, Industrial Area, P.O Box 30127-00100, Nairobi, Kenya. Table of contents Unit 1: Problem solving Revision Exercise 1 Unit 2: Investigative skills and techniques Use of glassware in heating substances Use of microscope Revision Exercise 2 Unit 3: Photosynthesis Parts of a leaf Adaptation of leaves for photosynthesis The process of photosynthesis Function of mineral elements involved in photosynthesis Types of pigments in leaves Importance of photosynthesis Plant products Revision Exercise 3 Unit 4: Transport in plants Tissues used for transport in plants Structure of xylem and its functions Transport of substances in the phloem Transpiration stream Factors affecting rate of transpiration Revision Exercise 4 Unit 5: Human Digestive System Nutrients in food Chemical composition of food substances Digestive enzymes Absorption of food substances Adaptation of the small intestines to their functions Functions of the large intestines Functions of the liver related to digestion Problems associated with the digestive system Revision Exercise 5 Unit 6: Human circulatory system Functions of the circulatory system Types of blood cells Role of haemoglobin in oxygen transport Structure of arteries, veins and capillaries Blood clotting process Structure of the heart How the heart works The lymphatic system Problems associated with the circulatory system Revision Exercise 6 Unit 7: Human respiratory system Tissue respiration How gaseous exchange takes place in the lungs and in the tissues How breathing is regulated Carbon monoxide poisoning First Aid for carbon monoxide poisoning Effects of smoking on the lungs Respiratory systems of fish Respiratory systems of insects Adaptations of gills, tracheal system and lungs Revision Exercise 7 Unit 8: Locomotion Types of skeletons Structure and functions of parts of bones Locomotory structures in fish and birds Adaptations of birds for locomotion Revision Exercise 8 Unit 9: Reproduction Structure and function of chromosomes Mitosis Meiosis Structure and function of the human reproductive systems Fertilisation and conception The role of hormones in the menstrual cycle The structure and function of the placenta Importance of breastfeeding Contraceptions Problems associated with reproduction Revision Exercise 9 Unit 10: Human diseases Diseases caused by bacteria Diseases caused by virus Diseases caused by fungi Diseases caused by protozoa Diseases caused by parasitic worms Control and preventive measures of diseases at household and community levels Revision Exercise 10 Unit 11: Human Population Population growth Population growth curves Factors affecting human population growth Problems related to rapid population growth Controlling problems associated with rapid population growth Revision Exercise 11 Unit 1 Problem solving Specific objectives By the end of this unit, you should be able to: (a) Identify problems from a given situation. (b) Suggest possible solutions to problems. (c) Test selected solutions and draw conclusions. (d) Evaluate evidence and make decisions. (e) Write reports. Introduction In Form One, we learnt about problem solving. We learnt that problem solving involves the use of scientific findings to come up with solutions to problems in our day-to-day lives. In Form Three, we will apply problem-solving skills to solve various problems in our surroundings. Problem solving process involves: (a) Identification of problem In Form One, we learnt that problems are identified by observing situations in our day-to-day lives. (b) Definition of the problem to be solved After a problem is identified, it must be put in form of a statement. For example, the problems identified in Hema village can be stated as follows: Poor health of children in Hema village. Deteriorating health of children of Hema village. Definition of the problems gives a reason why a given investigation should be carried out. (c) Gathering information This involves reading about the problem from books, magazines, journals and talking to experts to gather information about the problem. In this case, one can read on disease symptoms from books or visit the local health centre to ask the Health Officers to explain the problem in a professional way. In gathering information about the problem, one asks the following questions: Where else does the above problem occur or in which other place was such a problem observed? What other observations are made when such a problem occurs? What are the probable causes of the problem? What are the probable solutions to the problem? (d) Suggesting possible solutions After the information is gathered, it is analysed. In the analysis, information similar to the situation observed is identified. For instance, in which other area was such a problem observed? Possible solutions to the problem are then listed down. From the possible solutions, the ones more applicable to the situations observed are identified. (e) Testing the possible solutions Testing involves experimenting whether the solution suggested can really work to solve the problem. This is first done in a small group called a sample. If it works in a sample, then the solution can be implemented in the whole population. In testing the possible solutions the following are done: 1. Choose a sample population where the test will be carried out. Identify resources to be used in the test. For instance, educational materials, resource persons such as nurses and food items such as meat, milk and fish. 2. Identifying the variables to be used, what variable will be changed, what variable will be kept constant and what variable will be observed. 3. Choose the method to be used in the test. This is the systematic way of carrying out the test. (f) Conclusions From the data collected, a conclusion is made as to whether the suggested solution can really work. In the conclusion, the following are considered. Some factors that were not considered. Weaknesses of the method. Obstacles observed. (g) Evaluation of the evidence This involves evaluating the evidence in the conclusion to make decisions. The evaluation involves comparing the advantages and disadvantages of the solution. This is in terms of: How effective it can solve the problem. How expensive the method is. In this case, you consider whether funds can be available to solve the problem. The benefits of the solution. How was it received by the beneficiaries. (h) Making decisions This involves making a choice on whether to: - Accept the solution and implement it as it is. - Choose the solution but make some changes on the method before implementing it. - Reject the suggested solution and propose an alternative one. - After the decision is made, reasons as to why it is the best decision are given - Consequences of the solution are also stated. These include possible reactions after the solution is implemented. (i) Report writing After a decision has been made, a report is written. The report gives a guideline on how the problem was identified and tested. The report shows that the suggested solution can really work to solve the problem in the village. The report should have the following parts: (a) The aim of the investigation. (b) Description of the method used. (c) Presentation of results. (d) Drawing of conclusions. Activity 1.1: You are provided with the following sets of situations observed in Malawi. Study them and design a problem solving method. Under each case, write a report. Case 1 Most of Malawi was originally covered by forest but over the years, people have been cutting down the trees and burning them to open up areas for farming. This is commonly known as “slash and burn” agriculture. In the past, these areas were farmed for one to three years and then the farmers would move on, cut down some more forest and start all over again. When the population of malawi was small, the environment was able to recover as the trees would regenerate. Currently, the population of Malawi is estimated to be 15 million. This has increased pressure on available land. Today, more than 80% of Malawians live in the rural areas and are subsistence farmers. Case 2 Plastic bags are useful tools but they are becoming a menace in our urban centres. A solution is needed for this problem. Activity 1.2: Observe the environment in your school or around your school. Write down several problems that you can note. Choose one problem and design a problem solving mechanism. Write a report. Revision Exercise 1 1. Explain the importance of carrying out an investigation before implementing a suggested solution to a given problem. 2. Write an essay on important points to note in problem solving process. 3. Discuss obstacles that one faces when designing problem solving mechanisms. Unit 2 Investigative skills and techniques Specific objectives By the end of this unit, you should be able to: (a) Observe safety measures when doing investigations. (b) Use scientific equipment and materials safely. (c) Take measurements and make detailed drawings from observations. (d) Design investigations. (e) Do projects and write reports. Introduction In Form One, you learnt about the use of glassware in carrying out experiments. You also learned about using burners safely, safety measures in laboratory, using microscopes, taking measurements and carrying out investigations. In this unit, we will apply the skills learnt in Form One to carry out various activities. Use of glassware in heating substances Carry out the experiment below and answer question that follow. Activity 2.1: To determine whether chlorophyll dissolves in water Materials and apparatus Boiling tubes Test tubes Alcohol Water Bunsen burner Test-tube holders Wire gauze Leaf Procedure: 1. Take a 500 ml beaker and half fill it with water 2. Place it on the tripod stand and place a Bunsen burner with a good flame underneath. 3. Heat the water up to boiling. 4. Put alcohol into the boiling tube. 5. Put the leaf into the alcohol and ensure it is totally subemerged. 6. Dip the boiling tube with the alcohol and the leaf into the beaker with boiling water. 1. Heat for about 10 minutes. 2. Repeat the procedure using water instead of alcohol. 3. Record your observations. 4. Write a report on the above experiment. Questions 1. Suggest reasons why alcohol in the experiment was not heated directly. 2. What name is given to the hot water in the beaker that was used to heat the alcohol? 3. When removing the boiling tube from the hot water which equipment did you use? Discussion From the experiment, you may have realised that in heating substances using glassware such as beakers, a wire gauze is used. The beaker is placed on the wire gauze to protect the glass from direct flame. In heating alcohol, the boiling tube is dipped into a beaker with hot water. Alcohol is flammable and thus it cannot be heated directly. It can only be heated by use of a water bath. The beaker with hot water is called a water bath. From the experiment, you may have noted that when a leaf is boiled in alcohol, it becomes decolourised. However, when boiled with water, it does not decolourise. This shows that the green colour of leaves is soluble in alcohol but it is insoluble in water. Assembling glass apparatus in an experiment In some experiments, more than one glassware are used. The glass are connected together to form a set of apparatus. Tubes are used to connect one glassware to another. The mouth of glassware are closed by use of corks. Corks are made of soft materials such as rubber that fits tightly onto a glass without causing breakages. Tubes used to connect two glasses may be made of rubber or glass. Let us carry out an experiment to investigate production of gas in fermenting sugar. Activity 2.2: To investigate gas production when sugar is fermented Materials and apparatus Glass tubes Conical flasks Glucose solution Yeast Corks Lime water Boiling tube Thermometer Procedure 1. Put about 50cm3 of distilled water into a conical flask. Warm the water up to 37oC. Add 20 grams of glucose or sucrose and stir the mixture. 2. Add about 10 grams of yeast powder into the water. 3. Cork the conical flask with a cork having one hole. 4. Fit a U-shaped glass tube into the hole on the cork down into the conical flask. The glass tube should not touch the yeast solution in the conical flask. 5. Put limewater into a boiling tube upto half full mark. 6. Dip the other end of the glass tube from the conical flask into the limewater. 7. Allow the set to settle on a bench and observe for 30 minutes. The set-up is as shown below: Questions 1. What precautions did you carry out when setting the above apparatus? 2. What was the importance of using warm water instead of cold water? 3. What is yeast? 4. Explain the observations that were made from the experiment and write a report. Discussion In the experiment, warm water was used to increase the rate of activity. When temperatures are raised in chemical reactions, the rate at which chemical process take place also increases. Yeast is an organism. It is in the same group as fungi. It acts on sugar causing fermentation. As you may have noted, a gas was produced by the yeast in the conical flask. The gas rose up and passed through the glass tube into the lime water in the boiling tube. The gas escaped from the lime water in form of bubbles. As the gas was escaping, it caused the lime water to turn milky. This means that the gas produced was carbon dioxide. Use of microscope In this section, we will use a microscope to observe onion epidermal cells and organisms found in pond water. Activity 2.3: To examine cells from an onion epidermal cell. Materials and apparatus Microscope Slide Water Cover slips Scalpel Onion bulb Procedure 1. Remove a leaf from an onion bulb. Bend the leaf until it breaks. At the breaking point, check for a thin transparent structure (paper like) that appears. This is an epidermis. 2. Remove the epidermis using forceps, and place it on the slide. 3. Add a drop of water onto the tissue and cover carefully using a cover slip. 4. Mount the slide onto a microscope. 5. Focus the microscope and observe using low power lens and medium power lens. 6. Count the number of cells that are lined up from one end of the field of view to the other end when examined under: Low power lens Medium power lens 7. Draw two adjacent cells that you have observed. 8. Calculate the magnification of the observed images under: Low power lens Medium power lens 9. Write a report of your findings. Questions From the findings you have obtained from the experiment above, answer the following questions 1. Fill in the table below with the results you obtained from the experiment Objective lens used Lower power Medium power lens lens Magnification of the objective lens Magnification of the eye-piece lens Total magnification Number of cells counted across the field of view 2. Compare the total magnification with the number of cells observed across the diameter of the field of view Discussion From the experiment, you may have realised that at low magnification, many cells are seen across the field of view. As the magnification increases, the number of cells seen across the field of view are fewer. You may have noted that at low magnification, the cells do not show a lot of details. As the magnification is increased, more details are seen in a cell. Activity 2.4: To examine organisms in pond water Materials and apparatus Pond water Microscope Slides Cover slip Dropper Procedure 1. Place a drop of pond water onto a slide. 2. Lower the cover slip gently onto the drop of water. 3. Mount the slide onto a microscope. 4. Examine the specimen using low power and medium power objective lenses. 5. Identify organism that move and those that do not show any signs of movement. 6. Identify the structures used in movement by the organisms that you have observed. 7. Make a drawing of one organism you have observed. 8. Calculate the magnification of the image observed. Discussion From the activity, you may have realised that the microscope helps us to see tiny organisms that live in water that we cannot see with unaided eyes. Some of the organisms move freely in water by use of long whip like structures while others do not move. In the pond water, tiny plants such as spirogyra and algae are also observed. Taking measurement In this experiment, we will take measurements of different specimens. From the measurements, we will determine surface area, volume, length, force and mass. Activity 2.5: To determine volume and weight of different food materials of the same size Materials A piece of potato A piece of sugarcane Maize seeds Bean seeds Procedure 1. Design an experiment to compare the mass of 50 beans seeds with the mass of 50 maize seeds. 2. Write a report on your findings. Carrying out investigations In Form One, you learnt how to carry out scientific investigations. You learnt that every investigation has an aim, a hypothesis and a method to be used in the study. The results obtained from the observation are recorded and the data analysed to come up with a conclusion. After completion of an investigation, a report is written. Revision Exercise 2 Carry out the following activity A student noted that in their homestead, the kitchen was invaded by red ants at night. He decide to investigate and find out locally available substances that can be used to keep off red ants in the kitchen. He decided to use salt, naphthalene and ash. Instructions 1. Design an experiment that the student could use. Indicate the following: Aim of the investigation. Variable to be kept constant. Variables to be observed. Hypothesis. Procedure. Record expected observation. Present your observation in form of tables or graphs. Analyse the information and draw conclusions. 2. Write a report about your work. Unit 3 Photosynthesis Specific objectives By the end of this unit, you should be able to: (a) Label a diagram of a cross section of a leaf as seen through a light microscope. (b) State the functions of the parts of a leaf. (c) Explain the adaptations of leaves for photosynthesis. (d) Label a mesophyll cell as seen through an electron microscope. (e) Explain the functions of the parts of a plant cell. (f) Describe the process of photosynthesis. Review of photosynthesis In Form Two, you learnt that all green plants make their own food in a process called photosynthesis Photosynthesis is a process whereby green plants use light energy to make organic food substances by use of water and carbon dioxide. The process of photosynthesis takes place in all green plants and other organisms such as the green algae that contains chlorophyll in their cells. Parts of a leaf Activity 3.1: Observing a cross-section of a leaf Materials Pieces of carrot, scalpel or sharp blade, microscope slides, microscope cover slip, light microscope, water, dropper, mounting needle, dish for example petri-dishes or glass dishes, a fine brush, young leaf from a dicotyledon such as black jack, peas, beans and monocotyledons like maize, grass among others. Procedure 1. Take a carrot, wet it then slice it vertically halfway down the middle. 2. Insert the leaf blade into the slit made in the carrot. Make sure that the midrib of the leaf is placed vertically along the centre of the carrot. Trim off any protruding part of the leaf. 3. Hold the carrot in one hand, and cut several thin sections quickly and smoothly with a scalpel using the other hand. 4. Put the sections in water in a petri-dish. 5. Take a slide and put a drop or two drops of water on it using a dropper. 6. Select the thinnest section of a leaf preferably one that is cut through the midrib and place it into the 7. drop of water. Use a fine brush to transfer the sections to avoid damaging them. 8. Using a mounting needle, carefully place a cover slip on the section. Make sure no air bubbles are trapped. 9. Using tissue paper, wipe off excess water from the slide before observing it under the microscope. First use low power then medium power. 10. Try to identify the following layers of tissue: upper epidermis, palisade layer, spongy mesophyll, vascular bundles, lower epidermis and stomates. 11. Use a sharp pencil to draw the outline of the layers of tissues you have seen under low power magnification. Label the layers. 12. Under the medium power, examine and draw a cell from each of the following layers: upper epidermis, palisade layer, spongy mesophyll and lower epidermis. What do you notice about the shape of each cell? What do you notice about the arrangement of the cells in the layers? Are they closely packed or loosely packed? The transverse section of a leaf is as shown in Fig. 3.1 below. Discussion From activity 3.1, you may have identified the following parts of a leaf. Epidermis Mesophyll Air spaces Stomates Vascular bundles. We will now look at the functions of the above parts of a leaf. Functions of parts of a leaf Cuticle It covers the upper surface of most leaves and makes them appear shiny. The cuticle is thin and transparent to allow light to pass through. It has a waxy material that protects the leaf from attacks by bacteria. The waxy material also acts as a waterproof layer which prevents excessive loss of water from the leaf. Upper epidermis This is the layer of cells below the cuticle. It is usually one cell thick to allow light to pass through the cells easily. Cells of the epidermis do not have any chloroplasts, this allows them to remain transparent for light to pass through. The epidermis forms a protective layer over the cells that carry out photosynthesis. Fig. 3.1: Transverse section of a leaf. Palisade mesophyll This is a layer of cells located below the upper epidermis. Optimum photosynthesis takes place in the palisade mesophyll. Palisade cells are closely packed with few air spaces between them. The cells are elongated and lie at right angles to the leaf epidermis. They contain many chloroplasts. Their shape allows them to absorb most of the light falling on the leaf. They are close to the upper epidermis so as to absorb more light. The chloroplasts can move within the palisade cells to the side receiving the highest amount of light. Spongy mesophyll It is composed of cells located between the palisade mesophyll and the lower epidermis. Spongy mesophyll cells are also lined with moisture to facilitate uptake of oxygen and release of carbon dioxide gases. They have fewer chloroplasts than the palisade mesophyll cells. Suggest why the mesophyll is described as spongy. The cells are irregular in shape and are loosely arranged. They have large air spaces between them which allow air circulation and gaseous exchange between the cells and the air surrounding them. Vascular bundles The network of veins in the leaves is made up of vascular bundles. This tissue has vessels which supply water and mineral salts to the leaf. These are called xylem. Other vessels also take away manufactured food substances from the leaf to other parts of the plant. These are called phloem. Stomates Stomates are pores within guard cells. They are found on the upper or lower epidermis or both. They allow entry of carbon dioxide into the leaf to be used for photosynthesis. Adaptation of leaves to photosynthesis An adaptation is a characteristic that enables an organism to function properly. Let us carry out an experiment to investigate external features of a leaf and learn how they adapt the leaf to photosynthesis. Activity 3.2: To identify external features of a leaf that adapt it to photosynthesis Materials A variety of leaves for example grass, Hibiscus, bean leaves, Bougainvillaea among others. Procedure 1. Examine each leaf and list the observable characteristics that make it suitable for photosynthesis. As a guide for your observation, examine the following about the lamina. (a) Breadth. (b) Thickness. (c) Colour. (d) Structures that supply the leaf with water and minerals. (e) Structures that transport manufactured food from the leaf. (f) The firmness of the leaf 2. Draw and label the observable external features of the leaves. Questions 1. List the features that are common in all the leaves. 2. State how the features mentioned in (1) above adapt the leaves for photosynthesis. Discussion In your examination of the different types of leaves, you may have noticed that their shape is broad, thin and flat. The figure below shows the external parts of a leaf. Fig. 3.2: External parts of a leaf. Adaptations of leaves for photosynthesis From the previous activity, you may have realised that the leaf has the following adaptations. (i) The leaf is green in colour. This is because it contains a green colouring matter called chlorophyll. Chlorophyll traps light energy which is used in the process of photosynthesis. (ii) The leaf surface is broad. This increases the surface area for absorption of light and carbon dioxide for photosynthesis. (iii) Most leaves have thin blades. This reduces the distance across which carbon dioxide diffuses from the stomata to the photosynthetic cells. (iv) Leaves have a network of veins that have branches all over the leaf. The veins have vascular bundles. These vasular bundles transport water and mineral salts to the leaf cells and manufactured foods from the leaf cells to all the other parts of the plant. (v) The mesophyll cells are irregularly arranged leaving air spaces that allow efficient exchange of gases in the leaf cells. (vi) The cuticle covering the leaves is transparent. This allows light to penetrate into the leaf cells. (vii)The leaves have small pores on their surfaces. These allow air to move in and out of the leaf cells. Parts of a mesophyll cell and their functions Cell structure as seen under the electron microscope Figure 3.3 shows a mesophyll cell as seen under the electron microscope. It has the following parts. Protoplasm Protoplasm includes all the cell contents except the vacuoles and the material that is released or taken in by the cell. The cell membrane is also part of the protoplasm. The cell wall The cell wall is the outermost part of most plant cells. It is made of a chemical substance called cellulose. Cellulose is tough and resists stretching. The cell wall gives firmness and a fixed shape to a plant cell. This is due to the presence of cellulose. The cell wall has pores called plasmodesmata through which the cell exchanges materials with its environment. These pores also allow movement of substances between cells. The cell wall also protects the cell from bursting. This is similar to the leather covering of a football which prevents the inner tube from bursting due to high air pressure. Fig. 3.3: A mesophyll cell as seen under an electron microscope. The cell membrane The function of the cell membrane is to hold or enclose the contents of the cell. It also regulates the movement of materials in and out of the cell. Cytoplasm Cytoplasm is composed of all the cell contents except the nucleus. The fluid and semi-fluid part of the cytoplasm is called cytosol. Organelles and insoluble granules of various kinds are suspended in the cytosol. Cytoplasm also contains many dissolved substances like food nutrients, mineral ions, dissolved gases and vitamins. Cytoplasm keeps moving about in the cell. Many chemical processes are carried out in the cytoplasm by the different organelles suspended in it. Nucleus The nucleus is large and oval or spherical in shape. The nucleus is made up of nuclear material called nucleoplasm. It is enclosed by a nuclear membrane. The nucleus controls the activities of a cell and heredity (passing on traits from parents to offsprings). Ribosomes Ribosomes are small, spherical organelles. Some are attached to the rough endoplasmic reticulum and others float freely in the cytoplasm Ribosomes act as factories for making proteins. For example, in a bicycle factory, different parts like the seat, the tyres, the handle bars among others are put together or assembled to form the whole bicycle. Proteins are assembled in the ribosomes from amino acids in a similar way. Endoplasmic reticulum Endoplasmic reticulum is a system of canals or channels within the cytoplasm. It can appear either as rough or smooth when viewed under the electron microscope. The endoplasmic reticulum membranes are continuous with the outer nuclear membrane. The rough endoplasmic reticulum provides a surface for attachment of ribosomes. Proteins made by these ribosomes enter the rough endoplasmic reticulum to be transported to places where they are required. Smooth endoplasmic reticulum is not covered with ribosomes. Mitochondria The mitochondria are oval-shaped organelles. They produce energy for the cell. This energy is used in the cell for various activities. Mitochondria produces energy in a process called respiration. Vacuoles In plants they are large and centraly placed. They are filled with cell sap. They hold up the plant upright when filled up with water. They also store amino acids, salts and waste products. In animals, vacuoles are tiny and are located at the periphery. They help in elimination of excess water and waste products. Chloroplasts Plastids are small organelles in the cytoplasm of plant cells. They may contain coloured substances called pigments. The chloroplasts are an example of plastids with green pigment called chlorophyll. The chloroplast is an organelle in a plant cell where photosynthesis takes place. Chloroplasts are found in the cytoplasm of cells of palisade mesophyll, spongy mesophyll and guard cells. Cells that have chloroplasts are called photosynthetic cells. Each chloroplast is surrounded by two membranes. Inside each chloroplast are small units called grana (singular: granum). A granum consists of a number of discs placed on each other like a pile of coins. Each disc in the pile is a flat sac with a single membrane. One granum is connected to another by the intergranal lamellae. The remaining part of the chloroplast is filled with a fluid and is called stroma. The stroma contains enzymes involved in photosynthesis. The internal structure of a chloroplast is as illustrated below. Fig. 3.4: Internal structure of a chloroplast The process of photosynthesis In Form One, we learnt that the raw materials for photosynthesis are water and carbon dioxide and the products are glucose and oxygen. This may be summarised as follows: During the process, six molecules of oxygen are produced as a by product. Photosynthesis takes place in two stages; the light stage and dark stage. The light stage It takes place in the grana of the chloroplast. During this stage, chlorophyll absorbs light energy. This energy is used in two ways. Fig. 3.5: Summary of the process of photosynthesis. (i) Some is used to split up water molecules into hydrogen ions and oxygen. This is known as photolysis of water. The word photo means (light), lysis means (splitting). The hydrogen ions produced are used in the dark stage. Some of the oxygen formed is released from the leaf through the stomates. The rest is used up in the plant cells for respiration. (ii) Some of the absorbed sunlight energy is converted to ATP. The ATP is used in the dark stage. The dark stage The dark stage takes place in the stroma at the same time that the light stage is taking place in the grana. Carbon dioxide diffuses into the stroma from the cell cytoplasm. The hydrogen from the light stage combines with carbon dioxide to form glucose. This process uses the energy stored during the light stage. The manufacture of a carbohydrate (glucose) from carbon dioxide is called carbon dioxide fixation. The fate of glucose after photosynthesis As we mentioned earlier glucose is produced as the product during the dark stage. Glucose is the basic organic food substance. After photosynthesis, glucose is utilised in the following ways. 1. Respiration Some of the glucose is used by the plant cells for respiration to release energy. 2. Storage Most of the glucose produced in the leaf is usually converted to starch. Starch is stored in plant tissues. Some glucose is combined with fructose to form sucrose. Sucrose is stored in stems such as the stems of sugarcanes. 3. Formation of cellulose Some glucose is converted into a complex substance called cellulose. Cellulose is used in making of cell walls. 4. Converted into lipids Lipids are stored in plant tissues like the seed endosperm. 5. Glucose is combined with nitrates to form proteins Functions of mineral elements involved in photosynthesis What are mineral elements? How many mineral elements do you know? List down the mineral elements. Mineral elements are inorganic substances that are essential to all living organisms. This is because they are the basic units that form all substances making up the body of organisms. Mineral elements are also involved in chemical reactions taking place in the body of living organisms. In photosynthesis, mineral elements involved include. Carbon Pottassium Nitrogen Phosphorus Oxygen Sulphur Nitrogen Hydrogen Magnesium Iron. These minerals are involved in the following functions. 1. Formation of glucose Glucose is made of three minerals elements: carbon, hydrogen and oxygen. 2. Formation of chlorophyll Chlorophyll is a pigment found in green plants. Magnesium is an essential component in chlorophyll formation. Iron is involved in activating chemical reactions in a cell during chlorophyll formation. 3. Opening and closing of stomates Accumulation of potassium ions in the guard cells makes water to enter into the vacuole of the guard cells. This makes the guard cells to swell and become turgid. When guard cells becomes turgid, they open the stomates. 4. Enzyme formation The process of photosynthesis requires enzymes. Enzymes are organic substances that speed up or slow down chemical reactions. Formation of enzymes require mineral elements such as nitrogen and sulphur. 5. Energy changes Pottassium is involved in activation of enzymes involved in the process of photosynthesis. Phosphorus is involved in the formation of ATP in the light stage of photosynthesis. Types of pigments in leaves A pigment is any coloured substance found in cells. Leaves have three types of pigments. Plant pigments are mostly found in the chloroplast of the cells. They include; (i) Chlorophyll (ii) Carotene (iii) Xanthophylls To carry out this study, we will first carry out the following activity. Activity 3.3: To investigate types of pigments in plant leaves Requirements Green leaves Alcohol Filter paper Tape Beakers – small and large. Mortar and pestle Glass rod Procedure 1. Collect green leaves from plants in the school compound. 2. Chop the leaves into small pieces or crush them using pestle and mortar. 3. Place the crushed leaves into a small glass beaker. 4. Add alcohol to half fill the beaker. 5. Place the beaker into a large glass beaker half filled with water as shown below and then heat the large beaker. The large beaker with water is called a water bath. 6. Heat the beaker until the alcohol turns green. This shows that pigments in the leaves have dissolved into the alcohol. Note: Do not heat alcohol directly because it catches fire easily 7. Remove the alcohol mixture from the beaker. Filter it into another small beaker 8. Cut the small long strip of filter paper about 3 cm wide and 10 cm long. Stick it unto a glass rod using a sellotape as shown below. 9. Suspend the filter paper strip into a container with the coloured alcohol such that the tip of the paper touches the mixture. 10. Allow the strip to absorb the coloured alcohol for about 30 minutes. Observe the colours formed on the strip from the bottom to the top. Questions 1. Which colour was first visible as the alcohol was absorbed by the filter paper? 2. How many shades of colour can be observed on the filter paper strip? 3. Suggest the pigments that are represented by each of the colours. Discussion From the activity, you may have realised that the filter paper strip started to absorb the coloured alcohol. The absorbed substance rose up the paper forming different shades of colours. The first and prominent colour to appear was green. Then three other colours appeared that included light green, yellow and light yellow. The green colour represents the chlorophyll pigment. The yellow colour represents carotene pigment and the light yellow colour represents xanthophyll pigment. 1. Chlorophyll This is the main pigment found in plant leaves. It is green in colour. 2. Carotene Carotene is a pigment that is usually yellow, orange or red in colour. 3. Xanthophyll This is a pigment that is usually light yellow in colour. It absorbs lesser amount of green and blue light. Additional activity 1. Go to the field and collect leaves of different colours. In each leaf, suggest the pigments that they may contain. 2. Write a brief essay on how leaves change colour as they grow old. Importance of photosynthesis In your groups, discuss the importance of photosynthesis and write a report. From your discussion, check whether you have come up with the following points. 1. Production of food During photosynthesis, glucose is formed as a product. Glucose is a simple sugar hence it is a food substance used by living organisms to produce energy. Glucose is also used to make other food substances such as starch, sucrose, proteins and lipids. These food substances are used by plants for food. When animals feed on the plants, they obtain the food. Photosynthesis therefore is the main source of food for the whole universe. 2. Oxygen production For every molecule of glucose formed, six molecules of oxygen are released. The oxygen is released into the atmosphere. Oxygen is required by all living organisms for respiration. 3. Removal of carbon dioxide from the atmosphere Photosynthesis prevents accumulation of carbon dioxide in the atmosphere. Therefore, it prevents global warming. Plant products These are substances produced by plants. They are used by human beings in various ways. These substances include: 1. Food substances These include starch stored in roots, stems, leaves, seeds and fruits, oils stored in seeds, proteins stored in cotyledons and sucrose stored in stems. 2. Medicine These are substances stored in plant materials that are used for treatment of diseases. They include; (a) Quinine Quinine used in the treatment and control of malaria. It is found in the bark of Cinchona tree. (b) Codine This is a medicine for coughs. It contains caffeine which is a plant product. (c) Eucalyptus Eucalyptus tree produce chemicals which are used to relieve colds. 3. Industrial products These are substances stored in plant materials that are used by humans to manufacture industrial goods. Such products include tannins and rubber. (a) Tannins This is a group of complex substances that are acidic. They occur widely in plants dissolved in the cell sap. They are quite common in the bark of trees like wattle trees, unripe fruits and leaves. They are used in the conversion of hide (cow skin) into leather. They are also used in dyeing clothes, printing fabrics and in the manufacture of ink. (b) Rubber The primary source of natural rubber is the plant called Para rubber tree. It produces latex which is used to manufacture rubber products such as tyres, shoes among others. Revision Exercise 3 1. (a) Define the term photosynthesis. (b) Name the two stages of photosynthesis. 2. (a) State the site of photolysis in a plant cell. (b) State the role played by light energy during photolysis. 3. (a) Give the product and a by-product of photosynthesis in a green plant. (b) State the origin of the oxygen given out during photosynthesis. 4. The following organelle is found in plant cells. (a) Name the organelle. (b) State the functions of the organelle. (c) Identify the structures labelled: (i) J (ii) K (iii) M (d) Explain the processes that take place in structures labelled K and M. 5. Study the process below then answer the questions that follow. (a) Name substance X. (b) State the uses of substance X in animals. 6. Discuss the fate of glucose after photosynthesis. 7. Name three pigments found in the leaves. 8. What is the importance of photosynthesis? Unit 4 Transport in plants Specific objectives By the end of this unit, you should be able to: (a) Identify tissues that are used for transport in plants. (b) Describe the structural and functional differences between the xylem and phloem. (c) Describe the processes of diffussion, osmosis and active transport. (d) Explain the factors that affect the rate of diffussion. (e) Explain how substances are transported in the xylem and phloem. (f) Explain the significance of diffussion, osmosis and active transport. (g) Describe the transpiration stream. (h) Explain the importance of transpiration. (i) Explain the factors that affect the rate of transpiration. Introduction In this unit, we will learn about how substances move inside the plant. In the previous units, we learnt that plants use water, carbon dioxide and mineral salts during the process of photosynthesis. At the same time, substances such as glucose are formed during photosynthesis. All these substances need to be transported from one place to another in the plant. Meaning of transport in plants Transport is a process whereby substances move from one part of a plant to another. These substances include water, mineral salts and manufactured foods. These substances are transported by structures called vascular bundles. These are conducting structures found in the roots, stems and leaves. We will first study the structures of roots and stems to examine the structures involved in transport. Tissues used for transport in plants To investigate these tissues, let us carry out the following activities. Activity 4.1: To observe permanent slides of dicotyledonous and monocotyledonous roots Apparatus and materials Permanent slides of cross sections of dicotyledonous and monocotyledonous roots. Procedure 1. Place a prepared slide of a dicotyledon root under the microscope. Observe under low power and high power magnifications. 2. Note the different tissues present and their location. 3. Draw a plan diagram to show the position and layout of different layers of tissue. Do not draw any cells or shade. 4. Compare your diagram with that of the plan diagram of the dicotyledonous root section in Fig. 4.1 and use it to identify the tissues. Fig. 4.1: Transverse section of a dicotyledonous root. 5. Repeat this procedure for the monocotyledonous root and compare your drawing with that in Fig. 4.2. Fig. 4.2: Transverse section of a monocotyledonous root. Questions 1. Describe the pattern in the arrangement of xylem in relation to phloem in the: (a) Dicotyledonous root (b) Monocotyledonous root. 2. What is the difference in distribution of tissues between the dicot root and the monocot root ? 3. Where are the root hair cells located in the root? Discussion The distribution of tissues in a transverse section of the dicotyledonous root is not the same as that in the monocotyledonous root. In the dicotyledonous root, the xylem occupies the centre where it forms a star shape. The phloem is found in between the rays of the star. In the monocotyledonous root, the xylem and phloem are arranged to form a ring in which xylem tissue alternates with the phloem tissue. Activity 4.2: To observe permanent slides of dicotyledonous and monocotyledonous stems Apparatus and materials Permanent slides of dicotyledonous and monocotyledonous stems Procedure 1. Place a prepared slide of a dicotyledonous stem under the microscope. Observe under low power and high power magnifications 2. Note the different tissues present and their location. 3. Draw a plan diagram to show the position and layout of different layers of tissues. Do not draw any cells. 4. Compare your diagram with that of the plan diagram of the dicotyledonous stem in shown Fig. 4.3. Use it to identify the tissues. 5. Repeat this procedure for the monocotyledonous stem and compare your drawing with that shown in Fig. 4.4. Questions 1. Is the distribution of tissues in the cross (transverse) section of a dicotyledonous stem the same as that in a monocotyledonous stem? 2. Describe the pattern in the arrangement of xylem in relation to phloem in the: (a) Dicotyledonous stem (b) Monocotyledonous stem. 3. What is a vascular bundle? Discussion The distribution of tissues in a transverse section of the dicotyledonous stem is not the same as that in the monocotyledonous stem, In the dicotyledonous stem, the vascular bundles which contain both xylem and phloem are arranged to form a ring as shown below. Fig. 4.3: Cross section of a dicotyledonous stem. In the monocotyledon stem, the vascular bundles appear scattered in the stem as shown below. Fig. 4.4: Cross section of a monocot stem Most of the tissues in the root and stems are similar. This is because these tissue are continuous from the root into the stem. Common tissues to both the root and stem are the epidermis, cortex, endodermis, xylem and phloem. The stem has additional tissue known as pith. Note that there is no cambium in monocot stems. Some plant tissues are made of only one type of cell. Some tissues are found throughout the plant and some are found only in specific parts. Structure of xylem and its functions Xylem tissue is a specialised tissue which is modified to carry out two functions: (a) To transport water and mineral salts. (b) To give support to the plant. The xylem contains two types of modified cells namely tracheids and vessel elements. Tracheids Tracheids are empty dead cells. They are elongated and have tapering end walls. Water passes through them easily because they have no cellular contents that would otherwise cause obstruction. Their walls are reinforced by strengthening material known as lignin. The walls also have tiny pores known as pits. Tracheids are arranged end to end and also side by side. Water passes from one tracheid to another through the pits. Fig. 4.5: Tracheid element Vessel elements They are very long tube-like structures which are continuous from the root to the leaves of a plant. Several vessels are found side by side. They too are dead tissue and have no cell contents. This allows water to move through them freely. Their walls are reinforced with lignin which makes them strong and rigid to give support to plants. Fig. 4.6: Vessel element Structure of phloem Phloem tissue is made up of sieve tubes and companion cells. Fig. 4.7: Structure of the phloem Structural and functional differences between xylem and phloem Table 4.1: Structural differences between xylem and phloem Xylem Phloem Xylem is made of dead tissue. Phloem is made of living tissue. In xylem, cross walls between Cross walls are perforated into adjacent cells is absent. sieve pores. Xylem walls are lignified. Phloem walls are thin and not lignified. Xylem are made of hollow tubes. Phloem is made of sieve elements. Table 4.2: Functional differences between xylem and phloem Xylem Phloem Transports water and mineral Transports manufactured food salts. substances from the leaves to all other parts of the plant. Sieve tubes The sieve tubes are made up of cells called sieve elements arranged end to end with each other. Sieve elements are separated from each other by structures called sieve plates which have pores or perforations in them. Phloem sieve tubes, unlike xylem tissue are living tissue. They have fine cytoplasmic strands that run through from one element to another. However, the cytoplasm of sieve elements does not have a nucleus and several other cell organelles. Companion cells Each sieve element has an accompanying cell known as a companion cell. This cell has a dense cytoplasm, a prominent nucleus and other cell organelles. These take up the roles of those organelles that should have been in the sieve elements. Activity 4.3: The ringing experiment to show that substances are translocated through the phloem Apparatus and materials Tree or shrub with many branches. Sharp knife. Procedure 1. Remove completely a ring of bark with its phloem from two branches. The xylem tissue which makes up the bulk of the stem is left intact. 2. The set-up is left undisturbed for four weeks. Fig. 4.8: A ring of bark is removed from a woody shoot leaving the xylem Question 1. What observations were made in the stem after four weeks? 2. Explain these observations. Discussion When the ring of bark is removed, the phloem beneath it is also removed. After several weeks, a swelling above the cut ring is noted. This swelling is due to the accumulation of food substances that were being transported from the leaves but could not get across the debarked part of the stem. As a result, there is no swelling on the lower part of the ring. Fig.4.9: Swollen tissue above the cut part after four weeks. Note: This ringing procedure is sometimes employed to kill some unwanted trees before they are cut down. Do not use it on a tree that has economic value to your community. Discuss with your friends how the bark of medicinal trees can be harvested without killing these important trees. Transport of substances in the phloem As we mentioned earlier the phloem transports manufactured food substances from the leaf cells and the storage tissues to all other parts of the plant. This is by a process called translocation. The food substances move from the leaf cells into the sieve tubes by active transport. The energy for this process is produced by the companion cells. The substances accumulate in the sieve tubes of the phloem of the leaf. They draw water from adjacent cells by osmosis into the sieve tubes. This causes an increase in hydrostatic pressure in the sieve tubes resulting to the movement of materials to areas of storage or use. The materials can also be transported in the sieve tubes by cytoplasmic streaming. The materials are moved along the cytoplasmic filaments from one sieve tube cell to the next through the pores on the sieve plate. This process uses energy that is produced by the companion cells. How substances are transported in the xylem and the phloem In order to carry out the life processes, that we mentioned in Unit 3, a cell needs and takes in various substances. In carrying out these processes, the cell produces certain substances as well. Some of these are waste products and others are substances that the cell does not need but are useful to other cells in a tissue. Such useful substances are usually taken out of the cell to the other cells that need them. Therefore substances are always moving into and out of cells. The way substances move into or out of the cell depends on certain properties of the substances such as size of molecules and the type of substance. There are three main physiological processes by which substances move in and out of cells. These are diffusion, osmosis and active transport. Diffusion Let us carry out the following activity to show diffusion. Activity 4.4: To demonstrate diffusion using ink and perfume Materials Bottle of perfume or ink. Beaker Water and pipette Procedure 1. Fill a beaker half-full with water 2. Add a drop of ink into the water. What do you observe? Add another drop of ink. What do you observe? 3. Open a bottle of strong perfume. What do you observe after a few minutes? Discussion From the activity, you may have noticed that the ink spreads until the water becomes uniformly coloured. You also realised that after the perfume bottle was opened, you could smell the perfume after sometime. The particles of the ink spread from one point to the entire beaker until the water was uniformly coloured. In the same way, the particles of the perfume spread from the open bottle into the air in the whole class. This process where particles move from one point and spread to other regions is called diffusion. Activity 4.5: To demonstrate diffusion using potassium manganate (VII) crystals and ink in water Materials For each group; beakers, water, potassium manganate (VII) crystals, ink, pipette or dropper. Procedure 1. Put about 50 cm3 of water in a beaker. Select a large crystal of potassium manganate (VII) and drop it carefully in the water. Observe what happens. Questions 1. Explain the observation made when the crystal of potassium manganate (VII) is dropped in the water. Activity 4.6: To demonstrate diffusion using a visking tubing Materials Visking tubing, starch solution, dilute iodine solution, beakers, string. Procedure 1. Measure and cut a visking tubing 8 cm in length. 2. Tie up one end tightly with a string, about 1cm from the tip. Fill the visking tubing with starch solution and then tie up the other end. 3. Note the colour of the solution. Immerse this visking tubing into a beaker filled with dilute iodine solution. 4. Leave it to stand for 15 minutes. 5. Make observations of the colour changes and complete the table below. Before After 15 Conclusion minutes Colour of the starch solution inside the visking tubing. Colour of the iodine solution in the beaker. Questions 1. Why did the colour of the starch solution change after the visking tubing was immersed in a solution of iodine? 2. Why was there no change in the colour of the iodine solution in the beaker after 15 minutes? 3. The visking tubing seems to allow some molecules to pass through and not others. What is the name given to membranes that have this characteristic? Discussion The colour of the starch solution changed from white to blueblack because iodine molecules moved into the visking tubing which brought about the change. The colour of iodine remained brown because starch molecules did not move from the visking tube to the beaker. The membrane is permeable to the iodine molecules. Diffussion is the movement of particles of a substance from a region of high concentration of the particles to a region of low concentration of the particles. The difference in the concentration of particles in the two regions is called a concentration gradient or diffusion gradient. Diffusion occurs in liquids and gases. Diffusion does not take place in solids. Diffusion of certain particles also takes place through the cell membrane, cell walls and cell cytoplasm. When particles move from a region of high concentration to a region of low concentration, they are said to diffuse along a concentration gradient. As long as a concentration gradient is maintained, the movement of particles continues until they are evenly distributed in the available space. Examples of diffusion 1. When a drop of ink is placed into a glass of water, the ink particles spread in the water until all the water is uniformly coloured as seen in Activity 4.5. 2. We are able to smell perfume that other people have worn because the particles of perfume diffuse from them through the air to our organs of smell, the nose. Factors affecting the rate of diffusion The rate of diffusion of particles is the time taken for the particles to diffuse within an available (fixed) space until they are evenly distributed. Several factors affect the rate of diffusion. They include: Temperature When the temperature of particles is increased, their kinetic energy also increases and the particles move faster. Therefore, the higher the temperature, the faster the particles will diffuse. The lower the temperature the lower the rate of diffusion. Concentration of substance A greater difference in concentration of particles between two regions, results in a steeper concentration gradient and vice versa. Diffusion is faster when the concentration gradient is high. The difference in concentration of substances determines the rate of diffussion. Where the difference in concentration of substances is big, the rate of diffussion is high because this increases the concentration gradient. Size of molecules Small particles diffuse faster than large particles. This is bacause small particles are light and pass through air or water easily. Osmosis Let us carry out the activity below to learn about osmosis. Activity 4.7: To demonstrate osmosis using a visking tubing Materials Visking tubing, concentrated sucrose/sugar solution, beaker filled with distilled water, string. Procedure I 1. Measure and cut an 8cm length of visking tubing. 2. Tie one side of the visking tubing tightly with a string 1 cm from the end. 3. Fill the visking tubing about three-quarter full with the concentrated sugar solution. 4. Tie up the other end of the tubing tightly with a string. 5. Immerse the visking tubing with the sugar solution into a beaker full of distilled water. Note the level of water in the beaker. 6. Leave the setup to stand for several hours then note the amount of solution in the visking tubing and the level of water in the beaker. Fig. 4.10: The results of osmosis Questions 1. What happens to the following after several hours? (a) Contents of the visking tubing? (b) Level of the water in the beaker. Procedure II 1. Repeat procedure I, but this time, fill the visking tubing with distilled water then tie tightly until it is firm and bulging. 2. Immerse the visking tubing into a beaker of concentrated sugar solution and leave it to stand for several hours. Fig. 4.11: The results of osmosis Questions 1. What happens to the contents of the visking tubing after several hours? 2. Does it still feel firm to the touch? Does it still appear bulging? Explain your observations. In our activity of demonstrating osmosis, water also moves from a region where it is in high concentration to a region where it is in low concentration. Therefore, osmosis is a kind of diffusion. However, the term osmosis is used only when we are referring to diffusion of water molecules through a semi-permeable membrane. In our experiment, the visking tubing acts as a semi- permeable membrane. Osmosis is the movement of water molecules from a region of high concentration of water molecules to a region of low concentration of water molecules through a semi-permeable membrane. We have seen that diffusion is the movement of particles from a region of high concentration to a region of low concentration. Solute, solution and solvents When a solid is dissolved in water, we get a solution. The solid that is dissolved in this solution is called the solute. The liquid that dissolves the solid is known as the solvent. Thus: Solute + Solvent = Solution The concentration of a solution depends on the amount of solute dissolved. A dilute solution has more water (Solvent) molecules as compared to solute molecules. Fig. 4.12: A dilute solution A concentrated solution has more solute molecules than water molecules. Suppose, a dilute solution were separated from a concentrated solution by a semi-permeable membrane in a beaker. Water molecules will move from the dilute solution to the concentrated solution. Fig. 4.13: A concentrated solution This is because the dilute solution has more water molecules than the concentrated one. Water molecules pass easily through the channels or pores of the cell membrane mainly because they are very small. Solute molecules are too large to pass through the pores. Activity 4.8: To demonstrate osmosis in a living tissue Materials Fresh arrow roots/cassava, sweet potatoes, Mbatata, Irish potatoes, strong salt solution, distilled water, scalpel, large beaker or small basin. Procedure 1. Take a large Irish potato and peel it. You could also use arrowroot, cassava or sweet potato. 2. Cut off a piece so that it stands at least 6cm high. 3. Cut and scoop out a deep hollow portion in its middle and pour the strong salt solution halfway up the hollow portion. 4. Mark the level of the salt solution using a scalpel. 5. Place the potato in a beaker or basin containing distilled water and let it stand for several hours then note the level of solution in potato. 6. Repeat the experiment with boiled pieces of irish potato or cassava. Questions 1. Is the level of the strong salt solution still the same at the end of the experiment? 2. Explain what happens to cause the change in level of the salt solution. 3. Compare these results with those obtained when boiled Irish potato is used. Activity 4.9: To demonstrate osmosis using Irish potato tubes/cylinders Materials Irish potato, cork borer, ruler marked in millimetres, scalpel or razor blade, 3 test tubes, board to cut on, labels for beakers, concentrated sucrose solution, distilled water. Procedure 1. Select large Irish potatoes. 2. Using a cork borer, bore holes into the potato tuber and bore out long cylinders each 60mm in length. 3. Put concentrated sucrose solution in one test-tube and distilled water in another. Leave the third test tube empty. Label the test tubes. 4. Put the potato cylinders into the three test-tubes and leave them over- night. 5. Measure the change in length of the potato cylinders in each test-tube. 6. Record your results in a table like the one shown below. Test tube contents Sucrose Distilled water Empty Length of cylinder before experiment 30mm 30mm 30mm Length of cylinder after experiment Activity 4.10: To demonstrate osmosis using leaf petioles Materials Leaf petioles, salt solution, distilled water, petri-dishes, labels, scalpel. Procedure 1. Cut petioles from three leaves. 2. Split each petiole lengthwise and place the two strips from the same leaf in water, salt solution and an empty petri dish respectively. 3. Leave them for 30 minutes. 4. Draw the appearance of the petioles after 30 minutes in each solution. Questions Explain the appearance of each stalk/petiole in the solution it was immersed. Explain why it was necessary to cut a split on petioles in this experiment. Discussion The inner cut surface of a strip of petiole is described as the inner side. Its cells are exposed. The outer surface of the petiole is covered by a thick cuticle. When a strip of petiole is immersed in water, the cells in the inner side (exposed cells) take up water by osmosis much faster than the cells on the outer (covered by cuticle) surface. These cells expand more causing a curvature towards the outside (outwardly). When the strip is immersed in the salt solution, the cells on the inner side lose water much faster than the outside. The cells in the inner side shrink more and hence the strip of petiole curves inwardly. The strips in the empty petri dish do not change. Usually this is used for comparison purposes and is referred to as the control experiment. Active transport The type of movement of molecules and ions that we have considered so far are those where molecules move down or along a concentration gradient. These movements are called passive transport. Molecules and ions can also move from an area of low concentration to that of higher concentration. They are said to move against a concentration gradient. Such a process requires the use of energy and is called active transport. In active transport, the cell must use its own energy to move the molecules against a concentration gradient. Such energy is supplied by the mitochondria in the cell. Therefore active transport takes place only in living cells. Fig 4.14: Active transport across the cell membrane Active transport involves carrier protein molecules. They pick up molecules of a substance from one side of the cell membrane and transports them across. Activity 4.11: To show dyed water moving up the xylem Apparatus and materials Boiling tube, dilute solution of red ink or dye, cotton wool, bean seedling. Procedure 1. Set up the apparatus by placing a bean seedling in a dilute solution of red ink in boiling tube. Colour the solution with red ink or eosin stain. 2. Leave the set up for three or four hours then remove the plant and cut sections of the root, stem and leaf. 3. Examine under low power of the microscope or with a hand lens. Questions 1. Which tissues are stained red by the dye? 2. Suggest a control for this experiment. Discussion The sections of the root, stem and leaf will be seen to carry the dye or ink. If only the xylem vessels are stained red by the dye we can conclude that water travels up the plant in the xylem vessels. This conclusion would be strengthened if we cut sections of a second plant of the same species to confirm that the xylem vessels are not normally coloured. This would be the control experiment. We have already noted that water in the soil is absorbed by the roots and then transported through the xylem to all parts of a plant. In very tall trees, water is transported to the leaves which may be more than 100 metres above the level of the ground. This water moves against the force of gravity. A number of forces have been suggested to explain the movement of water in the xylem to such great heights. These forces include:- Capillarity Capillarity is the tendency of water to rise inside a narrow tube. This is because water molecules have the ability to cling to the surface of the tube. Water rises more in tubes with small diameters than in wide tubes. In plants, xylem forms narrow tubes through which water moves. Water rises in the xylem because of strong forces of attraction between the water molecules and the cell walls of the tubes or xylem vessels. However, capillarity can raise water up to a height of only a few centimetres. It is therefore not enough to raise water up to great heights in tall trees. Fig. 4.15: An experiment to demonstrate capillarity Cohesion and adhesion Water molecules have the ability to attract each other. This force by which water molecules attract each other is known as cohesion force. The water molecules also have a strong attraction to the walls of the xylem vessels or tubes. The forces of attraction between unlike molecules are called adhesion forces. Due to these two forces, water molecules next to the wall of the xylem tube “creep up” along the wall due to adhesion and pull along molecules that are not near the wall by cohesion. This is similar to the “tug of war” in which people on a line hold themselves firmly on the ground as they pull on the rope. Cohesive and adhesive forces prevent the column of water from breaking. Like capillarity, these two forces are not enough to raise water to the height of tall trees. Root pressure This is another force that moves water up the plant. Root pressure moves water from living root cells into the xylem. When the stem of a plant that is well supplied with water is cut off near the ground level, sap flows from the cut-stem. If a glass tube is attached to the cut end of the stem with a rubber tubing, the sap rises in the tube as shown in Fig. 4.15. The sap can rise up to a height of one metre. Fig. 4.16: An experiment to demonstrate root pressure The pressure that holds up the column of water against the pull of gravity is called root pressure. Root pressure is due to an increase of osmotic pressure. Such pressure is caused by accumulation of solutes in the xylem of roots. In the root, the endodermis moves solutes by active transport into the xylem. As a result, the osmotic pressure of sap in the root xylem increases. This causes water from the soil to diffuse into the root xylem by osmosis, causing an increase in pressure. This pressure in the root xylem causes water to move upwards in the xylem of the stem. As mentioned before, root pressure can only raise water to a height of about one metre. In addition, if a plant is growing in soil with little water, the maximum height that root pressure will raise water will be less than one metre. Therefore, root pressure alone cannot help to transport water to great heights in tall trees. Transpiration pull Transpiration pull is another force that transports water through the plant. This process begins in the leaf and sets up conditions that cause water to be drawn all the way from the roots in a continous stream. Water evaporates into the air spaces from the surrounding mesophyll cells in the leaf as shown in Fig 4.16. This causes the mesophyll cells next to the air space (cell B) to have a higher solute concentration than the cells further away from the air space (cell A). As a result, water moves by osmosis from cell A to cell B. The concentration in cell A therefore increases. If cell A is next to a xylem tube as shown in Fig. 4.16, then the water in the xylem moves by osmosis into the cell. Water in the xylem is drawn upwards by cohesion and adhesion forces to replace it. Fig. 4.17: To show detailed movement of water from xylem through cells to air spaces in the leaf. Significance of diffusion, osmosis and active transport (a) Absorption of water and mineral salts by plants Soil particles are usually surrounded by a film of water except when there is drought. Root hair cells absorb water from the soil by osmosis. The cell sap in the vacuole of the root hair cell has a high concentration of salts and sugars. It is therefore hypertonic to the water found between the soil particles. Due to this concentration gradient, water molecules move by osmosis from the soil through the semi-permeable membrane of root hair cells into the cell sap. The root hair cells will take up water as long as their concentration of salts is higher than that in the soil. The cell wall pressure is not large enough to prevent osmosis. (b) Active uptake of mineral salts As mentioned earlier, the concentration of salts in the cell sap of a root hair cell is higher than that of soil water. Soil water is a dilute solution of mineral ions such as potassium, magnesium, nitrates and phosphates. This means that a root hair must take up these mineral ions by active transport against their concentration gradients. Plants need minerals salts for normal growth and development. For example, they need magnesium to synthesise chlorophyll, and nitrate ions to synthesise proteins. (c) Transportation of manufactured foods Translocation is the transport of organic substances from one part of the plant to another. It takes place in the phloem tissue in plants. Transpiration stream Meaning of transpiration Transpiration is the evaporation of water from the plant surface mainly through the leaf. When water is lost from the surface of cell wall of spongy mesophyll cells, water is then drawn from the adjacent cells by osmosis. The adjacent cells continue to draw water from the xylem. This creates a sucking effect of water from the xylem which is known as suction. Continous flow of water from the xylem through the mesophyll cells and out of the stomata is called transpiration stream. Transpiration stream together with cohesion and adhesion forces draws water up the plant through the xylem. When the xylem in the root draws water from the adjacent cortex cells, the cortex cells draw water from the root hair cells by osmosis. This results in absorption of water from the soil through the root hair cells. Fig. 4.18: Transpiration stream Suction Have you ever taken a soft drink using a straw? What do you usually do? Do you usually push in air, or do you pull out or suck air out of the straw? How does the soda in the bottle behave when you suck air out of the straw? From the example probably you have noticed that as you suck air up a straw, the soft drink moves from the bottom of the bottle up the straw. This is called suction. Activity 4.12: To show that water is given off by the leaves during transpiration Apparatus and materials: Two potted plants (one with leaves, the other with its leaves removed) two polythene bags, strings. Procedure 1. Set up the potted plants and cover each with a polythene bag. Tie the polythene bag round the stem as shown in the following figures. Fig. 4.18: Set-up of experiment to show that water is given off by leaves during transpiration. 2. Leave the two set-ups in a sunny place. 3. Collect and test any liquid which collects in the plastic bag with anhydrous copper sulphate or anhydrous cobalt chloride paper. Questions 1. What observation was made in set-ups A and B after several hours? 2. What conclusion can be made from the above observations? 3. Which is the control experiment and why? 4. What changes are observed on the anhydrous cobalt chloride paper? Discussion The anhydrous dry cobalt chloride paper is used to confirm that the liquid collected is water. If the anhydrous cobalt chloride paper which is blue turns pink, then water is present which means transpiration is taking place. If water collects in the plastic bag in set-up A but not in set up B, we can conclude that water is given off mainly by the leaves of a plant. Set up B is the control. It proves that without leaves, no transpiration occurs. Activity 4.13: To compare transpiration on lower and upper leaf surfaces Apparatus and materials Anhydrous cobalt chloride papers, two glass slides, rubber bands or paper clips or cellotape. Plant that is still anchored in soil. Procedure 1. Identify an intact leafy shoot, a plant that is still growing will do. 2. Select a leaf, and attach equal size pieces of anhydrous cobalt chloride paper on both sides of the leaf. You can use the cellotape, slides and rubber bands or clips to secure the paper in position as shown in the figure below. 3. Record the time, and observe each piece of cobalt chloride paper. 4. Note how long it takes for each paper to change colour. 5. Make your conclusion. Fig. 4.20: Experiment to compare transpiration on lower and upper leaf surfaces. Questions 1. What is the colour of anhydrous cobalt chloride paper? 2. What is the colour change observed during the experiment? 3. Which paper took longer to change: the one on the upper surface of the leaf or the one on the lower surface? 4. Explain your answer in 3 above. Discussion Anydrous (dry) cobalt chloride paper is blue. Hydrated (wet) cobalt chloride paper is pink. Evaporation of water from the leaves (transpiration) occurs through the stomates located on the leaf surface. In a typical land plant, there are more stomates on the under side of the leaf than on the upper side. The more the number of stomates, the more the amount of water lost by transpiration. The cobalt chloride paper on the under surface of the leaf therefore changes colour from blue to pink faster/sooner than the cobalt chloride paper on the upper surface of the leaf. Importance of transpiration Transpiration is important to a plant for the following reasons: 1. Cooling of plants In hot climates or on hot days, direct sunlight causes the surface of the plants to heat up. In such situations, transpiration is important because it cools the plant as the water is evaporating, the plant is getting cooled. 2. Distribution of mineral salts throughout the plant When transpiration occurs, it causes water to flow through a plant. This is because when water evaporates through the stomata, more water is drawn from the leaf cells to replace it. This causes transpiration stream. As the water flows through the plant, it carries with it the mineral salts dissolved in it which are distributed throughout the plant. 3. Uptake of water The water lost by transpiration is replaced by water absorbed from the soil. Factors affecting the rate of transpiration Rate of transpiration is the speed at which a plant loses water through transpiration. The factors affecting transpiration rate include: Temperature Air movements Humidity Light intensity. Activity 4.14: Use of a potometer to investigate the effect of external conditions on the rate at which water enters a leafy shoot Apparatus and materials Leafy plant freshly uprooted or freshly cut for example a tomato plant, bean, or any other suitable plant, basin of water, scalpel, means of timing such as stopwatch or wristwatch, potometer, polythene bag. Procedure 1. Immerse the potometer in the basin of water making sure it is completely filled with water. Fig. 4.21(a): Cutting a leafy branch under water 2. Put the plant into the water and cut through the stem under water as shown below. 3. Attach the freshly cut end of the stem into the potometer still under water. Why is the stem of the plant cut under water? 4. Remove the plant and potometer from the water and mount them in a fixed position. The end of the capillary tube should rest in a beaker of water. 5. After setting up the potometer as shown, carry out the following activities and record the results in a table. (a) (i) Place the potometer with the shoot in a windy place outside the classroom. (ii) Introduce an air bubble into the capillary tube by removing the beaker of water at the end of the tube. What happens to the bubble? (iii) Measure the distance moved by the bubble for five minutes and record your results in a table like the one shown below. Conditions Distance moved by bubbles (cm) Time (min) Windy Still air Hot sun Cool place Table 4.1: External factors affecting transpiration (iv) Calculate the rate of water uptake as follows: Fig. 4.22(b): Experiment to show external factors affecting transpiration (b) Repeat the procedure but this time place the plant where the air is still for instance in the classroom. Record the distance covered by the bubble and calculate the rate of water uptake by the same shoot. (c) (i) Place the set-up outside in the hot sun and again in a cool place for example, inside the classroom. (ii) Calculate the rate of water uptake in each case using the procedure described in (a) above. (d) (i) Put the plant in a humid environment that is by covering the leaves with a polythene bag and leaving it without the polythene bag to compare. (ii) Calculate the rate of water uptake in each case. Questions 1. Explain what the potometer measures; (a) Directly. (b) Indirectly. 2. What conclusion can you draw from your results where the following environmental conditions were investigated (a) Wind. (b) Humidity. (c) Temperature. 3. Evaporation alone cannot account for the movement of water through a plant. What other forces might be involved? Discussion The potometer measures directly the rate of uptake of water. It also indirectly measures the rate of transpiration since evaporation of water from the leaf leads to the replacement of this water through uptake. In windy conditions, the rate of transpiration is higher than in still air. In hot environments, the rate of transpiration is higher than in cold conditions. In humid conditions, the rate of transpiration is lower than in dry conditions. When setting up the leafy shoot in the potometer, the stem is cut under water to prevent the introduction of an air bubble into the xylem tissue because it blocks the conduction of water. The plant uses up water for photosynthesis and it also loses water by transpiration. These two factors contribute to its uptake from the soil. Let us now discuss how the various factors affect transpiration. Temperature The temperature of the surrounding of a plant is an indication of the amount of heat present around the plant. On a hot day, Fig. 4.19(i), the temperature is high because there is a lot of heat in the atmosphere. This causes faster evaporation of water from the leaf and therefore transpiration is very fast. On cold days. Fig 4.19(ii), the temperature is low because there is very little heat in the environment. As a result of this, little evaporation of water from the leaf occurs causing transpiration to occur very slowly. Fig. 4.23: Effect of temperature on rate of transpiration Humidity Humidity is the amount of water vapour in the air. If this amount of water is a lot then humidity is high. If it is little, then the humidity is described as being low. When the humidity is high, the air becomes saturated with water vapour. Under these conditions there is very little space available in the air for water vapour from the leaf to occupy. This means that transpiration reduces or even stops. However, when the air is dry, that, is humidity is very low, there is plenty of space for water vapour being transpired from the leaf to occupy. Therefore the rate of transpiration is high. Fig. 4.24: Effect of humidity on rate of transpiration Air movement Moving air carries with it moisture that has evaporated from the leaf surface. This prevents the air surrounding the stomates from becoming saturated with water vapour from the leaf. Sometimes, the environmental conditions around the plant can be described as windy like when the air is moving quickly. The air can also be still when there is no wind. In a windy environment, increased movement of air prevents any moisture from accumulating on the surface of the leaf, Fig. 4.21(i). This creates room for vapour transpiring from the leaf. It also increases the diffusion gradient of water vapour between intercellular spaces in the leaf and the air. The rate of diffusion of water vapour from the leaf increases, thus increasing the rate of transpiration. When there is no wind, the air above the leaf surface quickly becomes saturated with water vapour hence reducing transpiration, Fig. 4.21(ii). The rate of transpiration decreases with increased humidity. Fig. 4.25: Effect of wind on rate of transpiration The situations just described can be compared with what happens in your class at break time. Everybody usually wants to leave the classroom. If the first students to leave just hang around the door, this will create crowding at the door and make it difficult for the others to leave. However, if those who leave first go away completely, then there is room for those behind to pass and everybody moves fast. Generally, the rate of transpiration is highest in hot, windy and dry conditions and it will be lowest in cold, non-windy and wet conditions. Light intensity Light intensity is the strength of light received by the earth’s surface. It varies in the course of the day depending on the position of the sun. For example, at dawn, there is very little light, and the light intensity is low. As the sun rises in the sky, the strength of sunlight increases. At noon, the sun’s intensity, is high and therefore light intensity is very high. As the sun sets, the strength of light decreases causing a low light intensity. Light intensity affects transpiration because it has an effect on the opening of stomates. The rate of transpiration is high when there is high light intensity because the stomates open more. When the intensity of light is low, the rate of transpiration is reduced because stomates open less. Stomates close in darkness, so at night very small amounts of water are lost. Revision Exercise 4 1. The diagram below represents a plant tissue. (a) (i) Identify the parts labelled G, H and K. (ii) State the functions of the vessel shown above. (b) Name food substances that are transported in plants by the above structures. 2. The diagram below represents a section through the stem of a plant. (a) Name the structures labelled K, P, and N. (b) State the class to which the plant belongs to giving reasons. 3. Explain the following observations. The freshly cut stump of a tree will continue releasing water for sometime after the tree has been cut down. 4. Explain why plants growing in an enclosed greenhouse have a lower rate of transpiration than plants growing in the open. 5. In an experiment, the bark of a stem was ringed as shown in figure M. Figure N shows the results after several weeks. (a) Account for the results obtained after several weeks. (b) (i) From the results of this experiments, explain why trees that have been ringed dry up after sometime. (ii) Suggest a possible aim of this experiment. 6. A student set up an experiment shown below using a freshly obtained leafy shoot. The shoot was cut and mounted on the capillary tube under water. (a) Suggest a possible aim of this experiment. (b) Explain why the leafy plant was cut and mounted under water. (c) Give the name for the apparatus used in the above experiment. (d) State the environmental factors that would influence the movement of the air bubble along the scale. 7. Define the following terms: (a) Osmosis. (b) Diffussion. Unit 5 Human digestive system Specific Objectives By the end of this unit, you should be able to: (a) State the functions of carbohydrates, proteins, lipids, vitamins, mineral salts and water in the body. (b) State the chemical composition of different food substances. (c) State different digestive enzymes and the food substances they work on. (d) List products of chemical digestion. (e) Describe properties of enzymes. (f) Carry out investigations in enzymes. (g) Describe the absorption of food substances. (h) Describe the adaptations of the small intestines to their function. (i) State the functions of the large intestines. (j) State the fate of digested food. (k) Explain problems associated with the digestive system. Introduction In Form Two, we learnt about food nutrients and what makes a balanced diet. In this unit, you will learn about the function of nutrients in the food, chemical digestion (molecular structure of food substances), enzymes (characteristics and properties of enzymes) absorption of food substances, assimilation, adaptations of the small intestine in its function and finally problems associated with the digestive system. We have seen that plants make their own food in a process called photosynthesis. Animals however do not make their own food. They depend on organic substances that make up the body of other organisms. Nutrients in food In Form Two, we discussed nutrients in food and what makes a balanced diet. In this section, we shall learn about functions of carbohydrates, proteins, lipids, vitamins, mineral salts and water in the body. Functions of carbohydrates Carbohydrates are important chemical compounds to all living things. They have several functions which include; 1. As a source of energy Some carbohydrates such as glucose are broken down to provide energy in the cell. Energy is used for various life processes such as locomotion, growth among others. 2. As part of the structure in plant cells A carbohydrate like cellulose forms part of the cell wall in plants. Cellulose makes the cell wall firm. 3. As roughage in humans Foods of plant origin such as vegetables and fruits are rich in cellulose or fibre. They provide bulk and resistance to the muscles in the alimentary canal. This allows easy movement and digestion of food in the gut and prevents constipation. Functions of proteins 1. Proteins are structural compounds of animal tissues Proteins form part of the structure of animal tissues. They are found in the form of: Keratin in hair, horns and feathers. Collagen in tendons and ligaments. Myosin in muscles. 2. Proteins are functional units in plants and animals Proteins perform many functions in animal bodies. Some functional proteins include: Enzymes These are proteins that speed up the reactions in plant and animal cells. Reactions like photosynthesis and respiration proceed with the help of enzymes. Haemoglobin This protein is found in red blood cells of vertebrates. Its function is to transport oxygen from the lungs to other parts of the body. Hormones These proteins regulate life processes in animals. An example of an animal hormone is insulin. It regulates the sugar level in the body. Antibodies These are proteins that provide the body with immunity against diseases. They assist the body to destroy disease-causing microorganisms. Fibrinogen This protein is important in the clotting of blood. Blood cannot clot without fibrinogen. Clotting of blood prevents excessive bleeding from wounds or injury. 3. Proteins are storage products Plants store their excess proteins in seeds. Such proteins are used by the seeds during germination. Mammals store some of their protein in form of casein in milk. This is a source of protein called albamen for their young ones. Similarly, a protein called a humen is stored in eggs and is used for the growth of the embryo. Functions of lipids Lipids are large, naturally occuring organic compounds. They are insoluble in water but they readily dissolve in alcohol. Fats are usually solids at room temperature. When heated, fats can melt to liquid. Oils can be turned to solids when cooled enough. 1. As a source of energy Lipids store energy just like carbohydrates. However, lipids store more energy than carbohydrates. 2. Lipids form part of the structure of the cell membrane Cell membrane is composed of a type of lipid called a phospholipid. 3. As a storage compound Plants store lipids as oil in seeds and in some fruits such as coconuts, avocado pear, castor seeds, groundnuts, simsim seeds, maize and macadamia nuts. Animals store lipids in the form of fat in a special tissue known as adipose tissue. Stored lipids can always be used by the plants and animals for respiration when need arises. 4. For insulation Lipids are used by animals for insulation against excessive heat loss. The insulatory lipid is in the form of solid, insoluble fat found beneath the skin in vertebrates. This layer helps to keep the animal warm. 5. As protective compounds The fat found around body organs like the heart and the kidneys acts as a protection against mechanical injury by cushioning the organs against physical impact. 6. As a source of metabolic water Chemical reactions that take place in cells are referred to as metabolic reactions. Some of these reactions produce water as a by-product. In the camel, the fat stored in the hump can be used as a source of water. This occurs when the fat is broken down during respiration to release energy. The water formed as a by product enables the camel to go for long periods without drinking water. 7. As a store of vitamins Vitamins A, D, E and K are soluble in fat. The human body is able to store these vitamins in fat found in the liver. Name of Use in the body vitamin A It maintains healthy epithelial growth It promotes growth It improves proper vision D It strengthens bone and teeth It enables absorption of calcium and phosphorus E Involved in cell metabolism It strengthens muscles K It helps in blood clotting by formation of prothrombin C Maintenance of healthy cells, tissues and blood vessels. It promotes absorption of iron It promotes healing of wounds and prevents infections For formation of cement and collagen fibres in the teeth B1 Involved in cell respiration It promotes nerve activity B2 It is needed for maintaining healthy tissues B3 Involved in energy release in cells Table 5.1: Vitamins and their uses in the body Functions of water The following are some ways in which our bodies use the water that we drink or take in with our food. 1. Digested food can only be absorbed when it is dissolved in the water in the ileum. 2. Excretory materials like carbon dioxide and urea are removed from the body when dissolved in water. 3. Oxygen is absorbed into the body after it dissolves in the moisture lining of the lungs. 4. Water gives organisms form. 5. Water dilutes harmful substances in our bodies and helps in their removal from the body. We must always take in a lot of fluids or water during any one of the following situations. when one has a high fever when one has diarrhoea when one is vomiting when one is sweating when one is bleeding Thirst is always a sign that the body needs water. 6. Water is used to transport substances in the cell and in the blood. Functions of mineral salts Table 5.2: Some important minerals, their sources, uses and deficiency signs Chemical composition of food substances (a) Carbohydrates These are chemical compounds made up of the elements carbon, hydrogen and oxygen. Their general formula is (CH2O)n. Common examples of carbohydrates are sugars and starch. Carbohydrates are classified into three main groups: Monosaccharides. Disaccharides. Polysaccharides. Monosaccharides. A monosaccharide is a single sugar unit. The general formula of a monosaccharide is (CH2O)n where n can be 3, 5 or 6. Some examples of monosaccharides include glucose, fructose, galactose. Monosaccharides They are single sugar units, their general formula is (CH2O)n. They are classified according to the number of carbon atoms present example trioses has three (3) carbon atoms and pentoses has five (5) carbon atoms. They are used in respiration to release energy and used in the manufacture of disaccharides and polysaccharides. They are reducing sugars. Disaccharides They are made by joining two monosaccharides in a process known as condensation. A molecule of water is removed in the process. Examples include: Lactose = glucose + galactose Sucrose = glucose + fructose They are sweet tasting. They are readily soluble in water. They have a general formula C12H22O11. Polysaccharides They are made by joining many monosaccharides. They are non-reducing sugars. They are not sweet, insoluble or slightly soluble in water. They are non-crystalline. They have a general formula; Cx(H2O)y (b) Proteins Proteins are complex compounds. They are made up of the elements carbon, hydrogen and oxygen. In addition, they have nitrogen. Some proteins may also contain sulphur or phosphorus. These elements make up units called amino acids. Amino acids are the building units of proteins. Several amino acids are joined together by peptide bonds to form proteins. There are about twenty different types of amino acids which occur naturally in plants and animals. These amino acids combine differently in a chain to form different types of protein. (c) Lipids Lipids are fats and oils. The elements found in lipids are the same as those found in carbohydrates namely, carbon, hydrogen and oxygen. However, lipids have much fewer oxygen atoms than hydrogen atoms as compared with carbohydrates. For example, in glucose which is a carbohydrate, there are two hydrogen atoms for every oxygen atom. In a lipid, there are twenty two hydrogen atoms for every oxygen atom. Fats are lipids commonly found in animal tissue. An exception is the whale which has oil. Oils are lipids commonly found in plants. The building units in a lipid molecule are fatty acids and glycerol. One glycerol molecule combines with three fatty acid molecules in a condensation reaction to form a lipid known as a triglyceride. Three water molecules are given out. This reaction can be reversed by hydrolysis where the triglyceride (lipid) is split to glycerol and three fatty acid molecules. There are different types of fatty acid molecules but only one type of glycerol. The type of lipid formed depends on the types of fatty acid molecules that it contains. Digestive enzymes Digestion is the process by which food substances from complex compounds are broken down to simple comp

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