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ContentsGetting the most from this book

About this book

Content GuidanceEnergy and living things

Respiration

Photosynthesis

Microbiology

Populations

Excretion

The nervous system

Responses in plants

Questions & AnswersQ1 ATP and respiration

Q2 Respiration

Q3 Photosynthesis

Q4 Microbiology

Q5 Populations

Q6 The kidney

Q7 The nervous system

Q8 The nitrogen cycle

Knowledge check answers

Glossary

Index

Getting the most from this bookExaminer tipsAdvice from the examiner on key points in the text to help you learn and recall unit content, avoidpitfalls, and polish your exam technique in order to boost your grade.

Knowledge checkRapid-fire questions throughout the Content Guidance section to check your understanding.

Knowledge check answersTurn to the back of the book for the Knowledge check answers.

Summaries• Each core topic is rounded off by a bullet-list summary for quick-check reference of what you

need to know.

Questions & Answers

About this bookThis guide will help you to prepare for BY4, the examination for WJEC A2 Biology Unit 4:Metabolism, Microbiology and Homeostasis. Your understanding of many of the principles in Unit 1may be re-examined here as well.

Content GuidanceThe Content Guidance section covers all the concepts you need to understand and facts you need toknow for the BY4 exam. It also includes examiner tips and knowledge checks to help you prepare forBY4.

The order in which topics appear in the guide follows the order of the specification with theexception of the detail of chemiosmosis, which is included in respiration and photosynthesis, ratherthan with ATP.

The concepts in each topic are presented first followed by details of the processes and adaptations ofthe various structures involved. You are advised to familiarise yourself with the key ideas beforeattempting to learn the associated facts.

The A2 biology course is more demanding than AS and includes stretch-and-challenge and synopticaspects.

Stretch and challenge: At A2 you have to develop a greater understanding of biological concepts anddemonstrate a greater ability to apply your knowledge and understanding (AO2). The ContentGuidance section contains boxes detailing investigations carried out on particular aspects of biology.The specification does not require you to know the details of these investigations, but they will giveyou an idea of the sort of information you could be provided with to assess AO2.

Synoptic element: You need to start piecing together the topics you have studied so far and try to seethe links between them; this is the synoptic element. In Unit BY1 you learnt the ‘core concepts’ inbiology — the fundamentals of biochemistry and cell biology. This knowledge underpins all aspectsof A2 biology. To ensure you have a good understanding of Unit BY4 it is essential that you revisitthese concepts. Synoptic links are highlighted throughout the Content Guidance section.

Questions and AnswersThis section will help you to:• familiarise yourself with the question styles you can expect in the unit test• understand what the examiners mean by terms such as ‘describe’ and ‘explain’• interpret the question material — especially any data that the examiners give you• write concise answers to the questions that the examiners set

It would be impossible to give examples of every kind of question in one book, but these should giveyou a flavour of what to expect. Two students, Student A and Student B, attempt each question in thissection. Their answers, along with the examiner comments, should help you to see what you need to

do to score a good mark — and how you can easily not score a mark even if you understand thebiology.

Content Guidance

Energy and living thingsKey concepts you must understand• Most energy available to living organisms is derived directly or indirectly from the sun.• Autotrophic organisms (e.g. plants) convert light energy into chemical energy during

photosynthesis. This chemical energy is locked up within organic molecules.• All organisms, both autotrophic and heterotrophic, break down these organic molecules during

respiration to produce adenosine triphosphate (ATP).• ATP is the only source of immediate energy within the cell for processes such as active transport,

muscle contraction and the synthesis of organic molecules, such as proteins.• ATP is often referred to as the ‘universal energy currency’ because it transfers energy for

biochemical reactions in the cells of all living organisms.

Adenosine triphosphate (ATP)The structure of ATP is shown in Figure 1. It is a free RNA nucleotide consisting of a ribose sugar,the base adenine and three phosphate groups (adenosine = ribose + adenine).

Figure 1 A molecule of ATP

Examiner tipSynoptic link to BY1: In Unit BY1 you studied the structure of nucleotides and the functions ofnucleic acids. Revisiting these topics will help you with aspects of energy transfer.

As energy is released when ATP is hydrolysed to adenosine diphosphate (ADP) and inorganicphosphate (Pi), it is an exergonic reaction (see Figure 2). This reaction is catalysed by the enzymeATPase and involves the removal of the terminal phosphate group. The reaction is always coupledwith an energy-requiring reaction (endergonic reaction) so that energy is transferred.

Figure 2 The interconversion of ATP, ADP and Pi

Examiner tipMake sure that you get the name right. You could be given a diagram similar to Figure 1 and asked toname it or label it. Many students get this wrong by labelling the base adenosine, or calling themolecule adenine triphosphate or adenosine triosephosphate.

ATP acts as an energy carrier and is suited to its function because:• the molecule is soluble and can be transported within the cell (but cannot leave the cell),

transferring chemical energy to energy-requiring processes

• the hydrolysis of ATP releases small quantities of energy (30.6kJmol−1) that are matched closelyto the energy required in the coupled reaction

• the energy is transferred quickly as the hydrolysis of ATP requires only one enzyme

Examiner tipThe ‘law of conservation of energy’ states that energy can neither be created nor destroyed.However, energy can be converted from one form to another. When answering questions relating tothe hydrolysis of ATP, you must refer to energy being released. You will not gain credit for statingthat energy is produced.

ATP is reformed from ADP and Pi by a condensation reaction. This requires the input of energy, i.e.it is an endergonic reaction. The energy required can come from cellular respiration or from thetransduction of light energy during photosynthesis. This reaction is catalysed by the enzyme ATPsynthase (also known as ATP synthetase).

Knowledge check 1(a) Give three examples of cellular activities that require ATP.(b) Describe three advantages of ATP for its function as the universal source of energy.

SummaryAfter studying this topic you should be able to:• understand the importance of chemical energy in biological processes• recognise the structure of ATP and describe its role as an energy carrier and its use in the

liberation of energy for cellular activity

RespirationKey concepts you must understand• Respiration is a process that occurs within the cells of all living organisms. It can be represented

by the following simple chemical equation:

C6H12O6 + 6O2 6CO2 + 6H2O

• Respiration releases chemical energy from the oxidation of organic molecules, such as glucose, tosynthesise ATP.

• There are three ways in which molecules can be oxidised or reduced:

Oxidation Reduction1 Gaining oxygen Losing oxygen2 Losing hydrogen Gaining hydrogen3 Losing electrons (e−) Gaining electrons (e−)

• Oxidation and reduction reactions always take place together because as one molecule is oxidisedanother molecule is reduced. These chemical reactions are called redox reactions.

• Figure 3 represents a typical step in the respiratory pathway. Note that the coenzyme (NAD) isreduced as the organic molecule is oxidised.

Figure 3 Oxidation of an organic molecule coupled with the reduction of a coenzyme

• The oxidation reactions involved in respiration are exergonic. The energy released from theorganic molecules is used to reduce the coenzymes NAD and FAD, as these reactions are coupled.During each oxidation reaction a small quantity of energy is, in effect, transferred to thesecoenzymes.

• During glycolysis, the link reaction and Krebs cycle, organic molecules are repeatedly oxidisedand therefore most of the energy contained within glucose is transferred to the reduced coenzymesNADH2 and FADH2. When the coenzymes are re-oxidised the stored energy is used to synthesiseATP via oxidative phosphorylation.

• The series of oxidation reactions in respiration brings about the gradual release of chemicalenergy from organic molecules in a series of small steps (as opposed to combustion which is theuncontrolled release of energy in a single step).

• Aerobic respiration occurs in the presence of oxygen. Respiration that takes place in the absenceof oxygen is called anaerobic respiration.

Examiner tipsIn biology it is more helpful to think about oxidation in terms of loss of hydrogen and loss ofelectrons and reduction in terms of gain of hydrogen and electrons.

Synoptic link to BY1: Respiration is a series of enzyme-catalysed reactions. Therefore factors thataffect enzymes affect the rate of respiration. The most important factor influencing the rate ofrespiration is temperature.

Aerobic respirationMitochondria are present in all eukaryotic cells and they are involved in synthesis of ATP duringaerobic respiration. Figure 4 shows the structure of a mitochondrion. The organelle is composed of adouble membrane enclosing a fluid-filled matrix. The inner membrane is highly folded to formcristae. This increases the surface area for the synthesis of ATP.

Figure 4 Structure of a mitochondrion

Figure 5 shows the location of the four stages of respiration. ATP is synthesised mainly in themitochondria.

Figure 5 Outline of the stages of aerobic respiration

Glycolysis (splitting of glucose)Glycolysis occurs in the cytoplasm — this is where the enzymes for glycolysis are located. The mainstages in the pathway are shown in Figure 6.

Figure 6 The steps involved in glycolysis

• Step 1: Two molecules of ATP are required for the phosphorylation of glucose to produce hexosebisphosphate. The energy from the hydrolysis of ATP activates glucose and makes the moleculemore reactive.

• Step 2: Hexose bisphosphate is split (lysis) producing two molecules of triose phosphate.• Step 3: Triose phosphate (TP) is oxidised via a dehydrogenation reaction into pyruvate. The

hydrogen removed is used to reduce the coenzyme NAD to reduced NAD (NADH2). This reactionis exergonic, and the energy released is used to synthesise four ATP molecules by substrate-levelphosphorylation.

Glycolysis results in a net gain of two ATP molecules (two ATP molecules are used initially and fourATP molecules are synthesised).

The link reactionGlycolysis links to the Krebs cycle via the link reaction (Figure 7). It takes place in themitochondrial matrix (where the enzymes involved in the link reaction are found).

Figure 7 The link reaction

Knowledge check 2Why is glycolysis referred to as being anaerobic?

Specific carrier proteins contained in the outer mitochondrial membrane transport pyruvate into thematrix. (Note that there are no glucose carrier proteins.) The pyruvate undergoes oxidativedecarboxylation to form acetate (a 2C molecule). This involves:• the removal of hydrogen (oxidation) to reduce the coenzyme NAD• the removal of a carboxyl group (decarboxylation) to form carbon dioxide

The acetate combines with coenzyme A to form acetyl coenzyme A.

Two molecules of pyruvate enter the link reaction, so the reactions shown in Figures 7 and 8represent only one-half of the reactions for the complete oxidation of one molecule of glucose.

Figure 8 The steps involved in the Krebs cycle

Examiner tipIt is important that you can state the precise locations within the cell where the different stages ofrespiration occur. You will not gain any credit for stating that the link reaction and Krebs cycle takeplace in the ‘mitochondria’ or in the ‘matrix’. You must say ‘mitochondrial matrix’.

Krebs cycleThe Krebs cycle occurs in the mitochondrial matrix (this is where the correct enzymes are located).It involves a series of decarboxylation reactions and dehydrogenation reactions. Carbon dioxide,ATP and reduced coenzymes are produced (see Figure 8).

Acetyl coenzyme A releases acetate (2C), which then combines with a 4C acid to form a 6C acid.This 6C acid is broken down in a series of oxidative decarboxylation reactions that regenerate the 4Cacid. These reactions involve:• the loss of hydrogen (dehydrogenation/oxidation) and the loss of carbon dioxide (decarboxylation)• the reduction of the coenzymes NAD and FAD to NADH2 and FADH2• the production of ATP from ADP and Pi by substrate-level phosphorylation

Examiner tipFigure 8 shows the names of some of the intermediate molecules involved in the Krebs cycle. Theymay appear in the exam, but you are not expected to know them. What is important is that youunderstand what is produced when these molecules are recycled.

Knowledge check 3Explain why carbon dioxide is not produced when glucose is added to a preparation of isolatedmitochondria.

The electron transport chain (ETC)The electron transport chain involves a chain of electron carriers located on the inner mitochondrialmembrane (cristae). The cristae have a large surface area so there are more electron carriers, whichincreases ATP synthesis. The reduced coenzymes, NADH2 and FADH2, produced during glycolysis,the link reaction and the Krebs cycle act as a source of electrons and protons.

Figure 9 shows the electron carriers at progressively lower energy levels. As electrons pass alongthe chain of carriers in a series of redox reactions, they release energy. This energy is used tosynthesise ATP by oxidative phosphorylation. Oxygen is the terminal electron acceptor. It combineswith protons (H+) and electrons (e−) and is reduced to water.

Figure 9 The electron transport chain

Figure 9 shows that for each NADH2 entering the chain three ATP molecules are produced; eachFADH2 only generates two ATP molecules.

Chemiosmotic theory of oxidative phosphorylationFigure 10 shows how ATP is synthesised during oxidative phosphorylation:• The energy released from the electrons during the redox reactions is used to pump protons (H+)

from the matrix through the inner mitochondrial membrane into the intermembrane space.• The protons accumulate so that steep concentration and electrochemical gradients are established

across the inner mitochondrial membrane.• The inner membrane is impermeable to protons, so they can only diffuse back into the matrix via the

stalked particles, which consist of a chemiosmotic channel protein attached to the enzyme ATPsynthase.

• The flow of protons through the ATP synthase provides the energy required to produce ATP fromADP and Pi.

Figure 10 Electron transport chain and chemiosmosis

Examiner tipSynoptic links to BY1: In Unit 1 you learnt that cyanide is a respiratory inhibitor. It is a non-competitive inhibitor of the enzyme cytochrome oxidase, which is associated with the final protonpump in the electron transport chain. When cyanide attaches to the enzyme the electron transport

chain cannot function and oxidative phosphorylation cannot occur.

Each molecule of NADH2 entering the chain results in three ATP molecules being synthesised asthree proton pumps are involved. Each molecule of FADH2 results in two ATP molecules beingsynthesised as only two proton pumps are involved.

Knowledge check 4Describe the role of oxygen in aerobic respiration.

Summary of aerobic respirationAerobic respiration involves the oxidation of glucose via a series of dehydrogenation reactions (seeFigure 11). Table 1 on p. 14 shows a summary of the molecules involved and the location of thedifferent stages in aerobic respiration.

Figure 11 Summary of aerobic respiration

Table 1 Molecules involved and the location of the different stages in aerobic respiration

Table 2 on p. 14 shows the number of ATP molecules produced via substrate-level phosphorylationand oxidative phosphorylation. The complete oxidation of one molecule of glucose can produce a

maximum of 38 molecules of ATP:• Four molecules of ATP (two from glycolysis and two from the Krebs cycle) are produced by

substrate-level phosphorylation.• Thirty-four molecules of ATP are produced by oxidative phosphorylation.

Table 2 The ATP tally

Examiner tipOn an exam paper, the reduced forms of NAD and FAD are likely to be written as reduced NAD andreduced FAD respectively (as in Figure 11). When answering questions it is perfectly acceptable touse NADH2 or NADH + H+ and FADH2 or FADH + H+.

Knowledge check 5Apart from oxygen and carbon dioxide, name:(a) two molecules that show a net movement into a mitochondrion(b) two molecules that show a net movement out of a mitochondrion

Knowledge check 6Explain what would happen to the production of ATP in an organism if:(a) its body temperature rose slightly(b) there was a reduced concentration of enzymes in the mitochondrial matrix

Anaerobic respirationAnaerobic respiration takes place in the cytoplasm of cells and occurs in the absence of oxygen. It isthe incomplete breakdown of glucose. Without oxygen the electron transport chain cannot occur andNADH2 and FADH2 are not oxidised. NAD and FAD become limiting factors (i.e. they run out) andtherefore the dehydrogenation reactions of the Krebs cycle and the link reaction can no longer occur.Glycolysis continues because the pyruvate enters a different pathway and is reduced, thereforeoxidising NADH2. The pyruvate is converted to lactate in animals and ethanol in plants and fungi.

Figure 12 shows the anaerobic pathway in animals. Pyruvate is reduced by NADH2 to form lactate.This recycles the NAD, which is then reused to oxidise triose phosphate, allowing ATP to besynthesised.

Figure 12 Anaerobic respiration in animals

Figure 13 shows the anaerobic pathway in plants and fungi. Pyruvate undergoes a decarboxylationreaction producing ethanal and carbon dioxide. The ethanal is then reduced by NADH2 to formethanol. This recycles NAD, which is then reused to oxidise triose phosphate.

Figure 13 Anaerobic respiration in plants and fungi

Knowledge check 7State the name of the molecule that acts as the terminal electron acceptor in the followingprocesses:(a) aerobic respiration

(b) anaerobic respiration in a muscle cell(c) anaerobic respiration in yeast

Comparison of energy yieldsNot all the energy of the glucose molecule is transferred to ATP. There is a loss of energy as heatenergy.

Aerobic respiration involves the complete breakdown of glucose to carbon dioxide and water. Itproduces 38 molecules of ATP per molecule of glucose. It is about 40% efficient.

Anaerobic respiration involves the incomplete breakdown of glucose and produces two moleculesof ATP per molecule of glucose. It is about 2% efficient. Energy still remains locked up inlactate/ethanol.

Alternative respiratory substratesUnder certain circumstances fats and proteins may be used as respiratory substrates. Individuals areable to survive for long periods without food because they can use their reserves of carbohydrate, fatand protein:• Glycerol is converted into triose phosphate and enters glycolysis. Long-chain fatty acid molecules

are split into 2C fragments, which enter the pathways as acetyl coenzyme A.• When a person is starving, tissue protein used as a source of energy. It is hydrolysed into its

constituent amino acids, which are deaminated (NH2 groups are removed). This leaves an organicacid that can enter the Krebs cycle.

Examiner tipAll organic molecules can be respired, but glucose is the main respiratory substrate. Most questionson respiration use glucose as the starting point.

SummaryAfter studying this topic you should be able to:• understand that all living organisms carry out respiration to provide energy in their cells• describe the process of glycolysis and the production of pyruvate, ATP and reduced NAD• describe the formation of acetyl coenzyme A during the link reaction• describe the Krebs cycle and the production of ATP, reduced NAD and reduced FAD with

release of carbon dioxide• explain the role of reduced NAD (and reduced FAD) as a source of electrons and protons for the

electron transport system• define the terms dehydrogenation, decarboxylation, substrate-level phosphorylation and

oxidative phosphorylation• explain the synthesis of ATP by means of a flow of protons through the enzyme ATP synthase by

chemiosmosis• describe and explain the breakdown of glucose under aerobic and anaerobic conditions• state how many molecules of ATP are produced in each of the four stages of respiration• understand that all organic molecules can be used as respiratory substrates and describe how

lipids and amino acids are utilised

PhotosynthesisKey concepts you must understand• Green plants are autotrophic organisms that, during photosynthesis, synthesise complex organic

molecules from simple inorganic molecules using light energy. Photosynthesis can be representedby the following simple chemical equation:

6CO2 + 6H2O C6H12O6 + 6O2

• During photosynthesis light energy is converted into chemical energy in the form of organicmolecules.

• Photosynthesis takes place in the chloroplasts of plant cells. The main site of photosynthesis is thepalisade tissue of the leaf.

• Chloroplasts are surrounded by a double membrane that encloses a fluid-filled stroma (see Figure14). Within the stroma is a series of flattened membrane-bound sacs called thylakoids, which formstacks called grana. These thylakoid membranes provide a very large surface area for theabsorption of light energy.

• There are two main stages in photosynthesis: the light-dependent stage and the light-independent stage. In the light-dependent stage:

– Photosynthetic pigments (e.g. chlorophyll) absorb light energy, which results in the loss ofelectrons. The electrons are transferred to an electron acceptor.

– The energy absorbed by the electrons is then released via a series of redox reactions and usedto synthesise ATP from ADP and Pi (photophosphorylation) and to reduce the coenzymeNADP to NADPH2.

– Light energy is converted into chemical energy within the organic molecules ATP andNADPH2.

Figure 14 The structure of a chloroplast

• The light-independent stage involves a series of enzyme-catalysed reactions in which carbondioxide is reduced to form a carbohydrate. This requires NADPH2 and energy released from the

hydrolysis of ATP.• Figure 15 shows an overview of the two main stages in photosynthesis occurring within a

chloroplast.

Figure 15 Outline of the stages of photosynthesis

Examiner tipsMost of life on Earth is dependent upon photosynthesis:• It converts light energy into chemical energy that can be used by other organisms.• It provides a source of complex organic molecules for heterotrophic organisms.• It releases oxygen which is necessary for aerobic respiration.

Synoptic link to BY1: Starch grains are found in chloroplasts. Within the stroma molecules ofglucose undergo condensation reactions to form starch. Therefore, there must be an enzyme in thestroma that catalyses these reactions.

Photosynthetic pigmentsThere are several photosynthetic pigments found in plants. They can be divided into two main groups,the chlorophylls and the carotenoids. The function of the pigments is to absorb light energy, therebyconverting it into chemical energy.

Chromatography can be used to separate out the leaf pigments so that they can be identified.

Absorption and action spectraThe absorption spectrum (Figure 16a) shows the percentage of light absorbed by a particularpigment at different wavelengths of light. The action spectrum (Figure 16b) shows the rate ofphotosynthesis at different wavelengths of light. Since the two graphs show a similar trend, it suggeststhat these pigments are those responsible for the absorption of the wavelengths of light used inphotosynthesis.

Figure 16 (a) Absorption spectra for photosynthetic pigments; (b) The action spectrum forphotosynthesis

From the absorption spectrum it can be seen that:• chlorophyll molecules absorb wavelengths of light in the blue–violet and red regions of the visible

spectrum• the peak absorptions for chlorophyll a and chlorophyll b differ slightly• the carotenoids (e.g. xanthophyll and carotene) absorb wavelengths of light in the blue–violet

region

Examiner tipAt A2 you are expected to use the correct terminology when answering questions. You will not gaincredit for vague statements such as ‘chlorophyll absorbs light’ or ‘chlorophyll absorbs blue–violetlight’. You must refer to chlorophyll absorbing either light energy or wavelengths of light in theblue–violet region of the spectrum.

PhotosystemsIt can be seen from Figure 17 that the photosynthetic pigments are arranged in clusters embedded inthe thylakoid membranes of the chloroplasts. These clusters are known as photosystems. A

photosystem consists of an antenna complex and a reaction centre (see Figure 18).

Figure 17 (a) The structure of a chloroplast; (b) A section through a single thylakoid; (c) Pigments ina thylakoid membrane

Figure 18 A photosystem

In a photosystem, chlorophyll a is the main photosynthetic pigment and is found in the reaction centre.Chlorophyll b and the carotenoids are accessory pigments found in the antenna complex. Themolecules in the complex are arranged so as to channel light energy to the reaction centre.

There are two types of photosystem:• In photosystem I, the reaction centre is called P700 as it contains a chlorophyll a molecule with a

maximum absorption at a wavelength of 700nm.• In photosystem II, the reaction centre is called P680 as it contains a chlorophyll a molecule with a

maximum absorption at a wavelength of 680nm.

Knowledge check 8Explain the advantage to a plant of having chloroplasts that contain several different light-absorbing pigments.

Knowledge check 9Explain why plants appear green.

Examiner tipSynoptic link to BY1: In BY1, you learnt about the structure of chloroplasts and mitochondria. Youmay be given electron micrographs or drawings of these organelles and asked to identify structuresor to indicate where the stages of photosynthesis or respiration occur.

Knowledge check 10State exactly where in the chloroplast you would expect to find photosystems.

Knowledge check 11Photosystems contain several pigments. State the location of the following pigments within aphotosystem:(a) chlorophyll a(b) chlorophyll b

The light-dependent stageThe light-dependent stage involves the formation of ATP by photophosphorylation and the reductionof the coenzyme NADP. These events are summarised in a diagram, known as the ‘Z-scheme’ shownin Figure 19.

Figure 19 The light-dependent stage (the ‘Z-scheme’)

(1) Photosynthetic pigments in the antenna complex of photosystem II absorb light energy. The energyis transferred to the reaction centre where it excites two electrons in the chlorophyll a molecule.

(2) The excited electrons are boosted to a higher energy level. They leave the chlorophyll a moleculeand are received by an electron acceptor.

(3) The electrons are passed along the electron transport chain (ETC) in a series of redox reactions tophotosystem I, which is at a lower energy level. The energy lost by the electrons is used toconvert ADP and Pi to ATP — this is photophosphorylation.

(4) Light absorbed by photosystem I boosts two electrons from the chlorophyll a molecule in thereaction centre to an even higher energy level. The electrons are received by another electronacceptor.

(5) Electrons (from the chlorophyll a molecule) and H+ (from the photolysis of water) are used toreduce NADP (the final electron acceptor) to NADPH2.

Examiner tipNADP is the coenzyme involved in photosynthesis. Do not confuse it with NAD, which is involvedin respiration. Remember that the letter ‘p’ occurs in both NADP and photosynthesis.

Photolysis of waterThis occurs at stage 1 of the light-dependent reaction. The electrons removed from the chlorophyll amolecule in photosystem II are replaced by electrons (e−) from a water molecule. The loss ofelectrons from the water molecule causes it to dissociate into protons (H+) and oxygen — this is

known as photolysis.

Examiner tipThe electrons lost from the chlorophyll a molecule in photosystem I are used to reduce NADP.These electrons are replaced by those lost by the chlorophyll a molecule in photosystem II, whichare in turn replaced by those lost by the water molecule during photolysis.

Chemiosmotic theory of photophosphorylationFigure 20 shows how ATP is synthesised during non-cyclic photophosphorylation. As the electronspass along the electron transport chain they lose energy. This energy is used to pump protons (H+)from the stroma, across the thylakoid membrane and into the thylakoid space. The protons accumulateso that steep concentration and electrochemical gradients are established between the thylakoid spaceand the stroma. These gradients are also maintained by:• the photolysis of water, which occurs in the thylakoid space and increases the H+ concentration• the reduction of NADP, which occurs in the stroma and decreases the H+ concentration

Figure 20 Chemiosmosis and non-cyclic photophosphorylation

The protons (H+) diffuse back into the stroma through the chemiosmotic protein channels where theenzyme ATP synthase is located. The flow of protons through ATP synthase provides the energyrequired to produce ATP from ADP and Pi.

Electrons from the chlorophyll a molecules in photosystem I are used to reduce NADP and arereplaced indirectly by electrons from the photolysis of water. This is known as non-cyclicphosphorylation and is represented by stages 1 to 5 in Figure 19.

You can see from Figure 19 that the electron acceptor at stage 4 is at the highest energy state. It ispossible for some of these excited electrons to return to the chlorophyll a molecule in photosystem Ivia the electron transport chain. This is known as cyclic phosphorylation and is represented by stages4, 6 and 3 in Figure 19.

Cyclic and non-cyclic photophosphorylation are compared in Table 3.

Table 3 Cyclic and non-cyclic photophosphorylation

Feature Cyclic photophosphorylation Non-cyclic photophosphorylationPhotosystems involved I only I and IIPhotolysis of water No YesElectron donor Chlorophyll a in photosystem I Chlorophyll a in photosystem ITerminal electron acceptor Chlorophyll a in photosystem I NADPProducts ATP ATP, NAPH2 and oxygen

The light-independent stage (Calvin cycle)The light-independent reactions take place in the stroma of the chloroplast because this is where theenzymes involved are located. During these reactions ATP and NADPH2 (from the light-dependentstage) are used to reduce carbon dioxide to triose phosphate (a 3C carbohydrate). Figure 21 showsthe main stages in the Calvin cycle.

Figure 21 The steps involved in the light-independent stage (the Calvin cycle)

(1) Carbon dioxide combines with ribulose bisphosphate (RuBP) to form two molecules of glycerate-3-phosphate (GP), which is the first product of photosynthesis. This reaction is a carboxylation(addition of carbon dioxide) reaction and is catalysed by the enzyme Rubisco (ribulosebisphosphate carboxylase). As carbon dioxide is converted from an inorganic form into anorganic molecule the process is also referred to as carbon fixation.

(2) NADPH2 is used to reduce the two molecules of glycerate-3-phosphate into two molecules oftriose phosphate (TP). The hydrolysis of ATP provides the energy for this reaction.

(3) Most of the TP (five out of six molecules) is converted by a series of reactions into RuBP. ATPsupplies the phosphate and energy required.

(4) Some of the TP (one of six molecules) is converted rapidly to glucose and other carbohydrates,amino acids, lipids and nucleic acids.

Examiner tipsBalancing the equation:

6CO2 + 6H2 O → C6H12O6 + 6O2

For every molecule of CO2 that enters the Calvin cycle two molecules of triose phosphate (TP) areproduced. Six molecules of CO2 entering the cycle results in the production of 12 molecules of TPof which two molecules leave the cycle and combine to form one molecule of glucose.

Synoptic links to BY1: Plants require magnesium to synthesise chlorophyll. Magnesium deficiencyleads to chlorosis and death. Plants also require nitrogen to synthesise amino acids and nucleic

acids from triose phosphate. Plants obtain their nitrogen as nitrates (NO3−) or ammonium ions(NH4+) from the soil. (This is also a synoptic link to the nitrogen cycle.)

Knowledge check 12Name the two products from the light-dependent stage of photosynthesis that are required for theCalvin cycle.

Investigating photosynthesis: Calvin’s lollipopMelvin Calvin was an American biochemist who investigated the pathway by which carbon dioxideis converted into organic compounds during photosynthesis. A suspension of the unicellular algaChlorella was placed in a flattened glass vessel that was called the ‘lollipop’ (see Figure 22a). Thesuspension of Chlorella was supplied with radioactive carbon dioxide. The lollipop wasilluminated and the algae allowed to photosynthesise. As the Chlorella photosynthesised, theradioactive carbon dioxide was ‘fixed’ and incorporated into organic molecules (the intermediatecompounds), which became radioactive. At specific time intervals samples of the Chlorella werereleased into boiling alcohol. This denatured enzymes, killed the Chlorella and stopped the light-independent reactions at a particular point in time.

Figure 22 (a) Calvin’s lollipop; (b) autoradiograms showing the different molecules synthesised

during Calvin’s experiments

Compounds that the radioactive carbon had reached at a particular moment were determined bychromatography and autoradiography (see Figure 22b). The order in which each compound isproduced was found by identifying the molecules and analysing the results. From these resultsCalvin discovered the metabolic pathway that is now known as the light-independent stage ofphotosynthesis.

SummaryAfter studying this topic you should be able to:• understand that chloroplasts are transducers that convert light energy into the chemical energy of

ATP• describe the absorption of various wavelengths of light by chlorophyll and associated pigments

and describe the relationship between absorption and action spectra• describe both the arrangement of photosynthetic pigments within photosystems and energy

transfer to reaction centres• describe the processes of cyclic and non-cyclic photophosphorylation (light-dependent stage)

including: – the source of electrons for the electron transport chain – photolysis of water as a source of electrons for photosystem II – the reduction of NADP by the addition of electrons and protons• explain the synthesis of ATP by means of a flow of protons through the enzyme ATP synthase by

chemiosmosis• describe the reactions occurring in the light-independent stage including: – the role of NADPH2 as a source of reducing power and ATP as a source of energy for the

reactions – the uptake of carbon dioxide and the role of Rubisco – the fate of triose phosphate• describe the role of magnesium and nitrogen in plant metabolism

MicrobiologyKey concepts you must understand• Microbiology is the study of organisms that are too small to be seen with the naked eye.• For thousands of years people have been manipulating microorganisms to produce various foods

and drinks including bread, cheese, yoghurt, beer and wine.• With increased knowledge and understanding of microorganisms, modern biotechnology is used to

produce other useful products, such as enzymes and antibiotics.• Genetic engineering has given the potential to produce a wide range of products including human

proteins such as insulin.• To produce useful products such as new antibiotics it is necessary to be able to: – culture microorganisms in the laboratory – have an understanding of a microorganism’s metabolism to provide the optimum conditions for

growth and the conditions necessary for the production of the useful product.

Bacterial classificationBacteria are classified by the shape of their cells and their reaction to the Gram stain.

Examiner tipSynoptic links to BY1: Look back at your notes on BY1 and construct a table to compare thestructure of a prokaryotic cell with a eukaryotic cell.

Classification by shapeThere are three main types of bacteria as classified by shape (see Figure 23). The shape of thebacteria is due to their rigid cell wall, which has a unique structure.

Figure 23 The classification of bacteria by shape

Classification by reaction to the Gram stainDifferent types of bacteria can also be identified using the Gram stain. The Gram-stain procedureand the colours that the bacteria are stained are shown in Figure 24.

Figure 24 The stages involved in the Gram staining technique

• Gram-positive bacterial stain purple because their cell walls retain crystal violet.• Gram-negative bacteria are stained red by the counterstain, as their cell walls do not retain crystal

violet.

Examiner tipWhen asked to explain the results of the Gram staining technique, a common error is to state thatGram-positive bacteria stain purple while Gram-negative bacteria do not. Both types stain purplebut Gram-positive bacteria retain the crystal violet stain whereas Gram-negative bacteria do not.

The Gram reaction reflects the more complex structure of Gram-negative cell walls.

Structure of bacterial cell wallsThere are two types of bacterial cell wall (Figure 25):• Gram-positive bacteria have a thick layer consisting of peptidoglycan.• Gram-negative bacteria have a thin layer consisting of peptidoglycan covered by a layer

containing lipopolysaccharides.

Figure 25 The difference in structure of the cell wall in Gram-positive and Gram-negative bacteria

The layer of lipopolysaccharides provides some protection against antibiotics and the enzymelysozyme, making Gram-negative bacteria more difficult to kill.

Knowledge check 13Staphylococcus aureus is a Gram-positive bacterium. Describe the shape of its cell and thestructure of its cell wall.

Culturing bacteriaTo study individual species of bacteria they have to be isolated and cultured. Bacterial cultures mustbe handled in such a way as to prevent their contamination by unwanted organisms. Of equalimportance is the need to ensure that no organisms escape to cause damage to health and theenvironment.

Aseptic techniques aim to prevent contamination of cultures and the environment. Figure 26 shows theaseptic techniques involved in the transfer of bacteria from a culture bottle to an agar plate.

Figure 26 Aseptic techniques and the safe transfer of microorganisms

Aseptic techniquesBefore culturing bacteria the equipment and media must be sterilised. Heat is normally used, with theequipment being placed in an autoclave at 121°C for 15 minutes.(1) Hands must be washed (before and after) with antibacterial soap.(2) The work surface must be cleaned (before and after) with disinfectant.(3) A sterile Petri dish containing sterile agar jelly is used. (Heat-labile plastics are irradiated.)(4) The plate is inoculated close to a Bunsen burner. This provides updrafts that move airborne

microorganisms away from the plates.(5) The lid of the culture bottle is removed and kept in the hand, away from the surface of the bench.

The mouth of the culture bottle is flamed to kill any unwanted microbes.(6) A flamed inoculating loop or sterile pipette is used to transfer the bacterial culture to the agar

plate.(7) When inoculating the plate, the lid of the Petri dish is opened only just wide enough to add the

sample.

(8) The plates are then secured with adhesive tape and incubated.

Examiner tipHuman pathogenic bacteria grow best under anaerobic conditions at 37°C. It is therefore importantnot to completely seal the inoculated plates with adhesive tape and to incubate them at a lowertemperature, usually 25°C.

Population growth

Patterns of growthWhen microorganisms are grown under laboratory conditions there are four phases of populationgrowth (Figure 27 on p. 28).

Figure 27 The population growth curve of a bacterial culture grown under laboratory conditions

During the lag phase there is a slow increase in the number of cells in the population. The bacteriaare adjusting to their new environment. This involves switching on genes and producing digestiveenzymes. As bacteria are saprophytic they secrete these enzymes and digest the mediumextracellularly. The soluble products of digestion are then absorbed. Time is also required for thereplication of DNA to occur before the cells can divide.

Examiner tipWhen answering questions about the phases of the growth curve avoid making vague statements andgive a detailed answer demonstrating your understanding of biology. You must be careful not torefer to ‘birth rate’ when talking about bacteria, as this will not gain credit.

During the log phase nutrients are in plentiful supply and the waste products that have been producedare in low concentrations. The rate of cell division is at a maximum and the population is growing atan exponential rate.

During the stationary phase, the carrying capacity (maximum population size that the environmentcan support indefinitely) has been reached and equilibrium is established. The number of new cellsbeing added to the population is equal to the number of cells dying. Limiting factors (nutrientdepletion and waste accumulation) prevent a further increase in population size.

During the death or decline phase, more cells are dying than are being produced. This is because ofshortage of nutrients and/or the accumulation of waste. The population begins to decrease.

Factors that affect growthEach species has optimum conditions for growth.

NutrientsThe nutrient broth or agar provided must contain certain nutrients to allow population growth.Microorganisms require:• a source of carbon, nitrogen and phosphorus• a respiratory substrate, which is normally glucose• vitamins and minerals to act as coenzymes• water — all metabolic reactions occur in aqueous solution

If a nutrient is depleted it will become a limiting factor and reduce the growth rate.

Knowledge check 14Explain why bacteria require a source of carbon, nitrogen and phosphorus.

TemperatureGrowth is coordinated by enzymes. Enzymes work most efficiently over a narrow range oftemperatures (around the optimum). If the temperature falls too low, the rate of enzyme-catalysedreactions becomes too low to sustain life. If the temperature is too high, the denaturation of enzymescauses cell death.

The optimum range of temperature for most microorganisms is between 20°C and 45°C. Thesebacteria are called mesophiles. Thermophiles have an optimum temperature range above 45°C;psychrophiles have an optimum temperature range below 20°C.

pHBacterial enzymes only work efficiently within a narrow range of pH. For most species this isbetween pH 5 and pH 7.5 (this is why stomach acid kills most species of bacteria). However,microorganisms in general can tolerate a wider range of pH than plant and animal cells. Some speciesgrow in very acidic environments (pH 2.5); others grow in very alkaline conditions (pH 9).

OxygenDifferent microorganisms have different oxygen requirements:• Microorganisms that require oxygen for growth at all times are called obligate aerobes.• Some microorganisms find oxygen toxic as it inhibits their respiration and they cannot grow in its

presence. These are called obligate anaerobes.• Populations of some bacteria grow rapidly in the presence of oxygen, but can survive without it,

although their population growth rate is slow. These are called facultative anaerobes.

Knowledge check 15The bacterium Escherichia coli is a Gram-negative facultative anaerobe. Describe what is meantby the terms ‘Gram negative’ and ‘facultative anaerobe’.

Measuring population growthA number of techniques can be used to estimate population. There are two types of populationestimate:• A total count includes both living and dead cells. It may be measured by haemocytometry. A

disadvantage of this type of count is that the numbers in the population can be overestimated.• A viable count includes only living cells. The method of dilution plating is used. This technique

involves growing bacteria to form distinct colonies that can be counted. It is based on theassumption that a single cell gives rise to a single colony. A disadvantage of this technique is thatnumbers can be underestimated because of clumping of cells when the plates are made, i.e. severalcells form a single colony.

Examiner tipSometimes serial dilutions involve the transfer of 1cm3 of the culture into 99cm3 of sterile nutrientmedium. This produces a 10−2 dilution. If 1cm3 of this dilution is transferred to 99cm3 of sterilenutrient medium it produces a 10−4 dilution. If asked in an exam to calculate the numbers of bacteriain the original sample make sure you read the information carefully: is it a 1 in 10 or a 1 in 100dilution and is 1cm3 or 0.1cm3 of the dilution being used to plate out?

Steps involved in dilution platingWhen cultures of bacteria are transferred to an agar plate it is impossible to distinguish individualcolonies. Therefore the original culture is diluted down, usually in ten-fold steps, to provide a finalnumber within a countable range. This is known as serial dilution (Figure 28 on p. 30).(1) 1cm3 of the original culture is transferred into 9cm3 of sterile nutrient medium (or sterile water).

This is then mixed to ensure an even distribution (×10 dilution).(2) 1cm3 of this mixture is transferred into 9cm3 of sterile nutrient medium and mixed (×100

dilution).(3) The process is then repeated a number of times to produce further 10-fold dilutions.(4) A known volume of bacterial culture (usually 1cm3 or 0.1cm3) is added to agar plates and

incubated.

Aseptic technique is required throughout.

Figure 28 The steps involved in dilution plating

Knowledge check 16Use the agar plate labelled E in Figure 28 to estimate the number of bacteria in 1cm3 of theoriginal culture.

Knowledge check 17What is the difference between a total count and a viable count? Explain why these methods canoverestimate or underestimate the size of a bacterial population.

It can be seen from Figure 28 that some plates have far too many colonies, many of which havemerged, meaning that it is impossible to count individual colonies. Other plates have only a fewcolonies, so they cannot be representative of the original culture.

To calculate the total viable cell count in the original culture, the number of colonies is multiplied bythe dilution factor.

Penicillin productionMicroorganisms can be grown on a large scale to produce substances that are useful to humans. Thesesubstances can be classified as primary metabolites or secondary metabolites.

Primary metabolites are required for the normal metabolism of the microorganism. Enzymes are anexample. To maximise production of primary metabolites continuous fermentation is used and themicrobes are kept in the exponential phase of growth.

Secondary metabolites are not required for the normal metabolism of the microorganism. Antibioticsare an example. Certain microbes produce these in overcrowded conditions when there is a shortageof nutrients. They are secreted to kill competing microbes. To maximise production of secondarymetabolites batch fermentation is used. This allows the microbes to reach the stationary phase,which is when secondary metabolites are produced.

Industrial fermentationThe fungus Penicillium notatum produces penicillin as a secondary metabolite. The antibiotic isproduced in the stationary phase, when glucose has become depleted. This reflects the need for thefree-living organism to reduce competition when food sources are depleted.

Examiner tipMost of this topic has involved the biology of bacteria. However, penicillin is produced by afungus. When answering questions about penicillin production it is important not to confuse thename of the antibiotic with the name of the fungus.

Penicillin is produced on a commercial scale using a batch culture fermenter (Figure 29). A batchfermenter is a closed tank made of stainless steel alloy to resist the corrosive effect of acidicfermentation products and to withstand pressure. The industrial production of penicillin uses batchfermenters that range from 40dm³ to 200dm³ in size.

Figure 29 A typical batch fermenter

A typical batch fermentation process involves the following:(1) The vessel is sterilised beforehand using high-pressure steam to kill any microbes that may infect

the batch.(2) A sterile nutrient medium is added.(3) A pure culture of Penicillium notatum is added.(4) The fungus is provided with suitable conditions for growth: (a) Sterile air is introduced into the fermenter to provide oxygen for aerobic respiration. The air

must be sterile to prevent contamination by airborne microbes. (b) The culture is heated to the optimum temperature for growth, which is then maintained — the

cooling water jacket removes any excess heat from fungal respiration. (c) The optimum pH is maintained by adding acid or alkali. (d) The culture and nutrients are thoroughly mixed by forced aeration using a sparging ring or

paddles.(5) Once the glucose has been depleted and the fungus enters the stationary phase, penicillin is

produced (see Figure 30).(6) The culture fluid is drained and filtered to remove the mycelium and the filtrate is purified to

extract the penicillin.

Figure 30 Graph to show the production of penicillin

Knowledge check 18Explain why the following features of a fermenter are important:(a) the cooling water jacket(b) the sparging ring(c) the air filters

Penicillin and bacterial cell walls• Penicillin prevents the formation of cross-links between peptidoglycan units during cell division

(bacterial reproduction). This weakens cell walls and leaves the bacteria susceptible to osmoticlysis.

• Penicillin is effective against Gram-positive bacteria because they have a thick layer ofpeptidoglycan in their cell walls.

• Penicillin is less effective against Gram-negative bacteria because they have only a thin layer ofpeptidoglycan and an additional layer of lipopolysaccharides in their cell walls.

SummaryAfter studying this topic you should be able to:• classify bacteria according to both their shape and their reaction to the Gram stain• describe how microorganisms are cultured in the laboratory and outline the principles of aseptic

technique• describe and explain the requirements for growth of microorganisms in relation to temperature,

pH, nutrients and oxygen• draw a generalised graph of the population growth curve of a microorganism cultured in the

laboratory and be able to describe and explain the four stages of the curve• state the difference between a total count and a viable count when monitoring population growth

• describe how a viable count is carried out using serial dilutions, plating and counting of colonies• describe and explain the principles involved in batch culture fermentation using penicillin

production as an example

PopulationsKey concepts you must understand• Populations are dynamic, constantly changing components of ecosystems.• When a species successfully invades a new area and establishes itself, the growth of its

population follows a predictable pattern.• The maximum population size that a particular environment can support indefinitely is called the

carrying capacity.• When the population of an organism increases and causes economic loss to humans it becomes a

pest.• The aim of pest control is to reduce the pest population to a level that does not result in an

economic loss for the farmer.• Pest populations can be controlled using either chemicals or biological control agents; there are

advantages and disadvantages to both methods.• All matter is recycled within ecosystems and decomposers play a crucial role in the breakdown

of organic molecules into inorganic molecules and ions.

A population is defined as a group of interbreeding individuals of the same species in a particularhabitat at a particular time. Figure 31 shows that the size of a population is determined by fourfactors:• Births and immigration increase population size.• Deaths and emigration decrease population size.

Figure 31 The factors that affect a population

Population growthpopulation growth = (births + immigration) – (deaths + emigration)

In the section on microbiology the population growth curve of a microorganism grown in thelaboratory was discussed. The curve is simplistic as it does not take into account immigration andemigration and is therefore based upon ‘birth rates’ and ‘death rates’ alone. However it is still a goodmodel to follow.

In the natural environment, when a species successfully invades and establishes itself in a new area,the growth of its population shows the following pattern (see Figure 32 on p. 34):• a period of slow growth, i.e. a lag phase• a period of rapid population growth, i.e. a log phase• a period of equilibrium when its numbers remain more or less constant, i.e. a stationary phase

Figure 32 Population growth under ‘natural’ conditions

Knowledge check 19Explain what causes the lag phase in a species that reproduces sexually (e.g. rats) and bacteria thatreproduce asexually.

During the lag phase the increase in population is slow. There are a limited number of individuals ofreproductive age in the population.

In the log/exponential phase there is a plentiful supply of nutrients. There is little intraspecificcompetition for nutrients and resources. The waste products that have been produced are in lowconcentrations. The birth rate is at a maximum and population growth is exponential.

Limits to population growthAs the population nears the carrying capacity of the environment the growth rate slows down. This isknown as environmental resistance and several factors will limit further population growth. Thesefactors can be either density dependent or density independent.

Knowledge check 20Explain what is meant by the following terms:(a) carrying capacity(b) exponential growth

Density-dependent factorsThe effect of a density-dependent factor increases as the density of the population increases, i.e. agreater percentage of the population is affected at higher population densities. Density-dependentfactors include the availability of food, competition, predation and disease.

For example, it is much easier for pathogens such as the influenza virus to spread from one animal toanother at higher population densities. Therefore a higher percentage of people within a populationwill have influenza as the population density increases. (This explains why influenza is more commonin the winter as people become crowded indoors.) In this way density-dependent factors slow downthe population growth rate.

Examiner tipWhen asked to define density-dependent and density-independent factors many students makereference to ‘the effect of the population on the factor’. This is incorrect — the reference should beto ‘the effect of the factor on the population’.

Most natural populations tend to fluctuate around the carrying capacity (i.e. a stationary/stable phase).The carrying capacity is dependent on the availability of resources — for example the concentrationof nutrients such as nitrates. If the population rises above the carrying capacity, then there will beincreased competition for nitrates. This will increase the death rate and/or decrease the birth rate —the death rate will exceed the birth rate and the population will decrease. If the population fallsbelow the carrying capacity then there will be decreased competition for nitrates. The birth rate willexceed the death rate and the population will increase.

Examiner tipExam questions on this topic are usually about a specific organism and you may not gain credit forusing vague terms such as food, resources or nutrients. Try to be specific with your answers. Forexample, carrot plants growing in a field will be competing with one another for light, water andnitrates; male tigers will be competing with each other for territories and females.

Density-independent factorsThe effect of a density-independent factor is the same regardless of the size of the population, i.e. thesame percentage of the population is affected, irrespective of the population density. Density-independent factors include the weather and climate, for example a sudden change in temperature.

Take small birds such as wrens as an example. They are susceptible to low temperatures and a

sudden freeze may kill a fixed percentage (say 70%) of the birds, irrespective of the populationdensity. Density-independent factors can, therefore, lead to a population crash.

Knowledge check 21State two density-dependent factors and one density-independent factor that can affect carryingcapacity.

Knowledge check 22Intraspecific competition for food can act in both a density-dependent manner and a density-independent manner. In an experiment, the number of eggs of the flour beetle added to a fixedvolume of flour was varied and the number of beetles dying monitored. Figure 33 shows some ofthe results. Use the graph to determine the type of competition shown by flour beetles and give areason for your answer.

Figure 33 The effect of competition on the survival of flour beetles

Knowledge check 23Explain why the introduction of grey squirrels into the British Isles has led to a decline in thenumber of native red squirrels.

Investigating competitionOrganisms compete with each other for limited resources, such as food, water, nesting sites orterritories, and mates.(a) Intraspecific competition is between individuals of the same species, for example carrot plants

competing for water and nitrates.(b) Interspecific competition is between individuals of different species, for example carrot plants

competing for water and nitrates with weeds such as dandelions.

The population growth of unicellular organisms, such as Paramecium, can be monitored within thelaboratory. They are grown in tubes of liquid medium which contains yeast cells. P. caudatum andP. aurelia feed on yeast cells suspended in the medium; P. bursaria feeds on yeast cells at the

bottom of the tubes. When the different species are grown separately each reaches a maximumpopulation size (see Figure 34 on p. 36).

Figure 34 Population growth in three species of Paramecium grown in isolation

Figure 35 (a) shows that when P. caudatum and P. aurelia are grown together only one speciesreaches its carrying capacity. The other population declines and eventually dies out. This is becausethey are in direct competition and P. aurelia outcompetes P. caudatum, which becomes locallyextinct.

Figure 35 Interspecific competition in Paramecium

Figure 35 (b) shows that when P. caudatum and P. bursaria are grown together, the species co-exist, but with smaller populations than when grown alone. They are able to co-exist because theyfeed in different areas. Their maximum populations are smaller because they are competing for thesame food source.

Pests and pest controlA pest is any unwanted organism that interferes, either directly or indirectly, with human activity.Most pests are associated with agriculture and food production.

The aim of pest control is to reduce the pest population to a level that does not result in an economicloss for the farmer, i.e. the costs of pest control must not exceed the additional economic value of thecrop.

Chemical control of pestsA pesticide is a chemical substance (poison) used for killing pests.

Advantages of chemical control• Fast acting — rapid eradication of a pest over a specified localised area before the population of

the pest has a chance to increase.• Relatively cost-effective — the cost of using the pesticide does not exceed the economic gains of

selling the crop.• The chemicals can be applied on a small scale.• Application does not require a high level of skill.

Disadvantages of chemical controlPesticides are non-selective. Many pesticides kill a wide range of organisms, not just the targetorganism. They kill beneficial organisms such as pollinating insects (e.g. bees) and natural predatorsof the pests. This may cause resurgence of the pest population as any pest individual that survives ormigrates into the treated area finds ideal conditions.

Pesticides are persistent. Many pesticides are not biodegradable (i.e. they cannot be broken downinto a harmless molecule using an enzyme released by a bacterium or fungus) and remain active in thesoil for many years. DDT is an example. Persistent pesticides tend to be fat-soluble and so they arenot excreted, but accumulate in fatty tissues. As an organism grows, the level of toxins in their bodiesincreases; this is known as bioaccumulation. As the pesticide passes from one trophic level to thenext it becomes more concentrated in the animals’ bodies; this is known as biomagnification.Animals higher up the food chain (including humans) will accumulate the highest concentrations andthis may prove fatal.

A few individuals in the pest population may contain alleles that give them genetic resistance to thepesticide. When an area is treated they will survive. With a lack of intraspecific competition theywill reproduce and pass on their resistance genes to their offspring. Natural selection will result inthe evolution of resistant populations.

Long-term exposure to pesticides (e.g. sheep dips) can cause harm to humans.

Biological control of pests

• Biological control exploits natural enemies of the pest species, using them to regulate the pestpopulation. The control agent could be the natural predators, parasites or disease-causingorganisms of the target pest species.

• The aim is to reduce the pest population below the economic damage threshold (see Figure 36).• It would be counter-productive to eradicate the pest species as then the control agent would die out.

Figure 36 Biological control

Advantages of biological control• Biological control methods tend to be specific, affecting only the target pest species.• The environment is not contaminated, so biological control is non-toxic to other species.• The pest species cannot develop resistance to the biological control agent.• Once a population of natural predators/parasites is established, control of the pest is often self-

perpetuating, i.e. a predator–prey cycle is established.• In the long term it is relatively inexpensive. (However, initial research costs are expensive.)• It is very effective in greenhouses.

Disadvantages of biological control• Slow — it takes time for the population of the control agent to increase to an effective level.• Expensive initially — it requires high levels of skill and research to implement.• An exotic control agent may itself become a pest. An example is the cane toad.• Repeated applications of the agent may be required to achieve long-term population balance.• It is of little use to individual producers — its introduction needs to be made on a large scale.

Examiner tipYou may be asked to compare the advantages/disadvantages of chemical and biological control.Read the following answers. Answer A: ‘biological control agents tend to be specific and only killthe pest species whereas chemical pesticides kill a wider variety of species.’ Answer B:‘biological control agents tend to be specific and only kill the pest species, however chemicalpesticides reduce the pest population faster.’ Answer A gains credit but answer B does not. It isessential to compare like with like and answer B (although not incorrect) does not do this.

Integrated pest managementEach crop and its pest are evaluated as an ecological system. Integrated pest management involvesthe use of biological control agents together with the minimal, well-targeted application of highlyselective pesticides.

Nutrient cyclesWhen organisms die decomposers (fungi and bacteria) break the tissues down. The organic moleculesin the tissues are broken down and the inorganic molecules are released into the environment. Thisenables nutrients to be recycled within ecosystems.

Examiner tipConsumers do more than just feed on other organisms. Once an organism has been eaten the largeinsoluble molecules they contain have to be digested. The small soluble products of digestion arethen absorbed and used to synthesise their own organic molecules (this is known as assimilation).

Figure 37 shows a basic nutrient cycle, with the three major feeding groups in an ecosystem(producers, consumers and decomposers) forming a triangle.

Figure 37 A simple nutrient cycle

The carbon cycleFigure 38 shows a simple carbon cycle:• Carbon is found in living organisms in carbohydrates, lipids, proteins and nucleic acids.• The carbon in the atmosphere and oceans is made available to organisms via photosynthesis. Green

plants and algae fix carbon dioxide and convert it into complex organic molecules, such as glucose.• Carbon dioxide is produced by all organisms during respiration and enters the atmosphere.

Figure 38 A simple carbon cycle

Human effects on the carbon cycleCarbon is also stored in wood and fossil fuels (coal, oil and gas). Fossil fuels were formed whenplants and animals died and their remains did not decompose but were fossilised. Coal is fossilised

wood; oil and gas result from the soft tissues of animals (hard tissues, such as bones, are preserved).The combustion of wood and fossil fuels releases the stored carbon into the atmosphere as carbondioxide.

Examiner tipWhen answering questions on the nitrogen cycle you must state clearly the form that nitrogen is in ateach stage. You are more likely to gain credit for an answer that refers to ‘proteins’ or ‘aminoacids’ when discussing plants and animals than one that uses a generic term such as ‘nitrogen-containing compounds’.

The nitrogen cycleIn living organisms, nitrogen is only found in certain molecules — for example amino acids andproteins, and nucleic acids. It is only available to producers as ammonium ions and nitrates in thesoil. Nitrogen leaves all organisms after death. It leaves consumers in indigestible matter (faeces) andthrough nitrogenous excretion (e.g. in urea).

Importance of bacteria in the nitrogen cycleBacteria are extremely important in the nitrogen cycle. They are responsible for several processes(see Figure 39):(1) Putrefaction/ammonification occurs when saprophytic bacteria metabolise the nitrogen-

containing organic matter in dead organisms, faeces and urine and release ammonium ions.(2) Nitrification — under aerobic conditions nitrifying bacteria convert: (a) ammonia to nitrites (Nitrosomonas) (b) nitrites to nitrates (Nitrobacter)

Nitrates are then taken up from the soil by producers.(3) Denitrification — denitrifying bacteria (Pseudomonas) convert nitrates and ammonium ions

into atmospheric nitrogen. This occurs in anaerobic conditions, for example in waterlogged soils.Denitrification removes nitrogen from ecosystems.

Figure 39 The importance of bacteria in the nitrogen cycle

(4) Nitrogen fixation — nitrogen-fixing bacteria convert atmospheric nitrogen into ammonium ions.Nitrogen fixation is brought about by Rhizobium and Azotobacter:

(a) Rhizobium is a genus of mutualistic bacteria found in the root nodules of leguminous plants,such as clover. The plant rapidly converts ammonium ions into nitrogen-containing organiccompounds. The bacterium benefits by gaining energy and nutrients, for example carbohydrates,from the plant.

(b) Azotobacter is found living free in the soil.

Knowledge check 24State two reasons why plants need to absorb nitrates.

Examiner tipAlthough it is unlikely that you would be asked to draw the nitrogen cycle, the ability to do so wouldhelp you answer questions based on it. Try to build up a flow diagram in stages:

Stage 1: the organic triangle (in black)

Stage 2: add the inorganic loop (in blue)

Stage 3: add in the names of the processes involved

Stage 4: add the names of the bacteria involved

Stage 5: add atmospheric nitrogen (in red); name the processes involved and give the names of thebacteria

Knowledge check 25Explain how the following farming practices affect the nitrogen cycle:(a) draining waterlogged fields(b) ploughing fields(c) planting leguminous crops such as clover(d) ploughing crops such as clover into the soil

SummaryAfter studying this topic you should be able to:• define the term population and describe the pattern of population growth• distinguish between density-dependent factors and density-independent factors and explain the

effect that these have on population growth• describe the principles involved in the chemical and biological control of pests and explain their

relative advantages and disadvantages• understand what is meant by organic breakdown and describe its importance to ecosystems in the

recycling of mineral nutrients• draw a labelled diagram of the carbon cycle linking the processes of photosynthesis, respiration,

decomposition, fossilisation and combustion• describe the role of bacteria in the nitrogen cycle and explain the significance of nitrogen to

living organisms• explain how human activities, such as ploughing and drainage, are important in the maintenance

of soil fertility

ExcretionKey concepts you must understand• Excretion is the removal of waste products of metabolism from the body.• If allowed to build up inside cells and tissues these waste products can be toxic.• The two main excretory products produced by animals are carbon dioxide, produced during

aerobic respiration, and nitrogenous waste from the breakdown of excess amino acids (andnucleic acids).

• To excrete nitrogenous waste safely, water is used to dilute it to non-toxic levels.• Nitrogenous waste is excreted in different forms (ammonia, urea or uric acid) depending upon the

availability of water within an animal’s habitat and the extent to which water loss is controlled bythe organism.

• The mammalian kidney performs two main functions: the excretion of urea and the regulation ofthe water potential of the blood (osmoregulation).

• Although mammals excrete urea, the mammalian kidney shows adaptations for controlling waterloss and there is a correlation between the availability of water in the environment and theanimal’s ability to conserve water.

Pre-existing knowledgeExcretion of nitrogenous waste• When proteins are digested, the amino acids absorbed into the body are used to synthesise other

proteins, for example enzymes and hormones.• Excess amino acids cannot be stored in the body.• The excess amino acids are broken down in the liver by a process called deamination (Figure

40).

Figure 40 Deamination of excess amino acids

• An amine group is removed from an amino acid and is then converted into ammonia.• The remainder of the amino acid (now an organic acid) is either respired or converted to

carbohydrate or lipid to be stored.

Different types of nitrogenous waste

AmmoniaAmmonia is a small, soluble and highly toxic substance. It must be excreted immediately. It cannot bestored and requires large volumes of water to dilute it to non-toxic levels for it to be excreted safely.

Freshwater fish have body fluids that have a lower water potential than their surroundings and sothey absorb large quantities of water by osmosis. Most of this water enters through their highlypermeable gills. Freshwater fish therefore excrete ammonia as they have to remove a large volume ofexcess water. Some of the ammonia is excreted via the kidneys. However, as it is highly soluble mostof it diffuses from the gills.

Examiner tipIf you are given a question about the different excretory products of animals you need to take intoaccount the type of animal involved (and if its embryo developed inside an amniote egg) and theavailability of fresh water in its environment. The lower the toxicity of the molecule, the longer itcan be stored and the less water is required to excrete it safely.

UreaMammals excrete urea. Urea is much less toxic than ammonia and therefore can be stored for aperiod of time in the tissues. As it is less toxic it requires less water to dilute it down to safe levels.

Although producing urea is energetically expensive it is an adaptation to life on land and helps toprevent dehydration.

Uric acidReptiles, birds and insects excrete uric acid. Uric acid has low toxicity and can therefore be storedfor long periods of time. Very little water is needed to store it and safely excrete it and it is removedfrom the bodies as a white paste. The low toxicity means that it can accumulate inside the eggs ofbirds and reptiles, without damaging the embryo.

Although producing uric acid is energetically expensive it allows these animals to survive in aridenvironments and reduces their mass, which is an advantage for flight.

Knowledge check 26Red-eared terrapins live in freshwater ponds and lakes. Giving a reason for your answer, statewhat type of nitrogenous waste they excrete.

Summary of nitrogenous waste typesFigure 41 summarises the different excretory products and the relative (energy) cost and volume ofwater needed to excrete them safely.

Figure 41 The excretion of different forms of nitrogenous waste

The mammalian kidneyIn mammals, deamination of amino acids occurs mainly in the liver. The ammonia produced isquickly converted to urea in the ornithine cycle. The urea then diffuses into the blood where it isremoved by the kidneys.

Gross anatomy of the urinary tract• The kidneys are located in the abdominal cavity.• They receive oxygenated blood from the renal artery.• The urine produced in the kidney travels to the bladder via the ureter.• The internal anatomy of the kidney is divided into two main regions — the inner medulla and the

outer cortex (Figure 42 on p. 44).

Figure 42 The gross structure of the kidney

Examiner tipSynoptic link to BY1: Unlike most organs, the kidneys are not protected by the skeleton. They areprotected by a thick layer of fat tissue that surrounds the delicate organ.

The nephronEach kidney has an estimated 1 million nephrons (Figure 43) providing a large surface area for theexchange of materials. The nephron has four functional regions:• the Bowman’s capsule and glomerulus• the proximal convoluted tubule• the loop of Henle• the distal convoluted tubule and collecting duct

Figure 43 A kidney nephron and associated blood vessels

Examiner tipThe nephron is the functional unit of the kidney. As the basic structure of all nephrons is the same thefunction of the kidney can be explained by looking at what happens along the length of one nephron.

The different structures of the nephron can be seen when sections of the kidney are viewed with anoptical microscope. Figure 44 (a) shows a section through the cortex. The Bowman’s capsule,glomerulus and the proximal and distal convoluted tubules are clearly visible. Figure 44 (b) shows asection through the medulla with the loop of Henle and collecting duct clearly visible.

Figure 44 Photomicrographs of sections through (a) the cortex of the kidney and (b) the medulla of thekidney

Examiner tipMany students have difficulty interpreting photomicrographs such as the ones shown in Figure 44.However, if you think about the biology you have learned you should be able to arrive at the correctanswer. You have probably drawn a nephron, including the boundary between the cortex and themedulla. You will remember that the Bowman’s capsule, glomerulus, proximal and distalconvoluted tubules are in the cortex and that the medulla contains only the loop of Henle andcollecting duct.

The Bowman’s capsule and glomerulusUltrafiltration is the removal of water and small solutes from the blood and the formation of fluidcalled the glomerular filtrate. It takes place between the capillaries of the glomerulus and theBowman’s capsule (see Figure 45 on p. 46).

Figure 45 Ultrafiltration in the glomerulus

Examiner tipAfferent means ‘to bring to’ and efferent means ‘to carry away’. Both terms are used in anatomy todescribe structures such as blood vessels and neurones. When describing ultrafiltration you willneed to use these terms. To help you get them the right way round think of the alphabet: a comesbefore e just as afferent comes before efferent.

The capillary walls of the glomerulus are made up of a single layer of endothelial cells with poresbetween them, called fenestrations, making the capillaries highly permeable.

The basement membrane surrounding the capillary acts as a molecular filter:• Water and small solutes are forced out of the blood and into the capsular space.• Blood cells and large solutes, such as plasma proteins, are prevented from leaving the capillaries.

The inner layer of the Bowman’s capsule is composed of cells called podocytes, which have ‘foot-like’ processes that wrap around the capillaries.

The hydrostatic pressure of the blood in the glomerulus is very high due to:• contraction of the left ventricle (the renal artery branches from the aorta)• the efferent arteriole leaving the glomerulus being narrower than the afferent arteriole, which

produces a ‘bottleneck’ effect

Knowledge check 27(a) Name three substances that are normally present in the filtrate.(b) Name an additional substance that may be present in the filtrate of a person with high blood

pressure.

The hydrostatic pressure forces fluid (water and small solutes) through the endothelium and the

basement membrane, and down through the spaces between the podocytes into the lumen of theBowman’s capsule. This fluid is now called the (ultra)filtrate.

The rate of ultrafiltration is high — about 125cm3 of filtrate is produced every minute.

Knowledge check 28Explain why the pressure in the glomerular capillaries is higher than capillaries found elsewhere inthe body.

The proximal convoluted tubuleThe proximal convoluted tubule is the longest region of the nephron and is composed of a single layerof epithelial cells adapted for the selective reabsorption of solutes (see Figure 46):• They possess microvilli to provide a large surface area for reabsorption via carrier proteins.• They contain many mitochondria to provide the ATP for active transport of solutes.

Figure 46 Selective reabsorption via an epithelial cell from the proximal convoluted tubule

The renal capillaries are close to the tubule, providing a short diffusion pathway for reabsorption.Over 80% of the filtrate is reabsorbed here, including all of the glucose and amino acids and most(approximately 85%) of the sodium ions. Some urea (about 50%) passes back into the blood byfacilitated diffusion. The reabsorption of these solutes, along with the presence of plasma proteins(which remained in the blood), lowers the water potential of the blood in the capillaries surroundingthe tubule. As a result, most of the water (about 85%) in the filtrate is reabsorbed by osmosis.

Examiner tipSynoptic link to BY1: Selective reabsorption of solutes involves active transport and diffusion; thereabsorption of water involves osmosis. It would be advisable to revisit your notes from section 1.3to ensure you fully understand cell membranes and transport. This will also help with your

understanding of nerve impulses.

Figure 46 shows that the plasma membrane nearest to the capillaries is highly folded to formnumerous intercellular spaces.

The numbers below refer to Figure 46.(1) Solutes (e.g. Na+) are actively transported out of the epithelial cells and into the intercellular

spaces.(2) This increases the concentration of the solutes in the intercellular spaces causing them to move

into the capillaries via facilitated diffusion. The solutes are then transported away in the plasma tothe renal vein.

(3) Due to the lower concentration of solutes inside the epithelial cell solutes move out of the filtrateby facilitated diffusion.

Knowledge check 29Describe two ways in which the cells from the proximal convoluted tubule are adapted forreabsorption.

The loop of Henle and collecting ductThe function of the loop of Henle is to conserve water. This results in the production of concentratedurine, i.e. urine that has a higher solute concentration and lower water potential than the blood.

The concentration of urine produced is related directly to the length of the loop of Henle, the longerthe loop the greater the volume of water reabsorbed (conserved).• The kidneys of semi-aquatic mammals, such as the beaver, have short loops of Henle and produce a

large volume of dilute urine.• The kidneys of desert-dwelling mammals, such as the kangaroo rat, have long loops of Henle and

produce a small volume of concentrated urine.

The structure of the loop of Henle can be divided into:• the descending limb, which is highly permeable to water (but relatively impermeable to ions)• the ascending limb, which is highly permeable to ions but relatively impermeable to water

Knowledge check 30Explain the difference in the lengths of the loops of Henle in an otter and a camel.

Walls of the capillaries surrounding the loop are freely permeable to both water and ions. Figure 47shows an overview of the structure and function of the loop of Henle.

Figure 47 The loop of Henle

Generation of water potential gradientsThe function of the ascending limb is to create a water potential gradient between the filtrate and thetissue fluid of the medulla. It achieves this by the active transport of sodium and chloride ions (fromthe filtrate) into the medulla (1 in Figure 47). This increases the concentration of ions, and lowers thewater potential of the tissue fluid of the medulla (2). This causes water to move out of the filtratefrom the descending limb (3) by osmosis (due to the ascending limb being impermeable to water).The water is reabsorbed into the capillaries and removed by the blood so that it does not dilute thetissues (i.e. increase the water potential) of the medulla.

Knowledge check 31Name the part of the nephron that provides the water potential gradient for the reabsorption ofwater from the collecting duct.

This process is repeated down the length of the loop so that:• As the filtrate moves down the descending limb it loses water, causing it to have a higher ion

concentration (i.e. become more concentrated) and lower water potential.• As the filtrate moves up the ascending limb it loses ions, causing it to have a lower ion

concentration (i.e. become more dilute) and higher water potential.• At the apex of the loop (4) the filtrate reaches a maximum ion concentration.

The loop of Henle acts as a countercurrent-multiplier system. Countercurrent refers to the fact thatthe filtrate flows in opposite directions in the loop, down the descending limb and up the ascendinglimb. The multiplier effect refers to the generation of the solute gradient (and therefore a waterpotential gradient) in the medulla. This gradient is caused by ions being pumped out of the ascendinglimb. The effect of the pump is multiplied due to ions being constantly removed from the filtrate in theascending limb and being replaced as the filtrate flows through the proximal convoluted tubule andthe descending limb.

The reabsorption of waterAs the filtrate enters the distal convoluted tubule it has a higher water potential than the tissues of themedulla.

As the filtrate moves down the collecting duct it once again meets tissue fluid with a lower waterpotential as a result of the pumping of solutes from the ascending limb of the loop of Henle.Therefore, water moves out of the filtrate by osmosis along the entire length of the collecting duct (5in Figure 47) and is reabsorbed into the blood. This produces urine that is greatly reduced in volumeand has a much lower water potential than blood (i.e. is concentrated).

The permeability of the walls of the distal convoluted tubule and collecting duct are affected byantidiuretic hormone (ADH). This allows mammals to control the volume of water reabsorbed intothe blood and so maintain the water potential of the blood at a relatively constant value. This is calledosmoregulation.

HomeostasisThe term homeostasis is used to describe the mechanisms that bring about the maintenance of aconstant internal environment. Homeostasis is important to ensure that the conditions inside cellsremain at the optimum for them to function effectively regardless of the external environmentalconditions. Homeostasis is brought about by both the nervous and endocrine systems. Features of anorganism’s internal environment have a set optimum level, also known as the norm.

Examiner tipIf you are asked to describe the process of homeostasis you need to make sure you make reference tothe following: the norm, the receptor/detector, the coordinator, the effector, the corrective actionsand negative feedback.

Figure 48 shows the homeostatic process involving:• a receptor (detector) — which detects a stimulus (i.e. change from the norm)• a coordinator — receives information from the receptor and initiates the corrective mechanism• an effector — carries out the corrective procedure to restore the norm

Figure 48 The principle of homeostasis

In most biological systems a negative feedback system operates to maintain the norm. Negativefeedback means that any change from the norm will bring about a corrective mechanism to restore thenorm.

An efficient homeostatic mechanism ensures that the factor being controlled shows minimalfluctuation around a set point.

OsmoregulationOsmoregulation is the control of the water content and solute composition of the blood. The bodymaintains the water potential of the blood at an approximately steady state by balancing water uptakefrom the diet with water loss by sweating, evaporation from the lungs, and in urine and faeces. Figure49 summarises the process of osmoregulation.

Figure 49 Osmoregulation

Osmoregulation is under endocrine control. The pituitary gland secretes the hormone ADH(antidiuretic hormone), which affects the reabsorption of water from the filtrate in the nephron. ADHincreases the permeability of the walls of the collecting duct and distal convoluted tubule to water.

The water potential of the blood falls when the diet contains too little water (or a high intake ofsolutes) or during periods of excessive sweating:(1) The decrease in water potential of the blood is detected by osmoreceptors located in the

hypothalamus.(2) Nerve impulses are generated and pass to the posterior pituitary gland, which responds by

secreting more ADH.(3) The ADH travels in the bloodstream and on reaching the kidneys brings about increased

permeability of the walls of the distal convoluted tubules and collecting ducts of the nephrons.(4) This results in more water being reabsorbed into the blood from the filtrate.(5) This restores the normal water potential of the blood and results in the production of a small

volume of hypertonic urine.

The water potential of the blood rises when the diet contains too much water (or a low intake ofsolutes):(1) The increase in water potential of the blood is detected by osmoreceptors located in the

hypothalamus.(2) Fewer nerve impulses are generated and the posterior pituitary gland responds by secreting less

ADH.(3) This decreases the permeability of the walls of the distal convoluted tubules and collecting ducts

of the nephrons.(4) This results in less water being reabsorbed into the blood from the filtrate.(5) This restores the normal water potential of the blood and results in the production of a large

volume of hypotonic urine.

Knowledge check 32Name the detector (receptor), coordinator and effector involved in osmoregulation.

SummaryAfter studying this topic you should be able to:• explain why different animals produce different excretory products• identify the different regions of the mammalian kidney, including the nephron• describe the functions of the mammalian kidney, including nitrogenous excretion and water

regulation• describe and explain the adaptations of the cells of the proximal convoluted tubule for

reabsorption• describe and explain the adaptations of the loop of Henle in animals in different environments• explain the concept of homeostasis and its importance in maintaining the body in a state of

dynamic equilibrium• explain what is meant by the term negative feedback and its importance in restoring conditions to

their original levels• describe the roles of the posterior pituitary gland and of antidiuretic hormone in regulating the

water potential of the blood

The nervous systemKey concepts you must understand• Organisms increase their chances of survival by responding to stimuli.• These stimuli include: – temperature – light intensity/wavelength/duration – chemicals — these may cause, for example, particular smells or pH changes• These stimulus–response control systems always follow the same basic pattern (see Figure 50).

Figure 50 The stimulus–response control system

The mammalian nervous systemThe individual components of the stimulus–response control system in a mammal are receptor,coordinator and effector.

Knowledge check 33Give the names of two different types of effector.

Receptors are specialised sensory cells that detect internal or external stimuli. They act astransducers by converting one form of energy into electrochemical energy. For example,photoreceptors in the retina convert light energy into electrochemical energy.

The coordinator is the central nervous system (CNS) and is composed of the brain and spinal cord. Itprocesses the information received from receptors and initiates the appropriate response bycommunicating with effectors.

Effectors are muscles or glands. They bring about the response.

The peripheral nervous system is composed of the individual neurones and ganglia that connect thereceptors and effectors to the CNS.

Neurones and nerve pathwaysNeurones are highly specialised cells that link together to form pathways (see Figure 51). There arethree basic types:• sensory neurones — transmit impulses from receptors to the CNS• motor neurones — transmit impulses from the CNS to effectors• relay neurones — link sensory and motor neurones in the CNS

Figure 51 The neural pathway involved in responding to a stimulus

Examiner tipA nerve and a neurone are not the same structure. A nerve is a bundle of neurones surrounded by aprotective sheath of fibrous tissue; a neurone is an individual nerve cell. Most of the content here isabout neurones. You must make sure that you refer to them correctly.

The motor neuroneFigure 52 shows the structure of a typical motor neurone. The cell body contains the nucleus and mostof the cell’s other organelles including ribosomes, which synthesise neurotransmitters, andmitochondria that provide ATP for the sodium and potassium ion (Na+/K+) pump. The dendritestransmit impulses to the cell body; the axon transmits impulses away from the cell body.

Figure 52 A myelinated motor neurone

Knowledge check 34What is the difference in function between an axon and a dendrite?

Some vertebrate neurones are surrounded and insulated by a fatty myelin sheath. This is formed bySchwann cells. During myelination, the membrane of the Schwann cell becomes extended and wrapsitself around the axon (see Figure 53). Schwann cells also secrete a fatty material called myelin. Thecombination of the membrane phospholipids and the myelin — the myelin sheath — insulates theneurone by preventing the movement of ions into and out of the axon. There are small gaps betweenthe Schwann cells called nodes of Ranvier. Here the axon is ‘exposed’, which allows the movementof ions into and out of the axon. Myelination occurs only in the vertebrate nervous system.

Figure 53 A Schwann cell wrapping around the axon during myelination

Pre-existing knowledgeMembrane structure• Membranes are composed mainly of phospholipids and proteins.• The phospholipids form a bilayer and the hydrophobic core prevents the passage of polar

molecules and ions.• Extrinsic proteins occur on only one side of the membrane; intrinsic proteins span the entire

membrane.• Intrinsic proteins are involved in the transport of polar molecules and ions across the membrane

via facilitated diffusion and active transport.• Facilitated diffusion involves the movement of molecules or ions down a concentration gradient

and is a passive process.• Active transport involves the movement of molecules or ions against a concentration gradient and

requires energy from the hydrolysis of ATP.

The nature of the nerve impulseNerve impulses are electrochemical in nature and are the result of a potential difference (pd) acrossthe membrane of the neurone. The nature of the nerve impulse was discovered by investigating thegiant axons of squid. By inserting microelectrodes into the axon and the fluid surrounding the axon,the potential difference can be measured. Changes in the potential difference can be monitored using acathode-ray oscilloscope (see Figure 54 on p. 54).

Figure 54 Changes in potential difference across neuronal membranes are detected usingmicroelectrodes and displayed on oscilloscopes

Structure of the axon membraneThe nerve impulse is based on the movement of sodium ions (Na+) and potassium ions (K+) acrossthe membrane of the axon. The Na+ and K+ ions are transported across the membrane by both activetransport and facilitated diffusion. Figure 55 shows the different types of intrinsic proteins involvedwith this:• Na+/K+ pump, which requires energy from the hydrolysis of ATP• K+ channels (relatively many)• voltage-gated Na+ channels• voltage-gated K+ channels

Figure 55 Structure of the neurone membrane

There are also Na+ channels. As they are relatively few in number, these are not shown in Figure 55.

The resting neuroneAxons have a differing permeability to Na+ and K+ ions. The membrane is more permeable to K+ions than Na+ ions. This is due to the relative abundance of K+ channel proteins.

Examiner tipNerve impulses, depolarisations and action potentials — what is the difference? The nervemembrane depolarises when the potential difference across the membrane rises from its polarisedstate (−70mV). An action potential is produced when the inside of the axon becomes positivelycharged.

The generation of the resting potential is illustrated in Figure 56. The Na+/K+ ion pump activelytransports Na+ ions out of the neurone and K+ ions into the neurone. For every three Na+ ionspumped out, only two K+ ions are pumped in. As a result, concentration gradients are establishedacross the membrane. The axon membrane is highly permeable to K+ ions, which diffuse rapidly outof the neurone. However, because the membrane is relatively impermeable to Na+ ions, they can onlydiffuse back into the neurone slowly.

Figure 56 The resting neurone

Knowledge check 35Describe how a resting potential is maintained in a neurone.

The overall outward movement of ions establishes a potential difference across the membrane. Theinside of the axon is negatively charged (relative to the outside) and the membrane is said to bepolarised. At rest, the potential difference is −70mV. This is called the resting potential.

The active neuroneFigure 57 illustrates how an action potential is generated. When the neurone is stimulated, thevoltage-gated Na+ ion channels open, which increases the permeability of the membrane to Na+ ions.Na+ ions diffuse rapidly into the axon, which depolarises the membrane. The potential differenceacross the membrane is briefly reversed, becoming positive (+40mV) on the inside. This change inpolarity (i.e. when the inside of the axon is positive) is known as an action potential.

Figure 57 The active neurone

Knowledge check 36Describe how the potential difference changes when a neurone is stimulated.

At the end of depolarisation the voltage-gated Na+ ion channels close and the permeability of themembrane to Na+ ions decreases. The membrane starts to repolarise and the voltage-gated K+ ionchannels open. This temporarily increases the permeability of the membrane to K+ ions, increasingthe rate of their outward diffusion (see Figure 58).

Figure 58 Repolarising the membrane

Due to the increased permeability to K+ ions the potential difference across the membrane‘overshoots’ the resting potential to −80mV. The neurone is hyperpolarised. The voltage-gated K+ion channels then close and the Na+/K+ ion pump restores the resting potential of the membrane.

Figure 59 shows the changes in potential difference and conductance (permeability) of the axonmembrane to Na+ and K+ ions during the passage of an action potential. It shows that thedepolarisation of the membrane coincides with the increase in permeability of the membrane to Na+ions. It can also be seen that the repolarisation coincides with an increased permeability to K+ ions.

Figure 59 The change in (a) potential difference and (b) the conductance of Na+ and K+ ions duringthe passage of an action potential

The thresholdFor an action potential to be generated the depolarisation must exceed a threshold value, which isusually 10–15mV above the resting potential (see Figure 59a). A threshold stimulus is a stimulus thatis sufficient to set up an action potential in a neurone. Any stimulus weaker than the threshold (a sub-threshold stimulus) will not generate an action potential. Above the threshold, the size or strength ofthe action potential is always the same irrespective of the size or strength of the stimulus. This isknown as the all-or-nothing law. The frequency of action potentials is directly related to the intensityof the stimulus, i.e. the greater the intensity of stimulus the greater is the frequency of actionpotentials.

The refractory periodFor a brief period following an action potential the voltage-gated Na+ channels are inactivated andthe inward movement of Na+ ions is prevented. Therefore another action potential cannot begenerated; this is known as the refractory period. The refractory period ensures that:• an action potential can be propagated in one direction only• a second action potential is separated from the first, which limits the frequency of action potentials

along a neurone

The transmission of a nerve impulseOnce an action potential has been set up it moves rapidly from one end of the neurone to the other.The movement is the result of local electrical currents set up by the ion movements of the actionpotential itself. This is illustrated in Figure 60 on p. 58.

Figure 60 Propagation of an action potential along a non-myelinated neurone

(1) When the neurone is stimulated, Na+ ions rush into the axon, depolarising the membrane (seesecond diagram).

(2) The action potential causes local electrical circuits to be established which cause gated Na+channels to open in the adjacent part of the membrane, which then depolarises (see third diagram).

(3) Behind the impulse, K+ ions begin to leave, causing the neurone to become repolarised behind theimpulse.

In this way a single action potential results in a step-by-step wave of action potentials along the entirelength of the neurone. It can also be seen that the gated Na+ channels are inactivated behind the

impulse (i.e. during the refractory period). By the time they are activated again (bottom diagram) theaction potential is too far down the axon to affect the area where the action potential originated. Thisexplains how the refractory period ensures that impulses travel in one direction only.

Speed of the impulseThe speed of the impulse varies from 0.5ms−1 to 120ms−1 depending upon the diameter of the axon,the temperature and the presence or absence of a myelin sheath.

In non-myelinated axons, the speed of transmission depends upon the longitudinal resistance of thecytoplasm in the axon (axoplasm). The resistance is related to the diameter of the axon — the largerthe diameter, the lower the resistance. This increases the length of the local electrical circuit alongthe membrane and therefore increases the speed of transmission.

Knowledge check 37Describe the effect of the following on the speed of transmission of a nerve impulse:(a) the presence of a myelin sheath(b) an increase in axon diameter

As the temperature increases to about 40°C, the speed of transmission increases. The generation ofnerve impulses is an active process and requires ATP from respiration. Therefore, anything thataffects the rate of respiration (such as temperature) affects the speed of transmission.

Examiner tipA nerve impulse refers to the movement of action potentials along a neurone. It is important thatwhen describing the transmission of a nerve impulse you state that the action potentials move alongthe neurone or that action potentials jump from node to node.

The myelin sheath acts as an electrical insulator. Ions cannot pass through it. The sheath isinterrupted by nodes of Ranvier. It is only at these points that local circuits are set up allowing themovement of Na+ and K+ ions across the membrane. This effectively increases the distance overwhich the local currents can bring about depolarisation. It causes the action potential to ‘jump’ fromnode to node and increases the speed of transmission. This process is known as saltatory conduction(Figure 61).

Figure 61 Saltatory conduction in a myelinated neurone

Examiner tipAs the Na+/K+ pumps are only active at the nodes of Ranvier, myelinated neurones use less ATPthan non-myelinated neurones.

SynapsesA synapse is a junction, 20nm in width, between two neurones. The transmission of the impulse fromone neurone to the next is by chemicals called neurotransmitters. There are two main types ofneurotransmitter:• acetylcholine (ACh), which is produced by motor neurones targeting muscle cells• noradrenalineOther neurotransmitters include dopamine and serotonin.

Structure of the synapseFigure 62 on p. 60 shows the structure of a synapse. At the end of the neurone, the axon swells to formthe synaptic knob. This contains synaptic vesicles containing the neurotransmitter and manymitochondria to produce ATP for the resynthesis of neurotransmitter molecules and the Na+/K+pumps.

Figure 62 The structure of a cholinergic synapse

Synaptic transmissionFigure 63 on p. 60 shows how an impulse is transmitted across a (cholinergic) synapse. The nerveimpulse reaches the synaptic knob (1 in Figure 63) and depolarises the presynaptic membrane causingvoltage-gated calcium ion channels to open (2). Calcium ions diffuse into the presynaptic neuronecausing the synaptic vesicles to move towards and fuse with the presynaptic membrane (3) wherethey release acetylcholine into the synaptic cleft via exocytosis (4).

Figure 63 Transmission of an impulse across a cholinergic synapse

The acetylcholine diffuses across the synaptic cleft (5) and binds with specific receptor proteins on

the postsynaptic membrane (6). The receptor proteins are attached to gated Na+ channels; the bindingof acetylcholine causes the protein to change shape and the Na+ channels to open. Na+ ions diffuseinto the postsynaptic neurone (7). If the threshold is reached an action potential is generated in thepostsynaptic neurone (8).

Once the acetylcholine has depolarised the postsynaptic neurone it is hydrolysed by the enzymeacetylcholinesterase, which is located on the postsynaptic membrane (9). This prevents successiveimpulses merging at the synapse. The resulting molecules (choline and ethanoic acid) diffuse backacross the synaptic cleft and are actively transported back into the synaptic knob of the presynapticneurone (10). Energy released from the hydrolysis of ATP is required to reform the acetylcholine,which is then stored in the synaptic vesicles.

Knowledge check 38Give two reasons why mitochondria are present in the synaptic knob.

Knowledge check 39Explain two reasons why nerve impulses can only travel in one direction.

Functions of synapses• Synapses transmit impulses in one direction only (unidirectional):• The neurotransmitter is released only from the presynaptic neurone.• The neurotransmitter receptors are found only on the postsynaptic membrane.• Neurotransmitter diffuses from the presynaptic neurone to the postsynaptic neurone.

Synapses filter out low-level stimuli. If the intensity of a threshold stimulus is low then the frequencyof impulses sent along a neurone is low. Only a small quantity of neurotransmitter is released and fewNa+ channels open on the postsynaptic membrane. This may be insufficient to exceed the thresholdvalue, in which case no action potential is generated in the postsynaptic neurone.

Synapses act as junctions and allow summation to occur. When several impulses arrive at a synapsea large quantity of neurotransmitter is released. This causes many gated Na+ channels to open. Thethreshold is exceeded and an action potential is generated in the postsynaptic neurone.

There are two types of summation:• Temporal summation — high frequency impulses arrive in quick succession from the same

presynaptic neurone and lead to an accumulation of neurotransmitter in the synapse• Spatial summation (Figure 64) — several impulses arrive at the same time from several different

presynaptic neurones

Figure 64 Spatial summation

Some synapses prevent the passage of impulses between neurones. Inhibitory synapses causehyperpolarisation to occur in the postsynaptic membrane. This prevents an impulse being generated inthe postsynaptic neurone. (See also the section on spinal reflexes on p. 62.)

The effect of drugs on synaptic transmissionMost drugs that affect the nervous system influence synaptic transmission. Drugs fall into two broadcategories: excitatory drugs (agonists) and inhibitory drugs (antagonists).

Excitatory drugs may have a similar shape to the neurotransmitter and bring about the same effect.They may inhibit the enzyme that breaks down the neurotransmitter, which results in theneurotransmitter remaining attached to the postsynaptic membrane. Examples are organophosphorusinsecticides — they inhibit cholinesterase so acetylcholine remains bound to the postsynapticmembrane.

The significance is that agonists bring about continuous stimulation of the postsynapticneurone/muscle. This can result in paralysis or death due to continuous contraction of cardiac andintercostal muscles.

Inhibitory drugs can bind to and block the receptors on the postsynaptic membrane. This prevents theneurotransmitter molecules from binding and prevents their action. Beta blockers are an example.

Inhibitory drugs can cause paralysis or death due to the inability of the muscles to contract.

Psychoactive drugs alter brain function and result in temporary changes in perception, mood,consciousness and behaviour. They include tobacco, cannabis, ecstasy, cocaine, heroin andamphetamines (speed).

Cocaine is an excitatory drug that works by preventing the normal re-uptake of the neurotransmitterdopamine. Dopamine accumulates in synapses, causing repeated action potentials in postsynapticneurones. Dopamine stimulates the pleasure centres of the brain, giving a sense of wellbeing andhappiness. Cocaine results in the over-stimulation of the pleasure centres and the user feels euphoric.

The molecule THC is the active ingredient in marijuana. THC is an inhibitory drug that binds toreceptors in the presynaptic membranes, thereby inhibiting the release of excitatory neurotransmitters.This prevents stimulation of the postsynaptic membranes, so a person intoxicated with marijuana islikely to feel relaxed and calm.

Knowledge check 40Organophosphorus insecticides are used to kill pests. Explain how the pesticide affects synaptictransmission.

Reflexes• Reflexes are not under the conscious control of the brain and are therefore involuntary.• The neurones forming the pathway taken by the nerve impulses in a reflex action make up a reflex

arc.• Reflexes generally have a protective function and increase an organism’s chances of survival.

Spinal reflexesThe position of the spinal cord within the vertebral column is shown in Figure 65. When the spinalcord is viewed in a transverse section it shows two distinct regions:• the central grey matter, which contains the cell bodies and relay neurones• the outer white matter, which contains myelinated axons that run up and down the spinal cord to

and from the brain

Figure 65 The position of the spinal cord inside the vertebral column

Sensory neurones transmit impulses from receptors to the CNS. They enter via the dorsal root andhave their cell bodies in a swelling called the dorsal root ganglion.

Motor neurones transmit impulses from the CNS to effectors. They leave via the ventral root.

Knowledge check 41Explain why white matter is white and grey matter is grey.

Figure 66 on p. 64 shows a withdrawal reflex.

Figure 66 A section through the spinal cord and the pathways formed by the neurones in thewithdrawal reflex

(1) Heat from the hot plate is detected by a heat receptor in the finger.(2) An action potential is generated and an impulse is transmitted along a sensory neurone to the

spinal cord via the dorsal root.(3) The sensory neurone synapses with a relay neurone in the grey matter of the spinal cord. The

release of neurotransmitter by the sensory neurone generates an action potential in the relayneurone.

(4) This is repeated at the synapse between the relay neurone and the motor neurone (also within thegrey matter of the spinal cord). The action potential is transmitted along the motor neurone, via theventral root, to the biceps muscle.

(5) The release of acetylcholine causes the biceps muscle to contract, resulting in an automaticwithdrawal of the hand from the hot plate.

Knowledge check 42Complete the table for the reflex shown in Figure 66.

Stimulus Receptor Coordinator Effector Response

Reflex arcs are not separate, and impulses can travel up and down the spinal cord. In the grey matter,the sensory neurone may synapse with another neurone transmitting impulses to the brain. Theinformation received may be stored in memory, which leads to learning. The brain may relate theinformation received with other sensory input — for example from the eyes — to modify theresponse.

For example, if a hot metal dish is picked up it will probably be dropped immediately (simple spinalreflex). However, if an equally hot glass dish is picked up it will probably be put down quickly, butgently. In this second situation if the glass dish is dropped it may break and cause further harm to theindividual. The brain makes a conscious decision and transmits impulses down the spinal cord vianeurones that terminate in inhibitory synapses, which prevents the dish from being droppedimmediately.

Examiner tipMany students have difficulty in completing a diagram to show the arrangement of neurones in areflex arc. The sensory neurone always enters via the dorsal root; this is the one with the ‘lump’ in it(the dorsal root ganglion) where the cell body of the sensory neurone is located. The synapsesalways occur in the grey matter. The motor neurone always leaves via the ventral root. Alwaysremember to label your diagrams.

Nerve netsHydra is a simple organism that belongs to the phylum Cnidaria, which also contains sea anemonesand jellyfish. They do not have a recognisable brain or true muscles. They have a nerve net (Figure67), which is a simple system compared with the mammalian nervous system. Photoreceptors andtouch-sensitive nerve cells are located in the body wall and tentacles. Therefore they respond tolimited stimuli — light intensity and touch. As a result, the numbers of effectors are small.

Figure 67 The structure of a nerve net in a Cnidarian jellyfish

The nerve net consists of simple nerve cells with short extensions joined to each other and branchingin a number of different directions. This means that the transmission of the nerve impulse is slow.

Knowledge check 43Complete the following table to compare the differences between the structure and function of theneurones found in the nerve net of a jellyfish (Cnidarian) and in a vertebrate.

Nerve net Vertebrate neuronesOnly one type of neurone Neurones are longerNeurones branched Neurones can transmit impulses in one

direction onlyImpulses pass in all directions from point ofstimulation

Neurones may be myelinatedMany synapses involved Slower transmission of impulses

SummaryAfter studying this topic you should be able to:

• understand that responding to a stimulus requires information from a receptor to be relayed to aneffector; in animals effectors are either muscles or glands

• draw a labelled diagram of a mammalian motor neurone and describe the function of the differentstructures

• describe and explain how a resting potential is established in a neurone• describe and explain how an action potential is generated and transmitted along a neurone• analyse oscilloscope traces• state and explain the factors that affect the speed of conduction of a nerve impulse• draw a labelled diagram of a synapse and describe the process of synaptic transmission• describe and explain the effect of chemicals such as organophosphates and psychoactive drugs

on transmission• draw a labelled diagram of a section through the spinal cord• describe and explain how a simple reflex arc acts as the basis of protective, involuntary actions• compare the structure and function of the mammalian nervous system with the nerve net of a

simple organism, such as Hydra

Responses in plantsKey concepts you must understand• Plants are responsive to the environment.• Plant responses are slow because coordination is achieved by chemicals (hormones) as they do

not have a nervous system.

PhotoperiodismThe relative length of day and night is known as the photoperiod and varies with the time of year intemperate latitudes. Photoperiodism describes the response of a plant to changes in the length of thephotoperiod and this can affect flowering as well as fruit and seed production.

Examiner tipDo not get confused; light intensity plays an important role in photosynthesis and transpiration buthas no role in the flowering of plants. This is controlled by the length of the photoperiod.

Experiments conducted in the 1920s demonstrated that the length of the photoperiod controlsflowering. Plants were classified according to the photoperiod in which they flower.• Long-day plants, for example lettuce, clover and many cereals, flower in late spring and early

summer when the light periods are long.• Short-day plants, for example chrysanthemums, poinsettia, tobacco and strawberries, flower

between late summer and early spring when the light periods are short.• In day-neutral plants, such as tomatoes and cucumber, the day length appears to have no effect on

flowering.

Figure 68 shows the effect of the photoperiod on the flowering of short-day and long-day plants.

Figure 68 Effect of photoperiod on flowering

Detection of the photoperiodPhytochrome is the photoreceptor responsible for absorbing light. It is a blue-green pigment andabsorbs light in the red part of the visible spectrum. Red light has a range of wavelengths from 625nmto 740nm.

Phytochrome exists in two interconvertible forms (see Figure 69):• Pr absorbs red light, with a wavelength of 660nm.• Pfr absorbs far-red light, with a wavelength of 730nm.

Figure 69 Interconversion of phytochrome

When Pr absorbs red light it is rapidly converted to Pfr; when Pfr absorbs far-red light it is rapidlyconverted to Pr. Sunlight contains more red light (660nm) than far-red light (730nm). Therefore,during the day Pr is converted to Pfr, which accumulates in cells. At night, the unstable Pfr slowlyreverts back to Pr.

Examiner tipBe sure you understand the difference between red and far-red light and the two different forms ofphytochrome.

The relative amounts of the two forms are used to measure the photoperiod. It is thought that the levelof Pfr is critical in the induction of flowering:• Low levels of Pfr induce flowering in short-day plants.• High levels of Pfr induce flowering in long-day plants.

Figure 70(a) shows the effect of the concentration of Pfr on the flowering of short-day and long-dayplants:• During the long day Pr is converted to Pfr; due to the short dark period little Pfr reverts back to Pr• High levels of Pfr are found in the plant• Long-day plants will flower• Short-day plants will flower• During the short day Pr is converted to Pfr; due to the long dark period most of the Pfr reverts back

to Pr• Low levels of Pfr are found in the plant• Long-day plants will not flower• Short-day plants will flower

Figure 70 (a) Effect of photoperiod on Pfr concentration and flowering

Knowledge check 44(a) Name one place in the plant where phytochrome is found.(b) State the effect of exposure to far-red light on Pfr.(c) State the effect of darkness on Pfr.(d) Describe how the effects of darkness on Pfr could be reversed.

Since the 1940s, experiments have shown that it is the length of the unbroken dark period that iscritical for flowering (see Figure 70b on p. 68):• Under normal conditions the long dark period will result in most of the Pfr reverting back to Pr;

however, the brief light period brings about the rapid conversion of Pr to Pfr resulting in highlevels of Pfr

• Long-day plants will now flower• Short-day plants will not flower

Figure 70 (b) Manipulation of flowering

Horticulturists can manipulate the flowering of plants:• In long-day (short-night) plants, a short dark period induces flowering. However, if these plants

are briefly exposed to light during a long dark period they will flower.• In short-day (long-night) plants, a long dark period induces flowering. However, if these plants are

briefly exposed to light during the long dark period they will not flower.

Examiner tipThe demand for flowers and flowering plants varies throughout the year. Being able to manipulatethe flowering of plants is an advantage because it allows horticulturists to prepare flowering plants

for particular dates such as Mother’s Day. Another advantage is that it allows plant breeders toproduce new hybrid varieties of plants by cross-pollinating flowers that would not usually flower atthe same time.

Investigating photoperiodismIn an experiment, one leaf of a short-day plant was exposed to daily routines of short days and longnights. The rest of the plant kept in constant darkness (see Figure 71a). The plant flowered. It wasconcluded that:• only one leaf needs to be exposed to the appropriate photoperiod for flowering to occur• the photoperiod is detected by the leaf• a chemical substance must be transported from the leaf to the bud to bring about flowering

Figure 71 (a) Detection of the photoperiod; (b) The chemical nature of flowering

In a second experiment, a short-day plant was grafted onto a long-day plant. Both plants wereexposed to daily routines of short days and long nights (see Figure 71b). Both plants flowered. Itwas concluded that:• the chemical produced in the short-day plant must be transported into the long-day plant• the chemical has the same effect in both types of plant

Other experiments have shown that phytochrome is present in the leaves and is responsible for thedetection of the photoperiod. A chemical messenger/hormone called florigen is thought to beresponsible for flowering:• In short-day plants, low levels of Pfr may stimulate the production of florigen.• In long-day plants, high levels of Pfr may stimulate the production of florigen.

Knowledge check 45Florigen is an organic molecule. Name the plant tissue that would be involved in the transport offlorigen from the leaves to the bud.

SummaryAfter studying this topic you should be able to:• understand that plants are responsive to their environment and that their responses are

coordinated by hormones• describe and explain how flowering in both short-day plants and long-day plants is brought about

by phytochromes and florigen

Questions & Answers

The unit testWhen exam papers are being prepared the examiner must try to ensure that all of the topics covered inthe unit are assessed, so you should expect to get questions on each topic. However, it would beimpossible to be asked a question on everything.

Examiners must also set questions that test the specific assessment objectives (these are described inthe WJEC Biology specification and also referred to in this Questions and Answers section) and youmay find it useful to understand the weighting of the assessment objectives that will be used.

There are 80 marks available in the BY4 exam. On the AS exams the marks available for AO1 (recallof knowledge and understanding) and AO2 (application of knowledge and understanding) are roughlyequal. However, at A2 there is much more emphasis on application of knowledge and understandingand almost two-thirds of the questions target AO2. There will always be a few marks targeting AO3(how science works) and these will probably relate to practical work you have carried out during thecourse.

The assessment objectives are weighted as follows:

About this sectionThis section contains questions on each of the topics. They are written in the style of the questions inBY4, so they will give you an idea of what you will be asked to do in the exam. After each questionthere are answers by two different students and then examiner comments on what they have written.

WJEC examination questions are used with permission from WJEC.

Examiner’s commentsEach question is followed by examiner tips explaining what you need to do to gain full marks (shownby the icon ). All student responses are then followed by examiner comments. These are precededby the icon , and highlight where credit is due. In the weaker answers, they also point out areas forimprovement, specific problems and common errors such as lack of clarity, irrelevance,misinterpretation of the question and mistaken meanings of terms.

Question 1 ATP and respiration(a) The diagram represents a molecule of ATP.

(i) Give the name of the reaction that has resulted in the bond X.

(1 mark) (ii) Name the pentose sugar Y.

(1 mark) (iii) Name the organic base Z.

(1 mark)(b) Describe how energy is released from ATP.

(3 marks)(c) The diagram shows the structure of a mitochondrion as seen using an electron microscope.

(i) Using the letters on the diagram identify the site of each of the following.

(3 marks) • Krebs cycle • oxidative phosphorylation • decarboxylation

(ii) What is the function of the circular DNA?

(1 mark) (iii) Explain how the pH in region Q becomes acidic.

(3 marks)

Total: 13 marks

Questions with a diagram of ATP are common. Here, the majority of the marks are testing recallwith understanding (AO1). Part (c)(iii) tests application of knowledge and understanding (AO2) and(a)(i) and (c)(ii) are synoptic marks. For a well-prepared student with a good grasp of the coreconcepts of biology, learnt in BY1, this question should pose no problems.

Student A(a) (i) phosphorylation a (ii) ribose b (iii) adenosine c(b) ATP is converted to ADP and Pi d when bond X is broken. e This exergonic reaction is

catalysed by ATP synthase.

(c) (i) Krebs cycle = S g; oxidative phosphorylation = Q h; decarboxylation = S. i. (ii) It allows the cell to replicate. j (iii) Aerobic respiration produces carbon dioxide as a waste product. This is an acidic gas and

dissolves in the mitochondria forming carbonic acid and lowering the pH. k

5/13 marks awarded a Correct — the formation of ATP from ADP and Pi is a phosphorylationreaction. b Correct. c This is incorrect. Adenosine refers to the combined ribose sugar and adeninebase. d Correct. e At A2, this statement is too vague. There should be reference to hydrolysis. f Thereis no mark for stating it is an exergonic reaction as the stem of the question refers to the release ofenergy. Student A has stated the wrong enzyme — ATP synthase catalyses the formation of ATP. gCorrect. h This is incorrect. ATP synthase is located on the inner mitochondrial membrane. i Correct.j This is incorrect. The DNA inside mitochondria allow the mitochondria to replicate but not the cell.k Although the student is correct in stating that carbon dioxide is both a waste product of respirationand an acidic gas, it is produced in the mitochondrial matrix (region S, not region Q) and does notaccumulate, It diffuses out of the mitochondria and out of the cell. As the context of the answer iswrong, it gains no credit.

Student B(a) (i) condensation a (ii) ribose b (iii) adenine c(b) ATP is hydrolysed d to ADP and Pi e. This reaction is catalysed by ATPase f and releases

30.6kJ of energy.

(c) (i) Krebs cycle = S g; oxidative phosphorylation = R h; decarboxylation = S. i. (ii) To synthesise mitochondrial proteins j (iii) NADH2 provides protons and electrons for oxidative phosphorylation. As the electrons pass

from carrier to carrier they provide the energy needed to pump protons from the matrix intothe intermembrane space (region Q). The accumulation of protons lowers the pH. k

12/13 marks awarded a–e Correct. f The student has stated the correct enzyme and gone on tostate the quantity of energy released. However, this cannot be awarded a mark because the student hasalready gained full marks for part (b). g–i Correct. j This is incorrect. The ribosomes inside themitochondria synthesise mitochondrial proteins; the mitochondrial DNA codes for the mitochondrialproteins. k This is an excellent response. The student has linked the intermembrane space with theirknowledge of chemiosmosis and gains all 3 marks.

If you are well prepared you should be able to score well on this type of question. Most of themarks gained by Student A are from his/her understanding of BY4. However, the student showsa poor understanding of some fundamental biology. The answers lack the detail and precise useof terminology that is expected at this level. Student A gains 5 marks (grade U). In contrast,Student B gains 12 marks (grade A) for demonstrating a good understanding of fundamentalaspects of biochemistry and cell biology. The information in the diagrams has been used toselect the appropriate biology to answer the questions. The answers are coherent and includethe terminology expected at this level.

Question 2 RespirationMuscle cells were broken up and separated into fractions. Samples of each fraction wereincubated with (i) glucose and (ii) pyruvate (pyruvic acid). Tests were carried out for theproduction of carbon dioxide and lactate (lactic acid) in each sample. The results are given inthe table below.

(a) Using your knowledge of respiration explain the results obtained with each cell fraction. (i) Ribosomes

(1 mark) (ii) Mitochondria

(2 marks) (iii) Cytoplasmic residue

(2 marks)(b) Explain why no carbon dioxide is evolved when cyanide is added to the mitochondrial

fraction prior to incubation.

(2 marks)

Total: 7 marks

This question tests mainly application of knowledge and understanding (AO2) and is typical ofmany questions on respiration. You must use the information given and pick out the aspects of biologyyou need. In this question the type of cell used is irrelevant, the parts of the cell (the ‘fractions’) areimportant as are the two different types of respiratory substrate given. You must link the correct stageof respiration with its location within the cell and the initial substrate involved. Part (b) has asynoptic element from BY1.

Student A(a) (i) Ribosomes synthesise enzymes involved in respiration. a (ii) No carbon dioxide is produced during glycolysis. It is only produced during the link reaction

and Krebs cycle. b (iii) Anaerobic respiration occurs in the cytoplasm producing lactate. c The mitochondria carry

out aerobic respiration and so do not produce lactate. d(b) Cyanide is a respiratory inhibitor e and prevents respiration from occurring. f

1/7 marks awarded a Although the student’s statement is correct it does not answer the questionbeing asked and simply recalls knowledge of ribosomes. b The student recalls the stages ofrespiration in which carbon dioxide is produced, but does not link this to the information in the table.c This attempt is better. The student links anaerobic respiration to the cytoplasm but does not statethat the lactate is produced from the glucose. d The reference to mitochondria is irrelevant and showsthat the student has struggled to interpret the data. e This is correct, for 1 mark, but the followingcomment f is too vague.

Student B(a) (i) Ribosomes synthesise proteins, they do not carry out respiration. a (ii) Glucose cannot enter the mitochondria as the mitochondrial membranes have no glucose

carriers and without the cytoplasm glucose cannot be broken down. b Pyruvate can enter themitochondria and so the link reaction and Krebs cycle will take place producing carbondioxide.c

(iii) Anaerobic respiration occurs in the cytoplasm and can produce lactate from glucose. d As nocarbon dioxide or lactate is produced in the cytoplasmic residue it implies that pyruvatecannot be broken down. e

(b) Cyanide inhibits the electron transport chain. f NAD cannot be regenerated and so the Krebscycle and the link reaction also stop. g

6/7 marks awarded a Correct. b, c The student has interpreted the information well anddemonstrates a clear understanding of the information provided, gaining both marks. d This is correct.The student links glucose to the production of pyruvate under anaerobic conditions. e The student hasused the information provided but the answer does not give an explanation. f,g Student B gains bothmarks for a concise answer showing full understanding.

Many students find interpreting experimental data difficult. You must ensure that you are wellprepared by answering as many past exam questions as you can. Make sure you read theinformation carefully and use a highlighter pen to pick out the important information. Thenthink about the biology you have learned and try to apply it to the situation given. Student Amakes little reference to the information provided. The only correct answer given is thatcyanide is a respiratory inhibitor, for 1 mark. In contrast, Student B gains 6 marks (grade A),which is an excellent mark demonstrating a good understanding of respiration and the ability tointerpret information. The only mark dropped was probably the hardest mark to gain. Ifpyruvate is produced during glycolysis, then NADH2 is also produced, which can convertpyruvate to lactate. However, in this experiment pyruvate is added to the cytoplasm and noNADH2 is available to convert it to lactate. To gain full marks on this question requires a goodunderstanding of respiration.

Question 3 Photosynthesis(a) The electron micrograph shows a section through a chloroplast.

Using the letters on the electron micrograph, complete the following table. You may use aletter once, more than once or not at all.

(5 marks)

Area Letter Where chlorophyll is found Where carbon dioxide is reduced to a hexose sugar Where oxygen is produced from water Where there is a high concentration of ribulose 1-5 bisphosphate carboxylase (Rubisco) Containing the polysaccharides amylose and amylopectin

(b) A representation of the light stage of photosynthesis is shown below.

(i) What is represented by the arrows labelled P?

(1 mark) (ii) What is the general name given to the structures X and Y?

(1 mark) (iii) Briefly explain what is happening in the process labelled G.

(1 mark) (iv) What is represented by R?

(1 mark) (v) What is represented by S?

(1 mark) (vi) What process is represented by the arrow labelled T?

(1 mark)(c) Atrazine is a commonly used weedkiller. It prevents non-cyclic photophosphorylation from

taking place. Using your knowledge of the light-independent stage of photosynthesis,explain why its use leads to the death of a plant.

(5 mark)

Total: 16 marks

In this well-structured question the questions get progressively more difficult. There are many easymarks to pick up in parts (a) and (b) which tests mainly recall with understanding (AO1). Thestructure of chloroplasts was covered and examined in BY1. A BY1 question may simply ask you toidentify the structure. This would be too simple for A2 and the questions asked here are morechallenging. However, if you have learnt the biology it should not prove too difficult. Part (c) testsapplication of knowledge and understanding (AO2). You must use the information in the stem of the

question to ensure you answer it correctly.

Student A(a) A; a B; b A; c B; d D; e(b) (i) light f (ii) electron acceptors g (iii) electrons are passing along a chain of carriers h (iv) photolysis of water i (v) NADH2 j (vi) electrons k(c) During the light-independent stage CO2 combines with RuBP to produce two molecules of GP.

The GP is then converted to TP using NADH2 and ATP from the light-dependent stage. Most ofthe TP is converted back to RuBP, but some is used to make glucose and other molecules. lAtrazine will stop this from happening. m

6/16 marks awarded a–d All correct. e Incorrect. f No mark awarded — at this level referencemust be made to light energy or photons of light. g,h Correct. i This is incorrect, although the studenthas recognised that photolysis of water is occurring. j The student makes a common error by choosingthe coenzyme used in respiration. k Electrons are passing at T. However, the question asks for a‘process’ so the answer is incomplete. l The student gives a good account of the light-independentstage but does not use it to answer the question. m This statement is far too vague and is a poorattempt to link the knowledge to the situation.

Student B(a) A; a B; b A; c B; d C; e(b) (i) light energy f (ii) electron acceptors g (iii) electrons are passing along a chain of carriers to the chlorophyll molecule in photosystem I h (iv) oxygen i (v) NADPH2 j (vi) the photolysis of water k(c) If atrazine prevents non-cyclic photophosphorylation then ATP and NADPH2 will not be

produced. l Therefore glycerate phosphate cannot be converted to triose phosphate m and theCalvin cycle will stop. n The plant can no longer photosynthesise and will die. o

14/16 marks awarded a–e Correct. f–j Correct. k This is an incomplete answer. The studentnames the process of photolysis of water but fails to mention that at T electrons are passing from thewater molecule to the chlorophyll a molecule in photosystem 11. l The student uses the informationgiven and understands the significance of the light-dependent stage on the light-independent stage mand makes specific reference to the reaction that would be prevented n and the effect onphotosynthesis. o Unfortunately the student does not make the final link with the previous statement.The plant will die as respiration can no longer take place because no respiratory substrate, for

example glucose, is being made.

If you are well prepared you should be able to gain most of the marks available on questions ofthis type. Student A lost some easy marks by giving answers that lack the detail and precise useof terminology that is expected at this level. It is clear that the student had learned the biologyrequired to pick up marks in part (c). However, the answers lack coherence. It is possible thatthe student did not read the question carefully. Student A gains 6 marks (grade U). In contrast,Student B gains 14 marks (grade A). This student has both learnt the biology and interpretedthe information provided in parts (a) and (b) giving precise answers. The answer to part (c) isconcise and well structured demonstrating an ability to apply knowledge and understanding toan unfamiliar situation.

Question 4 Microbiology(a) A swab from infected tissue was taken to a microbiology laboratory and stained to allow

the initial identification of the types of bacteria present. The results are shown in thediagram below.

(i) Bacteria can be classified according to their shape. Give the names of the types ofbacteria labelled A and B.

(2 marks) (ii) Although the bacteria labelled C and D have the same shape they are not the same type

of bacteria. Use the information in the diagram to identify the two types of bacteria.

(1 mark) (iii) Explain the reason for the difference in staining between the bacteria labelled C and D.

(2 marks)(b) Some bacteria are described as facultative anaerobes and some as obligate anaerobes.

State the difference between these two types of bacteria.

(2 marks)(c) Three fermenters were set up to study the population growth of E. coli in different sugar

solutions, 0.001M glucose, 0.001M lactose and a mixture of glucose and lactose both at0.001M. Samples were removed from the fermenter at timed intervals. The population sizein each fermenter was estimated. The results are shown in the graph below.

(i) Explain why there is a difference in population growth between the glucose and lactose.

(3 marks) (ii) Describe and explain the shape of the curve when the bacteria are grown in lactose and

glucose together.

(5 marks)

Total: 15 marks

This question combines classification of bacteria and aspects of the population growth curve. Parts(a) and (b) of this question are straightforward and there are some easy marks for simple recall withunderstanding (AO1). Part (c) is testing both AO1 and application of knowledge and understanding(AO2). There is also a synoptic element linking to biological molecules from BY1. Instead of simplygiving you a straightforward growth curve the examiner has presented a situation that requires you tothink and use the information provided.

Student A(a) (i) A = coccus a; B = bacillus b (ii) Gram-positive bacteria stain purple and Gram-negative bacteria stain pink c (iii) Gram-positive bacteria have a simple cell wall made of peptidoglycan. d Gram-negative

bacteria have a more complex cell wall with two layers. e(b) Facultative anaerobes can survive in both aerobic and anaerobic conditions. f Obligate

anaerobes are found in anaerobic conditions. g

(c) (i) Bacteria grow rapidly with glucose and reach the stationary phase after 120 minutes. h Thebacteria grow more slowly with lactose and only reach the stationary phase after 210minutes. i This is because the bacteria prefer glucose to lactose. j

(ii) The population of bacteria is increasing rapidly as there is plenty of food available k andtoxic waste products have not accumulated. The population then starts to level off l when theglucose runs out. m The bacteria then feed on the lactose and the population increases again.

n After 210 minutes it reaches the stationary phase o and the birth rate equals the death rate. p

6/15 marks awarded a Correct. b Incorrect. c Correct, although the student does not give the letterof the bacteria the link to the colour of the type of bacteria is made. d Correct. e The answer justmakes a comparison and does not give an explanation for the difference in colour. f,g Thesestatements are too vague. Reference to oxygen must be given. h,i The answer describes the curves; thequestion asks for an explanation. j This is a poor attempt at an explanation. k ‘Food’ is not acceptableand demonstrates poor use of terminology. l,m Correct. n The answer lacks detail. There is noreference to how the bacteria ‘feed’ on the lactose or that the increase in population is rapid at thisstage. o Correct. p Bacteria do not give birth! The student should refer to the rate of cell productionand the rate of cell death.

Student B(a) (i) A = coccus a; B = spirillum b (ii) C = Gram positive; D = Gram negative c (iii) Gram-positive bacteria have a cell wall made of a thick layer of peptidoglycan d which

retains the crystal violet stain. e Gram-negative bacteria have a more complex cell wall witha thin layer of peptidoglycan and an outer layer of lipopolysaccharides and do not retain thecrystal violet stain.

(b) Facultative anaerobes grow better in aerobic conditions but can also survive without oxygen. fObligate anaerobes are only found in anaerobic conditions as oxygen is toxic to them. g

(c) (i) When provided with glucose, the bacterial population grows rapidly as they do not have toproduce any enzymes to digest the glucose. h When provided with lactose, the bacterialpopulation grows slowly as it takes time to synthesise lactase i which is then secreted tohydrolyse the lactose. j The bacteria can then absorb the glucose and multiply.

(ii) After a relatively short lag phase, the bacteria enter the exponential phase of growth k asthere is plenty of glucose l. The population starts to level off m but after 150 minutes thepopulation starts to rise again. This is because the bacteria are now secreting lactase n so thelactose is being hydrolysed into glucose o so the bacteria can continue to grow. After 210minutes the bacteria reach the stationary phase p when all of the lactose has run out. q

13/15 marks awarded a–c Correct. d, e Correct. This is a good answer that explains how thestructure of the cell wall accounts for the difference in the colour of the bacteria. f,g Correct. h Thestudent is on the right lines with this answer and would have gained the mark for stating that theglucose can therefore be absorbed quickly and used in respiration. i, j The student demonstrates thebasic difference between glucose and lactose and explains why this causes the difference in thecurves. Student B gains 2 marks even though galactose is not named as the other product of lactosedigestion. k–n,p,q These are all valid points and the student gains full marks for this part-question. oThis would not gain a mark as galactose has not been mentioned. However, the maximum mark hasbeen achieved.

A large proportion of the marks for this question — particularly for an A2 question — testrecall with understanding (AO1). Therefore, if you are well prepared you should be able to gainmost of the marks available. Student A has not read the question carefully and has provided

answers that lack the detail and precise use of terminology expected at this level. Student Agains 6 marks (grade U). In contrast, Student B gains 13 marks (grade A). This student hasspent time learning the terminology in this section of the course. The answers given are wellstructured, detailed and include precise biological terminology.

Question 5 Populations(a) Two species of the single-celled organism Paramecium, P. aurelia and P. caudatum, were

grown together in a single culture of the bacterium Bacillus pyocyaneus, on which they bothfeed. Their population densities were measured every 2 days and the results are shown inthe graph below.

(i) For P. aurelia, on which day of the experiment did population growth enter thestationary phase?

(ii) On which days of the experiment is the population of P. caudatum in the death phase?(b) The experiment shows both interspecific and intraspecific competition. (i) Which type of competition is most likely to have caused the population of P. caudatum to

decrease after day 2? (ii) What is the carrying capacity for P. aurelia in this experiment? (iii) How might the carrying capacity have been increased in this experiment? (iv) Suggest, with an explanation, what would happen to the numbers of P. caudatum if P.

aurelia became infected with a parasitic microorganism at day 8.

(c) (i) Distinguish between the terms density-dependent and density-independent factors. (ii) Name one density-independent factor that could have changed in the experiment.

Total: 10 marks

This is a straightforward question about competition and its effects on population growth. It testsboth recall with understanding (AO1) and application of knowledge and understanding (AO2).Population growth is commonly assessed through structured questions like this. It is important to learnall of the key terms in this section and to be able to distinguish between density-dependent anddensity-independent factors. There are 3 marks for reading data from the graph and, if you are careful,you should earn all 3 marks.

Student A(a) (i) Day 10 a (ii) Days 2 to 14 b(b) (i) Intraspecific competition c (ii) 240 d (iii) Carry out the experiment in a bigger container so they have more space e. (iv) The numbers would increase f as the P. aurelia start to die and there is more food for P.

caudatum. g.(c) (i) Density-dependent factors — larger populations have more of an effect on the factor than a

smaller population h and with a density-independent factor the size of the population isirrelevant. i

(ii) A natural disaster such as an earthquake. j

4/10 marks awarded a, b Correct. c Incorrect. d The student has read the correct value from thegraph but loses the mark by not including the units. e Incorrect. If the experiment could be carried outwith animals such as lions feeding on zebra, it would be easier to appreciate that the maximumnumber of lions is dependent upon the number of zebra and that making the ‘enclosure’ bigger will notincrease the number of lions (it will just make it harder for them to catch the zebra!). f, g Correct. h, iThis shows confusion and implies that the population is affecting the factor, when it is the factor thataffects the population. j The student has not read the question carefully and has given an implausibleanswer.

Student B(a) (i) Day 10 a (ii) Days 2 to 14 b(b) (i) Interspecific competition c (ii) 240 per cm3 d (iii) Provide the paramecium with more food (B. pyocyaneus). e (iv) The numbers would increase f as there would be fewer P. aurelia and therefore less

competition. g.(c) (i) The effect of a density-dependent factor on the population increases with increasing

population density. h The effect of a density-independent factor on the population is the sameregardless of the population density. i

(ii) Temperature j

9/10 marks awarded a–f Correct. g The statement is correct but incomplete as there is noreference to food/bacteria. h, i Correct. In both definitions the student states clearly that it is the factorthat affects the population. j Correct. in many respects this is the best example of a density-independent factor. It is equally applicable to the natural environment and experimental situations.

If you are well prepared you should be able to gain most of the marks available on this type ofquestion. Student A’s answers show a lack of preparation. The student has not learnt the keyterms and easy marks have been lost for not including units and by not reading the question

carefully. Student A gains 4 marks (grade U). In contrast, Student B gains 9 marks (grade A).The information provided has been used correctly and the precise answers demonstrate thattime has been spent learning the biology.

Question 6 The kidney(a) The diagram below is of a kidney nephron.

(i) In which region of the kidney do the Bowman’s capsules occur?

(1 mark) (ii) Explain why the urea concentration is higher in the region labelled Y than it is in the

region labelled X.

(1 mark) (iii) Match any of the letters A–G from the diagram with the statements shown in the

following table.

(4 marks)

Statement Letter(s) Main site of selective reabsorption Areas involved in ultrafiltration Sodium ions actively pumped from this region Antidiuretic hormone acts on this region

(iv) The water potential of the blood changes as it passes through the glomerulus and thewater potential of the filtrate changes as it passes along the nephron. Using a tick [✓]complete the table below to show if the water potential increases or decreases as itpasses along each of the regions labelled in the diagram.

(4 marks)

Flowing along region Water potential increasing Water potential decreasingA

C D E

(b) The kangaroo rat, Dipodomys spectabilis, is found in desert regions of North America. Itdoes not drink water and feeds on dry seeds and other dry plant material. It produces verylittle urine.

(i) Explain how the kidney of this mammal is adapted to reduce the volume of urineproduced.

(2 marks) (ii) Suggest how desert animals are able to obtain water from dry seeds.

(2 marks)

Total: 14 marks

Questions on the kidney tend to be structured, although you might be asked a 10-mark question onthe topic, particularly with regard to osmoregulation. Part (a) tests recall with understanding (AO1)and there are some easy marks to gain. However part (a)(iv) could prove tricky as students getconfused with water potential. Part (b) tests both AO1 and AO2 (application of knowledge andunderstanding) and there is a synoptic element to the question, with a link to BY1.

Student A(a) (i) Medulla. a (ii) This is because water is being reabsorbed from the filtrate. b (iii)

Statement Letter(s)Main site of selective reabsorption C cAreas involved in ultrafiltration B dSodium ions actively pumped from this region D and E eAntidiuretic hormone acts on this region G f

(iv)

Flowing along region Water potential increasing Water potential decreasingA ✓ g C ✓ hD ✓ i E ✓ j

(b) (i) It would have a larger loop of Henle k, which is an adaptation to living in the desert. l (ii) When the seeds are digested, water is released and absorbed into the blood. m

4/14 marks awarded a Incorrect. b, c Correct. d No mark awarded as both the glomerulus (A) andBowman’s capsule (B) are involved in ultrafiltration. e E is correct, but D is wrong so no mark isgiven. f Although ADH acts on both the distal convoluted tubule (F) and the collecting duct (G) thefocus on the specification is the collecting duct and the wording of the question implies one region,therefore the student gains the mark. g Incorrect. h Correct. i, j Incorrect. k The student states that theloop of Henle is ‘larger’. This is not the same as longer so the student fails to score. l The answer istoo vague and is a poor attempt at an explanation. m The student demonstrates a complete lack ofunderstanding of some basic biology.

Student B(a) (i) Cortex a (ii) This is because water is being reabsorbed from the filtrate and the urea is concentrated into a

smaller volume of filtrate. b (iii)

Statement Letter(s)Main site of selective reabsorption C cAreas involved in ultrafiltration A and B dSodium ions actively pumped from this region E eAntidiuretic hormone acts on this region F and G f

(iv)

Flowing along region Water potential increasing Water potential decreasingA ✓ gC ✓ hD ✓ iE ✓ j

(b) (i) As the rat lives in arid conditions it has longer loops of Henle. k This creates a high soluteconcentration in the tissues of the medulla l allowing more water to be reabsorbed into theblood. m

(ii) The seeds contain lipids n which can be used as a respiratory substrate o by the rat. Thisreleases metabolic water. p

14/14 marks awarded a, Correct. b This is an excellent answer that explains fully the increase inconcentration. c–f Correct. g–j Correct. The student understands the change in water potential as thefiltrate passes through the nephron. k–m The student has linked structure to function and given a fullexplanation linking the length of the loop with the effect on the solute concentration and the volume ofwater reabsorbed. n Student B uses knowledge from BY1 o and links this to knowledge of respirationp to explain one of the functions of lipids. This is an excellent answer.

The function of the kidney relies heavily on the principles of movement of molecules across

membranes. Solutes are actively transported out of the filtrate lowering the water potential —therefore water follows by osmosis. This question demonstrates the need to have a fullunderstanding of these aspects of biology. If you are well prepared, and have reviewed yournotes from BY1 you should be able to gain most of the marks available. Student A has gainedsome of the marks available for simple recall about the different regions of the nephron. Themark in part (a) (iv) was probably gained by chance. There is no evidence that the student hasspent time learning the detail involved in kidney function or the underlying principles involved.Student A gains 4 marks (grade U). In contrast, Student B gains full marks (grade A) for linkingstructure to function and cause and effect. The student has spent time learning the biology andshows an excellent understanding of the fundamental principles involved in kidney function.

Question 7 The nervous system(a) A diagram of a section of the spinal cord is shown.

(i) Identify structures A–D.

(4 marks) (ii) On the diagram above, draw a sensory neurone, a relay neurone and a motor neurone.

The sensory neurone should enter at one side of the spinal cord and the motor neuroneshould exit on the other side. The neurones should link the receptor to the effector.Label each neurone.

(3 marks)(b) Pacinian corpuscles are receptors found in the skin and consist of a single sensory neurone

surrounded by connective tissue. They respond to changes in pressure. The Paciniancorpuscle was stimulated and the electrical activity across the membrane of the sensoryneurone was recorded using a microelectrode as shown.

(i) Explain the change in potential difference shown by the microelectrode after lightpressure was applied.

(3 marks) (ii) Explain the change in potential difference across the membrane shown by the

microelectrode when heavy pressure was applied.

(6 marks)

Total: 16 marks

Questions on the nervous system involving spinal reflexes are common. Students often lose markswhen asked to complete diagrams showing the neurones involved. The spinal cord can be shownorientated in different ways and you must make sure that you use any information given in the text.Part (a) of this question tests both recall with understanding (AO1) and application of knowledge andunderstanding (AO2). However, these are relatively easy marks to obtain if you are well prepared.Part (b) tests AO2 and gives a scenario about receptor cells (which you probably have not beentaught about because it is not required on this specification). However on careful inspection of theinformation given it becomes apparent that the question is simply testing knowledge of nerveimpulses. Note that part (b)(ii) is worth 6 marks and so you need to make sure that you provide adetailed answer.

Student A(a) (i) A = motor neurone a; B = central canal b; C = sensory neurone c; D = peripheral nervous

system d (ii) e, f and g.

(b) (i) When light pressure was applied the potential difference rose from −60mV to about −50mVh. This is due to sodium ions entering the axon i and then the membrane repolarises back to−60mV. j

(ii) This is a threshold stimulus. k The heavy pressure causes the sodium channels to open l andthe sodium ions are actively transported into the axon. m The inside of the axon is now about+45mV. n The membrane then repolarises o due to the potassium ions diffusing out. p

8/16 marks awarded a Incorrect. b Correct. c, d Incorrect. e–g The student has ignored theinformation provided and attempted to draw a normal reflex pathway. However the sensory neuroneand motor neurone are entering and leaving the spinal cord via the wrong roots and no credit can begiven for the relay neurone as there are no labels. h The student describes the trace i and gives acorrect explanation for the depolarisation. j The concept of threshold is being tested here and as thestudent makes no reference to this so only 2 marks are awarded. k, l and n–p The student gives aconcise answer and correctly links the change in potential difference with the movement of sodiumand potassium ions, for 5 marks. m The student loses a mark for incorrectly stating that activetransport is involved in the inward movement of sodium ions.

Student B(a) (i) A = dorsal root ganglion a; B = central canal b; C = ventral root c; D = dorsal root d (ii) e, f and g.

(b) (i) When the receptor is stimulated with light pressure the membrane starts to depolarise h assodium ions diffuse into the axon. i However the threshold is not reached j and an actionpotential not generated. k

(ii) An action potential is produced. l The heavy pressure causes the gated sodium channels toopen m and the sodium ions rapidly diffuse into the axon. n The inside of the axon is nowpositively charged. o The sodium channels then close and the potassium channels open. p Themembrane then repolarises due to the potassium ions diffusing out at a faster rate. q

15/16 marks awarded a–c Correct. d Both students failed to recognise this as a spinal nerve. e–gThe student has used the information provided and correctly drawn the position of the three neuronesand labelled them. h–k This is an excellent answer. The data from the graph are linked clearly withknowledge of the nerve impulse and the concept of the threshold. The student earns all 3 marks l–qThis is a well-structured answer using knowledge of the nerve impulse to explain the data from thegraph. Student B gains all 6 marks.

There will always be questions that contain information about unfamiliar situations. If you takeyour time reading the information you should be able to pick out the relevant biology and beable to answer the question. If you spend time looking in different textbooks at drawings ofreflex arcs, being asked to complete one in the exam should not pose a problem. Student Ashows a complete lack of understanding of the structure of reflex pathways and an inability tofollow the instructions provided. However in part (b) the student scores 7 of the 9 marksavailable with good knowledge of the biology of nerve impulses. Overall, student A scores 8marks (grade E). In contrast. Student B gains 15 marks (grade A) for analysing the data andproviding detailed, coherent answers. The student has clearly spent time learning the biology.

Question 8 The nitrogen cycleDescribe the nitrogen cycle. Include the form that nitrogen takes in each part and the role ofbacteria (giving names wherever possible). Any diagrams included in your answer must be fullyannotated.

Total: 10 marks

The last question on every WJEC exam paper is an essay-style question worth 10 marks. There arealways two alternatives and you are required to answer one. Although the subject matter of thequestions will differ, on the whole these questions are testing recall with understanding (AO1).Therefore, if you have revised the whole of the unit and are well prepared you should gain most of themarks on this type of question. You may include diagrams within your answer and you are stronglyadvised to do this. Biology is a visual subject and your notes will probably contain many diagrams tohelp you understand the biology. You will not gain any credit for just drawing a diagram. However, ifyou annotate it then the annotations will gain credit. Drawing diagrams will also help you to constructa coherent answer.

Student A• Nitrogen is an important element and is found in proteins and nucleic acids in plants and animals.

a• When plants and animals die decomposers b release ammonia into the soil. c• The ammonia is then converted into nitrates by bacteria. d• The plants can then absorb the nitrates to make more proteins. e• Other bacteria change nitrates into nitrogen gas f which enters the atmosphere. g• Some plants, such as peas, can absorb the nitrogen directly. h

3/10 marks awarded The diagram of the nitrogen cycle is incomplete and there are six statementsabout the cycle, demonstrating a lack of preparation. a, c g These points gain marks for correctlyidentifying the locations of different forms of nitrogenous compound. b, d, e, f There is no referencemade to any of the processes involved (putrefaction, nitrification, denitrification and nitrogenfixation) or to the names of the bacteria involved. h This statement is incorrect and shows a lack ofunderstanding.

Student BPlan:

Essay:

Plants absorb nitrates from the soil a to provide a source of nitrogen for the synthesis of amino acidsand proteins. b Animals consume the plants, digest the proteins into amino acids, which they then useto make proteins.

During decomposition by saprophytic bacteria the proteins within the dead plants and animals arebroken down c and ammonium ions are released d into the soil. These ions are converted to nitritese by Nitrobacter f and then to nitrates g by Nitrosomonas. h Under anaerobic conditions,denitrifying bacteria convert nitrates into atmospheric nitrogen gas i which is unavailable to plants.Nitrogen-fixing bacteria can convert nitrogen gas into ammonium ions. j Rhizobium are symbioticbacteria that live in the root nodules of legumes. k The plant uses the ammonium ions to make aminoacids. Azotobacter is another type of nitrogen-fixing bacteria that is free-living in the soil. l

10/10 marks awarded The student gains full marks even though the answer is incomplete andcontains a couple of errors. The answer is well-structured and includes detail of the processes andthe names of the bacteria involved although f and h Nitrosomonas and Nitrobacter are mixed up. Theanswer demonstrates that the student is well prepared and has taken the time to learn the detail of thecycle.

If you are well prepared you should be able to gain most of the marks available on this type ofquestion. There are usually about 15 marking points on the mark scheme, which allows you toachieve maximum marks without knowing ‘everything on the topic’; as demonstrated byStudent B who gains 10 marks (grade A). Quality of written communication is important andyou must ensure that you give a coherent answer that is detailed and contains precise biologicalterminology. Student A’s answers lack the detail expected at this level and the student gainsonly 3 marks (grade U).

Many students shy away from practising the 10-mark essay questions from past papers.However it is worth bearing in mind the following:1 The grade boundaries for this unit tend to be 5 or 6 marks apart. Therefore the quality of your

answer to the 10-mark question can have a big effect on your overall grade.2 The majority, if not all, of the marks are for recall of knowledge with understanding (AO1).3 You can gain full marks without getting everything right.

As part of your revision, when you have finished reviewing each topic you are advised to answerat least one appropriate past paper essay question. This will allow you to assess howappropriate your knowledge is. If you are well prepared you will get most, if not all, of themarks available!

Knowledge check answers 1 (a) Active transport, muscle contraction, synthesis of organic molecules, e.g. proteins,

mitosis/meiosis. (b) ATP is soluble and easily transported around the cell; ATP is an immediate source of energy

as only one enzyme is needed to hydrolyse it; ATP releases energy in useable amounts that arematched closely to the energy required in a coupled reaction.

2 The metabolic pathway does not involve oxygen. 3 The glucose molecules cannot enter the mitochondria and as the mitochondria have been isolated

from the cell there are no enzymes present to convert the glucose into pyruvate. 4 Oxygen is the terminal electron acceptor in the electron transport chain. It accepts electrons and

protons and is reduced to water; this allows the coenzymes NAD and FAD to be regenerated. 5 (a) Any two molecules from pyruvate, NADH2, ADP or inorganic phosphate (Pi) (b) Any two molecules from water, NAD or ATP 6 (a) The rate of ATP synthesis would increase. The enzymes involved in glycolysis, the link

reaction and Krebs cycle would have more kinetic energy and would therefore increase therate of the reactions involved. Therefore, more NADH2 and FADH2 would be produced,leading to more ATP being produced via oxidative phosphorylation. More ATP would also beproduced via substrate-level phosphorylation during glycolysis and the Krebs cycle.

(b) The rate of ATP synthesis would fall. Less ATP would be produced via substrate-levelphosphorylation during the Krebs cycle. Less ATP would be produced via oxidativephosphorylation as less NADH2 and FADH2 would be produced during the link reaction andKrebs cycle.

7 (a) oxygen (b) pyruvate (c) ethanal 8 By containing several pigments the chloroplast is able to absorb a greater range of wavelengths of

light, therefore absorbing more light energy and increasing the rate of photosynthesis. 9 Plants appear green because the photosynthetic pigments reflect the green wavelengths of light.10 On the thylakoid membranes11 (a) the reaction centre (b) the antenna complex12 ATP and NADPH213 Its cells are spherical and it has a simple cell wall composed of a thick layer of peptidoglycan.14 Carbon: to be able to synthesise organic molecules Nitrogen: to be able to synthesise amino acids, proteins and nucleic acids Phosphorus: to be able to synthesise nucleotides for the production of ATP and nucleic acids,

DNA and RNA.15 Gram negative: it has a complex cell wall composed of a thin layer of peptidoglycan and an outer

layer of lipopolysaccharides. These bacteria stain pink as the cell wall does not retain the crystalviolet stain. Facultative anaerobe: they grow better in the presence of oxygen but can survive inanaerobic conditions.

16 19 colonies counted from a dilution factor of 10−4 therefore 19 × 104 = 190000/1.9 × 105bacteria

17 A total count includes both living and dead cells and can therefore overestimate the population. Aviable count includes only living cells and is based on the assumption that one cell gives rise toone colony. However, if clumping of cells occurs then the colonies will merge, giving anunderestimate of the population.

18 (a) The cooling water jacket prevents overheating due to microbial respiration. (b) The sparging ring provides oxygen, allowing aerobic respiration to occur. As the oxygen

bubbles through the culture it mixes the microbes with the nutrients. (c) The air filters prevent contamination from airborne microbes.19 The lag phase in the rat population is caused by a low number of individuals of reproductive age.

The lag phase in the bacterial population is due to the switching on of genes and the synthesis ofthe enzymes necessary for the bacteria to utilise their culture medium.

20 (a) Carrying capacity is the maximum population size that the environment can supportindefinitely.

(b) Exponential growth is the doubling of the population per unit time.21 Density-dependent factors include competition for food/water/mates, predation, disease and the

accumulation of toxic waste. Density-independent factors include climatic factors such astemperature.

22 It is acting in a density-independent manner as 20% of the beetles are dying at each populationdensity.

23 In broad-leafed woodland the two species are in direct competition and the grey squirrels haveout-competed the red squirrels, which have become locally extinct.

24 The nitrates provide a source of nitrogen to synthesise amino acids/proteins andnucleotides/nucleic acids.

25 (a) Draining waterlogged fields reduces anaerobic conditions and therefore reducesdenitrification by denitrifying bacteria.

(b) Ploughing fields aerates the soil so that nitrifying bacteria can convert ammonium ions intonitrates for plant growth.

(c) Planting leguminous crops such as clover allows symbiotic nitrogen-fixing bacteria to convertatmospheric nitrogen to ammonium ions.

(d) Ploughing crops such as clover into the soil encourages putrefaction by saprophytic bacteria,which increases the concentration of ammonium ions (and ultimately nitrate ions) in the soil.

26 Although they have access to fresh water, they excrete uric acid. They are reptiles and thereforethey lay eggs. The low toxicity of uric acid means that it can accumulate inside the eggs withoutdamaging the embryos.

27 (a) Any three from water, glucose, amino acids, urea, fatty acids, glycerol, small proteins,inorganic ions (e.g. Na+)

(b) Large plasma proteins28 The efferent arteriole has a narrower lumen than the afferent arteriole, which causes a build-up of

blood in the capillaries, which increases the hydrostatic pressure.29 The microvilli provide a large surface area for absorption and they contain many mitochondria to

provide ATP for active transport.

30 The otter is a mammal that lives in and around fresh water. Therefore it has short loops of Henlebecause it does not need to produce urine with a low water potential. The camel is a mammal thatlives in deserts. It has long loops of Henle, which enable the generation of a high soluteconcentration in the tissues of the medulla so that the camel can reabsorb more water.

31 Loop of Henle32 The detectors (receptors) are the osmoreceptors in the hypothalamus; the coordinator is the

posterior lobe of the pituitary gland (and ADH); the effectors are the walls of the distalconvoluted tubule and collecting ducts of the nephron.

33 Muscles and glands34 An axon transmits impulses away from the cell body; a dendrite transmits impulses towards a cell

body.35 The Na+/K+ pump actively transports three Na+ ions out of the axon in exchange for two K+. The

membrane of the axon is more permeable to K+ ions so they diffuse out more rapidly than the Na+ions. This creates an uneven distribution of charge and the membrane is polarised with the insideof the axon being negatively charged.

36 When the neurone is stimulated the sodium channels open, increasing the permeability of themembrane to Na+. The Na+ ions diffuse rapidly into the axon, depolarising the membrane; theinside of the axon, becomes positively charged.

37 (a) The myelin sheath provides electrical insulation to the neurone by preventing ion exchangeacross the membrane. This increases the length of local circuits and the action potential movesby saltatory conduction. This speeds up the transmission of a nerve impulse.

(b) The increase in diameter reduces the longitudinal resistance of the cytoplasm in the axon. Thisincreases the length of the local electrical circuits in the axon, increasing the speed oftransmission.

38 The mitochondria provide ATP for the resynthesis of neurotransmitter and for the Na+/K+ pump.39 Synapses transmit impulses in only one direction because the neurotransmitter is found on the

presynaptic side and the neurotransmitter receptors are found on the postsynaptic membrane. Therefractory period prevents the nerve impulse from travelling in two directions along an axon.

40 The insecticide is an excitatory drug. It inhibits acetylcholinesterase and therefore acetylcholinecannot be broken down. The acetylcholine remains in the synapse and causes continuousstimulation of the postsynaptic membrane.

41 White matter contains myelinated axons; the grey matter is composed of cell bodies.42

Stimulus Hot surfaceReceptor Thermoreceptors and pain receptors in the skinCoordinator CNS (spinal cord)Effector Arm (biceps) musclesResponse Contraction of muscle to remove the hand/arm

43

Nerve net Vertebrate neurones

One type of neurone Three types of neuroneNeurones are shorter Neurones are longerNeurones branched Neurones not branchedNeurones can transmit impulses in bothdirections

Neurones can transmit impulses in one direction only

Impulses pass in all directions from pointof stimulation

Impulses follow a unidirectional pathway from thepoint of stimulation

Neurones are non-myelinated Neurones may be myelinatedMany synapses involved Few synapses involvedSlower transmission of impulses Faster transmission of impulses

44 (a) The leaf/stem/bud (b) It would cause the rapid conversion to Pr (c) It would cause the slow conversion to Pr (d) Expose the plant to sunlight/white light/red light45 Phloem

GlossaryDecarboxylation reactions involve the removal of a carboxyl group from a molecule resulting in theproduction of CO2. They are catalysed by decarboxylase enzymes.

Dehydrogenation reactions involve the removal of hydrogen from a molecule. They are catalysed bydehydrogenase enzymes.

Exponential growth means that the number of individuals in the population doubles per unit time.

Oxidative phosphorylation involves the synthesis of ATP using energy released from redoxreactions.

Photophosphorylation involves the synthesis of ATP using light energy.

A reflex is a rapid, automatic (involuntary) response to a particular stimulus.

A stimulus is a change in an organism’s internal or external environment.

Substrate-level phosphorylation involves the synthesis of ATP using energy released from thebreakdown of a high-energy substrate molecule.

IndexNote: Bold page numbers indicate defined terms

Aabsorption spectrum 18acetylcholine (ACh) 59acetyl coenzyme A (acetyl coA) 9, 10–11, 13, 14action potentials 54, 55–57

generation of 55propagation of 58–59and withdrawal reflex 63

action spectrum 18active neurones 55–56active transport 47, 48, 54, 55adenine 6adenosine triphosphate (ATP) 6–7

questions & answers 71–72and respiration 8–15synthesis 11–12, 20–21

ADH (antidiuretic hormone) 49, 50aerobic respiration 8–9

electron transport chain 11–12energy yield 15glycolysis 9–10Krebs cycle 10–11link reaction 10summary 12–14

all-or-nothing law 57ammonia 42, 43ammonification 40anaerobic conditions, denitrification 40anaerobic respiration 14–15

energy yield 15animals

anaerobic respiration 14, 15excretory products 42–43

nervous system 52–64antibiotics 24, 25, 30

penicillin production 31–32antidiuretic hormone (ADH) 49, 50aseptic techniques 26–27, 30assessment objectives 70ATP (adenosine triphosphate) 6–7ATPase 6ATP synthase/synthetase 7, 12, 21autoradiograms 23autotrophic organisms 6, 16axons 52, 54

action potentials 55–56membrane structure 54myelinated 52, 53, 59, 62and nerve impulse transmission 57–58non-myelinated 57–58permeability of 55

Azotobacter 41

Bbacteria

cell walls 26, 32classification 24–26culturing 26–27denitrifying 40Gram-negative 25, 26Gram-positive 25, 26nitrifying 40and nitrogen cycle 40–41nitrogen-fixing 41penicillin’s effect on 32population growth 27–30saprophytic 40

batch fermentation 30bioaccumulation 37biological control of pests 37–38

biomagnification 37birth rates/births 33, 34–35Bowman’s capsule 45–46

CCalvin cycle 22–23Calvin’s lollipop 23carbon cycle 39carbon fixation 22carotenoids 17, 18carrying capacity 28, 33, 34–35cell bodies, neurones 52, 59, 62, 64cell walls, bacteria 26, 32chemical control of pests 37chemiosmotic channel protein 12, 21chemiosmotic theory

oxidative phosphorylation 12photophosphorylation 20–21

chlorophylls 17, 18, 19, 20chloroplasts 16, 17, 19cholinergic synapse 60chromatography 17, 23classification of bacteria 24–26cocaine 62coenzymes 8, 9, 10, 11, 13collecting duct 44–45, 49, 50–51competition 34–36, 81–82continuous fermentation 30coordinators 49, 51, 52countercurrent-multiplier system 48culturing bacteria 26–27cyclic phosphorylation 21

Dday-neutral plants 66deamination 42death/decline phase 28

death rates/deaths 33, 34–35decarboxylation 10dehydrogenation reactions 10dendrites 52, 60denitrification 40denitrifying bacteria 40density of populations 34–35diameter of axon and impulse speed 57dilution plating 29–30Diplococci 25distal convoluted tubule 44–45, 49, 50dopamine 62drugs, effect on synaptic transmission 61–62

Eeffectors 49, 51, 52, 64electrochemical nature of nerve impulses 53–54electron transport chain 11–12, 14endergonic reaction 7energy carrier, ATP 6–7energy yield, respiration 15excitatory drugs 62excretion 41–42

homeostasis 49–51mammalian kidney 43–49nitrogenous waste 42–43

exergonic reaction 6, 10exponential growth 28

Ffacultative anaerobes 29far-red light 66–67fermentation 30–32florigen 69flowering of plants

effect of photoperiod 66–67manipulation of 68–69

fossil fuels and carbon cycle 39freshwater fish, ammonia excretion 42fungi

anaerobic respiration 15use in penicillin production 31–32

Ggenetic resistance 37glomerulus 45–46glycolysis 9–10Gram stain procedure 25–26

Hhexose bisphosphate 9, 13homeostasis 49–51Hydra 64hydrostatic pressure 46hyperpolarisation 56, 61hypothalamus 50

Iindustrial fermentation, penicillin 31–32inhibitory drugs 62inhibitory synapses 61insecticides 62integrated pest management 38intercellular spaces 47interspecific competition 35, 36intraspecific competition 35, 36involuntary (of reflexes) 62

Jjellyfish, nerve net 64–65

Kkidneys

gross anatomy 43–44nephron 44–49questions & answers 83–85

Krebs cycle 10–11, 13

Llag phase, population growth 27–28light-dependent stage, photosynthesis 20–21light-independent stage, photosynthesis 22–23limiting factors, growth 28link reaction 10, 13, 14log phase, population growth 28long-day plants 66, 67–69loop of Henle 44, 45, 47–49

Mmammals

kidneys 43–49nervous system 52–64

marijuana 62membrane structure 53microbiology 24

bacterial classification 24–26culturing bacteria 26–27penicillin production 30–32population growth 27–30questions & answers 78–80

microvilli 46mitochondrial matrix 10mitochondrion, structure of 8molecular filter, ultrafiltration 45motor neurones 52–53myelinated axons 62myelin sheath 53, 58

N

negative feedback 49nephron 44–45

Bowman’s capsule 45–46collecting duct 48–49glomerulus 45–46loop of Henle 47–49proximal convoluted tubule 46–47

nerve impulses 53–57transmission of 57–59

nerve nets, jellyfish 64–65nervous system 51

jellyfish 64–65mammalian 52–64questions & answers 86–88

neural pathway 52neurones 52–53, 59–64neurotransmitters 59, 61–62, 63nitrification 40nitrifying bacteria 40nitrogen cycle 40–41

questions & answers 89–90nitrogen fixation 41nitrogen-fixing bacteria 41nitrogenous waste, excretion of 41–43nodes of Ranvier 52, 53, 58–59non-competitive inhibitor 12non-cyclic phosphorylation 20–21non-myelinated neurones 57–58non-selective pesticides 37noradrenaline 59nucleus 52nutrient cycles 39–41nutrients and growth 28

Oobligate aerobes 29obligate anaerobes 29

optimum conditions, bacterial growth 28–29organophosphorus insecticides 62ornithine cycle 43oscilloscope readings 54osmoreceptors 50, 51osmoregulation 50–51osmosis 46–47, 48, 49oxidation 7–8, 10oxidative phosphorylation 11, 12–13oxygen and bacterial growth 29

PParamecium 35–36penicillin production 30–32Penicillium notatum 31peripheral nervous system 52permeability

ADH affecting 49, 50of axon membranes 55, 56loop of Henle, nephron 47–48

persistent pesticides 37pest control 33

biological 37–38chemical 37integrated 38

pesticides 37pests 33, 37–38phases of population growth, bacteria 27–28pH and bacterial growth 29phosphorylation 9

oxidative 11–13substrate-level 10

photolysis of water 20, 21photoperiod 66

detection of 66–69photoperiodism 66–69photophosphorylation 16, 20

chemiosmotic theory 20–21photoreceptors 64photosynthesis 16–17

light-dependent stage 20–21light-independent stage 22–23photosynthetic pigments 17–19questions & answers 75–77

photosynthetic pigments 17–19photosystems 18–21phytochrome 66–67, 69pigments, photosynthetic 17–19pituitary gland 50, 51plants

anaerobic respiration 14–15photoperiodism 66–69photosynthesis 16–24

podocytes 45–46population growth

measuring 29–30optimum conditions 28–29phases of 27–28

populations 33growth of 27–30, 33–34limits to growth 34–36nutrient cycles 39–41pests and pest control 37–38questions & answers 81–82

posterior pituitary gland 50potential difference 53–54, 55, 56primary metabolites 30proximal convoluted tubule 44, 45, 46–47psychoactive drugs 62putrefaction 40pyruvate (pyruvic acid) 9, 10, 13, 14–15

Rreabsorption of water 49

receptors 49, 51, 52, 62, 63red light 66, 67redox reactions 7–8reduction 7–8reflex arc 62, 63, 64reflexes 62–64refractory period 57relay neurones 52, 63, 64, 86, 88repolarisation, axon membranes 56respiration 7–8

aerobic 8–14anaerobic 14–15questions & answers 73–74

respiratory substrates, alternative 15responses in plants 65–69resting neurones 55resting potential 54, 55, 56, 57Rhizobium 41ribose 6Rubisco 22

Ssaltatory conduction 58–59saprophytic bacteria 27–28, 40Schwann cells 53, 59secondary metabolites 30, 31selective reabsorption 46, 47sensory neurones 52, 62, 63, 64serial dilution 29–30short-day plants 66spatial summation 61speed of nerve impulses 57–59spinal cord 62–63, 64, 86–87spinal reflexes 62–64Staphylococcus aureus 25stationary phase, population growth 28stimuli 51

stimulus–response control systems 51Streptococci 25substrate-level phosphorylation 10, 12summation, synapses 61synapses 59

effect of drugs 61–62functions of 61transmission via 59–61

synaptic knob 59synaptic vesicles 59

Ttemperature

and impulse speed 58and population growth 29, 35

temporal summation 61threshold, action potentials 57thylakoid membranes 18, 19, 20–21total count, population growth 29touch-sensitive nerve cells 64triose phosphate 9–10, 13, 14–15, 22

Uultrafiltration 45–46universal energy currency, ATP 6urea, excretion of 42, 43uric acid 43urinary tract anatomy 43–44

Vviable count 29

Wwater, photolysis of 20, 21water potential of blood, control of 50–51water potential gradients, loop of Henle 48–49

wavelengths of light 18, 66

ZZ-scheme 20