Biology SL IB Assessment

8/2/2019 Biology SL IB Assessment

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1.1.1 State that error bars are graphical representations of thevariability of dataError bars are graphical representations of the variability of data.1.1.2 Calculate the mean and standard deviation of a set of

values.See video (coming soon)1.1.3 State that the term standard deviation is used to summarizethe spread of values around the mean, and that 68% of the valuesfall within one standard deviation of the mean.The term standard deviation is used to summarize the spread ofvalues around the mean, and that 68% of the values fall within onestandard deviation of the mean. This rises to about 95% for +/- 2standard deviations.

1.1.4 Explain how the standard deviation is useful for comparingthe means and spread of data between two or more samples.The standard deviation is used to show how the values are spreadabove and below the mean. A low standard deviation means that thevalues are closely grouped around the mean whereas a high standarddeviation means that the values are widely spread. About 68% of thevalues fall within one standard deviation of the mean. This rises toabout 95% for +/- 2 standard deviations.We can use the standard deviation to decide weather the differences

between two means is significant. If the difference between the twomeans is larger than that of the standard deviations then thedifference between the two means is significant. If the differencebetween the two means is smaller than that of the standard deviationthen the differences between the two means are insignificant.1.1.5 Deduce the significance of the differences between two setsof data using calculated values for t and the appropriate tables.See video (coming soon)1.1.6 Explain that the existence of a correlation does not

establish that there is a casual relationship between twovariables.Correlation often shows a casual relationship between two variablessuch as height and weight. Taller people tend to be heavier and so wecan see a correlation in these two sets of data. However, somevariables may show correlation when in fact there is no casualrelationship between them. The results may be correlated by chance.This means that even the correlation is a useful tool for studying data,it is not always reliable.


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2.1 Cell Theory2.1.1 Outline the cell theory.The cell theory states that:

All living organisms are composed of cells. Multicellular organisms (example:humans) are composed of many cells while unicellular organisms (example:bacteria) are composed of only one cell. Cells are the basic unit of structure inall organisms.

Cells are the smallest unit of life. They are the smallest structures capable ofsurviving on their own.

Cells come from pre-exsisting cells and cannot be created from non-livingmaterial. For example, new cells arise from cell division and a zygote (the veryfirst cell formed when an organism is produced) arises from the fusion of anegg cell and a sperm cell.

2.1.2 Discuss the evidence for the cell theory.When scientists started to look at the structures of organisms underthe microscope they discovered that all living organisms where madeup of these small units which they proceeded to call cells. When thesecells were taken from tissues they were able to survive for someperiod of time. Nothing smaller than the cell was able to liveindependently and so it was concluded that the cell was the smallestunit of life. For some time, scientists thought that cells must arise fromnon-living material but it was eventually proven that this was not thecase, instead they had to arise from pre-exsisting cells. An experimentto prove this can be done as follows:

Take two containers and put food in both of these Sterilize both of the containers so that all living organisms are killed Leave one of the containers open and seal the other closed

What will happen is that in the open container mold will start to growbut in the container that was sealed no mold will be present. Thereason for this is because in the open container, cells are able to enter

the container from the external environment and start to divide andgrow. However, due to the seal on the other container no cells will beable to enter and so no mold will develop, proving that cells cannotarise from non-living material.2.1.3 State that unicellular organisms carry out all the functionsof life.Unicellular organisms carry out all the functions of life includingmetabolism, response, homeostasis, growth, reproduction andnutrition.


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2.1.4 Compare the relative sizes of molecules, cell membranethickness, viruses, bacteria, organelles and cells, using theappropriate SI unit.Remember:

1 millimeter (mm) = 10-3

meters1 micrometer (m) = 10-3 millimeters1 nanometer (nm) = 10-3 micrometers

A molecule = 1 nmThickness of cell membrane = 10 nmViruses = 100 nmBacteria = 1mOrganelles = up to 10 m

Eukaryotic cells = up to 100 m2.1.5 Calculate the linear magnification of drawings and theactual size of specimens in images of known magnification.

Take a measurement of the drawing (width or length) Take this same measurement of the specimen Remember to convert units if needed to Place your values into the equation Magnification = length of drawing / length of actual specimen

You can also calculate the length of the specimen if this is unknown:length of the drawing / magnification.Conversion of units:1 centimeter = 10-2 meters1 millimeter = 10

-3meters

1 micrometer = 10-6 meters1 nanometer = 10

-9meters

2.1.6 Explain the importance of the surface area to volume ratioas a factor limiting cell size.Many reactions occur within the cell. Substances need to be taken intothe cell to fuel these reactions and the wast products of the reactionsneed to be removed. When the cell increases in size so does itschemical activity. This means that more substances need to be takenin and more need to be removed. The surface area of the cell is vitalfor this. Surface area affects the rate at which particles can enter andexit the cell (The amount of substances that it takes up from theenvironment and excretes into the environment), whereas the volumeaffects the rate at which material are made or used within the cell,hence the chemical activity per unit of time.


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Every cell in a multicellular organisms contains all the genes of thatorganism. However, the genes that are activated vary from cell to cell.The reason we have different types of cells in our body (the cells inyour eyes are not the same as the ones that make up your hair) is

because different genes are activated in different cells. For example,the gene that produces keratin will be active in hair and nail cells.Keratin is the protein which makes up hair and nails. Genes encodefor proteins and the proteins affect the cells structure and function sothat the cell can specialize. This means cells develop in different ways.This is called differentiation. Differentiation depends on geneexpression which is regulated mostly during transcription. It is anadvantage for multicellular organisms as cells can differentiate to bemore efficient unlike unicellular organisms who have to carry out all of

the functions within that one cell.2.1.9 State that stem cells retain the capacity to divide and havethe ability to differentiate along different pathways.Adults have stems cells in the tissues in their bodies that need to befrequently replaced such as the skin. Stem cells have the ability toproduce a wide range of cells which means that they are pluripotent.They retain their ability to divide and produce many different cells bycell division and the process of differentiation. For example, one typeof stem cells in the bone marrow produce a variety of red and white

blood cells.2.1.10 Outline one therapeutic use of stem cells.Bone marrow transplants are one of the many therapeutic uses ofstem cells. Stem cells found in the bone marrow give rise to the redblood cells, white blood cells and platelets in the body. These stemcells can be used in bone marrow transplants to treat people whohave certain types of cancer.When a patient has cancer and is given high doses of chemotherapy,the chemotherapy kills the cancer cells but also the normal cells in the

bone marrow. This means that the patient cannot produce blood cells.So before the patient is treated with chemotherapy, he or she canundergo a bone marrow harvest in which stem cells are removed fromthe bone marrow by using a needle which is inserted into the pelvis(hip bone). Alternatively, if stem cells cannot be used from the patientthen they can be harvested from a matching donor. After thechemotherapy treatment the patient will have a bone marrowtransplant in which the stem cells are transplanted back into thepatient through a drip, usually via a vein in the chest or the arm. These

transplanted stem cells will then find their way back to the bone


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marrow and start to produce healthy blood cells in the patient.Therefore the therapeutic use of stem cells in bone marrowtransplants is very important as it allows some patients with cancer toundergo high chemotherapy treatment. Without this therapeutic use of

stem cells, patients would only be able to take low doses ofchemotherapy which could lower their chances of curing the disease.

2.2.1 Draw and label a diagram of the ultrastructure ofEscherichia coli (E. coli) as an example of a prokaryote.

Figure 2.2.1 - Annotated drawing of Escherichia coli2.2.2 Annotate the diagram from 2.2.1 with the functions of eachnamed structure.Cell wall: Protects the cell from the outside environment andmaintains the shape of the cell. It also prevents the cell from bursting ifinternal pressure rises.

Plasma membrane: Semi-permeable membrane that controls thesubstances moving into and out of the cell. It contains integral and

peripheral proteins. Substances pass through by either active orpassive transport.


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Cytoplasm: Contains many enzymes used to catalyze chemicalreactions of metabolism and it also contains the DNA in a regioncalled the nucleoid. Ribosomes are also found in the cytoplasm.

Pili: Help bacteria adhere to each other for the exchange of geneticmaterial.

Flagella (singular flagellum): Made of a protein called flagellin. Helpsbacteria move around by the use of a motor protein that spins theflagellum like a propeller.

Ribosomes: They are the site of protein synthesis. Contributes to

protein synthesis by translating messenger RNA.

Nucleoid: Region containing naked DNA which stores the hereditarymaterial (genetic information) that controls the cell and will be passedon to daughter cells.2.2.3 Identify structures from 2.2.1 in electron micrographs of E.coli.

Figure 2.2.2 - Electron micrograph of a prokaryotic cell2.2.4 State that prokaryotic cells divide by binary fission.


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Prokaryotic cells divide by binary fission. Binary fission is a method ofasexual reproduction involving the splitting of the parent organism intotwo separate organisms.

Eukaryotic Cells2.3.1 Draw and label a diagram of the ultrastructure of a liver cellas an example of an animal cell.

Figure 2.3.1 - Annotated drawing of an animal cell2.3.2 Annotate the diagram from 2.3.1 with the functions of eachnamed structure.

Ribosomes: Found either floating free in the cytoplasm or attached tothe surface of the rough endoplasmic reticulum and in mitochondriaand chloroplast. Ribosomes are the site of protein synthesis as theytranslate messenger RNA to produce proteins.Rough endoplasmic reticulum: Can modify proteins to alter theirfunction and/or destination. Synthesizes proteins to be excreted fromthe cell.Lysosome: Contains many digestive enzymes to hydrolyzemacromolecules such as proteins and lipids into their monomers.


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Golgi apparatus: Receives proteins from the rough endoplasmicreticulum and may further modify them. It also packages proteinsbefore the protein is sent to its final destination which may beintracellular or extracellular.

Mitochondrion: Is responsible for aerobic respiration. Convertschemical energy into ATP using oxygen.Nucleus: Contains the chromosomes and therefore the hereditarymaterial. It is responsible for controlling the cell.2.3.3 Identify structures from 2.3.1 in electron micrographs ofliver cells.

Figure 2.3.2 - Electron micrograph of an animal cell2.3.4 Compare prokaryotic and eukaryotic cells.

1. Prokaryotic cells have naked DNA which is found in the cytoplasm in a region

named the nucleoid. On the other hand, eukaryotes have chromosomes thatare made up of DNA and protein. These chromosomes are found in thenucleus enclosed in a nuclear envelope.

2. Prokaryotes do not have any mitochondria whereas eukaryotes do.3. Prokaryotes have small ribosomes (70S) compared to eukaryotes which have

large ribosomes (80S).4. In prokaryotes there are either no or very few organelles bounded by a single

membrane in comparison to eukaryotes which have many of them includingthe Golgi apparatus and the endoplasmic reticulum.

2.3.5 State three differences between plant and animal cells.


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1. Animal cells only have a plasma membrane and no cell wall. Whereas plantcells have a plasma membrane and a cell wall.

2. Animal cells do not have chloroplasts whereas plant cells do for the process ofphotosynthesis.

3. Animal cells store glycogen as their carbohydrate resource whereas plants

store starch.4. Animal cells do not usually contain any vacuoles and if present they are small

or temporary. On the other hand plants have a large vacuole that is alwayspresent.

5. Animal cells can change shape due to the lack of a cell wall and are usuallyrounded whereas plant cells have a fixed shape kept by the presence of thecell wall.

2.3.6 Outline two roles of extracellular components.The plant cell wall gives the cell a lot of strength and prevents it from

bursting under high pressure as it is made up of cellulose arranged ingroups called microfibrils. It gives the cell its shape, preventsexcessive water up take by osmosis and is the reason why the wholeplant can hold itself up against gravity.The animal cell contains glycoproteins in their extracellular matrixwhich are involved in the support, movement and adhesion of the cell.

Membranes2.4.1 Draw and label a diagram to show the structure ofmembranes.


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Figure 2.4.1 - Annotated drawing of a cell membrane

2.4.2 Explain how the hydrophobic and hydrophilic properties ofphospholipids help to maintain the structure of cell membranes.Phospholipid molecules make up the cell membrane and arehydrophilic (attracted to water) as well as hydrophobic (not attracted to

water but are attracted to other hydrophobic tails). They have ahydrophilic phosphate head and two hydrophobic hydrocarbon tails.Cell membranes are made up of a double layer of these phospholipidmolecules. This is because in water the hydrophilic heads will face thewater while the hydrophobic tails will be in the center because theyface away from the water. The phospholipid bilayer makes themembrane very stable but also allows flexibility. The phospholipid inthe membrane are in a fluid state which allows the cell to change it sshape easily.

2.4.3 List the functions of membrane proteins.Membrane proteins can act as hormone binding sites, electroncarriers, pumps for active transport, channels for passive transportand also enzymes. In addition they can be used for cell to cellcommunication as well as cell adhesion.2.4.4 Define diffusion and osmosis.Diffusion is the passive movement of particles from a region of highconcentration to a region of low concentration.Osmosis is the passive movement of water molecules, across a

partially permeable membrane, from a region of lower soluteconcentration to a region of higher solute concentration.


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2.4.5 Explain passive transport across membranes by simplediffusion and facilitated diffusion.Membranes are semi-permeable which means that they allow certainmolecules through but not others. The molecules can move in and out

through passive transport which is a method that does not require anyinput of outside energy. It can either be done by simple diffusion orfacilitated diffusion. Molecules will go from a region of highconcentration to a region of low concentration as they move randomlyand eventually become evenly distributed within the system if they arepermeable to the membrane. Simple diffusion involves the diffusion ofmolecules through the phospholipid bilayer while facilitated diffusioninvolves the use of channel proteins embedded in the membrane. Thecell membrane is hydrophobic inside so hydrophobic (lipid soluble)

molecules will pass through by simple diffusion whereas hydrophilicmolecules and charged particles will use facilitated diffusion. Watermoves through by osmosis which is also by passive transport.Osmosis involves the movement of water molecules from a region oflow solute concentration, to a region of high solute concentration. So ifthe solute concentration is higher inside the cell than outside the cell,water will move in and vice versa.2.4.6 Explain the role of protein pumps and ATP in activetransport across membranes.

Active transport involves the movement of substances through themembrane using energy from ATP. The advantage of active transportis that substances can be moved against the concentration gradient,meaning from a region of low concentration to a region of highconcentration. This is possible because the cell membrane has proteinpumps embedded it which are used in active transport to movesubstances across by using ATP. Each protein pump only transportscertain substances so the cell can control what comes in and whatgoes out.

2.4.7 Explain how vesicles are used to transport materials withina cell between the rough endoplasmic reticulum, Golgi apparatusand the cell membrane.After proteins have been synthesized by ribosomes they aretransported to the rough endoplasmic reticulum where they can bemodified. Vesicles carrying the protein then bud off the roughendoplasmic reticulum and are transported to the Golgi apparatus tobe further modified. After this the vesicles carrying the protein bud offthe Golgi apparatus and carry the protein to the plasma membrane.

Here the vesicles fuse with the membrane expelling their content (the


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modified proteins) outside the cell. The membrane then goes back toits original state. This is a process called exocytosis. Endocytosis is asimilar process which involves the pulling of the plasma membraneinwards so that the pinching off of a vesicle from the plasma

membrane occurs and then this vesicle can carry its content anywherein the cell.2.4.8 Describe how the fluidity of the membrane allows it tochange shape, break and re-form during endocytosis andexocytosis.The phospholipids in the cell membrane are not solid but are in a fluidstate allowing the membrane to change its shape and also vesicles tofuse with it. This means substances can enter the cell via endocytosisand exit the cell via exocytosis. The membrane then returns to its

original state. In exocytosis the vesicles fuse with the membraneexpelling their content outside the cell. The membrane then goes backto its original state. Endocytosis is a similar process which involves thepulling of the plasma membrane inwards so that a vesicle is pinchedoff it and then this vesicle can carry its content anywhere in the cell.

Cell Divison2.5.1 Outline the stages in the cell cycle, including interphase,

(G1, S, G2), mitosis and cytokinesis.The first stage of cell division is interphase which is divided into 3phases; G1, S and G2. The cell cycle starts with G1 (Gap phase 1)during which the cell grows larger. This is followed by phase S(synthesis) during which the genome is replicated. Finally, G2 (gapphase 2) is the second growth phase which separates the newlyreplicated genome and marks the end of interphase.The fourth stage is mitosis which is divided into prophase, metaphase,anaphase and telophase. During mitosis the spindle fibers attach tothe chromosomes and pull sister chromatids apart. This stageseparates the two daughter genomes. Finally, cytokinesis is the laststage during which the cytoplasm divides to create two daughter cells.In animal cells the cell is pinched in two while plant cells form a platebetween the dividing cells.2.5.2 State that tumors (cancers) are the result of uncontrolledcell division and that these can occur in any organ or tissue.Tumors are formed when cell division goes wrong and is no longercontrolled. This can happen in any organ or tissue.


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2.5.3 State that interphase is an active period in the life of a cellwhen many metabolic reactions occur, including proteinsynthesis, DNA replication and an increase in the number ofmitochondria and/or chloroplast.

Interphase is an active period in the life of a cell during which manymetabolic reactions occur such as protein synthesis, DNA replicationand an increase in the number of mitochondria and/or chloroplast.2.5.4 Describe the events that occur in the four phases of mitosis(prophase, metaphase, anaphase, telophase).During prophase the spindle microtubules grow and extend from eachpole to the equator. Also chromosomes super coil and become shortand bulky and the nuclear envelope breaks down.During metaphase the chromatids move to the equator and the spindle

microtubules from each pole attach to each centromere on oppositesides.During anaphase the spindle microtubules pull the sister chromatidsapart splitting the centromeres. This splits the sister chromatids intochromosomes. Each identical chromosome is pulled to opposite poles.During telophase the spindle microtubules break down and thechromosomes uncoil and so are no longer individually visible. Also thenuclear membrane reforms. The cell then divides by cytokinesis toform two daughter cells with identical genetic nuclei.

2.5.5 Explain how mitosis produces two genetically identicalnuclei.Mitosis is divided into four stages; prophase, metaphase, anaphaseand telophase. During prophase, the chromosomes become visibleunder a light microscope as they super coil and therefore they getshorter and more bulky. The nuclear envelope disintegrates and thespindle microtubules grow and extend from each pole to the equator.At metaphase the chromatids move to the equator. The sisterchromatids are two DNA molecules formed by DNA replication and are

therefore identical. These sister chromatids are then separated inanaphase as the spindle microtubules attaches to centromere andpulls the sister chromatids to opposite poles. As the sister chromatidsseparate they are called chromosomes. This means that each polehas the same chromosomes (same genetic material). Finally themicrotubules break down, the chromosomes uncoil and the nuclearmembrane reforms. The cell then divides into two daughter cells withgenetically identical nuclei.2.5.6 State that growth, embryonic development, tissue repair

and asexual reproduction involve mitosis.


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Growth, embryonic development, tissue repair and asexualreproduction involve mitosis.

3.1 Chemical Elements and Water3.1.1 State that the most frequently occurring chemical elementsin living things are carbon, hydrogen, oxygen and nitrogen.Carbon, hydrogen, oxygen and nitrogen are the most frequentlyoccurring chemical elements in living things.3.1.2 State that a variety of other elements are needed by livingorganisms, including sulfur, calcium, phosphorus, iron and

sodium.A variety of other elements are needed by living organisms, includingsulfur, calcium, phosphorus, iron and sodium.3.1.3 State one role for each of the elements mentioned in 3.1.2.Sulfur: Needed for the synthesis of two amino acids.Calcium: Acts as a messenger by binding to calmodulin and a fewother proteins which regulate transcription and other processes in thecell.Phosphorus: Is part of DNA molecules and is also part of the

phosphate groups in ATP.Iron: Is needed for the synthesis of cytochromes which are proteinsused during electron transport for aerobic cell respiration.Sodium: When it enters the cytoplasm, it raises the soluteconcentration which causes water to enter by osmosis.3.1.4 Draw and label a diagram showing the structure of watermolecules to show their polarity and hydrogen bond formation.3.1.5 Outline the thermal, cohesive and solvent properties ofwater.

Thermal properties of water include heat capacity, boiling and freezingpoints and the cooling effect of evaporation.Water has a large heat capacity which means that a considerableamount of energy is needed to increase its temperature. This is due tothe strength of the hydrogen bonds which are not easily broken. Thisis why the temperature of water tends to remain relatively stable. It isbeneficial for aquatic animals as they use water as a habitat.Water has a high boiling and freezing point. It boils at 100 C becausethe strong hydrogen bonds. All these hydrogen bonds between the

water molecules need to break for the liquid to change to gas. Waterbecomes less dense as it gets closer to the freezing point and so ice


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always forms on the surface first. The high boiling point of water isvital for life on earth as if water boiled at a lower temperature the waterin living organisms would start to boil and therefore these organismswould not survive.

The fact that water becomes less dense as it freezes is beneficial toorganisms as ice will always form at the surface of lakes or seas andby doing so it insulates the water underneath, maintaining a possiblehabitat for organisms to live in.Water can evaporate at temperatures below the boiling point.Hydrogen bonds need to break for this to occur.Cohesion is the effect of hydrogen bonds holding the water moleculestogether. Water moves up plants because of cohesion. Long columnsof water can be sucked up from roots to leaves without the columns

breaking. The hydrogen bonds keep the water molecules sticking toeach other.The solvent properties of water mean that many different substancescan dissolve in it because of its polarity.3.1.6 Explain the relationship between the properties of water andits uses in living organisms as a coolant, medium for metabolicreactions and transport medium.Water can evaporate at temperatures below the boiling point.Hydrogen bonds need to break for this to occur. The evaporation of

water cools body surfaces (sweat) and plant leaves (transpiration) byusing the energy from liquid water to break the hydrogen bonds. Thesolvent properties of water mean that many different substances candissolve in it because of its polarity. This allows substances to becarried in the blood and sap of plants as they dissolve in water. It alsomakes water a good medium for metabolic reactions.

Carbohydrates, Lipids and Proteins3.2.1 Distinguish between organic and inorganiccompounds.Organic compounds are compounds that are found in living organisms and contain carbon.Inorganic compounds are the ones that dont contain carbon. Although, there are a fewcompounds found in living organisms which also contain carbon but are considered asinorganic compounds. These include carbon dioxide, carbonates and hydrogen carbonates.3.2.2 Identify amino acids, glucose, ribose and fatty acids from diagrams showingtheir structure.


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3.2.3 List three examples of each monosaccharides, disaccharides andpolysaccharides.

Glucose, galactose and fructose are all monosaccharides. Maltose, lactose and sucrose are all disaccharides.

Starch, glycogen and cellulose are all polysaccharides.3.2.4 State one function of glucose, lactose and glycogen in animals, and of fructose,sucrose and cellulose in plants.In animals, glucose is used as an energy source for the body and lactose is the sugar foundin milk which provides energy to new borns until they are weaned. Finally, glycogen is usedas an energy source (short term only) and is stored in muscles and the liver.In plants, fructose is what makes fruits taste sweet which attracts animals and these then eatthe fruits and disperse the seeds found in the fruits. Sucrose is used as an energy source forthe plant whereas cellulose fibers is what makes the plant cell wall strong.3.2.5 Outline the role of condensation and hydrolysis in the relationships betweenmonosaccharides, disaccharides and polysaccharides; between fatty acids, glyceroland triglycerides; and between amino acids and polypeptides.

3.2.6 State three functions of lipids.

Lipids can be used for energy storage in the form of fat in humans and oil in plants. Lipids can be used as heat insulation as fat under the skin reduces heat loss. Lipids allow buoyancy as they are less dense than water and so animals can float in

water.3.2.7 Compare the use of carbohydrates and lipids in energy storage.Carbohydrates and lipids can both be used as energy storage however carbohydrates areusually used for short term storage whereas lipids are used for long term storage.Carbohydrates are soluble in water unlike lipids. This makes carbohydrates easy to transportaround the body (from and to the store). Also, carbohydrates are a lot easier and more

rapidly digested so their energy is useful if the body requires energy fast. As for lipids, theyare insoluble which makes them more difficult to transport however because they areinsoluble, lipids do not have an effect on osmosis which prevents problems within the cells inthe body. They also contain more energy per gram than carbohydrates which makes lipids alighter store compared to a store of carbohydrates equivalent in energy.


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DNA Structure

3.3.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose),base and phosphate.

A nucleotide is made of the sugar deoxyribose, a base (which can be eitheradenine, guanine, cytosine or thymine) and a phosphate group. Below is arepresentation of a nucleotide.3.3.2 State the names of the four bases in DNA.Adenine, Guanine, Cytosine and Thymine.

3.3.3 Outline how DNA nucleotides are linked together by covalent bonds into a singlestrand.Below is a diagram showing how nucleotides are linked to one another to form a strand. Acovalent bond forms between the sugar of one nucleotide and the phosphate group of anothernucleotide.

3.3.4 Explain how a DNA double helix is formed using complementary basepairing and hydrogen bonds.DNA is made up of two nucleotide strands. The nucleotides are connectedtogether by covalent bonds within each strand. The sugar of one nucleotide formsa covalent bond with the phosphate group of another. The two strandsthemselves are connected by hydrogen bonds. The hydrogen bonds are foundbetween the bases of the two strands of nucleotides. Adenine forms hydrogenbonds with thymine whereas guanine forms hydrogen bonds with cytosine. This iscalled complementary base pairing. Below is a digram showing the molecularstructure and bonds within DNA.3.3.5 Draw and label a simple diagram of the molecular structure of DNA.

DNA Replication3.4.1 Explain DNA replication in terms of unwinding the doublehelix and separation of the strands by helicase, followed byformation of the new complementary strands by DNApolymerase.

DNA replication is semi-conservative as both of the DNA moleculesproduced are formed from an old strand and a new one. The firststage of DNA replication involves the unwinding of the double strandof DNA (DNA double helix) and separating them by breaking thehydrogen bonds between the bases. This is done by the enzymehelicase. Each separated strand now is a template for the newstrands. There are many free nucleotides around the replication forkwhich then bond to the template strands. The free nucleotides formhydrogen bonds with their complimentary base pairs on the template

strand. Adenine will pair up with thymine and guanine will pair up withcytosine. DNA polymerase is the enzyme responsible for this. The


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new DNA strands then rewind to form a double helix. The replicationprocess has produced a new DNA molecule which is identical to theinitial one.3.4.2 Explain the significance of complementary base pairing in

the conservation of the base sequence of DNA.Complementary base pairing is very important in the conservation ofthe base sequence of DNA. This is because adenine always pairs upwith thymine and guanine always pairs up with cytosine. As DNAreplication is semi-conservative (one old strand an d one new strandmake up the new DNA molecules), this complementary base pairingallows the two DNA molecules to be identical to each other as theyhave the same base sequence. The new strands formed arecomplementary to their template strands but also identical to the other

template. Therefore, complementary base pairing has a big role in theconservation of the base sequence of DNA.3.4.3 State that DNA replication is semi-conservative.DNA replication is semi-conservative.

Transcription and Translation

3.5.1 Compare the structure of RNA and DNA.DNA and RNA both consist of nucleotides which contain a sugar, abase and a phosphate group. However there are a few differences.Firstly, DNA is composed of a double strand forming a helix whereasRNA is only composed of one strand. Also the sugar in DNA isdeoxyribose whereas in RNA it is ribose. Finally, both DNA and RNAhave the bases adenine, guanine and cytosine. However DNA alsocontains thymine which is replaced by uracil in RNA.3.5.2 Outline DNA transcription in terms of the formation of an

RNA strand complementary to the DNA strand by RNApolymerase.DNA transcription is the formation of an RNA strand which iscomplementary to the DNA strand. The first stage of transcription isthe uncoiling of the DNA double helix. Then, the free RNA nucleotidesstart to form an RNA strand by using one of the DNA strands as atemplate. This is done through complementary base pairing, howeverin the RNA chain, the base thymine is replaced by uracil. RNApolymerase is the enzyme involved in the formation of the RNA strand

and the uncoiling of the double helix. The RNA strand then elongatesand then separates from the DNA template. The DNA strands then


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reform a double helix. The strand of RNA formed is called messengerRNA.3.5.3 Describe the genetic code in terms of codons composed oftriplets of bases.

A triplet of bases (3 bases) forms a codon. Each codon codes for aparticular amino acid. Amino acids in turn link to form proteins.Therefore DNA and RNA regulate protein synthesis. The genetic codeis the codons within DNA and RNA, composed of triplets of baseswhich eventually lead to protein synthesis.3.5.4 Explain the process of translation, leading to polypeptideformation.Translation is the process through which proteins are synthesized. Ituses ribosomes, messenger RNA which is composed of codons and

transfer RNA which has a triplet of bases called the anticodon.The first stage of translation is the binding of messenger RNA to thesmall subunit of the ribosome. The transfer RNAs have a specificamino acid attached to them which corresponds to their anticodons.A transfer RNA molecule will bind to the ribosome however itsanticodon must match the codon on the messenger RNA. This is donethrough complementary base pairing.These two form a hydrogen bond together. Another transfer RNAmolecule then bonds. Two transfer RNA molecules can bind at once.

Then the two amino acids on the two transfer RNA molecules form apeptide bond. The first transfer RNA then detaches from the ribosomeand the second one takes its place.The ribosome moves along themessenger RNA to the next codon so that another transfer RNA canbind. Again, a peptide bond is formed between the amino acids andthis process continues. This forms a polypeptide chain and is the basisof protein synthesis.3.5.5 Discuss the relationship between one gene and onepolypeptide.

A polypeptide is formed by amino acids liking together through peptidebonds. There are 20 different amino acids so a wide range ofpolypeptides are possible. Genes store the information required formaking polypeptides. The information is stored in a coded form by theuse of triplets of bases which form codons. The sequence of bases ina gene codes for the sequence of amino acids in a polypeptide. Theinformation in the genes is decoded during transcription andtranslation leading to protein synthesis.


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Enzymes3.6.1 Define enzyme and active site.Enzymes: Globular proteins which act as catalysts of chemical

reactions.Active site: Region on the surface of an enzyme to which substratesbind and which catalyses a chemical reaction involving the substrates.3.6.2 Explain enzymesubstrate specificity.The active site of an enzyme is very specific to its substrates as it hasa very precise shape. This results in enzymes being able to catalyzeonly certain reactions as only a small number of substrates fit in theactive site. This is called enzyme-substrate specificity. For a substrateto bind to the active site of an enzyme it must fit in the active site and

be chemically attracted to it. This makes the enzyme very specific toits substrate. The enzyme-substrate complex can be compared to alock and key, where the enzyme is the lock and the substrate is thekey.3.6.3 Explain the effects of temperature, pH and substrateconcentration on enzyme activity.Enzyme activity increases with an increase in temperature and usuallydoubles with every 10 degrees rise. This is due to the moleculesmoving faster and colliding more often together. However at a certain

point the temperature gets to high and the enzymes denature and stopfunctioning. This is due to the heat causing vibrations within theenzyme destroying its structure by breaking the bonds in the enzyme.Enzymes usually have an optimum pH at which they work mostefficiently. As the pH diverges from the optimum, enzyme activitydecreases. Both acid and alkali environments can denature enzymes.Enzyme activity increases with an increase in substrate concentrationas there are more random collisions between the substrate and theactive site. However, at some point, all the active sites are taken up

and so increasing the substrate concentration will have no more effecton enzyme activity. As long as there are active sites available, anincrease in substrate concentration will lead to an increase in enzymeactivity.3.6.4 Define denaturation.Denaturation is changing the structure of an enzyme (or other protein)so it can no longer carry out its function.3.6.5 Explain the use of lactase in the production of lactose-freemilk.

Lactose is the sugar found in milk. It can be broken down by theenzyme lactase into glucose and galactose. However some people


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lack this enzyme and so cannot break down lactose leading to lactoseintolerance. Lactose intolerant people need to drink milk that has beenlactose reduced. Lactose-free milk can be made in two ways. The firstinvolves adding the enzyme lactase to the milk so that the milk

contains the enzyme. The second way involves immobilizing theenzyme on a surface or in beads of a porous material. The milk is thenallowed to flow past the beads or surface with the immobilized lactase.This method avoids having lactase in the milk.

Cell Respiration3.7 Cell Respiration3.7.1 Define cell respirationCell respiration is the controlled release of energy from organiccompounds in cells to form ATP.3.7.2 State that, in cell respiration, glucose in the cytoplasm isbroken down by glycolysis into pyruvate, with a small yield ofATP.In cell respiration, glucose in the cytoplasm is broken down byglycolysis into pyruvate with a small yield of ATP.

3.7.3 Explain that, during anaerobic cell respiration, pyruvate canbe converted in the cytoplasm into lactate, or ethanol and carbondioxide, with no further yield of ATP.In anaerobic cell respiration the pyruvate stays in the cytoplasm and inhumans is converted into lactate which is the removed from the cell. Inyeast the pyruvate is converted into carbon dioxide and ethanol. Ineither case, no ATP is produced.3.7.4 Explain that, during aerobic cell respiration, pyruvate can bebroken down in the mitochondrion into carbon dioxide and waterwith a large yield of ATP.If oxygen is available, the pyruvate is taken up into the mitochondriaand is broken down into carbon dioxide and water. A large amount ofATP is released during this process.


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Photosynthesis3.8.1 State that photosynthesis involves the conversion of lightenergy into chemical energy.

Photosynthesis involves the conversion of light energy into chemicalenergy.3.8.2 State that light from the Sun is composed of a range ofwavelengths (colours).The light from the sun is composed of a range of wavelengths(colours).3.8.3 State that chlorophyll is the main photosynthetic pigment.Chlorophyll is the main photosynthetic pigment.3.8.4 Outline the differences in absorption of red, blue and green

light by chlorophyll.Chlorophyll can absorb red and blue light more than green.Chlorophyll cannot absorb green light and so instead reflects it makingleaves look green.3.8.5 State that light energy is used to produce ATP, and to splitwater molecules (photolysis) to form oxygen and hydrogen.Light energy is used to produced ATP and to split water molecules(photolysis) to form oxygen and hydrogen.3.8.6 State that ATP and hydrogen (derived from the photolysis of

water) are used to fix carbon dioxide to make organic molecules.ATP and hydrogen derived from photolysis of water are used to fixcarbon dioxide to make organic molecules.3.8.7 Explain that the rate of photosynthesis can be measureddirectly by the production of oxygen or the uptake of carbondioxide, or indirectly by an increase in biomass.Photosynthesis can be measured in many ways as it involves theproduction of oxygen, the uptake of carbon dioxide and an increase inbiomass. For example, aquatic plants release oxygen bubbles during

photosynthesis and so these can be collected and measured. Theuptake of carbon dioxide is more difficult to measure so it is usuallydone indirectly. When carbon dioxide is absorbed from water the pH ofthe water rises and so this can be measured with pH indicators or pHmeters. Finally, photosynthesis can be measured through an increasein biomass. If batches of plants are harvested at a series of times andthe biomass of these batches is calculated, the rate increase inbiomass gives an indirect measure of the rate of photosynthesis in theplants.

3.8.8 Outline the effects of temperature, light intensity and carbondioxide concentration on the rate of photosynthesis.


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As temperature increases, the rate of photosynthesis increases moreand more steeply until the optimum temperature is reached. Iftemperature keeps increasing above the optimum temperature thenphotosynthesis starts to decrease very rapidly.

As light intensity increases so does photosynthesis until a certainpoint. At a high light intensities photosynthesis reaches a plateau andso does not increase any more. At low and medium light intensity therate of photosynthesis is directly proportional to the light intensity.As the carbon dioxide concentration increases so does the rate ofphotosynthesis. There is no photosynthesis at very low levels ofcarbon dioxide and at high levels the rate reaches a plateau.

Chromosomes, genes, alleles and

mutations4.1.1 State that eukaryote chromosomes are made of DNA andproteins.Eukaryote chromosomes are made of DNA and proteins.4.1.2 Define gene, allele and genome.Gene: a heritable factor that controls a specific characteristic.Allele: one specific form of a gene, differing from other alleles by one or a few bases only andoccupying the same gene locus as other alleles of the gene.Genome: the whole of the genetic information of an organism.4.1.3 Define gene mutation.Gene mutation: a change to the base sequence of a gene.4.1.4 Explain the consequence of a base substitution mutation in relation to the processesof transcription and translation, using the example of sickle-cell anemia.Sickle cell anaemia is a genetic disease that affects red blood cells inthe body. It is due to a mutation on the Hb gene which codes for apolypeptide of 146 amino acids which is part of haemoglobin(haemoglobin is an important protein component in red blood cells). In

sickle cell anaemia the codon GAG found in the normal Hb gene ismutated to GTG. This is called a base substitution mutation asadenine (A) is replaced by thymine (T). This means that when themutated gene is transcribed, a codon in the messenger RNA will bedifferent. Instead of the normal codon GAG, the messenger RNA willcontain the codon GUG. This in turn will result in a mistake duringtranslation. In a healthy individual the codon GAG on the messengerRNA matches with the anticodon CUC on the transfer RNA carryingthe amino acid glutamic acid. However, if the mutated gene is present

then GUG on the messenger RNA matches with the anticodon CACon the transfer RNA which carries the amino acid valine. So the base


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substitution mutation has caused glutamic acid to be replaced byvaline on the sixth position on the polypeptide. This results inhaemoglobin S being present in red blood cells instead of the normalhaemoglobin A. This has an effect on the phenotype as instead of

normal donut shaped red blood cells being produced some of the redblood cells will be sickle shaped. As a result these sickle shaped redblood cells cannot carry oxygen as efficiently as normal red blood cellswould. However, there is an advantage to sickle cell anemia. Thesickle cell red blood cells give resistance to malaria and so the alleleHbson the Hb genewhich causes sickle cell anemia is quite common inparts of the world where malaria is found as it provides an advantageover the disease.Summary of important steps:

Normal Gene Mutated gene

Codon GAG GTG

Transcription GAG on mRNA GUG on mRNA

Translation Anticodon CUC and amino acidglutamic acid on tRNA.

Anticodon CAC and amino acidvaline on tRNA.

Haemoglobin HbA HbS

Phenotype Normal donut shaped red bloodcells.

Sickle cell shaped red bloodcells.

Effects Carry oxygen efficiently but areaffected by malaria.

Do not carry oxygen efficientlybut give resistance to malaria.

4.2.1 State that meiosis is a reduction division of a diploid

nucleus to form haploid nuclei.Meiosis is a reduction division of a diploid nucleus to form haploidnuclei.4.2.2 Define homologous chromosomes.Homologous chromosomes: chromosomes with the same genes aseach other, in the same sequence but do not necessarily have thesame allele of those genes.4.2.3 Outline the process of meiosis, including pairing ofhomologous chromosomes and crossing over, followed by two

divisions, which results in four haploid cells.


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Meiosis involves two divisions. Meiotic cells have an interphase stagebefore the start of meiosis I which is similar to mitosis. It includes G1,S and G2 phases. (See notes on mitosis) After meiosis I there isanother brief interphase stage which is followed by meiosis II.

Meiosis IThe first stage of meiosis I is prophase I. In prophase I thechromosomes pair up so that the chromosomes in each pair arehomologous. Once the homologous chromosomes are paired up,crossing over occurs. Crossing over is the exchange of geneticmaterial between non-sister chromatids. The nuclear membrane alsostarts to break down and the spindle microtubules stretch out fromeach pole to the equator.

The second stage is metaphase I. Here the paired up homologouschromosomes line up at the equator and the spindle fibbers attach tothe chromosomes in a way that ensures that for each homologouspair, one chromosome moved to one pole and the other moves to theopposite pole.The third stage is anaphase I. This is the stage where the homologouschromosomes are separated and pulled to opposite poles. This halvesthe chromosome number however each chromosome is stillcomposed of two sister chromatids. The cell membrane starts to

prepare for its separation at the equator to form two cells.The fourth stage is telophase I. Here each chromosome from thehomologous pair are found at opposite poles and the nuclearmembrane reforms around each daughter nucleus. The membranethen divides through citokinesis.There is a brief interphase stage before the start of meiosis II. Thisstage does not include the S phase.

Meiosis II

The first stage of meiosis II is prophase II. Here the cell has dividedinto two daughter haploid cells however the process does not endhere as these two cells immediately start to divide again. The spindlemicrotubules stretch out from each pole again and the nuclearmembrane breaks down as in prophase I.The second stage is metaphase II. Here the chromosomes in each cellline up at the equator and the spindle microtubules attach to thecentromere of each chromosome.The third stage is anaphase II. Here the centromere devised as a

result of the spindle microtubules pulling each sister chromatid to


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opposite poles in both cells. Each sister chromatid then becomes achromosome.The fourth stage is telophase II. Here the nuclear membrane reformsaround the four sets of daughter chromosomes. Cytokinesis then

follows to divide the cytoplasm of the two cells and so the result is fourdaughter cells each with a haploid set of chromosomes.Below is a summary of meiosis (click on images to expand them).


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4.2.4 Explain that non-disjunction can lead to changes inchromosome number, illustrated by reference to Down syndrome(trisomy 21).A number of problems can arise during meiosis. A common problem isnon-disjunction. This is when the chromosomes do not separateproperly during meiosis, either in meiosis I (in anaphase I) or meiosisII (in anaphase II). This leads the production of gametes that eitherhave a chromosome too many or too few. Gametes with a missingchromosome usually die quite fast however gametes with an extrachromosome can survive. When a zygote is formed from the


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fertilization of these gametes with an extra chromosome, threechromosomes of one type are present instead of two. An example ofthis is Down syndrome. Down syndrome is a disease in which thechromosomes failed to separate properly during meiosis leading to

three chromosomes of type 21 instead of two. A person with thecondition therefore has a total of 47 chromosomes instead of 46. Thenon-disjunction can take place either in the formation of the egg or thesperm. Down syndrome leads to many complications and also the riskof having a child with the condition increases with age.Below is a diagram illustrating a non-disjunction:

4.2.5 State that, in karyotyping, chromosomes are arranged inpairs according to their size and structure.In karyotyping, chromosomes are arranged in pairs according to theirsize and structure.4.2.6 State that karyotyping is performed using cells collected bychorionic villus sampling or amniocentesis, for pre-nataldiagnosis of chromosome abnormalities.Karyotyping is performed using cells collected by chorionic villussampling or amniocentesis, for pre-natal diagnosis of chromosomeabnormalities.4.2.7 Analyse a human karyotype to determine gender andwhether non-disjunction has occurred.Karyotyping can be used to determine gender of a fetus and look for

chromosome abnormalities such as non-disjunction. The gender canbe deduced by looking at the sex chromosomes. Females will have


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two X chromosomes while males have one X and one Y. We candistinguish this on with karyotyping as the Y chromosome is smallerthan the X. As for non-disjunctions we can see if a chromosome ismissing or if their is an extra one by looking at the number of

chromosomes. If There should only be two of each chromosome.Each 23 chromosomes should have a pair resulting in 46chromosomes in total. For example, if we notice that there are threechromosomes 21 then we can conclude that a non-disjunctionoccurred. In this case, the non-disjunction results in Downs syndrome.(trisomy 21)Below are two images of a karyotype. The first one is of a normalhealthy male patient as on the karyotype there are two chromosomesfor each chromosome number and a Y chromosome is present. The

second image shows the karyotype a person with Downs syndromewould get.Karyotype 1:

Karyotype 2:


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Theoretical genetics4.3.1 Define genotype, phenotype, dominant allele, recessiveallele, codominant alleles, locus, homozygous, heterozygous,carrier and test cross.Genotype: the alleles of an organism.Phenotype: the characteristics of an organism.Dominant allele: an allele that has the same effect on the phenotype whether it is present in the

homozygous or heterozygous state.Recessive allele: an allele that only has an effect on the phenotype when present in thehomozygous state.Codominant alleles: pairs of alleles that both affect the phenotype when present in aheterozygote.Locus: the particular position on homologous chromosomes of a gene.Homozygous: having two identical alleles of a gene.Heterozygous: having two different alleles of a gene.Carrier: an individual that has one copy of a recessive allele that causes a genetic disease inindividuals that are homozygous for this allele.Test cross: testing a suspected heterozygote by crossing it with a known homozygous recessive.

4.3.2 Determine the genotypes and phenotypes of the offspring of

a monohybrid cross using a Punnett grid.


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4.3.3 State that some genes have more than two alleles (multiplealleles).Some genes have more than two alleles (multiple alleles).

4.3.4 Describe ABO blood groups as an example of codominanceand multiple alleles.The ABO blood group is a good example of codominance and multiple alleles. There are threeallele that control the ABO blood groups. If there are more than two allele of a gene then they arecalled multiple allele. The allele IA corresponds to blood group A (genotype IAIA) and the alleleIB corresponds to blood group B (genotype IBIB). Both of these are dominant and so if IA and IB arepresent together they form blood group AB (genotype IAIB). Both allele affect the phenotype sincethey are both codominant. Codominant allele are pairs of allele that both affect the phenotypewhen present together in a heterozygote. The allele i is recessive to both IA and IB so if you havethe genotype IA i you will have blood group A and if you have the genotype IB i you will have bloodgroup B. However if you have the genotype ii then you are homozygous for i and will be of bloodgroup O. Below is a table to summaries which genotypes give which phenotypes.

Phenotype Genotype

A IAIA or IAiB IBIB or IBiAB IAIBO ii

Below are diagrams illustrating the inheritance of the ABO blood groups.
http://www.ibguides.com/images/biology/4.3.1.png


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4.3.5 Explain how the sex chromosomes control gender by

referring to the inheritance of X and Y chromosomes in humans.There are two chromosomes which determine gender. These are called the sex chromosomesand there are two types, the X and the Y chromosome. Females have two X chromosomeswhereas males have one X and one Y chromosome. The X chromosome is relatively largecompared to the Y (which is much smaller) and contains many genes. The Y chromosome on theother hand only contains a few genes. The female always passes on to her offspring the Xchromosome from the egg (female gamete). The male can pass on either the Y or the Xchromosome from the sperm (male gamete). If the male passes on the X chromosome then thegrowing embryo will develop into a girls. If the male passes on the Y chromosome then thegrowing embryo will develop into a boy. Therefore gender depends on whether the sperm whichfertilizes the egg is carrying an X or a Y chromosome.4.3.6 State that some genes are present on the X chromosome and absent from the shorterY chromosome in humans.Some genes are present on the X chromosome and absent from the shorter Y chromosome inhumans.

4.3.7 Define sex linkage.Sex linkage: when the gene controlling the characteristic is located on the sex chromosome andso we associate the characteristic with gender.4.3.8 Describe the inheritance of colour blindness and hemophilia as examples of sexlinkage.Most of the time sex-linked genes are carried on the X chromosome. Since females have two Xchromosomes they have two copies of the sex-linked gene whereas males only have one sincethey only have one X chromosome. Hemophilia and colour blindness are both examples of sexlinkage.HemophiliaXH is the allele for normal blood clotting and is dominate over Xh which is recessive and causeshemophilia. If a mother is heterozygous she is a carrier of the disease but does not have


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hemophilia as the dominate allele is present. She can however pass the disease on to heroffspring. Below is a punnett showing how a carrier mother and an unaffected father can pass thedisease on to their offspring.

From our four possible outcomes we can see that a female child cannot get hemophilia but canbe a carrier. This is because the father will always pass on the dominate allele (XH) on the Xchromosome in females. Depending on whether the mother passes on the dominant or recessiveallele will determine if the female child is a carrier or is unaffected by the hemophilia. If the child isa boy then the father has passed on the Y chromosome which does not contain the allele of thegene. So whether the child has the disease or is unaffected depends on which allele the motherhad passed on. If she has passed on the recessive allele (Xh) then the male child will havehemophilia, however if she has passed on the dominate allele (XH) then the child will beunaffected.So there is a 50% chance of the child having hemophilia if it is male as half of the eggs produced

by the mother will carry the recessive allele. The chance of a female offspring having hemophiliais 0% since the father always passes on the dominant allele on the X chromosome. Finally thereis a 25% chance overall that the offspring will be affected.

4.3.9 State that a human female can be homozygous orheterozygous with respect to sex-linked genes.A human female can be homozygous or heterozygous with respect to sex-linked genes.

4.3.10 Explain that female carriers are heterozygous for X-linkedrecessive alleles.Female carriers for X-linked recessive alleles are always heterozygous since they require adominant allele and a recessive allele to be carriers. They inherit the recessive allele from oneparent and the dominate allele from the other. For example hemophilia is a sex-linked disease. Ifa carrier mother and an unaffected father have offspring then the unaffected father will alwayspass on his dominate allele to his female offspring. The carrier mother can either pass on the


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dominate or recessive allele. If she passes on the recessive allele to her female offspring than thefemale offspring will be a carrier as well.

4.3.11Predict the genotypic and phenotypic ratios of offspring ofmonohybrid crosses involving any of the above patterns ofinheritance.

Crossing between two heterozygous individuals gives a 3:1 ratio if one of the alleles is dominate

and the other is recessive.


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4.3.12 Deduce the genotypes and phenotypes of individuals inpedigree charts. Squares represent males Circles represent females Shaded symbols represent affected individuals

Unshaded symbols represent unaffected individuals.

If most of the males in the pedigree are affected the disorder is X-linked. If it is a 50/50 ratio between men and women the disorder is autosomal. If the disorder is dominant, one of the parents must have the disorder. If the disorder is recessive than neither of the parents has to have the disorder as they can be

heterozygous.

Genetic engineering and biotechnology4.4.1 Outline the use of polymerase chain reaction (PCR) to copy and amplify minutequantities of DNA.Polymerase chain reaction is used to copy and amplify minute quantities of DNA. It can be usefulwhen only a small amount of DNA is available but a large amount is required to undergo testing.We can use DNA from blood, semen, tissues and so on from crime scenes for example. The PCRrequires high temperature and a DNA polymerase enzyme from Thermus aquaticus (a bacteriumthat lives in hot springs).


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4.4.2 State that, in gel electrophoresis, fragments of DNA move in an electric field and areseparated according to their size.In gel electrophoresis, fragments of DNA move in an electrical field and are separated accordingto their size.4.4.3 State that gel electrophoresis of DNA is used in DNA profiling.Gel electrophoresis of DNA is used in DNA profiling.

4.4.4 Describe the application of DNA profiling to determine paternity and also in forensicinvestigations.The DNA is copied using PCRand then cut up into small fragments using restriction enzymes. Gelelectrophoresis separates fragmented pieces of DNA according to their size and charge. Thisgives a pattern of bands on a gel which is unlikely to be the same for two individuals. This iscalled DNA profiling. DNA profiling can be used to determine paternity and also in forensicinvestigations to get evidence to be used in a court case for example.4.4.5 Analyse DNA profiles to draw conclusions about paternity or forensic investigations.For a suspect look for similarities between the DNA found at the crime scene and the suspect.For a paternity test, look for similarities between the child and the possible father.4.4.6 Outline three outcomes of the sequencing of the complete human genome. It is now easier to study how genes influence human development.

It helps identify genetic diseases. It allows the production of new drugs based on DNA base sequences of genes or the

structure of proteins coded for by these genes. It will give us more information on the origins, evolution and migration of humans.4.4.7 State that, when genes are transferred between species, the amino acid sequence ofpolypeptides translated from them is unchanged because the genetic code is universal.When genes are transferred between species, the amino acid sequence of polypeptidestranslated from them is unchanged because the genetic code is universal.4.4.8 Outline a basic technique used for gene transfer involving plasmids, a host cell(bacterium, yeast or other cell), restriction enzymes (endonucleases) and DNA ligase.The human gene that codes for insulin can be inserted into a plasmid and then this plasmid canbe inserted into a host cell such as a bacterium. The bacterium can then synthesis insulin whichcan be collected and used by diabetics. This is done as follows. The messenger RNA whichcodes for insulin is extracted from a human pancreatic cell which produces insulin. DNA copiesare then made from this messenger RNA by using the enzyme reverse transcriptase and theseDNA copies are then given extra guanine nucleotides to the end of the gene to create stickyends. At the same time, a selected plasmid is cut using restriction enzymes which cut the DNA atspecific base sequences. Then extra cytosine nucleotides are added to create sticky ends. Oncewe have both the plasmid and the gene ready, these are mixed together. The two will link bycomplementary base pairing (between cytosine and guanine) and then DNA ligase is used tomake the sugar phosphate bonds. The plasmids with the human insulin gene (called recombinantplasmids) can then be mixed with host cells such as bacterium. The bacterium will take in theplasmid and start producing insulin which can then be collected and purified.4.4.9 State two examples of the current uses of genetically modified crops or animals. The transfer of a gene for factor IX which is a blood clotting factor, from humans to sheep so

that this factor is produced in the sheeps milk. The transfer of a gene that gives resistance to the herbicide glyphosate from bacterium to

crops so that the crop plants can be sprayed with the herbicide and not be affected by it.4.4.10 Discuss the potential benefits and possible harmful effects of one example ofgenetic modification.It is quite common to see genetic modifications in crop plants. An example of this is the transfer ofa gene that codes for a protein called Bt toxin from the bacterium Bacillus thuringiensisto maizecrops. This is done because maize crops are often destroyed by insects that eat the corn and soby adding the Bt toxin gene this is prevented as the toxin kills the insects. However this is verycontroversial as even though it has many positive advantages, it can also have some harmfulconsequences. The table below summarizes the benefits and possible harmful effects ofgenetically modifying the maize crops.


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Benefits Harmful Effects

Since there is less damage to themaize crops, there is a higher cropyield which can lessen foodshortages.

We are not sure of theconsequences of humans andanimals eating the modified crops.The bacterial DNA or the Bt toxin

itself could be harmful to human aswell as animal health.

Since there is a higher crop yield,less land is needed to grow morecrops. Instead the land canbecome an area for wild lifeconservation.

Other insects which are notharmful to the crops could bekilled. The maize pollen willcontain the toxin and so if it isblown onto near by plants it can killthe insects feeding on theseplants.

There is a reduction in the use ofpesticides which are expensive

and may be harmful to theenvironment, wild life and farmworkers.

Cross pollination can occur whichresults in some wild plants being

genetically modified as they willcontain the Bt gene. These plantswill have an advantage over othersas they will be resistant to certaininsects and so some plants maybecome endangered. This willhave significant consequences onthe population of wild plants.

4.4.11Define clone.Clone: a group of genetically identical organisms or a group of genetically identical cells derivedfrom a single parent cell.

4.4.12 Outline a technique for cloning using differentiated animal cells.Dolly the sheep was cloned by taking udder cells from a donor sheep. These cells were thancultured in a low nutrient medium to make the genes switch off and become dormant. Then anunfertilized egg was taken from another sheep. The nucleus of this egg cell was removed byusing a micropipette and then the egg cells were fused with the udder cells using a pulse ofelectricity. The fused cells developed like normal zygotes and became embryos. These embryoswere then implanted into another sheep whos role was to be the surrogate mother. One lambwas born successfully and called Dolly. Dolly was genetically identical to the sheep from whichthe udder cells were taken.4.4.13 Discuss the ethical issues of therapeutic cloning in humans.There are many ethical issues involving therapeutic cloning in humans. Below is a tablesummarizing the arguments for and against therapeutic cloning in humans.

Arguments for Arguments against

Embryonic stem cells can beused for therapies that save

lives and reduce pain forpatients. Since a stem cell candivide and differentiate into anycell type, they can be used to

replace tissues or organsrequired by patients.

Every human embryo is apotential human being and

should be given the chance ofdeveloping.


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Cells can be taken fromembryos that have stopped

developing and so these cellswould have died anyway.

More embryos are generallyproduced than are needed and

so many are killed.

Cells are taken at a stage whenthe embryos have no nerve cells

and so they cannot feel pain.

There is a risk of embryonicstem cells developing into

tumour cells.

Communities and ecosystems5.1.1 Define species, habitats, populations, community,

ecosystems and ecology.Species: a group of organisms that can interbreed and produce fertile offspring.Habitat: the environment in which a species normally lives or the location of a living organism.Population: a group of organisms of the same species who live in the same area at the sametime.Community: a group of populations living and interacting with each other in an area. Ecosystem: a community and its abiotic environment.Ecology: the study of relationships between living organisms and between organisms and theirenvironment.5.1.2 Distinguish between autotroph and heterotroph.Autotrophs are organisms that synthesize their organic molecules from simple inorganicsubstances whereas heterotrophs are organisms that obtain organic molecules from otherorganisms.

5.1.3 Distinguish between consumers, detritivores and saprotrophs.Consumer: an organism that ingests other organic matter that is living or recently killed.Detritivore: an organism that ingests non-living organic matter.Saprotroph: an organism that lives on or in non-living organic matter, secreting digestiveenzymes into it and absorbing the products of digestion.5.1.4 Describe what is meant by a food chain, giving three examples, each with at leastthree linkages (four organisms).A food chain shows the direction of energy flow from one species to another. For example, anarrow from A to B means that A is being eaten by B and therefore indicates the direction of theenergy flow.


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Figure 5.1.1 - Example of a food chain

5.1.5 Describe what is meant by a food web.A food web is a diagram that shows all the feeding relationships in a community with arrowswhich show the direction of the energy flow.5.1.6 Define trophic level.Trophic level: the trophic level of an organism is its position in the food chain. Producers, primaryconsumers, secondary consumers and tertiary consumers are examples of trophic levels.5.1.7 Deduce the trophic level of organisms in a food chain and a food web.Plants or any other photosynthetic organisms are the producers. Primary consumers are thespecies that eat the producers. Secondary consumers are the species that eat the primaryconsumers and tertiary consumers in turn eat the secondary consumers.5.1.8 Construct a food web containing up to 10 organisms, using appropriate information.


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Figure 5.1.2 - Food web

5.1.9 State that light is the initial energy source for almost all communities.Light is the initial energy source for almost all communities.5.1.10 Explain the energy flow in a food chain.

Producers receive their energy from light energy (the sun) by means of photosynthesis. After this,the energy in organic matter flows from producers to primary consumers to secondary consumersto tertiary consumers. This is because producers will be eaten by primary consumers which inturn will be eaten by secondary consumers and so on. However, between these trophic levels,energy is always lost. All of the trophic levels lose energy as heat through cell respiration. Also,as the organic matter passes from one trophic level to the next, not all of it is digested and so wehave loss of energy in organic matter through feces. This energy then passes on to thedetritivores and saprotrophs. Another energy loss occurs through tissue loss and death which canhappen at any trophic level. Once again, this energy would be passed on to detritivores andsaprotrophs as they digest these. Detritivores and saprotrophs in turn lose energy as heatthrough cell respiration.

Summary:

1. Energy flows from producers to primary consumers, to secondary consumers, to tertiaryconsumers...

2. Energy is lost between trophic levels in the form of heat through cell respiration, faeces,tissue loss and death.

3. Some of this lost energy is used by detritivores and saprotrophs. These in turn also loseenergy in the form of heat through cell respiration.

5.1.11 State that energy transformations are never 100% efficient.Energy transformations are never 100% efficient.5.1.12 Explain that energy enters and leaves ecosystems, but nutrients must be recycled.Energy is not recycled. It is constantly being supplied to ecosystems through light energy and

then flows through the trophic levels. As it flows through the trophic levels energy is lost in feces,tissue loss and death. This energy from these losses is passed on to detritivores andsaprotrophs. However the energy is then lost from the ecosystem as the remaining energy in thetrophic levels and the energy in the saprotrophs and detritivores is lost through cell respiration inthe form of heat. As a result, energy needs to be constantly supplied to the ecosystems. Nutrientson the other hand are different as they constantly have to be recycled. Carbon, nitrogen andphosphorus are all examples of nutrients. There is only a limited supply of these as they are notresupplied to the ecosystems like energy. Therefor they have to be recycled over and over. Theyare absorbed from the environment, used by living organisms and then returned to theenvironment.

Summary:

1. Energy is not recycled. Constantly being supplied to the ecosystem through light energy.2. Energy is lost from the ecosystem in the form of heat through cell respiration.3. Nutrients must be recycled as there is only a limited supply of them.4. They are absorbed by the environment, used by organisms and then returned to the

environment.5.1.12 State that saprotrophic bacteria and fungi (decomposers) recycle nutrients.Saprotrophic bacteria and fungi (decomposers) recycle nutrients.


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5.2.1 Draw and label a diagram of the carbon cycle to show the processes involved.

5.2.2 Analyse the changes in concentration of atmospheric carbon dioxide using historicalrecords.See video (coming soon)

5.2.3 Explain the relationship between rises in concentrations of atmospheric carbondioxide, methane and oxides of nitrogen and the enhanced greenhouse effect.The earths mean average temperature is regulated by a steady equilibrium which exists betweenthe energy reaching the earth from the sun and the energy reflected by the earth back into space.The incoming radiation is short wave ultraviolet and visible radiation. Some of the radiation will beabsorbed by the atmosphere and some of it will be reflected back from the earths surface intospace. The radiation that is reflected back into space is infrared radiation which has a longerwavelength. Green house gases such as carbon dioxide, methane, and oxides of nitrogen tend toabsorb some of the reflected infrared radiation and re-reflect it back towards the earth. This iswhat causes the greenhouse effect and it results in an increase in average mean temperature onearth. It is a natural phenomenon. However, since there has been an increase in the green housegases in the past century, this has resulted in an increase of the green house effect leading tohigher than normal average temperatures which could lead to disastrous consequences in thefuture.

Summary:

1. The incoming radiation from the sun is short wave ultraviolet and visible radiation.2. Some of this radiation is absorbed by the earths atmosphere.3. Some of the radiation is reflected back into space by the earths surface.4. The radiation which is reflected back into space is infrared radiation and has a longer

wavelength.5. The greenhouse gases in the atmosphere absorbe some of this infrared radiation and re-

reflect it back towards the earth.6. This causes the green house effect and results in an increase in average mean temperatures

on earth.


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7. A rise in greenhouse gases results in an increase of the green house effect which can bedisastrous for the planet.

5.2.4 Outline the precautionary principle.The precautionary principle holds that, if the effects of a human-induced change would be verylarge, perhaps catastrophic, those responsible for the change must prove that it will not do harmbefore proceeding. This is the reverse of the normal situation, where those who are concerned

about the change would have to prove that it will do harm in order to prevent such changes goingahead.5.2.5 Evaluate the precautionary principle as a justification for strong action in response tothe threats posed by the enhanced greenhouse effect.There is strong evidence that shows that green house gases are causing global warming. This isvery worrying as global warming has so many consequences on ecosystems. If nothing is done,and the green house gases are in fact causing the enhanced green house effect, by the time werealize it, it will probably be too late and result in catastrophic consequences. So even thoughthere is no proof for global warming, the strong evidence suggesting that it is linked with anincrease in green house gases is something we can not ignore. Global warming is a globalproblem. It affects everyone. For these reasons, the precautionary principle should be followed.Anyone supporting the notion that we can continue to emit same amounts or more of the green

house gases should have to provide evidence that it will not cause a damaging increase in thegreen house effect.5.2.6 Outline the consequences of a global temperature rise on arctic ecosystems.Global warming could have a number of disastrous consequences largely affecting the arcticecosystems:

The arctic ice cap may disappear as glaciers start to melt and break up into icebergs. Permafrost will melt during the summer season which will increase the rate of decomposition

of trapped organic matter, including peat and detritus. This in turn will increase the release ofcarbon dioxide which will increase the green house effect even further.

Species adapted to temperature conditions will migrate north which will alter food chains andhave consequences on the animals in the higher trophic levels.

Marine species in the arctic water may become extinct as these are very sensitive totemperature changes within the sea water.

Polar bears may face extinction as they loose their ice habitat and therefore can no longerfeed or breed as they normally would.

Pests and diseases may become quite common with rises in temperature. As the ice melts, sea levels will rise and flood low lying areas of land. Extreme weather events such as storms might become common and have disastrous effects

on certain species.

Populations

5.3.1 Outline how population size is affected by natality,immigration, mortality and emigration.Natality: increases population size as offspring are added to the population. Immigration: increases population size as individuals have moved into the area from somewhereelse and so this adds to the population.Mortality: decreases the population as some individuals get eaten, die of old age or get sick. Emigration: decreases the population as individuals have moved out of the area to go livesomewhere else.

5.3.2 Draw and label a graph showing a sigmoid (S-shaped)population growth curve.


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5.3.3 Explain the reasons for the exponential growth phase, theplateau phase and the transitional phase between these twophases.The sigmoid graph showing the population growth of a species hasthree phases which are; the exponential phase, the transitional phaseand the plateau phase. At the start of the sigmoid curve we can seethe exponential phase. This is where there is a rapid increase inpopulation growth as natality rate exceeds mortality rate. The reasonfor this is because there are abundant resources available such as

food for all members of the population and diseases as well aspredators are rare. As time passes, the population reaches thetransitional phase. This is where the natality rate starts to fall and/orthe mortality rate starts to rise. It is the result of a decrease in theabundance of resources, and an increase in the number of predatorsand diseases. However, even though population growth hasdecreased compared to the exponential phase, it is still increasing asnatality rate still exceeds mortality rate. Finally, the population reachesthe plateau phase. Here, the population size is constant so no more

growth is occurring. This is the result of natality rate being equal tomortality rate and is caused by resources becoming scarce as well as


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an increase in predators, diseases and parasites. These are thelimiting factors to the population growth. If natality rate starts to dropthen mortality rate will drop too as more resources become available.As natality rate starts to increase again so does mortality rate as

resources become scarce. This keeps the population numberrelatively stable. If a population is limited by a shortage of resourcesthen we say that it has reached the carrying capacity of theenvironment.Summary:Exponential phase:1. Rapid increase in population growth.2. Natality rate exceeds mortality rate.3. Abundant resources available. (food, water, shelter)4. Diseases and predators are rare.

Traditional phase:1. Natality rate starts to fall and/or mortality rate starts to rise.2. There is a decrease in the number of resources.3. An increase in the number of predators and diseases.4. Population still increasing but at a slower rate.

Plateau phase:1. No more population growth, population size is constant.2. Natality rate is equal to mortality rate.3. The population has reached the carrying capacity of the environment.4. The limited resources and the common predators and diseases keep the population numbers

constant.

5.3.4 List three factors that set limits to population increase.

1. Shortage of resources (e.g. food)2. Increase in predators3. Increase in diseases and parasites

Evolution5.4.1 Define evolutionEvolution is the cumulative change in the heritable characteristics of apopulation.5.4.2 Outline the evidence for evolution provided by the fossilrecord, selective breeding of domesticated animals andhomologous structures.Fossils, selective breeding and homologous structures have providedscientists with evidence that support the theory of evolution. As theystarted to study fossils they realised that these were not identical but

had similarities with existing organisms. This suggested thatorganisms changed over time. Selective breeding of domesticated


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animals also provides this evidence as the domestic breeds havesimilar characteristics to the wild ones and can still breed with them.As selected wild individuals with desirable characteristics were bred,over time this resulted in a more desirable species from a human point

of view. An example of this is the taming of wild wolves and theirselective breeding in order to produce the domestic dogs we knowtoday. This suggests that not only have these animals evolved butalso that they can evolve rapidly. Finally scientists have found anumber of homologous structures within different species. Manybones in the limbs are common to a number of species and thereforesuggests that these have evolved from one common ancestor.5.4.3 State that populations tend to produce more offspring thanthe environment can support.

Populations tend to produce more offspring than the environment cansupport.5.4.4 Explain that the consequence of the potential overproduction of offspring is a struggle for survival.If the mortality rate remains lower than the natality rate then apopulation will keep growing. As more offspring are produced, therewill be less resources available to other members of the population. Ifthere is an over production of offspring this will result in a struggle forsurvival within the species as the resources become scarce and

individuals in the population will start to compete for these. This resultsin an increase in mortality rate as the weaker individuals in thepopulation will lose out on these vital resources that are essential fortheir survival.5.4.5 State that members of a species show variation.Members of a species show variation.5.4.6 Explain how sexual reproduction promotes variation in aspecies.Sexual reproduction is important for promoting variation as even

though mutations form new genes or alleles, sexual reproductionforms a new combination of alleles. There are two stages in sexualreproduction that promote variation in a species. The first one is duringmeiosis during which a large variety of genetically different gametesare produced by each individual. The second stage is fertilisation.Here, alleles from two different individuals are brought together to formone new individual.5.4.7 Explain how natural selection leads to evolution.Individuals in a population differ from each other. Some individuals will

have characteristics that make them well adapted to their environment


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whereas others will have characteristics that make them less adaptedto their environment. The better adapted individuals are the ones thatare more likely to survive and produce offspring while the less adaptedones are more likely to die. This is called natural selection. Natural

selection results in the better adapted individuals to pass on theircharacteristics to more offspring as the lesser adapted ones are morelikely to die before they reproduce. Over time, this result accumulatesand a new generation is created with the favourable characteristicsthat makes this species better adapted to its environment. Naturalselection has lead to the species evolving.5.4.8 Explain two examples of evolution in response toenvironmental change; one must be antibiotic resistance inbacteria.

Antibiotic resistance in bacteria is a common problem. It results fromthe transfer of a gene that gives resistance to a specific antibioticusually by means of a plasmid to a bacterium. Some bacteria will thenhave this gene and become resistant to the specific antibiotic whileothers will lack the gene and so will die if exposed to the antibiotic.Over time, the non-resistant ones will all die off as doctors vaccinatepatients, but the resistant ones will survive. Eventually, the resistantones will be the only ones left as a result of natural selection and so anew antibiotic must be created. However, this has to be done on a

regular basis as the bacteria keep evolving and become resistant tomultiple antibiotics.The Peppered Moth is another example of evolution in response toenvironmental change. There are two types of these moths, onespecies has a light colour while the other one is darker. When Britainbegun industrialising, the soot from the factories would land on tre

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Biology SL IB Assessment