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Common themes IB Biology

RespirationTypical question- describe the process by which energy is released from food during respiration

Points to include:

Glycolysis is the gradual breakdown of a glucose (6C) molecule to produce pyruvate

2 molecules of pyruvate per molecule of glucose

This process takes place in the cytoplasm

It requires an investment of 2 ATP molecules to phosphorylate the glucose.

The reaction results in the formation of several intermediary molecules

4 molecules of ATP are released (pay off)

NAD is reduced to form NADH+H

Pyruvate X 2 (3C) are produced

The net products of this reaction are 2 molecules of pyruvate and 2 molecules of ATP along

with 2 molecules of NADH + H+.

If oxygen is present, then the pyruvate will enter the mitochondrion

The link reaction

Takes place in the matrix

Pyruvate (3C) is converted to acetyl CoA (2)

NAD is reduced to NADH+H

Carbon dioxide is given off (decarboxylation)

The oxidised acetyl group bonds with CoA to form acetyl CoA

Output per pyruvate- 1 CO2, 1Acetyl CA, 1 NADH+H

Krebs cycle

Takes place in the matrix

Acetyl CoA joins with oxaloacetate (4C) to form citrate (6C)

Citrate 6C releases CO2 by decarboxylation


-ketoglutaric acid (5C) is produced

5C carbohydrate releases CO2 through decarboxylation


ATP is formed from ADP + Pi

FAD is reduced to FADH2

Another NAD is reduced to NADH+H

Output per cycle= 2 CO2, 3 NADH+H, 1FADH2, 1 ATP

Electron transport chain Takes place in the cristae of the inner mitochondrial membrane

Common themes IB Biology Electron carriers NADH+H and FADH2 needed The reduced electron carrier donates electrons to the proteins (NAD Dehydrogenase) Electrons move along the electron transport chain Hydrogen ions are pumped in to the inter-membrane space by proton pumps The electrons move through the chain and release energy as they pass between molecules

(Cytochromes) This causes more H+ ions to be pumped in to the inter-membrane space of the mitochondria. A concentration of H+ builds up in the intermembrane space of the mitochondria Oxygen is the terminal electron acceptor ½ O2 +H2 ----------- H2O

Chemiosmosis H+ ions diffuse down a concentration gradient from the intermembrane space of the mitochondrion

to the matrix H+ ions move through ATP synthase located In the inner membrane Energy changes cause the phosphorylation of ADP and Pi to form ATP (oxidative phosphorylation) This is reversible

Anaerobic respiration in muscles In the absence of oxygen pyruvate is converted to lactic acid Electron transport chain stops as no terminal electron acceptor Pyruvate not absorbed by the mitochondrion Incomplete oxidation of glucose forms lactic acid NADH oxidised to NAD Lactic acid converted to lactate by lactate dehydrogenase Lactate moves to heart where it is used for respiration (lactate shuttle)

Anaerobic respiration in yeast (fermentation) Pyruvate converted to ethanol CO2 is removed from pyruvate (decarboxylation) NADH is oxidised to NAD This is irreversible

Aerobic respiration Glucose + oxygen-------------- Water + carbon dioxide + energy

Anaerobic respiration in muscle cells Glucose------------------- Lactic acid + some energy

Anaerobic respiration in yeast Glucose ------------------- Ethanol + carbon dioxide +some energy


Photosynthesis

Typical question – describe the steps involved in photosynthesis

Points to include:

Groups of pigments are found in the thylakoid membrane we call these photosystems. The thylakoid membrane is folded to bring the photosystems closer together.

Takes place in the chloroplast Light dependent and light dependent reactions Photosynthesis is the production of carbon compounds using light energy. It is very important as it fixes the carbon from carbon dioxide forming organic molecules

such as glucose, we call this carboxylation. 6H2O +6CO2 C6H12O6 + 6O2, The glucose can then be converted in to starch, fructose, cellulose and amino acids.

Light dependent reaction Takes place in the thylakoid membrane of the chloroplast Made up of photosystem II and photosystem I

Photosystem II (P680- light 680nm)- takes place in the thylakoid membrane Photons of light are absorbed by pigments and the energy is passed to electrons in the

reaction centre (P680) of the photosystem, Electrons become excited (photo-activated) Electrons jump from one orbital to another. The excited electrons are then passed along a transport chain of carrier molecules (electron

transport chain) This flow of electrons causes protons to be pumped in to the thylakoid. At the end of the chain electrons are removed from water leaving behind oxygen and H+

ions (photolysis) which takes place in the thylakoid These electrons replace those electrons lost from the pigment P680 at the reaction centre in

PSII. The electrons that passed along the electron chain are now transported to PSI where more

light is absorbed

Photosystem I - P700 (700nm)- takes place in the thylakoid membrane Electrons are received from PSII via an electron carrier (plastoquinone Pq and plastocyanin

Pc), Additional photons of light activate the electrons further. Two electrons then pass on through the photosystem until- along with hydrogen from the stroma,

they are used to reduce a molecule of NADP to form NADPH, The electron carrier ferredoxin Fd and the enzyme NADP reductase assist with this This process may be cyclic (without water) or non-cyclic (with water) Cyclic phosphorylation refers to the fact that the electrons that are lost from Photosystem 1 return

PS1 and there is no need for water to be used as an electron acceptor, In cyclic phosphorylation no NADPH is generated. In non-cyclic water acts as an electron acceptor, oxygen gas is produced and NADP is reduced to form

NADPH The electrons lost from PS1I are replaced by those removed by the photolysis of water. Cyclic is more common in bacteria and non-cyclic in green plants. Chemiosmosis- take place in the thylakoid membrane

Common themes IB Biology A H+ concentration gradient forms between thylakoid and the stroma H+ ions move down the concentration gradient from the thylakoid through ATP synthase-

embedded in the thylakoid membrane and into the stroma. This movement causes the formation of ATP from ADP and inorganic phosphate

(photophosphorylation)

Light independent reaction- takes place in the stroma

The Calvin cycle is part of the light independent reaction

Takes place in the stroma of the chloroplast.

The purpose of the reaction is to produce glycerate-3-phosphate (GP) a triose sugar that is reduced to form triose phosphate (TP) and then glucose.

The enzyme rubisco is used to fix (carboxylation) a molecule of carbon dioxide onto an existing molecule of ribulose bisphosphate (RuBP) (5C)

This forms an unstable 6C compound

The 6C compound is quickly broken down to form two molecules of phosphoglycerate (PGA),

These molecules are then converted to GP and then TP - this requires an energy investment of 2ATP and the use of 2 NADPH.

Some of these triose phosphate molecules are used to form glucose. Others are converted back in to RuBP an additional ATP molecule needs to be invested at this point.

It is important to note that a single cycle will not produce the required products of glucose and RuBP, several cycles are taking place at the same time and the products of 6 cycles are combined to produce the glucose (1/6 of product) and the RuBP (5/6 of product).

Note- an additional energy investment is needed to reform the RuBP. Although this reaction does not require light, it is dependent upon the presence of ATP and NADPH, both of which are produced in the light dependent reaction.


Membrane structure and transport

Draw the structure of a plasma membrane, explain the arrangement of proteins in the membrane and discuss how substances move in and out of the cell

An integral protein is one that provides structure for the cell membrane. They are embedded into the hydrophobic part of the phospholipid bi-layer.

Transmembrane proteins pass all the way through the cell membrane. These proteins may be channel proteins or carrier proteins to allow facilitated diffusion (passive) or act as pumps for active transport. They may also act as hormone binding sites eg for insulin. Some transmembrane proteins allow the passage of water and are called aquaporins. Some proteins also act as a pump for sodium and potassium.

Peripheral proteins are on the outer surface of the membrane they may act as immobilized enzymes such as sucrase.

Membrane proteins (glycoproteins) are also important for cell recognition, adhesion and communication.

Other membrane bound proteins act as receptors for neurotransmitters, and as electron carriers.

Proteins may contain both polar and non-polar amino acids. Integral proteins are found embedded in the membrane; it is the properties of the non-polar amino acids that cause this embedding in the hydrophobic lipid part of the membrane bi-layer. Peripheral proteins are found on the outer part of the membrane; polar amino acids cause the protein to protrude from the surface. Transmembrane proteins have both polar and non-polar amino acids, as a result these proteins can pass through the membrane and can act as transport channels.

Diffusion is a passive process in which particles move down a concentration gradient. This means that the particles move from a high concentration to low concentration, this movement is random and due to the kinetic energy of the particles, it will continue until the point of even distribution eg oxygen moving into a cell

Facilitated diffusion is a form of passive diffusion that requires a protein channel. Potassium ions are an example of a substance that will pass out of a nerve cell in this way.

Osmosis is the diffusion of water through a partially permeable membrane. Water moves from a dilute (high water potential) to a concentrated (low water potential) high solute solution. This is a passive process and water moves down a water potential gradient. Water may also move by facilitated diffusion through protein channels called aquaporins.

Active transport is the movement of a solute across a cell membrane against its concentration gradient. This process requires the use of energy in the form of ATP which is hydrolyzed to form ADP and inorganic phosphate and a specific protein pump is also needed eg glucose transport in the villi.

Common themes IB BiologySubstances may move in to the cell by endocytosis and out of the cell by exocytosis

EnzymesEnzymes have many functions in the organism including digestion, respiration and photosynthesis. All enzymes have an active site which is a specific shape to fit a specific substrate it may also have an electrical charge which attracts the substrate. This active site is found on the surface of the enzyme and it is here where the substrate binds and the reaction is catalyzed.

Allosteric enzymes have an additional binding point/s which is different to the active site and which other substances may bind to. This may change the shape (conformation) of the active site making the enzyme active or inactive. Chemicals needed to activate the enzyme are called co-enzymes.

All chemicals reactions need energy to start them off. An enzyme will reduce the amount of activation energy needed- without the enzymes reactions would take far too long to occur at body temperature. The enzyme brings the substrates closer together and in the correct orientation. The substrate binds with the active site forming an enzyme substrate complex, this puts the substrate under stress causing a slight change in its structure and weakening molecular bonds. We call this a conformational change.

One theory of enzyme activity is that of the lock and key theory, the substrate has a complimentary shape to the rigid shape of the active site and the two fit exactly to form the enzyme-substrate complex. The second theory of enzyme activity used by the IB is the theory that the enzyme is not rigid and will change the shape of its active site to fit the substrate, we call this the induced fit model. The active site and the substrate do not match exactly, the enzyme will mold its active site to compliment the shape of the substrate as the enzyme substrate complex is formed. The enzyme and substrate may have opposite electrical charges to assist with this. This change in configuration of the enzyme puts strain on the bonds of the substrate and weakens them, lowering the activation energy needed to break the bonds. The enzyme therefore catalyses the reaction. Once the reaction has occurred the products will leave the active site, leaving the enzyme unchanged and available for the next substrate. Some enzymes are able to act upon more than one substrate and this is explained by the induced fit method.

Enzymes work best at an optimum temperature, as temperature increase towards the optimum, the rate of reaction will also increase due to an increase in motion of the molecules and an increase in the number of collisions between the enzyme and substrate, more complexes will be formed and more of the substrate will obtain their activation energy. Above this temperature the active site will change shape and the tertiary structure of the protein is altered, due to the breaking of hydrogen bonds (due to vibrations inside the enzyme disrupting the intermolecular forces). The substrate will no longer be able to bind with the active site and the enzyme substrate complex will not form we say that the enzyme has denatured. Each enzyme also has an optimum pH (note for pepsin this is pH 2 for amylase pH8) outside of which there will be lower activity and if the pH is too high or low the enzyme may go through a change in shape or may denature, this may be as a result of ionization or changes in electric charge due to the acidity. The concentration of the substrate will also affect the activity of the enzyme, the increased concentration will mean a greater number of collisions between the enzyme and substrate and therefore more reactions take place. The rate of reaction is directly proportional to substrate concentration, however this will eventually plateau as all the active sites on the enzymes are utilized, therefore further increases in concentration will see no

Common themes IB Biologyfurther increase. Salivary amylase produced in the salivary glands found in the mouth breaks starch down in to maltose, this enzyme prefers a pH or around 7. Pepsin in the stomach breaks protein in to amino acids and prefers a pH of 2.

When an enzyme denatures the bonds between the different amino acids that maintain the 3D shape will break. This means that the active site loses its specific shape. We say that the active site has gone through a conformational change. This change is irreversible and will prevent the enzyme from working.

In a metabolic pathway the end product (metabolites) in the chain may switch off an allosteric enzyme found earlier in the chain by binding to its allosteric site- some enzymes have multiple allosteric sites. The metabolites act as allosteric inhibitors. This causes a temporary change in the shape (conformational change) of the active site preventing the enzyme from working, it is a form of non-competitive inhibition as the metabolite does not compete for the active site. It is a form of negative feedback as the build-up of the product switches off the process. It is reversible as when the volume of the product falls the metabolite will detach from the allosteric site and the enzyme will once again activate. We call this end product inhibition (allosteric inhibition), allostery is the process when the binding of substance to one part of an enzyme causes a conformational change to another part of the enzyme. An example of end-product inhibition is the regulation of ATP formation by phosphofructokinase (PFK) an enzyme in glycolysis that is inhibited by ATP binding to its allosteric site.

Sometimes a chemical may have a similar shape to the substrate and will temporary bind to the active site preventing the substrate from binding, this results in competitive inhibition An example of competitive inhibition could be malonic acid which competes with succinate for active sites of succinic dehydrogenase, an important enzyme in the Krebs cycle.

Some substances such as a heavy metals (eg mercury and silver) may bind with a part of the enzyme that is not the active site or with the allosteric site, this causes an irreversible conformational change (change in shape) at the active site and stops the enzyme from working we call this non-competitive inhibition An example of non-competitive inhibition could be cyanide (or potassium cyanide – KCN) which inhibits Cytochrome C Oxidase responsible for the transfer of hydrogen atoms during cellular respiration preventing the production of ATP.


Osmoregulation

TranspirationDescribe the structure of a leaf and discuss the process of transpiration, describe the adaptations that can be seen in xerophytes to reduce transpiration.

Transpiration is the loss of water from the leaf through evaporation

It is an inevitable consequence of gas exchange

Water vapour evaporates from the surface of the mesophyll cells and diffuses through the air spaces and out of the stomata

Common themes IB BiologyStomata open and close- this is controlled by the guard cells which open during the day and close at night

Roots are branched and contain root hairs, this provides a large surface area for the plant to absorb water through osmosis

Water will move from the soil to the root down a water potential gradient

To maintain the water potential gradient, the plant will actively take up solutes to reduce the water potential inside the root

Water will move through the root cortex towards the xylem located in the stele. Water may move by the symplastic or the apoplastic method. As it reaches the waterproof Casparian strip it is directed by the symplastic route towards the xylem

The xylem is made from a hollow lignified narrow tube made up of tracheids. Through which a continuous column of water flows

Water moves through the xylem due to capillary action, adhesion and cohesion. This combined with root pressure and the transpiration pull mean that the transpiration stream can occur over large distances

Transpiration can be modelled using capillary tubes to measure the uptake of water by the xylem due to its adhesive and cohesive properties, blotting paper can be used to imitate the flow of water through the xylem and porous pot can also be used to show the evaporation of water from the leaf.

Transpiration can occur over long distances and is a passive process using the energy from the sun. Factors that can influence the rate of transpiration include

Temperature- the higher the temperature the greater evaporation the faster transpiration, the higher temperature also speeds up the diffusion of the water molecules as they leave through the stomata of the leaf they move further away from the leaf due to their kinetic energy,

light- as light intensity increases a greater number of stomata are opened resulting in faster transpiration, wind speed- more wind means faster evaporation and the lower the humidity, therefore faster transpiration.

Humidity- the higher the humidity the more saturated the air is and the slower transpiration is

Xerophytes

A xerophyte is a plant that is adapted to survive in very dry conditions such as a desert- (eg prickly pear cactus) plants adapted to dry conditions, these have a thick waxy cuticle, few or small stomata, stomata that only open at night, modified leaves (spines) with a small surface area, water storage tissue in the stem/leaves/roots, A deep and extensive root system for greater water uptake. Marram grass is another xerophyte, this plant has rolled leaves to increase humidity, stomata arranged in pits to trap moist air. Small hairs on the leaf surface also reduce air movement and water loss.


Protein structureDescribe the stages involved protein synthesis and relate the protein structure to protein function.

Specific genes in the DNA code for specific proteins

Transcription is the production of a complimentary strand of mRNA for a section of DNA (gene) it is a similar process for both prokaryotes and eukaryotes.

mRNA as a template to make a protein.

Primary proteins are a long chain of amino acids it is the order of these amino acids that determine the function of the protein, they are synthesised using ribosomes bound to the rough endoplasmic reticulum and free ribosomes in the cytoplasm. Note that free ribosomes synthesise protein for use inside the cell, ribosomes on the RER synthesise proteins for export.

Secondary proteins have hydrogen bonds between C=O and N-H groups of different amino acids (every 4 amino acids may form a H bond) which then form an alpha helix and beta pleated sheet these proteins tend to be structural such as collagen.

Tertiary proteins also have disulfide bonds (bridges), hydrogen bonds and ionic bonds, they have a 3D shape (globular) eg enzymes, the 3D shape is due in part to the hydrophobic and hydrophilic interactions, hydrophobic amino acids will face inwards, causing folds in the molecule, this helps to form the shape of the active site in an enzyme polar and non-polar amino acids also add to the shape of the active site.

Quaternary proteins are formed from more than one peptide or it may be conjugated by the addition of a prosthetic group (haemoglobin and iron). Haemoglobin is a quaternary protein formed from multiple peptide chains (2α chains and 2β). It is a globular protein; it also has a prosthetic heme group (Fe2+) to which oxygen will bind. Each molecule of haemoglobin therefore has four protein chains and each heme group will bind to one oxygen molecule (O2) therefore each molecule of haemoglobin will bind with 4 molecules of oxygen.


DNA replication

Explain how DNA is replicated and discuss what is meant by semi-conservative replication

DNA is made up from nucleotides

DNA nucleotides consist of a phosphate group, deoxyribose sugar and genetic base

DNA has 4 bases adenine, guanine, cytosine and thymine

The bases form complimentary pairs on antiparallel DNA strands A-T and C-G

DNA is replicated in the S phase of the cell cycle.

Each strand of existing DNA is replicated semi-conservatively through complimentary base pairing. This means that in each DNA molecule there will be one of the original strands and one new complimentary strand.

We can use radioactive isotopes of nitrogen to show the semi-conservative replication

DNA replication is an enzyme led process and requires the presence of Helicase, RNA Primase and Polymerase.

Leading strand. DNA helicase unwinds the DNA double helix by breaking the hydrogen bonds between complimentary bases. Each DNA strand will act as a blueprint for a complimentary new strand. RNA Primase adds an RNA nucleic acid (primer) to the point where replication will begin- this primer will later be removed. DNA Polymerase III binds to the DNA and moves along it from the 3’ to the 5’ direction BUT builds the new complimentary strand of DNA in the 5’ and 3’ direction. Free nucleotides form complimentary base pairs by forming hydrogen bonds (A-T, C-G). The nucleotides are covalently linked by the polymerase III and the new DNA strand forms. Once formed the double helix will again form the result is an identical copy of the original DNA strands.

Lagging strand. The DNA is replicated in sections- still reading in the 3’5’ direction. The fragments are called Okazaki fragments these fragments are still formed by Polymerase III, however need to be joined together by DNA ligase once the RNA primers have been removed.

DNA Polymerase III elongates the DNA by adding a new nucleotide to the 3’ end of the chain building the DNA in the 5’3’dorection. This enzyme also proofreads the new DNA strand.

DNA Polymerase I removes the RNA primer and replaces it with DNA nucleotide

DNA Polymerase II is believed to repair damaged DNA

DNA Gyrase acts to relieve the strain on the DNA strand as it unwinds.

Helicase- unwinds the double helix by breaking the hydrogen bonds between complimentary base pairs.

Ligase- an enzyme that joins together the Okazaki fragments from the lagging strands, forms sugar phosphate bonds between the nucleotides.

Single stranded binding proteins are used to stabilize the unwound single strands of DNA and prevents them from annealing (rejoining to form a double strand).


TranscriptionDescribe the process of transcription, compare the structure of DNA with that of RNA

The molecules of DNA are long and only certain sections are expressed in certain cells.

The molecules are too large to pass through the pores in the nuclear membrane.

Therefore, shorter chains of mRNA are used for gene expression, these molecules are produced through the process of transcription.

During transcription RNA polymerase controls the process, it unwinds the DNA exposing the bases and separates the DNA strands by breaking the hydrogen bonds.

RNA polymerase uses one strand as a template (antisense).

It starts complimentary base pairing at a binding site called the promoter and will build the mRNA molecule in the 5’ 3’ whilst reading the antisense strand in the 3’ 5’ direction.

As the mRNA is constructed hydrogen bonds form between the complimentary bases.

RNA polymerase will stop transcribing the DNA when it reaches the terminator codon.

After which the mRNA and RNA polymerase will detach completely from the DNA and be released.

The DNA double helix will reform (anneal).

mRNA will be modified before it leaves the nucleus, encoding introns are removed and the remaining exons are spliced together

The 5’end of the mRNA is capped and a tail is added to the 3’end

Energy is required for transcription and this is attained by the hydrolysis of phosphate compounds (eg ATP).

Note that the sense strand is the DNA strand that has the correct code for the gene. The antisense strand which has the complimentary base pairs acts as the template so that the copy of the gene can be transcribed. Remember that in RNA Thymine is substituted by Uracil.


Translation

Describe the process of translation and discuss what is meant by the genetic code

Mature mRNA needs to be translated in order for proteins to be produced.

The mRNA is a copy of a gene and will leave the nucleus by passing through the nuclear pores.

Once in the cytoplasm it will locate a ribosome.

Ribosomes are made from two subunits- large and small which consist of proteins and ribosomal RNA.

The 5’ end (cap) of the mRNA will bind to a binding site on the small subunit of the ribosome.

The small subunit of the ribosome will move along the mRNA towards the 3’ end until it locates the start triplet codon AUG.

At this point the large subunit of the ribosome will bind along with a specific molecule of tRNA forming a ribosome complex- we call this initiation.

The ribosome subunits will then move along the mRNA molecule to the next triplet codon/base triplet (translocation)

Here a specific tRNA molecule with the correct anticodon will pair with the triplet codon on the mRNA forming hydrogen bonds.

The enzyme peptidyl transferase will bind the amino acids that are attached to each molecule of tRNA to form peptide bonds (condensation reaction) note that a peptide bond is a type of covalent bond and that two tRNA molecules will be bound to the ribosome with one of them at the P site holding the peptide chain.

This process will continue forming a primary protein chain (polypeptide)- we call this elongation.

When the ribosomal complex reaches a stop codon (UAG, UGA and UAA), a release factor will bind with the codon and the peptide chain will be released from the ribosomes (termination).

The peptide chain will then go through further modifications until it has the correct structure.

Note that the large subunit of the ribosome has three binding points E P and A, the tRNA first binds to the A site and then is moved to the P site during translocation and finally the E site where it exits, leaving behind the amino acid. Remember that the mRNA is read in the 5’ 3’ direction and the polypeptide is constructed from the amino to the carboxyl terminus.

The genetic code

All living organisms use the same genetic code. The genetic code is therefore universal and determines the expression of the DNA base sequence and therefore the amino acid sequence during the synthesis of peptides.

Each base sequence forms a triplet codon (3 bases), many of these codons code for a specific amino acid. There are 64 possible triplet codons, as a result each amino acid has more than on triplet codon we say that the genetic code is degenerate.


Polymerase Chain reaction (PCR)Discuss the importance of the PCR and state how this assists with genetic profiling

The polymerase chain reaction (PCR) is used to replicate DNA from a small sample eg blood, semen, hair.

This is particularly useful when examining DNA from a crime scene as this is often in very short supply.

A sample of DNA is heated in order for the double helix to be unwound (denatured)

The separated DNA strands then act as a template for the production of new strands the enzyme Taq polymerase is used in this process, in the presence of nucleotides,

The DNA is then cooled and a RNA primer is bound to the DNA strands, Taq polymerase will the join nucleotides to form a complimentary strand.

The double helix will then reform (annealing)

The DNA goes through temperature cycles with the help of a thermos-cycler.

Over a four-hour period, a single strand of DNA can be replicated over a million times (exponential replication)

Genetic profiling is a procedure used to analyse the DNA in a sample

It can be used to identify if a person was involved in a crime

It can be used prove paternity- DNA is taken and tested form the child, mother and or father.

It can be used to examine a person’s DNA to see if they are predisposed to genetic diseases.

It can be used in disputes to prove inheritance beneficiaries.

Gel electrophoresis is a technique that is used to examine a genetic profile.

Electrophoresis works on the principle that the phosphate backbone of DNA has a negative charge.

Samples of DNA are placed into wells in a gel and an electric current is passed through the gel.

The voltage pushes the DNA fragments along the gel- the smaller the fragment the further it will move.

Completed gel plates can be compared for similarities.

Restriction enzymes are enzymes that are extracted from bacterial and will cut DNA in particular places

These enzymes are used to cut DNA into fragments at specific places for electrophoresis.

A sample of DNA can be taken from saliva, leucocytes, skin or hair.

The DNA is then put through the PCR to amplify the DNA content.

Repeating satellite units from the centromeres are used in the profiling.

The DNA is exposed to restriction enzymes in order to produce the DNA fragments the fragments then go through electrophoresis.

The fragments separate according to length.

The bands of DNA that are produced can be compared to other samples. This allows forensic scientists to determine if a suspect has committed a crime such as rape, to prove paternity if at least half the bands of DNA match between the two samples, and family relationships and to see if an organism is a clone, also it can be used to track individuals.


Water

Discuss the biological importance of water and relate this to the chemical properties of water

Water is made from one oxygen and two hydrogen atoms, it is a polar molecule (di-pole) which means that one end (Hydrogen) has a positive charge and the other (Oxygen) has a negative charge. It therefore forms a ‘V’ shape. A water molecule will form an attractive bond (hydrogen bond- each water molecule can form 3 H bonds) with neighbouring water molecules, we call this cohesion, it will also form hydrogen bonds with other polar molecules which we call adhesion. These properties are essential for water transport in the xylem (Transpiration stream- a continuous column of water molecules).

Water is an essential component of life and as a habitat.

It is important due to its properties as a solvent

Water is a polar molecule with polar properties

Many polar substances will dissolve in water making it good for transport and as a place for metabolic reactions.

For temperature regulation (sweat) water is important due to a high specific heat capacity (amount of energy needed to raise water by 1oC).

Hydrogen bonds form between the O- and H+ of adjacent molecules, each water molecule can form 3 H bonds.

Cohesion between water molecules provide surface tension for transpiration and also adhesion between the water molecules and the xylem.

The cohesive properties of water allow organisms to move on the surface of water.

This cohesion also gives water a high latent heat of vaporization (high boiling point) - meaning it takes a high amount of energy to break the hydrogen bonds and it is therefore good as a coolant whist at the same time a high boiling point means that the water is unlikely to boil in a natural habitat

Its SHC means that it will maintain the temperature of the environment.

As water reaches its maximum density at 4oC and the fact that ice float means that it can act as an insulator allowing life to survive in cold waters.

The transparency of water allows light to penetrate allowing the survival of hydrophytes and allows animals to see under water.

One vital property of water is that it can be found in its liquid form over a wide global temperature range therefore water provides a thermally stable habitat.


Immunity

Discuss how the body responds to foreign antigens that have entered the blood stream

Innate barriers such as the skin, mechanical and chemical are breached

Mast cells release histamine, this widens the blood vessels allowing the migration of white

blood cells to the site of infection, we call this the inflammatory response

Macrophages engulf the pathogen and present antigens (Antigen Presenting Cells) to T

lymphocytes

Major histocompatibility proteins (MHC) found on the membranes of these cells present the

antigens/fragment to specific naïve T lymphocytes which bind to the antigen on the APC

cells (the CD4 glycoprotein on the surface of the T cell identifies the antigen)

When the correct T cell has been identified (Clonal selection) the APC will release the

cytokine interleukin 1.

The naïve/helper T cells are activated and produce interleukin II which activate specific T

cells and specific B lymphocyte cells causing them to divide (proliferation by clonal

expansion).

The T cells may also bind to inactive B lymphocytes cell and activate them.

The T cells differentiate to form Cytotoxic T cells (CD8 cells)- which locate the pathogens and

release the chemical perforin which destroys the invading cells through cellular lysis

(apoptosis) and Memory T cells which remain in the blood in case the antigen is encountered

again.

Once activated Naïve B cells will clone and either become memory B cells- in preparation for

a secondary encounter with a pathogen, or grow and become antibody producing Plasma

cells.

The plasma cells proliferate rapidly producing large quantities of the specific antibody

(immunoglobulin) which leave the cell by exocytosis and flood the body- the RER and Golgi

apparatus is quite extensive in these cells.

Note- there are many different types of T and B lymphocytes, -lymphocytes are white

blood cells that are initially produced and stored in the lymph nodes before migrating to

the bone marrow and thymus gland where they mature.


Vaccination

Discuss the use of vaccinations and discuss the successes and failures of vaccination programmes

Vaccinations ae a form of artificial immunity

The patient is exposed to weak or inactive antigens which evoke an immune response and the production of antibodies.

Macrophages will ingest the antigen and present it to the T-helper cells which bind to the macrophages and are activated.

Alternatively, the antigen may bind to receptors on the B cells,

T-cells will then bind to the B-cells and activate them.

The B-cells undergo proliferation (mitosis) and produce specific antibodies to fight the pathogen.

The memory B cells will then be able to produce the correct antibodies should the person ever be exposed to the real disease,

This will initiate a secondary immune response and will result in the proliferation of the memory B lymphocytes cells allowing rapid antibody production.

Thus the process gives a person immunity or resistance to a specific disease.

Some viruses and bacteria evolve quickly or exist in multiple forms and as a result it is difficult to produce a vaccination.

Equally some diseases attack our organs and therefore are difficult for the immune system to get to.

Booster injections may be needed to top up the immune system.

The elimination of smallpox from the globe was a phenomenal success for the vaccination

programme.

The last reported case was in Somalia in 1977.

Since then organizations have tried to repeat the success with other vaccines such as polio and

prevent epidemics and pandemics,

This has not always been possible due to global conflict, reluctance amongst people to take the

vaccine, the remoteness of communities and the rate at which the disease mutates.

MMR is a mixture of vaccines that is given to infants to reduce the chances of getting mumps,

measles and rubella.

Exposure to the vaccine reduces the chances of the diseases causing deafness, blindness and heart

problems.

There is some resistance amongst groups to accept vaccine as it is believed that too many vaccines

may actually weaken the immune system if exposed to other diseases.

There may be side effects from vaccinations for example the whooping cough vaccine is suspected of

causing encephalitis.

Common themes IB BiologySome people may have allergic reactions to the vaccine and others are sensitive to the preservatives

(eg mercury) of the vaccine.

In some cases, the vaccine has not been properly prepared and as a result people have actually

contracted the disease from the vaccine.

Vaccines do increase life expectancy and protect vulnerable groups such as the young and elderly.

Economic benefits will result as the population becomes more disease resistant.


Digestion

Explain why food substances need to be digested and discuss the process of digestion, discuss the importance of villi

Large and complex molecules cannot be absorbed directly in to the blood, they therefore need to be

broken down into simpler soluble substances that can be absorbed

Enzymes are needed for this to happen

Amylase hydrolyses starch in to maltose

Maltase hydrolyses maltose to form glucose

Pepsin hydrolyses protein in to amino acids

Lipase hydrolyses triglycerides to form fatty acids and glycerol

The digestive system consists if the following structures: Mouth, esophagus, stomach, small intestine

(duodenum / ilium), large intestine (Colon, caecum), Rectum and anus.

The accessory organs of the teeth, tongue, pancreas, gall bladder and the liver also play important

roles in the digestive process.

Throughout the digestive system smooth muscle can be found. This muscle consists of circular and

longitudinal muscle and is important for the pushing of substances through the esophagus and the

intestines through the process of peristalsis.

In the mouth food is mechanically broken down by the teeth, it is mixed with saliva and mucus, alpha

amylase

Amylase starts the breakdown of starch in to maltose

The food is rolled in to a bolus by the tongue.

Food passes through the esophagus by the process of peristalsis.

Once in the stomach the food is churned and mixed with the gastric juices containing hydrochloric

acid and pepsin.

As food (chyme) leaves the stomach it enters the small intestine (duodenum) where additional

enzymes are added along with bicarbonate ions from the pancreas and bile from the gall bladder.

The digestive process continues and the products are then absorbed in to the blood through the villi-

these are small structures that line the small intestine (ilium).

As the remaining substances enter the large intestine (colon) water is absorbed into the blood.

The waste is then stored in the rectum until it is released from the anus (egestion).

Villi are thin small structures (one layer thick) located in the jejunum and the ilium.

They are small protrusions with a large SA/Vol ratio, they are responsible for the absorption of

digested material.

Each villus is folded to form microvilli- structures which increase the surface area of the small

intestine many times over.

Common themes IB Biology Each villus is formed from a single layer of epithelial cells which reduces the distance for molecules to

diffuse.

Villi contain a lacteal, which is part of the lymph system

Capillaries are found close to the surface of the villi and are connected to a rich blood supply.

Some substances such as lipids are absorbed by the lacteal

Others such as glucose and amino acids are absorbed into the blood (capillaries).

The villi have protein channels to allow facilitated diffusion and carrier proteins for active transport,

They also contain many mitochondria to produce the ATP needed for active transport.

Tight junctions between cells assist with the controlling of absorption,

Pinocytosis is also utilized by the cells of the villi.


Mitosis

Discuss the process of mitosis and explain how mitotic index is calculated

Mitosis is a type of cell division that involves somatic (non sex) cells.

Mitosis is important for growth and tissue repair, it is also important in the early stages of spermatogenesis and oogenesis

During interphase cell organelles are synthesised (G1 phase) DNA is replicated (S phase) DNA is checked for errors

Normal cell activity also occurs during interphase eg transcription and respiration

During DNA replication semi conservative copies are made of the chromosomes, the copies are joined at the centromere- whilst joined they are referred t as chromatids

Mitosis occurs in recognisable stages

The stages are:

Prophase which sees the breakdown of the nuclear membrane, the supercoiling of chromosomes, the migration of the centrioles to the poles and the formation of micro-filaments.

Metaphase which involves the attachment of the microfilaments to the centromeres and the alignment of the chromatids on the equator.

Anaphase sees the contraction of the microfilaments, the splitting of the centromeres and the pulling of the chromosomes to the poles.

Telophase involves the breakdown of the microfilaments, the formation of a nuclear membrane and the uncoiling of the chromosomes.

The cell membrane will then be cleaved due to the contraction of actin and myosin fibres using ATP.

The end product of mitosis is two identical diploid daughter cells (2n).

Mitotic index is a measure of the rate of mitosis. It looks at the ratio between cells involved in mitosis and cells that are not involved in mitosis in a tissue sample.

Meiosis

Describe the process of meiosis and discuss how this leads to greater variation

Meiosis is important for the production of gametes.

Gametes are produced by reduction division

Meiosis produces 4 non-identical haploid gametes (n)

During prophase 1- the nuclear membrane is broken down, centrioles divide and migrate to the poles and spindle fibres develop. The chromosomes supercoil and homologous pairs of chromosomes form bivalents (tetrads) through the process of synapsis, at this point crossing over (recombination) occurs.

Common themes IB BiologyDuring metaphase 1- spindle fibres attach to the centromere and the bivalent align independently on the equator of the cell.

During anaphase 1- The bivalents are separated and the chromatids are pulled to opposite poles of the cell (half go to one pole half to the other).

During telophase 1- the cell membrane is cleaved (cytokinesis) and the nuclear membrane reforms, the spindle fibres break down and the chromosome begin to un coil. (note at this point the cells are considered to be haploid).

Prophase 11- the nuclear membrane breaks down, centrioles migrate to the poles, spindle fibers reforms and the chromosomes supercoil.

Metaphase 11- the spindle fibers attach to the centromeres and the chromatids orientate randomly on the equator (further enhancing variation).

Anaphase 11- the spindle fibers contract and the chromatids separate to form chromosomes which move to opposite poles of the cell.

Telophase 11- the nuclear membrane reforms, the spindle fibers break down and the cell starts to divide.

Crossing over occurs during prophase 1 when tetrads form during the process of synapsis.

DNA is exchanged between homologous non sister chromatids.

During crossing over the points at which the homologous chromatids cross are called chiasma this is evident by an X shaped structure which helps to hold the bivalents together until anaphase I.

The recombinant paternal and maternal chromosomes then align on the equator during metaphase 1- the chromosomes orientate independently of other chromosomes (independent assortment).

As the chromatids are pulled apart in anaphase 1 they break at the same point and the genetic material between them will be exchanged forming recombinants.

This leads to increase variation giving a possibility of 223 gametes.

Note- when talking about variation through meiosis for other species quote 2n to represent the number of possible gametes.

Crossing over and random assortment of chromosomes increase variation during meiosis


Muscle contraction

Discuss the process of muscle contraction and emphasise the role that calcium plays in the process.

Motor neurons rapidly carry impulses from the CNS to the muscles (effectors) and stimulate

them to contract resulting in movement.

The neurons control the timing of the contraction

When the action potential of the neuron reaches the muscle, the neuron will release the

neurotransmitter acetylcholine

This then activates the sarcoplasmic reticulum to release calcium ions with is an important

step in muscle contraction.

Calcium ions bind to troponin

Troponin changes position and moves tropomyosin from the actin binding sites

The heads on the myosin filaments will bind to sites on the actin filaments,

This creates a cross-bridge which allows the filaments to create a force.

There are many heads on each myosin and many binding sites on each actin filament,

They are spaced at regular intervals so that many cross-bridges can form at the same time.

In order to form the cross-bridge, ATP must be used (hydrolysis):

The myosin heads form cross-bridges with the binding sites of the actin filaments,

The myosin heads swivel towards the centre of the sarcomere, moving the actin along

slightly

ATP binds to the myosin heads and the cross-bridges are broken

ATP is hydrolyzed to form ADP and Pi, these bind to the myosin causing the heads to

change position (swivel) by moving away from the centre of the sarcomere (cocked)- in this

position they store potential energy

The heads then attach to binding sites on the actin that are further away forming a new

cross bridge

The ADP and Pi detach and the myosin heads push the actin towards the centre of the

Sarcomere (power stroke)

The process repeats as long as signals are received from a motor neuron

The sliding of these filaments causes the contraction of the muscle. In a contracted muscle

the Z lines can be seen closer together.

When muscles contract, the actin and myosin filaments will slide over each other and the

sarcomeres become smaller


Nerve impulse

Draw an label a typical motor neuron, discuss how a neuron generates an action potential and what happens after the axon has depolarised, explain what is meant by the all or none response

A nerve cell generates an electrochemical impulse

This is referred to as an action potential,

An action potential is caused by the change in electrical charge across the membrane of a

nerve cell.

Electrically charged ions are able to pass across the semipermeable membrane of the nerve

cell.

At rest the inside of the cell has a negative charge (-70mV) compared to the outside, we call

this the resting potential

The sodium potassium pump maintains this- note the resting potential is the point when the

neuron does not conduct an impulse and there is a Net negative charge (-70mV) across the

membrane.

At rest we say the nerve is polarized.

When the nerve is activated by a stimulus such as pressure, an action potential (impulse) is

generated.

The stimulus causes the membrane of the nerve to open voltage gated sodium channels

increasing the cells permeability to Na+

Common themes IB Biology The electrical charge across the membrane moves towards -55mv due to the influx of Na+

ions and we say that the nerve is depolarized.

At -55mV more sodium channels open and there is an increased influx of Na+ ions.

This causes the electrical charge to move towards +30mV and an electrochemical signal

passes along the axon.

The sodium channels then close and the voltage gated potassium channels reopen

potassium ions flood out of the cell (efflux) moving down a concentration gradient.

This causes the electrical charge to fall (repolarization),

Initially the charge falls too much -90mV and we say that the cell is hyperpolarized.

To return to the resting potential the cell actively pumps K+ ions into the cell and pumps Na+

out (2K+ in 3Na+ out using ATP) using the sodium-potassium pump.

Eventually the membrane potential will get back to -70mV

The time taken to get back to the resting potential is the refractory period.

As the depolarization wave moves along the nerve the process is repeated many times in

adjacent parts of the neuron.

In cells that are myelinated, the action potential jumps from node of Ranvier to node of

Ranvier, we call this salutatory conduction.

An action potential will only be generated at -55mV, this means that any changes in charge

below -55mV will not generate an impulse, we say that this level is a threshold potential and

only when we reach the threshold do we get a response (all or none).


Genetic engineering

Discuss how bacteria can be used to produce human insulin. Discuss with an example how reverse transcriptase can also be used in genetic engineering

Insulin is essential to regulate blood glucose levels.

It is produced by the β cells in the pancreas

Some people suffer from diabetes and are unable to produce sufficient amounts of insulin and therefore need to inject it.

In order to meet the global demand for insulin it can be produced by using genetically modified bacteria.

In this procedure the DNA from healthy donor pancreas cells is removed and the gene for insulin removed through the use of restriction enzymes.

Sticky ends are then added to the gene and it is inserted in to a plasmid, which is a small loop of DNA, that has been extracted from an e coli bacterium and cut with the same restriction enzymes

Complimentary sticky ends are added to the plasmid

Ligase enzymes are used to splice (join) the host and donor DNA sections together.

The recombinant plasmid is then reinserted in to the bacterium and a culture of clones is grown in a fermenter

The insulin is then harvested and purified.

Alternatively reverse transcriptase, can be used to make copies of required DNA

The mRNA that codes for a specific protein is extracted from a cell

The enzyme reverse transcriptase is extracted from a virus

Reverse transcriptase transcribes mRNA back into a strand of DNA which can be inserted in to a plasmid using restriction enzymes, sticky ends and ligase.

The plasmid can be reinserted in to a bacterium which can then be cloned in a fermenter, the product can then be harvested.

The clotting factor IX needed to treat hemophilia can be produced in this way, by removing the mRNA for the clotting factor from healthy cells.


Conservation

Discuss the process of conservation and identify why in situ is more favourable than ex situ conservation

Every species has the right to life regardless of its use to humans, therefore we need to conserve endangered species as they are at risk of becoming extinct,

Loss of the organism from its habitat may impact upon the ecosystem especially if they are keystone species.

Conservation programs have been set up to help protect endangered species. These programs may be on site in an organisms natural habitat (in situ) eg a nature reserve, which is a defined as a region in a protective framework, usually in the

organisms natural habitat Example: Pelican island in Florida set up to protect the brown pelican or national parks and

nature reserves, other examples of in situ would be the conservation of tigers in India Or they may be offsite (ex situ) eg a breeding program in a zoo, botanical garden or seed

bank. It is better to set up in situ programs as this reduces the stress caused to the organism by

being moved. In situ projects tend to be less expensive, They allow the continuation of evolution and adaptation in the natural environment. In situ also allows greater genetic variation, It allows for natural learned behavior and interactions within a community, Avoids the risk of injury and stress that is often caused by capture. The organisms can also maintain their natural diet and establish territories and adapt according to their natural habitat, they can maintain symbiotic relationships and are more likely to breed. Exposure to new disease and pathogens can be reduced. In situ conservation preserves whole ecosystems, not just a single species and provides a

location for the reintroduction of species. In situ is not always possible due to small numbers of destruction of habitat. Ex situ are often set up as the last resort if the habitat of the organism is under threat from

poachers or from development, zoos are a good example of ex situ conservation, they establish breeding programs and use technology and assisted breeding techniques such

as artificial insemination and DNA extraction for gene banks to maintain the species. Botanical gardens allow for the protection of a plant species from climate change and the

manipulation of environmental conditions to provide a plant with its requirements. An example of an organism that has become extinct in recent years, would be the

Tasmanian Tiger which became extinct due to habitat destruction, interspecific competition with the dingo and hunting by humans.

Plants that have become extinct include the St Helena olive which became extinct to habitat destruction.

International collaboration and funding can help with conservation projects.


Nature reserves

Discuss the need for nature reserves and explain some of the problems encountered within them

Nature reserves are established when there is a need to actively manage a natural environment.

The setting up of these reserves often results in the reversal of damage that has been caused by human activity.

By restricting access to sensitive ecological areas the exploitation of natural resources is reduced eg through poaching.

The establishment of these areas allows the recovery of threatened species, the control of alien species and the establishment of protective buffer zones around an area that is under threat. Some people still try to exploit the resources of the reserve and measures need to be put in

place to reduce this. Education of the biological importance of the area, regular monitoring, legislation and the use of wardens to patrol the area are measures that are frequently used. Enforced international bans in the trade of animal products such as tiger bones. Where fragmentation has occurred in nature reserves habitat corridors can be established

between the fragments. Example of a nature reserve would be Pelican island in Florida set up to protect the brown pelican

Some reasons why nature reserves are needed

Ethical all life needs to be protected, protection of the indigenous people’s way of life and culture. Maintains species richness for future generations.

Ecological native species might be replaced by alien species which could become invasive, extinction of one species may lead to the extinction of many, deforestation leads to erosion and flooding

Economical medical pharmaceuticals not yet discovered, ecotourism, wild alleles may improve crop plants through genetic modification


Carbon cycle

Autotrophs fix inorganic material in to useable organic material using energy from the sun.

Plants are autotrophs and convert water and carbon dioxide in to glucose and oxygen.

The process requires CO2, H2O, energy from the sun and chlorophyll in the leaf.

0.04% of the earth’s atmosphere is carbon dioxide.

The only way to ‘fix’ the carbon is through photosynthesis.

Carbon dioxide will move in to the leaf through the stomata by diffusion.

It will then move through the air spaces in the leaf and is absorbed in to the palisade cells.

Plants also release CO2 when they respire,

Once the carbon has been converted into glucose it may be stored by the plant as starch,

converted to amino acids or cellulose or it may be used by the plant during respiration.

The organic compounds are consumed by herbivores and will pass through the food chain.

The carbon is released back in to the environment as carbon dioxide during respiration.

When the organisms die they will be decomposed by bacteria which also respire.

Decomposers will release carbon compounds and inorganic ions from decaying material-

fewer decomposers mean that the nutrients will be recycled more slowly.

Detritivores will also use the carbon compounds from decaying material.

Combustion of fossil fuels also returns carbon to the atmosphere as CO2.

Some of the carbon may be trapped and fossilised as fossil fuel.


Nitrogen cycle

Discuss the importance of the nitrogen cycle, describe the impact of human activity on the nitrogen cycle

Nitrogen is essential for the formation of proteins and the formation of DNA. Nitrogen exists in the atmosphere as nitrogen gas (78%) this gas cannot be used by the majority of organisms and first needs to be converted to the nitrogen compound nitrates.

This activity is done by bacteria: Azobacter- free living nitrogen fixing bacteria in soil Rhizobium- nitrogen fixing bacteria in rhizomes of legumes (mutualistic symbiosis)Putrefying bacteria convert urea to ammonia (ammonification)Nitrosomas bacteria convert ammonia to nitrites (nitrification)Nitrobacter nitrifying bacteria convert nitrite to nitrateP denitrificans- anaerobic bacteria that converts nitrates to nitrogen gas.

A plant lacking nitrogen will have poor growth and yellow leaves. Humans are able to influence the nitrogen cycle adding natural and synthetic fertilizer to soil

which increases the nitrogen content in the ecosystem, Planting legumes helps add nitrates to the land, Ploughing the land aerates the soil allowing nitrogen fixing bacteria to obtain oxygen and

provides drainage- reducing the activity of anaerobic denitrifying bacteria. Leaving field fallow for a season will help reduce di-nitrification. Motor vehicles add nitrogen

compounds to the atmosphere, although there are some bacteria that can convert these emissions in to nitrates, these nitrous oxides often combine with sulphur dioxide from acid rain when this falls in rivers it lowers pH, this causes the leaching of toxic aluminum, resulting in the death of fish. The dumping of sewage in rivers increase nitrogen content due to the activity of decomposers.

Abiotic factors that favour nitrification include warmer temperatures- as this increases enzyme activity


Natural selection

Discuss how new species may arise due to natural selection

Natural selection is the concept that organisms which are better adapted to the environment are more likely to survive and pass on their characteristics.

As populations tend to overproduce offspring, they will compete and those that have the traits most suited the environment will survive to sexual maturity.

These advantageous traits will be passed on to more offspring and will gradually become more prevalent over time and the species evolves.

The adaptations are therefore determined by genetics.

Types of natural selection include stabilizing selection, directional selection and diversifying (disruptive) selection.

In stabilizing selection, a particular range becomes the preferred trait eg birth weight in humans.

In directional selection one end of the range for particular characteristic is favoured due to environmental pressure eg the peppered moth and pollution.

This leads to a progressive change in the genotypes of the population towards the favoured direction, over time the species may change enough to achieve reproductive isolation and therefore will be classed as a new species. Isolated populations will often be subjected to directional selection.

In disruptive selection different forms of the organism become more prevalent in different niches, the extremes of the characteristic are favoured over the intermediates (which disappear), the extremes will be adapted to different niches and the two types may then evolve reproductive barriers leading to speciation, eg the pocket rock mouse has light and dark forms.

Organisms experience different selective pressures that affect the fitness of a species and lead to natural selection. These pressures may be abiotic such as climate and pH etc., or biotic such as predation and disease.

Variation refers to differences between individuals in a population, variation within a population is advantageous as it enables some individual to survive environmental changes.


Populations and species distribution

Discuss the factors that may affect the distribution of a species and explain how a transect can be used to monitor plant distribution

Factor ExplanationBreeding site Some organisms require a specific breeding siteFood supply Organisms are more abundant closer to the food supply, larger populations when food is

abundantTerritory Some organisms have a large territory and will defend it keeping competitors out, there may

be limited space so those without territory are unable to breed or access food and water, leads to intraspecific and interspecific competition

Competition Availability of space and food effects the number of organisms that can be supported

Predation/ herbivory

Controls the numbers of organisms in the lower trophic level

Climate Must be within the range of tolerance for the organismWater Organisms need access to a water supply to surviveGeographic barriers

Controls the distribution of organisms by limiting inhabitable space

Temperature

Homeothems may a wider distribution, SA/Vol ration determines habitat due to heat loss and need for metabolism, poikilothems are more dependent on environmental temperature, only organisms with special adaptations can survive in extreme temperature

Pollution May have a detrimental effect on species distribution, some organisms are sensitive to pollution

A transect is a way of comparing the distribution of plants in relation to an abiotic factor. For example to observe the changes in species distribution with the incline of a hill we would

use a transect. There are two types of transect, a line transect, a tape measure is laid out over an area of

study- the actual positioning must be random to avoid bias. Every 5m or so the number of organisms touching the line or within a quadrat is counted.

The data for the whole transect is studied. A belt transect involves laying our two lines a short distance apart. Every 5m or so the

number of organisms between the two lines is counted. The data from either transect can be plotted on to a kite diagram. This diagram allows us to compare species distribution in relation to the abiotic factor It also allows us to see the effect of zonation and succession.


Simpson’s diversity index

Discuss how species diversity may be calculated

We use the Simpsons diversity index to calculate diversity, The test allows us to compare the species richness and abundance (diversity) in different

areas, Or it can be used to measure the impact of a disturbance on biodiversity or the success of a

conservation program. A high index suggests a more stable or ancient community. It can also be used to measure the impact of pollution over time A low index suggests that there is high pollution. The higher the value attained the more diversity there is in a community. A low index may also indicate recent colonization or that the land is managed or agricultural. he index is usually used to measure plant diversity, but may also be used for animal diversity

with samples being taken at regular intervals to monitor changes.

Biotic indexDiscuss how biotic indices can be used to monitor the impact of pollution on an ecosystem

All aquatic invertebrates are given a biotic value depending upon their tolerance to pollution. The higher the value the less tolerant to a pollutant they are. Mayfly nymph may have a value of 10 as they require a high oxygen environment, Midge larvae may have a value of 1 as they have a high tolerance to pollution. The biotic index is an indication of the pollution level or impact of an abiotic factor on an ecosystem. The richness or absence of a particular indicator species suggest the level of pollution in the

environment. The abundance of these organisms allows us to calculate a biotic index. To calculate the biotic index for an aquatic system, organisms are collected in a kick sample and then

counted. We then multiply the number of individuals by their biotic value and add all the totals together. We will then divide by 10. The higher the value the cleaner the water The lower the value the lower the diversity of organisms and the more polluted the water is. To gain a full picture using the biotic index, we need to know the number of different species and the

number of individuals within them, We also need to know the pollution tolerance rating for each organism. Only organisms that are known to have a pollution tolerance level will be counted. The same method can be used to calculate diversity within a lichen community. Changes in numbers of these indicator species allows us to monitor the impact of the pollutant. Note that the absence of a species may not necessarily indicate pollution. Keys may need to be used to identify organisms. Regular samples should be taken and compared Seasonal changes may affect populations and changes in numbers may not necessarily indicate

pollution

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