1 - Statistical AnalysisState that error bars are a graphical representation of the variability of dataError bars show the spread of measurements around a central tendency (e.g. mean) on a graphError bars will usually show either:The range of data (all the samples)The standard deviation (68% of the sample)The 95% confidence intervals (CI)

Calculate the mean and standard deviation of a set of valuesData can be measured in one of three ways:

Mean:The average of a group of data entriesCalculation method = sum of data entries / total number of data entries

Standard Deviation:Measures the variability (spread) in a set of data that is normally distributed (Gaussian)In order to ensure normal distribution of the standard deviation, certain outliers may need to be excluded from the data set (with justification)

State that the term standard deviation is used to summarise the spread of values around the mean, and that 68% of the values fall within onestandard deviation of the meanData which is normally distributed will exhibit a bell-shaped (Gaussian) curve which is symmetrical around a central mean68% of data values will fall within one standard deviation of the mean95% of data values will fall within two standard deviations of the mean99.8% of data values will fall within three standard deviations of the mean

Explain how the standard deviation is useful for comparing the means and spread of data between two or more samplesData sets that have the same mean may not have the same degree of variation in dataAs standard deviation measures the spread of data, it can be used to compare data setsThe diagram below shows three sets of data with identical means but different spreads of dataA higher standard deviation means there is greater variation in the data set, whereas a lower standard deviation means there is less

Graphical Representations of Data Spread

Deduce the significance of the difference between two sets of data using calculated values for t and the appropriate tablesA t-test is a statistical test used to compare two means (e.g. between a control group and an experimental group)It can be used to compare a null hypothesis against an alternative hypothesis:Null hypothesis:There is no significant difference between the meansAlternative hypothesis:There is a significant difference between two meansA t-test can be either one-tailed (difference only) or two-tailed (difference and direction)

Conducting a t-test:A t-test is a statistical formula that calculates at valuebased on a set of dataIf there is a significant difference between the two means, then the t value will be greater than a determined value for a given degree of probabilityScientists generally consider data to be statistically significant when there is less than a 5% probability the results are due to chanceThis 95% confidence level is a probability value ofp allele B > allele C)

4.3.4 Describe ABO blood groups as an example of codominance and multiple allelesWhen assigning alleles for codominance, the convention is to use a common letter to represent dominant and recessive and use superscripts to represent the different codominant allelesI stands for immunoglobulin (antigenic protein on blood cells)A and B stand for the codominant variants

The ABO gene has three alleles: IA, IBand iIAand IBare codominant, wherease i is recessive (no antigenic protein is produced)Codominance means that both IAand IBalleles will be expressed within a given phenotype

The genotypes and phenotypes of the ABO blood groups are:

The ABO Blood Group System

4.3.5 Explain how sex chromosomes control gender by referring to the inheritance of X and Y chromosomes in humansHumans have 23 pairs of chromosomes for a total of 46 (excluding instances of aneuploidy)The first 22 pairs are autosomes - each chromosome pair possesses the same genes and structural featuresThe 23rd pair of chromosomes are heterosomes (or sex chromosomes) and determine genderFemales are XX - they possess two X chromosomesMales are XY - they posses one X chromosome and a much shorter Y chromosome

The Y chromosome contains the genes for developing male sex characteristic - hence the father is always responsible for determining genderIf the male sperm contains the X chromosome the growing embryo will develop into a girlIf the male sperm contains a Y chromosome the growing embryo will develop into a boyIn all cases the female egg will contain an X chromosome (as the mother is XX)

Because the X and Y chromosomes are of a different size, they cannot undergo crossing over / recombination during meiosisThis ensures that the gene responsible for gender always remains on the Y chromosome, meaning that there is always ~ 50% chance of a boy or girl

4.3.6 State that some genes are present on the X chromosome and absent from the shorter Y chromosome

The Y chromosome is much shorter than the X chromosome and contains only a few genesIncludes the SRY sex-determination gene and a few others (e.g. hairy ears gene)

The X chromosome is much longer and contains several genes not present on the Y chromosomeIncludes the genes for haemophilia and red-green colour blindness

In human females, only one of the X chromosomes remains active throughout lifeThe other is packaged as heterochromatin to form a condensed Barr bodyThis inactivation is random and individual to each cell, so heterozygous women will be a mosaic - expressing both alleles via different cells

4.3.6 Define sex linkageSex linkage refers to when a gene controlling a characteristic is found on a sex chromosome (and so we associate the trait with a predominant gender)Sex-linked conditions are usually X-linked, as very few genes exist on the shorter Y chromosome

4.3.7 Describe the inheritance of colour blindness and haemophilia as examples of sex linkageColour blindness and haemophilia are both examples of X-linked recessive conditionsThe gene loci for these conditions are found on the non-homologous region of the X chromosome (they are not present of the Y chromosome)As males only have one allele for this gene they cannot be a carrier for the conditionThis means they have a higher frequency of being recessive and expressing the traitMales will always inherit an X-linked recessive condition from their motherFemales will only inherit an X-linked recessive condition if they receive a recessive allele from both parents

When assigning alleles for sex-linked traits the convention is to write the allele as a superscript to the sex chomosome (usually X)Haemophilia:XH= unaffected ; Xh= affectedColour Blindness:XA= unaffected ; Xa= affected

Male and Female Genotypes for a Sex-Linked Condition

4.3.8 State that a human female can be homozygous or heterozygous with respect to sex-linked genesAs human females have two X chromosomes (and therefore two alleles for any given X-linked gene), they can be either homozygous or heterozygousMales only have one X chromosome (and therefore only one allele) and are hemizygous

4.3.9 Explain that female carriers are heterozygous for X-linked recessive allelesAn individual with a recessive allele for a disease condition that is masked by a normal dominant allele is said to be a carrierCarriers are heterozygous and can potentially pass the trait on to the next generation, but do not suffer from the defective condition themselvesFemales can be carriers for X-linked recessive conditions because they have two X chromosomes - males (XY) cannot be carriersBecause a male only inherits an X chromosome from his mother, his chances of inheriting the disease condition from a carrier mother is greater

4.3.10 Predict the genotypic and phenotypic ratios of offspring of monohybrid crosses involving any of the above patterns of inheritanceAutosomal Dominance / RecessiveChoose a letter where the upper and lower case forms are easily distinguishable (e.g. E/e, A/a, B/b)Use the capital letter for the dominant allele and the lower case letter for the recessive alleleExample:

CodominanceChoose a letter to denote the general trait encoded by the gene (capital = dominant, lower case = recessive)Use different superscript letters (capitals) to represent the different codominant allelesExample:

X-linked RecessiveUse a capital "X" to denote the X chromosomeChoose a superscript letter to represent the trait (capital = dominant, lower case = recessive)Example:

4.3.11 Deduce the genotype and phenotype of individuals in pedigree chartsA pedigree is a chart of the genetic history of a family over several generationsMales are represented as squares, while females are represented as circlesShaded symbols means an individual is affected by a condition, while an unshaded symbol means they are unaffectedA horizontal line between a man and woman represents mating and resulting children are shown as offshoots to this line

Autosomal DominanceAll affected individualsmusthave at least one affected parentIf two parents are unaffected, all offspringmustbe unaffected (homozygous recessive)If two parents are affected, theymayhave offspring who are unaffected (if parents are heterozygous)

Autosomal RecessiveIf two parents show a trait, all childrenmustalso show the trait (homozygous recessive)An affected individualmayhave two normal parents (if parents are both heterozygous carriers)

X-Linked RecessiveIf a female shows the trait, so must all sons as well as her fatherThe disorder is more common in males

Identifying Modes of Inheritance

4.4 - Genetic Engineering and Biotechnology

4.4.1 Outline the use of polymerase chain reaction (PCR) to copy and amplify minute quantities of DNA

PCR is a way of producing large quantites of a specific target sequence of DNAIt is useful when only a small amount of DNA is avaliable for testingE.g. crime scene samples of blood, semen, tissue, hair, etc.

PCR occurs in a thermal cycler and involves a repeat procedure of 3 steps:Denaturation:DNA sample is heated to separate it into two strandsAnnealing:DNA primers attach to opposite ends of the target sequenceElongation:A heat-tolerant DNA polymerase (Taq) copies the strands

One cycle of PCR yields two identical copies of the DNA sequenceA standard reaction of 30 cycles would yield 1,073,741,826 copies of DNA (230)

4.4.2 State that, in gel electrophoresis, fragments of DNA can move in an electric field and are separated according to their sizeGel electrophoresis is a technique which is used to separate fragments of DNA according to size

Samples of fragmented DNA are placed in the wells of an agarose gelThe gel is placed in a buffering solution and an electrical current is passed across the gelDNA, being negatively charged (due to phosphate), moves to the positive terminus (anode)Smaller fragments are less impeded by the gel matrix and move faster through the gelThe fragments are thus separated according to sizeSize can be calculated (in kilobases) by comparing against a known industry standard

4.4.3 State that gel electrophoresis of DNA is used in DNA profilingDNA profiling is a technique by which individuals are identified on the basis of their respective DNA profilesWithin the non-coding region of an individual's genome, there exists satellite DNA - long stretches of DNA made up of repeating elements called short tandem repeats (STRs)These repeating sequences can be excised to form fragments, by cutting with a variety of restriction endonucleases (which cut DNA at specific sites)As individuals all have a different number of repeats in a given sequence of satellite DNA, they will all generate unique fragment profilesThese different profiles can be compared using gel electrophoresis

DNA Profiling Using STR Analysis

4.4.4 Describe the application of DNA profiling to determine paternity and also in forensic investigationA DNA sample is collected (blood, saliva, semen, etc.) and amplified using PCRSatellite DNA (non-coding) is cut with specific restriction enzymes to generate fragmentsIndividuals will have unique fragment lengths due to the variable length of their short tandem repeats (STR)The fragments are separated with gel electrophoresis (smaller fragments move quicker through the gel)The DNA profile can then be analysed according to need

Two applications of DNA profiling are:Paternity testing (comparing DNA of offspring against potential fathers)Forensic investigations (identifying suspects or victims based on crime-scene DNA)

4.4.5 Analyse DNA profiles to draw conclusions about paternity or forensic investigationsPaternity Testing: Children inherit half of their alleles from each parent and thus should possess a combination of their parents allelesForensic Investigation: Suspect DNA should be a complete match with the sample taken from a crime scene if a conviction is to occur

Paternity TestForensic Investigation

4.4.6 Outline three outcomes of the sequencing of the complete human genomeThe Human Genome Project (HGP) was an international cooperative venture established to sequence the 3 billion base pair (~25,000 genes) in the human genomeThe outcomes of this project include:Mapping:We now know the number, location and basic sequence of human genesScreening:This has allowed for the production of specific gene probes to detect sufferers and carriers of genetic disease conditionsMedicine:With the discovery of new proteins and their functions, we can develop improved treatments (pharmacogenetics and rational drug design)Ancestry:It will give us improved insight into the origins, evolution and historical migratory patterns of humans

With the completion of the Human Genome Project in 2003, researcher have begun to sequence the genomes of several non-human organisms

4.4.7 State that, when genes are transferred between species, the amino acid sequence of polypeptides translated from them is unchanged because thegenetic code is universalThe genetic code is universal, meaning that for every living organism the same codons code for the same amino acids (there are a few rare exceptions)This means that the genetic information from one organism could be translated by another (i.e. it is theoretically transferable)

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 ligaseDNA ExtractionA plasmid is removed from a bacterial cell (plasmids are small, circular DNA molecules that can exist and replicate autonomously)A gene of interest is removed from an organism's genome using a restriction endonuclease which cut at specific sequences of DNAThe gene of interest and plasmid are both amplified using PCR technology

Digestion and LigationThe plasmid is cut with the same restriction enzyme that was used to excise the gene of interestCutting with certain restriction enzymes may generate short sequence overhangs ("sticky ends") that allow the the two DNA constructs to fit togetherThe gene of interest and plasmid are spliced together by DNA ligase creating a recombinant plasmid

Transfection and ExpressionThe recombinant plasmid is inserted into the desired host cells (this is called transfection for eukaryotic cells and transformation for prokaryotic cells)The transgenic cells will hopefully produce the desired trait encoded by the gene of interest (expression)The product may need to subsequently be isolated from the host and purified in order to generate sufficient yield

Treating Haemophilia via the Isolation of Human Factor IX Clotting Protein from Transgenic Sheep Milk

4.4.9 State two examples of current uses of genetically modified crops or animalsCropsEngineering crops to extend shelf life of fresh produceTomatoes (Flavr Savr) have been engineered to have an extended keeping quality by switching off the gene for ripening and thus delaying the natural process of softening of fruitEngineering of crops to provide protection from insectsMaize crops (Bt corn) have been engineered to be toxic to the corn borer by introducing a toxin gene from a bacterium (Bacillus thuringiensis)

AnimalsEngineering animals to enhance productionSheep produce more wool when engineered with the gene for the enzyme responsible for the production of cysteine - the main amino acid in the keratin protein of woolEngineering animals to produce desired productsSheep engineered to produce human alpha-1-antitrypsin in their milk can be used to help treat individuals suffering from hereditary emphysema

4.4.10 Discuss the potential benefits and potential harmful effects of one example of genetic modificationExample:Maize introduced with a bacterial gene encoding a toxin to the European Corn Borer (i.e. Bt Corn)

Potential BenefitsAllows for the introduction of a characteristic that wasn't present within the gene pool (selective breeding could not have produced desired phenotype)Results in increased productivity of food production (requires less land for comparable yield)Less use of chemical pesticides, reducing the economic cost of farmingCan now grow in regions that, previously, may not have been viable (reduces need for deforestation)

Potential Harmful EffectsCould have currently unknown harmful effects (e.g. toxin may cause allergic reactions in a percentage of the population)Accidental release of transgenic organism into the environment may result in competition with native plant speciesPossibility of cross pollination (if gene crosses the species barrier and is introduced to weeds, may have a hard time controlling weed growth)Reduces genetic variation / biodiversity (corn borer may play a crucial role in local ecosystem)

4.4.11 Define cloneA clone is a group of genetically identical organisms or a group of cells derived from a single parent cell

4.4.12 Outline a technique for cloning using differentiated animal cellsSomatic Cell Nuclear Transfer (SCNT) is a method of reproductive cloning using differentiated animal cellsA female animal (e.g. sheep) is treated with hormones (such as FSH) to stimulate the development of eggsThe nucleus from an egg cell is removed (enucleated), thereby removing the genetic information from the cellThe egg cell is fused with the nucleus from a somatic (body) cell of another sheep, making the egg cell diploidAn electric shock is delivered to stimulate the egg to divide, and once this process has begun the egg is implanted into the uterus of a surrogateThe developing embryo will have the same genetic material as the sheep that contributed the diploid nucleus, and thus be a clone

Different Uses of Cloning

4.4.13 Discuss the ethical issues of therapeutic cloning in humansRefer to Topic 2.1.10 for an outline of uses for therapeutic cloning in humans

Arguments for Therapeutic CloningMay be used to cure serious diseases or disabilities with cell therapy (replacing bad cells with good ones)Stem cell research may pave the way for future discoveries and beneficial technologies that would not have occurred if their use had been bannedStem cells can be taken from embryos that have stopped developing and would have died anyway (e.g. abortions)Cells are taken at a stage when the embryo has no nervous system and can arguably feel no pain

Arguments Against Therapeutic CloningInvolves the creation and destruction of human embryos (at what point do we afford the right to life?)Embryonic stem cells are capable of continued division and may develop into cancerous cells and cause tumorsMore embryos are generally produced than are needed, so excess embryos are killedWith additional cost and effort, alternative technologies may fulfil similar roles (e.g. nuclear reprogramming of differentiated cell lines)

5.1 - Communities & Ecosystems

5.1.1 Define species, habitat, population, community, ecosystem and ecologySpecies:A group of organisms that can interbreed and produce fertile, viable offspringHabitat:The environment in which a species normally lives or the location of a living organismPopulation:A group of organisms of the same species who live in the same area at the same timeCommunity:A group of populations living and interacting with each other in an areaEcosystem:A community and its abiotic environmentEcology:The study of relationships between living organisms and between organisms and their environment

5.1.2Distinguish between autotroph and heterotrophAutotroph:An organism that synthesises its organic molecules from simple inorgance substances (e.g. CO2and nitrates) - autotrophs areproducersHeterotroph:An organism that obtains organic molecules from other organisms - heterotrophs areconsumers

5.1.3Distinguish between consumers, detritivores and saprotrophsConsumer:An organism that ingests other organic matter that is living or recently killedDetritivore:An organism that ingests non-living organic matterSaprotroph:An organism that lives on or in non-living organic matter, secreting digestive enzymes into it and absorbing the products of digestion

5.1.4Describe what is meant by a food chain, giving three examples, each with at least three linkages (four organisms)A food chain shows the linear feeding relationships between species in a communityThe arrows represent the transfer of energy and matter as one organism is eaten by another (arrows point in the direction of energy flow)The first organism in the sequence is the producer, followed by consumers (1, 2, 3, etc.)

Examples of Food Chains

5.1.5Describe what is meant by a food webA food web is a diagram that shows how food chains are linked together into more complex feeding relationships within a communityThere can be more than one producer in a food web, and consumers can occupy multiple positions (trophic levels)

5.1.6Define trophic level An organism's trophic level refers to the position it occupies in a food chainProducers always occupy the first trophic level, while saprotrophs would generally occupy the ultimate trophic level of a given food chain or food webThe trophic levels in a community are:

5.1.7Deduce the trophic levels of organisms in a food web and food chainThe trophic level of an organism can be determined by counting the number of feeding relationships preceding it and adding one (producer always first)Trophic Level = Number of arrows (in sequence) before organism + 1In food webs, a single organism may occupy multiple trophic levels

5.1.8Construct a food web containing up to 10 organisms, using appropriate informationHint:When constructing a food web, always try to position an organism relative to itshighesttrophic level (to keep all arrows pointing in same direction)

Food web (trophic levels in red)

5.1.9State that light is the initial energy source for almost all communitiesAll green plants, and some bacteria, are photo-autotrophic - they use light as a source of energy for synthesising organic moleculesThis makes light the initial source of energy for almost all communitiesSome bacteria are chemo-autotrophic and use energy derived from chemical processes (e.g. nitrogen-fixating bacteria)

5.1.10Explain the energy flow in a food chainEnergy enters most communities as light, where it is absorbed by autotrophs (e.g. plants) and converted into chemical energy via photosynthesisEnergy then gets passed to the primary consumer (herbivore) when they eat the plant, and then gets passed to successive consumers (carnivores) as they are eaten in turnOnly ~10% of energy is passed from one trophic level to the next, the rest is lostBecause ~90% of energy is lost between trophic levels, the number of trophic levels are limited as energy flow is reduced at higher levels

Summary of Energy Flow in a Food Chain

5.1.11State that energy transformations are never 100% efficientWhen energy transformations take place in living organisms the process is never 100% efficientTypically, energy transformations in living things are ~10% efficient, with about 90% of the energy lost between trophic levelsThis energy may be lost as heat, be used up during cellular respiration, be excreted in faeces or remain unconsumed as the uneaten part of food

5.1.12Explain the reason for the shape of pyramids of energy

A pyramid of energy is a graphical representation of the amount of energy of each tropic level in a food chainThey are expressed in units of energy per area per time (e.g. kJm2year-1)Pyramids of energy will never appear inverted as some of the energy stored in one source is always lost when transferred to the next sourceThis is an application of the second law of thermodynamicsEach level of the pyramid of energy should be approximately one tenth the size of the level preceding it, as energy transformations are ~10% efficient

5.1.13Explain that energy enters and leaves ecosystems, but nutrients must be recycledThe movement of energy and matter through ecosystems are related because both occur by the transfer of substances through feeding relationshipsHowever, energy cannot be recycled and an ecosystem must be powered by a continuous influx of new energy from an external source (e.g the sun)Nutrients refer to material required by an organism, and are constantly being recycled within an ecosystem as food (either living or dead)The autotrophic activities of the producers (e.g. plants) produce organic materials from inorganic sources, which are then fed on by the consumersWhen heterotrophic organisms die, these inorganic nutrients are returned to the soil to be reused by the plants (as fertiliser)Thus energy flows through ecosystems, while nutrients cycle within them

5.1.14State that saprotrophic bacteria and fungi (decomposers) recycle nutrientsIn order for organisms to grow and reproduce, they need a supply of the elements of which they are madeThe saprotrophic activity of decomposers (certain bacteria and fungi), free inorganic materials from the dead bodies and waste products of organisms, ensuring a continual supply of raw materials for the producers (which can then be ingested by consumers)Thus saprotrophic bacteria and fungi play a vital role in recycling nutrients within an ecosystem

5.2 - The Greenhouse Effect

5.2.1 Draw and label a diagram of the carbon cycle to show the processes involvedThere are four main 'pools' of carbon in the environment: Atmosphere Biosphere Sediments Ocean

There are a number of processes by which carbon can be cycled between these pools:Photosynthesis:Atmospheric carbon dioxide is removed and fixed as organic compounds (e.g. sugars)Feeding:In which organic carbon is moved from one trophic level to the next in a food chainRespiration:All organisms (including plants) metabolise organic compounds for energy, releasing carbon dioxide as a by-productFossilization:In which carbon from partially decomposed dead organisms becomes trapped in sediment as coal, oil and gas (fossil fuels)Combustion:During the burning of fossil fuels and biomassIn oceans, carbon can be reversibly trapped and stored as limestone (storage happens more readily at low temperatures)

The Carbon Cycle

5.2.2Analyse the changes in concentration of atmospheric carbon dioxide using historical recordsRecent Trends:Atmospheric carbon dioxide concentrations have been measured at the Mauna Loa atmospheric observatory in Hawaii from 1958 and has since been measured at a number of different locations globallyThe data shows that there is an annual cycle in CO2concentrations which may be attributable to seasonal factors, but when data from the two hemispheres is incorporated, it suggests that atmospheric CO2levels have risen steadily in the past 30 years

Long Term Estimates:Carbon dioxide concentration changes over a long period of time have been determined by a variety of sources, including analysing the gases trapped in ice (and thus providing a historical snapshot of atmospheric concentrations)Data taken from the Vostok ice core in Antarctica shows that fluctuating cycles of CO2concentrations over thousands of years appear to correlate with global warm ages and ice agesIt is compelling to note that CO2levels appear to be currently higher than at any time in the last 400,000 years

Recent and Long-term Changes in Carbon Dioxide Concentration

Mauna Loa CO2Data (last 50 years) Vostok Ice Core Data - CO2vs Temperature (last 400,000 years)

5.2.3Explain the relationship between the rises in concentrations of atmospheric carbon dioxide, methane and oxides of nitrogen and the enhancedgreenhouse effectThe greenhouse effect is a natural process whereby the earth's atmosphere behaves like a greenhouse to create the moderate temperatures to which life on earth has adapted (without the greenhouse effect, temperatures would drop significantly every night)The incoming radiation from the sun is short-wave ultraviolet and visible radiationSome of this radiation is reflected by the earth's surface back into space as long-wave infrared radiationGreenhouse gases absorb this infrared radiation and re-reflect it back to the earth as heat, resulting in increased temperatures (the greenhouse effect)The Greenhouse Effect

Theenhanced greenhouse effectrefers to the suggested link between the increase in greenhouse gas emissions by man and changes in global temperatures and climate conditionsThe main greenhouse gases are water vapour, carbon dioxide (CO2), methane (CH4) and oxides of nitrogen (e.g. NO2)While these gases occur naturally, man is increasing greenhouse gas emissions via a number of processes, including: Deforestation (less trees) Industrialisation (more combustion) Increased farming / agriculture (more methane)With increases in greenhouse gas emission, it is thought that the atmospheric temperature may increase and threaten the viability of certain ecosystems, although this link is still being debated

5.2.4Outline the precautionary principleThe precautionary principle states that when a human-induced activity raises a significant threat of harm to the environment or human health, then precautionary measures should be taken even if there is no scientific consensus regarding cause and effectBecause the global climate is a complex phenomena with many emergent properties, and is based on time frames well beyond human lifespans, it is arguably impossible to provide appropriate scientific evidence for enhanced global warming before consequences escalate to potentially dire levelsAccording to the precautionary principle, the onus falls on those contributing to the enhanced greenhouse effect to either reduce their input or demonstrate their actions do not cause harm - this makes it the responsibility of governments, industries, communities and even the individualThe precautionary principle is the reverse of previous historical practices whereby the burden of proof was on the individual advocating action

5.2.5Evaluate the precautionary principle as a justification for strong action in response to the threats posed by the enhanced greenhouse effectArguments for ActionRisks of inaction are potentially severe, including increased frequency of severe weather conditions (e.g. droughts, floods) and rising sea levelsHigher temperatures will increase the spread of vector-borne diseasesLoss of habitat will result in the extinction of some species, resulting in a loss of biodiversityChanges in global temperature may affect food production, resulting in famine in certain regionsThe effects of increased temperatures (e.g. rising sea levels) could destroy certain industries which countries rely on, leading to povertyAll of these consequences could place a far greater economic burden on countries than if action were taken nowThese factors would increase competition for available resources, potentially leading to increased international tensions

Arguments for InactionCutting greenhouse emissions may delay economic growth in developing countries, increasing poverty in these regionsVery difficult to police - what level of action would be considered sufficient on a global scale in the current absence of scientific consensus?Boycotting trade with non-compliant countries could negatively effect economies and create international tensionsNo guarantee that human intervention will be sufficient to alter global climate patternsMoney and industrial practices that may be used to develop future technologies may be lost due to restrictions imposed by carbon reduction schemesCarbon reduction schemes will likely result in significant job losses from key industries, retraining workers will require significant time and money

5.2.6Outline the consequences of a global temperature rise on arctic ecosystemsIncreases in global temperature pose acrediblethreat to arctic ecosystems, including:Changes in arctic conditions (reduced permafrost, diminished sea ice cover, loss of tundra to coniferous forests)Rising sea levelsExpansion of temperate species increasing competition with native species (e.g. red fox vs arctic fox)Decomposition of detritus previously trapped in ice will significantly increase greenhouse gas levels (potentially exacerbating temperature changes)Increased spread of pest species and pathogens (threatening local wildlife)Behavioural changes in native species (e.g. hibernation patterns of polar bears, migration of birds and fish, seasonal blooms of oceanic algae)Loss of habitat (e.g. early spring rains may wash away seal dens)Extinction and resultant loss of biodiversity as food chains are disrupted

5.3 - Populations

5.3.1 Outline how population size is affected by natality, immigration, mortality and emigrationThe change in population size over a given period of time can be summarised by the following equation: Population Size = (N+I) - (M+E)

Natality:Increases to population size through reproduction (i.e. births)Immigration:Increases to population size from external populationsMortality:Decreases to population size as a result of death (e.g. predation, senescence)Emigration:Decreases to population size as a result of loss to external populations

5.3.2 Draw and label a graph showing the sigmoid (S-shaped) population growth curvePopulation Growth Curve

5.3.3 Explain reasons for the exponential growth phase, the plateau phase and the transitional phase between these two phasesInitially, population growth may be slow, as there is a shortage of reproducing individuals which may be widely dispersedAs numbers increase and reproduction gets underway, three stages of population growth are seen:

Exponential Growth PhaseThere is a rapid increase in population size / growth as the natality rate exceeds the mortality rateThis is because there is abundant resources (e.g. food, shelter and water) and limited environmental resistance (disease and predation uncommon)

Transitional PhaseAs the population continues to grow, eventually competition increases as availability of resources are reducedNatality starts to fall and mortality starts to rise, leading to a slower rate of population increase

Plateau PhaseEventually the increasing mortality rate equals the natality rate and population size becomes constantThe population has reached the carrying capacity (K) of the environmentLimited resources, predation and disease all contribute to keeping the population size balancedWhile the population size at this point may not be static, it will oscillate around the carrying capacity to remain relatively even (no net growth)

5.3.4 List three factors that sets limits to population increaseEvery species has limits to the environmental conditions it can endure and must remain within appropriate levels for population growth to occurSome of these factors are density-dependent, while others are unrelated to the density of the population

Factors affecting population growth:

5.4 - Evolution

5.4.1 Define evolutionEvolution is the cumulative change in the heritable characteristics of a population

5.4.2 Outline the evidence for evolution provided by the fossil record, selective breeding of domesticated animals and homologous structuresSomething provides evidence for evolution when it demonstrates a change in characteristics from an ancestral form

The Fossil Record

A fossil is the preserved remains or traces of any organism from the remote pastFossil evidence may be either:Direct (body fossils): Bones, teeth, shells, leaves, etc.Indirect (trace fossils): Footprints, tooth marks, tracks, burrows, etc.

The totality of fossils (both discovered and undiscovered) is known as thefossil recordThe fossil record reveals that, over time, changes have occurred in features of organisms living on the planet (evolution)Moreover, different kinds of organisms do not occur randomly but are found in rocks of particular ages in a consistent order (law of fossil succession)This suggests that changes to an ancestral species was likely responsible for the appearance of subsequent species (speciationvia evolution)Furthermore, the occurrence oftransitional fossilsdemonstrate the intermediary forms that occurred over the evolutionary pathway taken within a single genus

While fossils may provide clues regarding evolutionary processes and ancestral relationships, it is important to realise that the fossil record is incompleteFossilization requires a unusual combination of specific circumstances to occur, meaning there are many gaps in the fossil recordOnly the hard parts of an organism are preserved and often only fragments of fossilized remains are discovered

Selective Breeding

Selective breeding of domesticated animals is an example of artificial selection, which occurs when man directly intervenes in the breeding of animals to produce desired traits in offspringAs a result of many generations of selective breeding, domesticated breeds can show significant variation compared to the wild counterparts, demonstrating evolutionary changes in a much shorter time frame than might have occurred naturallyExamples of selective breeding include:Breeding horses for speed (race horses) versus strength and endurance (draft horses)Breeding dogs for herding (sheepdogs), hunting (beagles) or racing (greyhounds)Breeding cattle for increased meat production or milkBreeding zebras in an attempt to retrieve the colouration gene from the extinct Quagga

Homologous StructuresComparative anatomy of groups of animals or plants shows certain structural features are basically similar, implying a common ancestryHomologous structures are those that are similar in shape in different types of organisms despite being used in different waysAn example is the pentadactyl limb structure in vertebrates, whereby many animals show a common bone composition, despite the limb being used for different forms of locomotion (e.g. whale fin for swimming, bat wing for flying, human hand for manipulating tools, horse hoof for galloping, etc.)This illustrates adaptive radiation (divergent evolution) as a similar basic plan has been adapted to suit various environmental nichesThe more similar the homologous structures between two species are, the more closely related they are likely to be

Homologous Structures (Pentadactyl Limb)

5.4.3 State that populations tend to produce more offspring than the environment can supportThe Malthusian dilemma states that populations tend to multiply geometrically, while food sources multiply arithmeticallyHence populations tend to produce more offspring than the environment can support

5.4.4 Explain that the consequence of the potential overproduction of offspring is a struggle for survival

When there is an abundance of resources, a population can achieve a J-curve maximum growth rate (biotic potential)However, with more offspring there will be less resources available to other members of the population (environmental resistance)This will lead to competition for available resources and a struggle for survivalIntraspecific competition occurs when members of the same species compete for the same resources in an ecosystem (e.g. light, food, water)It is density dependent, as the available resources must be shared among members of the speciesCompetition that occurs between different species for resources is interspecificThe result of this competition will be an increase in the mortality rate, leading to an S-curve growth rate as the population approaches the carrying capacity (K)

5.4.5 State that members of a species show variationMembers of a species show variation, which can manifest itself in one of two forms:Discontinuous variation: A type of variation usually controlled by a single gene, which leads to distinct classes (e.g. ABO blood group in humans)Continuous variation: A type of variation controlled by many genes, which leads to a range of characteristics (e.g. skin pigmentation in humans)

There are three primary sources of variation within a given populationGene mutations (a permanent change to the genetic composition of an individual)Gene flow (the movement of genes from one population to another via immigration and emigration)Sexual reproduction (the combination of genetic materials from two parental sources)

5.4.6 Explain how reproduction promotes variation within a speciesThere are three primary ways by which sexual reproduction promotes variation within a species:

Independent AssortmentDuring metaphase I, when homologous chromosomes line up at the equator, the paired chromosomes can randomly arrange themselves in one of two orientations (paternal left / maternal rightOR maternal left / paternal right)When the chromosomes separate in anaphase I, the final gametes will differ depending on whether they got the maternal or paternal chromosomeIndependent assortment of chromosomes creates 2ndifferent gamete combinations (n = haploid number of chromosomes)

Crossing OverDuring prophase I, when homologous chromosomes pair up as bivalents, genetic information can be exchanged between non-sister chromatidsThe further apart two genes are on a chromosome, the more likely they are to recombineCrossing over greatly increases the number of potential gamete variations by creating new genetic combinations

Random FertilisationFertilisation results from the fusion of gametes from a paternal and maternal source, resulting in offspring that have a combination of paternal and maternal traitsBecause fertilisation is random, offspring will receive different combinations of traits every time, resulting in near infinite genetic variability

5.4.7 Explain how natural selection leads to evolutionThe theory of natural selection was postulated by Charles Darwin (and also independently by Alfred Wallace) who described it as 'survival of the fittest'There is genetic variation within a population (which can be inherited)There is competition for survival (populations tend to produce more offspring than the environment can support)Environmental selective pressures lead to differential reproductionOrganisms with beneficial adaptations will be more suited to their environment and more likely to survive to reproduce and pass on their genesOver generations there will be a change in allele frequency within a population (evolution)

5.4.8 Explain two examples of evolution in response to environmental change; one must be antibiotic resistance in bacteriaExample 1:Staphylococcus aureus(associated with a variety of conditions, including skin and lung infections)Variation: Antibiotic resistance (some strains have a drug-resistant gene ; other strains do not)Environmental change:Exposure to antibiotic (methicillin)Response: Methicillin-susceptibleS. aureus(MSSA) die, whereas methicillin-resistantS. aureus(MRSA) survive and can pass on their genesEvolution:Over time, the frequency of antibiotic resistance in the population increases (drug-resistant gene can also be transferred by conjugation)

Example 2:Peppered Moth (Biston betularia)Variation:Colouration (some moth have a light colour, while others are a darker melanic colour)Environmental change: Pollution from industrial activities caused trees to blacken with soot during the Industrial RevolutionResponse:Light coloured moths died from predation, whereas melanic moths were camouflaged and survived to pass on their genesEvolution:Over time, the frequency of the melanic form increased (with improved industrial practices, the lighter variant has become more common)

5.5 - ClassificationWednesday, July 30, 201411:46 AM5.5.1 Outline the binomial system of nomenclatureThe binomial system of nomenclature was devised by Carolus Linnaeus as a way of classifying organisms that was globally recognised and could demonstrate evolutionary relationships between organisms (and thus allow for the prediction of features closely related organisms may share)According to the binomial system of nomenclature, every organism is designated a scientific name with two parts:Genus is written first and is capitalised (e.g.Homo)Species follows and is written in lower case (e.g.Homo sapiens)Some species may also have a sub-species designation (e.g.Homo sapiens sapiens)Conventions:When typing, the name should be initalics; whereas when hand writing, it should beunderlined

5.5.2 List the seven levels in the hierarchy of taxa - kingdom, phylum, class, order, family, genus and species - using an example from two different kingdoms for each levelWhen classifying living things, organisms are grouped according to a series of hierarchical taxa - the more similar their characteristics, the closer the groupingClassification of Animals and Plants

5.5.3 Distinguish between the following phyla of plants, using simple external recognition features: bryophyta, filicinophyta, coniferophyta and angiospermophyta

5.5.4 Distinguish between the following phyla of animals, using simple external recognition features: porifera, cnidaria, platyhemlnthes, annelida, mollusca and arthropoda

5.5.5 Apply and design a key for up to eight organismsA dichotomous key is a method of identification whereby a group of organisms are sequentially divided into two categories until all are identified

Example of a Dichotomous Key:Organism is a plant ...................................................................................... Go to Q2Organism is not a plant (animal) ................................................................ Go to Q5Has no 'true' leaves or roots ....................................................................... BryophytaHas leaves and roots ................................................................................... Go to Q3Has no seeds (sporangia) .......................................................................... FilicinophytaHas seeds ..................................................................................................... Go to Q4Has no flowers ............................................................................................. ConiferophytaHas flowers ................................................................................................... AngiospermophytaAsymmetrical body plan ............................................................................. PoriferaSymmetrical body plan ............................................................................... Go to Q6Has radial symmetry ................................................................................... CnidariaHas bilateral symmetry ............................................................................... Go to Q7Has no anus ................................................................................................. PlatyhelminthesHas an anus ................................................................................................. Go to Q8Has a segmented body .............................................................................. Go to Q9Has no visible body segmentation ........................................................... MolluscaHave an exoskeleton ................................................................................. ArthropodaHave no exoskeleton ................................................................................. Annelida

Dichotomous Key as a Flowchart

7.1 - DNA Structure

7.1.1 Describe the structure of DNA, including the antiparallel strands, 3' - 5' linkages and hydrogen bonding between purines and pyrimidinesThe carbon atoms in deoxyribose are numbered, with the nitrogenous bases attach to C1and the phopshate group is attached to C5Nucleotides are joined by a covalent phosphodiester bond between the C5phosphate group and the C3hydroxyl groupHence one nucleotide strand runs 5' - 3'The nitrogenous bases interact via hydrogen bonding (complementary base pairing)Adenine (A) and thymine (T) share 2 hydrogen bondsGuanine (G) and cytosine (C) share 3 hydrogen bondsIn order for the bases to associate (i.e. face each other), one strand must run antiparallel to the other (this antiparallel strand runs 3' - 5')Double stranded DNA forms a double helix, with 10 nucleotides per turn and the structure containing both major and minor grooves

Structural Organisation of DNA

7.1.2 Outline the structure of nucleosomes

The DNA double helix contains major and minor grooves on its outer diameter, which expose chemical groups that can form hydrogen bondsThe DNA of eukaryotes associates with proteins called histonesDNA is wound around an octamer of histones (146 bases and 1.65 turns of the helix per octamer)The octamer and DNA combination is secured to a H1 histone, forming a nucleosome

7.1.3 State that nucleosomes help to supercoil DNA and help to regulate transcriptionNucleosomes serve two main functions:They protect DNA from damageThey allow long lengths of DNA to be packaged (supercoiled) for mobility during mitosis / meiosisWhen supercoiled, DNA is not accessible for transcriptionCells will have some segments of DNA permanently supercoiled (heterochromatin) and these segments will differ between different cell types

7.1.4 Distinguish between unique or single copy genes and highly repetitive sequences in nuclear DNA

7.1.5 State that eukaryotic genes contain introns and exonsIntron:A non-coding sequence of DNAwithina gene (intervening sequence) that is cut out by enzymes when RNA is made into mature mRNAExon:The part of the gene which codes for a protein (expressing sequence)Eukaryotic DNA contains introns but prokaryotic DNA does not

7.2 - DNA Replication

7.2.1 State that DNA replication occurs in a 5' - 3' directionDNA replication is semi-conservative, meaning that a new strand is synthesised from an original template strandDNA replication occurs in a 5' - 3' direction, in that new nucleotides are added to the C3 hydroxyl group such that the strand grows from the 3' endThis means that the DNA polymerase enzyme responsible for adding new nucleotides moves along the original template strand in a 3' - 5' direction

Direction of DNA Replication

7.2.2 Explain the process of DNA replication in prokaryotes, including the role of enzymes (helicase, DNA polymerase, RNA primase and DNA ligase), Okazaki fragments and deoxynucleoside triphosphatesDNA replication is semi-conservative and occurs during the S phase of interphaseHelicaseunwinds and separates the double stranded DNA by breaking the hydrogen bonds between base pairsThis occurs at specific regions (replication origins), creating a replication fork of two polynucleotide strands in antiparallel directionsRNA primasesynthesises a short RNA primer on each template strand to provide an attachment and initiation point for DNA polymerase IIIDNA polymerase IIIaddsdeoxynucleoside triphosphates(dNTPs) to the 3' end of the polynucleotide chain, synthesising in a 5' - 3' directionThe dNTPs pair up opposite their complementary base partner (adenine pairs with thymine ;guanine pairs withcytosine)As the dNTPs join with the DNA chain, two phosphates are broken off, releasing the energy needed to form a phosphodiester bondSynthesis is continuous on the strand moving towards the replication fork (leading strand)Synthesis is discontinuous on the strand moving away from the replication fork (lagging strand) leading to the formation ofOkazaki fragmentsDNA polymerase Iremoves the RNA primers and replaces them with DNADNA ligasejoins the Okazaki fragments together to create a continuous strand

Overview of DNA Replication

7.2.3 State that DNA replication is initiated at many points in eukaryotic chromosomesBecause eukaryotic genomes are (typically) much larger than prokaryotic genomes, DNA replication is initiated at many points simultaneously in order to limit the time required for DNA replication to occurThe specific sites at which DNA unwinding and initiation of replication occurs are called origins of replication and form replication bubblesAs replication bubbles expand in both directions, they eventually fuse together, two generate two separate semi-conservative double strands of DNAOrigins of Replication

7.3 - Transcription

7.3.1 State that transcription is carried out in a 5' - 3' directionTranscription is carried out in a 5' - 3' direction (of the new RNA strand)

7.3.2 Distinguish between the sense and antisense strands of DNADNA consists of two polynucleotide strands, only one of which is transcribed into RNATheantisense strandis transcribed into RNAIts sequence will be complementary to the RNA sequence and will be the "DNA version" of the tRNA anticodon sequenceThesense strandis not transcribed into RNAIts sequence will be the "DNA version" of the RNA sequence (identical except for T instead of U)

7.3.3 Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates and theterminatorA gene is a sequence of DNA which is transcribed into RNA and contain three main parts:Promoter:Responsible for the initiation of transcription (in prokaryotes, a number of genes may be regulated by a single promoter - this is an operon)Coding Sequence:The sequence of DNA that is actually transcribed (may contain introns in eukaryotes)Terminator:Sequence that serves to terminate transcription (mechanism of termination differs between prokaryotes and eukaryotes)

Transcription is the process by which a DNA sequence (gene) is copied into a complementary RNA sequence and involves a number of steps:RNA polymerase binds to the promoter and causes the unwinding and separation of the DNA strandsNucleoside triphosphates (NTPs) bind to their complementary bases on the antisense strand (uracil pairs with adenine, cytosine pairs with guanine)RNA polymerase covalently binds the NTPs together in a reaction that involves the release of two phosphates to gain the required energyRNA polymerase synthesises an RNA strand in a 5' - 3' direction until it reaches the terminatorAt the terminator, RNA polymerase and the newly formed RNA strand both detach from the antisense template, and the DNA rewindsMany RNA polymerase enzymes can transcribe a DNA sequence sequentially, producing a large number of transcriptsPost-transcriptional modification is necessary in eukaryotes

Overview of Transcription

7.3.4 State that eukaryotic RNA needs the removal of introns to form mature mRNAEuakaryotic genes may contain non-coding sequences called introns that need to be removed before mature mRNA is formedThe process by which introns are removed is called splicingThe removal of exons (alternative splicing) can generate different mRNA transcripts (and different polypeptides) from a single gene

7.4 - Translation

7.4.1 Explain that each tRNA molecule is recognised by a tRNA-activating enzyme that binds a specific amino acid to the tRNA using ATP for energy

Each different tRNA molecule has a unique shape and chemical composition that is recognised by a specific tRNA-activating enzymeThe enzyme (aminoacyl-tRNA synthetase) first binds the amino acid to a molecule of ATP (to form an amino acid-AMP complex linked by a high energy bond)The amino acid is then transferred to the 3'-end of the appropriate tRNA, attaching to a terminal CCA sequence on the acceptor stem and releasing the AMP moleculeThe tRNA molecule with an amino acid attached is thus said to be 'charged' and is now capable of participating in translationThe energy in the bond linking the tRNA molecule to the amino acid will be used in translation to form a peptide bond between adjacent amino acids

7.4.2 Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA binding sites and mRNA bindingsitesRibosomes are made of protein (for stability) and ribosomal RNA (rRNA - for catalytic activity)They consist of two subunits:The small subunit contains an mRNA binding siteThe large subunit contains three tRNA binding sites - an aminacyl (A) site, a peptidyl (P) site and an exit (E) siteRibosomes can be either found freely in the cytosol or bound to the rough ER (in eukaryotes)Ribosomes differ in size in eukaryotes and prokaryotes (eukaryotes = 80S ; prokaryotes = 70S)

7.4.3 State that translation consists of initiation, elongation, translocation and terminationTranslation occurs in four main steps:Initiation:Involves the assembly of an active ribosomal complexElongation:New amino acids are brought to the ribosome according to the codon sequenceTranslocation:Amino acids are translocated to a growing polypeptide chainTermination:At certain "stop" codons, translation is ended and the polypeptide is released

7.4.4 State that translation occurs in a 5' - 3' directionThe start codon (AUG) is located at the 5' end of the mRNA sequence and the ribosome moves along it in the 3' directionHence translation occurs in a 5' - 3' direction

7.4.5 Draw and label a diagram showing the structure of a peptide bond between two amino acids

7.4.6 Explain the process of translation, including ribosomes, polysomes, start codons and stop codonsPre-Initiation:Specific tRNA-activating enzymes catalyse the attachment of amino acids to tRNA molecules, using ATP for energy

Initiation:The small ribosomal subunit binds to the 5' end of mRNA and moves along it until it reaches the start codon (AUG)Next, the appropriate tRNA molecule binds to the codon via its anticodon (according to complementary base pairing)Finally, the large ribosomal subunit aligns itself to the tRNA molecule at its P-site and forms a complex with the small ribosomal subunit

Elongation:A second tRNA molecule pairs with the next codon in the ribosomal A-siteThe amino acid in the P-site is covalently attached via a peptide bond to the amino acid in the A-site

Translocation:The ribosome moves along one codon position, the deacylated tRNA moves into the E-site and is released, while the tRNA bearing the dipeptide moves into the P-siteAnother tRNA molecules attaches to the next codon in the newly emptied A-site and the process is repeatedThe ribosome moves along the mRNA sequence in a 5' - 3' direction, synthesising a polypeptide chainMultiple ribosomes can translate a single mRNA sequence simultaneously (forming polysomes)

Termination:Elongation and translocation continue until the ribosome reaches a stop codonThese codons do not code for any amino acids and instead signal for translation to stopThe polypeptide is released and the ribosome disassembles back into subunitsThe polypeptide may undergo post-translational modification prior to becoming a functional protein

Overview of the Process of Translation

7.4.7 State that free ribosomes synthesise proteins for use primarily within the cell, and that bound ribosomes synthesise proteins primarily for secretionor for lysosomesRibosomes floating freely in the cytosol produce proteins for use within the cellRibosomes attached to the rough ER are primarily involved in producing proteins to be exported from the cell or used in the lysosomeThese proteins contain a signal recognition peptide on their nascent polypeptide chains which direct the associated ribosome to the rough ER

7.5 - Proteins

7.5.1 Explain the four levels of protein structure, indicating the significance of each levelPrimary (1) StructureThe order / sequence of the amino acids of which the protein is composedFormed by covalent peptide bonds between adjacent amino acidsControls all subsequent levels of structure because it determines the nature of the interactions between R groups of different amino acids

Secondary (2) StructureThe way the chains of amino acids fold or turn upon themselvesHeld together by hydrogen bonds between non-adjacent amine (N-H) and carboxylic (C-O) groupsMay form an alpha helix, a beta-pleated sheet or a random coilSecondary structure provides a level of structural stability (due to H-bond formation)

Tertiary (3) StructureThe way a polypeptide folds and coils to form a complex molecular shape (e.g. 3D shape)Caused by interactions between R groups; including H-bonds, disulphide bridges, ionic bonds and hydrophilic / hydrophobic interactionsTertiary structure may be important for the function of the enzyme (e.g. specificity of active site in enzymes)

Quaternary (4) StructureThe interaction between multiple polypeptides or prosthetic groups that results in a single, larger, biologically active proteinA prosthetic group is an inorganic compound involved in protein structure or function (e.g. the heme group in haemoglobin)A protein containing a prosthetic group is called a conjugated proteinQuaternary structure may be held together by a variety of bonds (similar to tertiary structure)

Levels of Protein Organisation

7.5.2 Outline the difference between fibrous and globular proteins, with reference to two examples of each protein type

7.5.3 Explain the significance of polar and non-polar amino acidsPolar amino acids have hydrophilic R groups, whereas non-polar amino acids have hydrophobic R groupsFor water soluble proteins, non-polar amino acids tend to be found in the centre of the protein (stabilising structure) while polar amino acids are found on the surface (capable of interacting with water molecules)For membrane-bound proteins, non-polar amino acids tend to be localised on the surface in contact with the membrane, while polar amino acids line interior pores (to create hydrophilic channels)For enzymes, the active site specifically depends on the location and distribution of polar and non-polar amino acids as hydrophobic and hydrophilic interactions can play a role in substrate binding to the active site

7.5.4 State four functions of proteins, giving a named example of eachStructure: Support for body tissue (e.g. collagen, elastin, keratin)Hormones:Regulation of blood glucose (e.g. insulin, glucagon)Immunity:Bind antigens (e.g. antibodies / immunoglobulins)Transport:Oxygen transport (e.g. haemoglobin, myoglobin)Movement:Muscle contraction (e.g. actin / myosin, troponin / tropomyosin)Enzymes:Speeding up metabolic reactions (e.g. catalase, lipase, pepsin)

7.6 - Enzymes

7.6.1 State that metabolic pathways consist of chains and cycles of enzyme-catalysed reactionsMost chemical changes in a cell results from chains and cycles of reactions, with each step controlled by a separate specific enzymeThis allows for a far greater level of control and regulation of metabolic pathways (such as photosynthesis and cell respiration)

7.6.2 Describe the induced fit modelWhen enzymes and substrates bind, the active site is not completely rigid and may undergo a conformational change in shape to better fit the substrateThis conformational change may increase the reactivity of the substrate and be necessary for the enzyme's catalytic activityThe induced fit model explains how an enzyme may be able to bind to, and catalyse, several different substrates (broad specificity)

The Induced Fit Model

7.6.3 Explain that enzymes lower the activation energy of the chemical reactions that they catalyseEvery reaction requires a certain amount of energy to proceed - this is the activation energy (Ea)Enzymes speed up the rate of a biochemical reaction by lowering the activation energyIf more energy is in the products than the reactants, energy is lost from the system (endergonic)These reactions are usually anabolic (building things up), as the energy is being used up in bond formation between two substrate moleculesIf more energy is in the reactants than the products, excess energy is released into the system (exergonic)These reactions are usually catabolic (breaking things down), as the energy is released from the broken bonds within molecules

Reaction Pathway of a Typical Exergonic / Exothermic Reaction

7.6.4 Explain the difference between competitive and non-competitive inhibition, with reference to one example of eachCompetitive InhibitionA molecule (inhibitor) which is structurally / chemically similar to the substrate and binds to the active site of the enzymeThis serves to block the active site and thus prevent substrate binding (competes for the active site)Its effect can be reduced by increasing substrate concentrationExample:Relenza is a competitive inhibitor of neuraminidase (influenza virus enzyme), preventing the release of virions from infected cells

Non-competitive InhibitionA molecule (inhibitor) which is not structurally or chemically similar to the substrate and binds to a site other than the active site (allosteric site)This causes a conformational change in the active site, meaning the substrate cannot bindIts effect cannot be reduced by increasing substrate concentration as it isnot competing for the active siteExample:Cyanide (CN-) inhibits enzymes (cytochrome oxidase) in the electron transport chain by breaking disulphide bonds within the enzyme

Competitive versus Non-competitive Inhibition

7.6.5 Explain the control of metabolic pathways by end-product inhibition, including the role of allosteric sitesEnd-product inhibition is a form of negative feedback in which increased levels of product decrease the rate of product formationBecause metabolic pathways usually consist of chains (e.g. glycolysis) or cycles (e.g. Krebs cycle), the product can regulate the rate of its own production by inhibiting an earlier enzyme in the metabolic pathwayThe product binds to an allosteric site of an enzyme, causing a conformational change in the active site (non-competitive inhibition)As the enzyme can not currently function, the rate of product formation will decrease (and with less product there is less enzyme inhibition)

End-Product Inhibition

An example of end-product inhibition is the regulation of ATP formation by phosphofructokinase (an enzyme in glycolysis)ATP inhibits phosphofructokinase, so that when ATP levels are high, glucose is not broken down (but instead can be stored as glycogen)When ATP levels are low, phosphofructokinase is activated and glucose is broken down to make more ATP

8.1 - Cell Respiration

8.1.1 State that oxidation involves the loss of electrons from an element, whereas reduction involves a gain of electrons and that oxidation frequentlyinvolves gaining oxygen or losing hydrogen, whereas reduction frequently involves losing oxygen or gaining hydrogen

Redox (reduction-oxidation) reactions are chemical reactions that involve the transfer of electrons (gain or loss) between speciesMnemonics for redox reactions include:OIL RIG:OxidationIsLoss (of electrons),ReductionIsGain (of electrons)ELMO:ElectronLossMeansOxidationLEO goes GER:Loss ofElectrons isOxidation,Gain ofElectrons isReductionIn metabolic reactions, a species that has been reduced has the ability to reduce other species (this is the predominant role of hydrogen carriers)

The differences between oxidation and reduction can be summarised by the following table:

8.1.2 Outline the process of glycolysis, including phosphorylation, lysis, oxidation and ATP formation

Glycolysis is the first stage of cell respiration and involves the breakdown of glucose into two molecules of pyruvateIt is an anaerobic reaction (does not require the presence of oxygen) and occurs in the cytoplasm

There are four main parts in glycolysis (not including intermediary steps):Phosphorylation:A hexose sugar is phosphorylated by two ATP to become hexose biphosphateLysis:The hexose biphosphate splits into two triose phosphates (3C sugars)Oxidation:Hydrogen removed from the triose phosphates via oxidation (NAD is reduced to NADH + H+)ATP Formation:Four ATP molecules are released as the triose phosphates are converted into pyruvateOverall:One molecule of glucose results in 2 pyruvate, 2 (NADH + H+) and 2 ATP (net gain)

8.1.3 Draw and label a diagram showing the structure of a mitochondrion as seen in electron micrographs2D Representation 3D Representation Electron Micrograph

8.1.4 Explain aerobic respiration, including the link reaction, the Krebs cycle, the role of NADH + H+, the electron transport chain and the role of oxygenAerobic respiration takes place in the mitochondria, using the pyruvate produced via glycolysisIt produces large amounts of ATP in the presence of oxygen via three main processes:

The Link ReactionPyruvate is transported from the cytosol to the mitochondrial matrix in a reaction that produces (one) NADH + H+via oxidationThe pyuvate loses a carbon (as CO2) and the remaining two carbons are complexed with coenzyme A (CoA) to form acetyl CoA

The Krebs CycleIn the matrix, acetyl CoA combines with a 4C compound to form a 6C compoundOver a series of reactions the 6C compound is broken back down into the original 4C compoundThese reactions result in the formation of 2 CO2molecules, 1 ATP molecule and multiple hydrogen carriers, specifically 3 (NADH + H+) and 1 FADH2

The Electron Transport ChainThe hydrogen carriers (NADH + H+and FADH2) provide electrons to the electron transport chain on the inner mitochondrial membraneAs the electrons cycle through the chain they lose energy, which is used to translocate H+ions to the intermembrane space (creating a gradient)The hydrogen ions return to the matrix through the transmembrane enzyme ATP synthase, producing multiple ATP molecules (via chemiosmosis)Oxygen acts as a final electron acceptor for the electron transport chain, allowing further electrons to enter the chainOxygen combines the electrons with H+ions to form water moleculesThe electron transport chain produces the majority of the ATP molecules produced via aerobic respiration (~32 out of 36 ATP molecules)

8.1.5 Explain oxidative phosphorylation in terms of chemiosmosisOxidative phosphorylation describes the production of ATP from oxidised hydrogen carriers (as opposed to substrate level phosphorylation)When electrons are donated to the electron transport chain, they lose energy as they are passed between successive carrier moleculesThis energy is used to translocate H+ions from the matrix to the intermembrane space against the concentration gradientThe build up of H+ions creates an electrochemical gradient, or proton motive force (PMF)The protons return to the matrix via a transmembran enzyme called ATP synthaseAs they return they release energy which is used to produce ATP (from ADP and Pi)This process is called chemiosmosis and occurs in the cristaeThe H+ions and electrons are combined with oxygen to form water, allowing the process to be repeated anew

Overview of Chemiosmosis

8.1.6 Explain the relationship between the structure of the mitochondria and its functionInner membrane:Folded into cristae to increase surface area for electron transport chainIntermembrane space:Small space between inner and outer membranes for accumulation of protons (increases PMF)Matrix:Contains appropriate enzymes and a suitable pH for the Krebs cycle to occurOuter membrane:Contains appropriate transport proteins for shuttling pyruvate into the mitochondria

8.2 - Photosynthesis

8.2.1 Draw and label a diagram showing the structure of a chloroplast as seen in an electron micrograph2D Representation 3D RepresentationElectron Micrograph

8.2.2 State that photosynthesis consists of the light-dependent and light-independent reactionsPhotosynthesis is a two-step process:The light dependent reactions convert the light energy into chemical energyThe light independent reaction uses the chemical energy to make organic molecules

Overview of Photosynthesis

8.2.3 Explain the light dependent reactionsThe light dependent reactions occur on the thylakoid membrane and may occur by either cyclic or non-cyclic processesIn both processes, light excites chlorophyll (clustered in photosystems) which release electrons that pass through an electron transport chain, making ATP (photophosphorylation)

Non-Cyclic PhotophosphorylationChlorophyll in photosystems I and II absorbs light, which triggers the release of high energy electrons (photoactivation)The electrons from photosystem II pass along a series of carriers (electron transport chain), producing ATP via chemiosmosisThe electrons from photosystem I reduce NADP+to generate NADPH + H+Electrons lost from photosystem I are replaced by electrons from photsystem IIElectrons lost from photosystem II are replaced by electrons generated by the photolysis of water (oxygen is produced as a by-product)

Cyclic PhotophosphorylationOnly photosystem I is involved in cyclic photophosphorylationThe high energy electrons released by photoactivation pass along an electron transport chain (producing ATP) before returning to photosystem ICyclic photophosphorylation does not produce NADPH + H+, which is needed for the light independent reactionsThus while cyclic photophosphorylation can make chemical energy (ATP) from light, it cannot be used to make organic molecules

Non-Cyclic versus Cyclic Photophosphorylation

8.2.4 Explain photophosphorylation in terms of chemiosmosisAs the electrons (released from chlorophyll) cycle through the electron transport chains located on the thylakoid membrane, they lose energyThis free energy is used to pump H+ions from the stroma into the thylakoidThe build up of protons inside the thylakoid creates an electrochemical gradient (or proton motive force)The H+ions return to the stroma via the transmembrane enzyme ATP synthase, which uses the potential energy from the proton motive force to convert ADP and an inorganic phosphate (Pi) into ATPThis process is called chemiosmosis

Photophosphorylation via Chemiosmosis

8.2.5 Explain the light independent reactionThe light independent reaction occurs in the stroma and uses the ATP and NADPH + H+produced by the light dependent reaction (non-cyclic)The light independent reaction is also known as the Calvin cycle and occurs via three main steps:

Carbon FixationThe enzyme rubisco (RuBP carboxylase) catalyses the attachment of CO2to the 5C compound ribulose bisphosphate (RuBP)The unstable 6C compound that is formed immediately breaks down into two 3C molecules called glycerate-3-phosphate (GP)

ReductionEach GP molecule is then phosphorylated by ATP and reduced by NADPH + H+This converts each GP molecule into a triose phosphate (TP) called glyceraldehyde phosphate

Regeneration of RuBPFor every six molecules of TP produced, only one may be used to form half a sugar molecule (need two cycles to form a complete glucose)The remaining TP molecules are reorganised to regenerate stocks of RuBP in a reaction that involves ATPWith RuBP regenerated, this cycle will repeat many times and be used to construct chains of sugars (e.g. sucrose) for use by the plant

The Light Independent Reaction (Calvin Cycle)

8.2.6 Explain the relationship between the structure of the chloroplast and its functionThylakoids:Small lumen means small changes in proton concentration have a large effect on the proton motive forceGrana:Thylakoids arranged in stacks to greatly increase surface area available for light absorption (chlorophyll located in thylakoid membrane)Stroma:Contains appropriate enzymes and suitable pH for the light independent reaction to occur

8.2.7 Explain the relationship between the action spectrum and absorption spectrum of photosynthetic pigments in green plantsPigments absorb light as a source of energy for photosynthesisThe absorption spectrum indicates the wavelengths (frequency) of light absorbed by each pigmentThe action spectrum indicates the rate of photosynthesis for each wavelength / frequencyThere is a strong correlation between the cumulative absorption spectrum of all photosynthetic pigments and the action spectrumBoth display two main peaks - a larger peak at ~450 nm (blue) and a smaller peak at ~670 nm (red) with a decrease in between (green)

Absorption Spectrum versus Action Spectrum

8.2.8 Explain the concept of limiting factors in photosynthesis, with reference to light intensity, temperature and concentration of carbon dioxideThe law of limiting factors states that when a chemical process depends on more than one essential condition being favourable, its rate will be limited by the factor that is nearest its minimum valuePhotosynthesis is dependent on a number of favourable conditions, including:

Light IntensityLight is required for the light dependent reactions (photoactivation of chlorophyll and photolysis of water molecules)Low light intensities results in insufficient production of ATP and NADPH + H+(both needed for the light independent reaction)

TemperaturePrimarily affects the light independent reaction (and to a lesser extent the light dependent reactions)High temperatures will denature essential enzymes (e.g. rubisco), whereas insufficient thermal energy will prohibit reactions from occurring

Concentration of Carbon DioxideCarbon dioxide is required for the light independent reaction to occur (carbon fixation of RuBP by rubisco)At low levels, carbon fixation will occur very slowly, whereas at higher levels the rate will peak as all rubsico are being used

Factors Affecting the Rate of Photosynthesis

9.1 - Plant Structure and Growth

9.1.1 Draw and label plan diagrams to show the distribution of tissues in the stem and leaf of a dicotyledonous plantStem Tissue Leaf Tissue

9.1.2 Outline three differences between the structures of dicotyledonous and monocotyledonous plants

9.1.3 Explain the relationship between the distribution of tissues in the leaf and the function of these tissuesUpper EpidermisFunction:Main function is water conservation (secretes cuticle to create a waxy outer boundary)Distribution:On top of leaves where light intensity and heat are greatest

Palisade MesophyllFunction:Main photosynthetic tissue (cells contains many chloroplasts)Distribution:Upper half of leaf where light intensity is greatest (upper epidermal cells are transparent)

Spongy MesophyllFunction:Main site of gas exchange (made of loosely packed cells with spaces)Distribution:Lower half of leaf, near the stomatal pores (where gases and water are exchanged with the atmosphere)

Vascular TissueFunction:Transport of water (xylem) and the products of photosynthesis (phloem)Distribution:Found in middle of leaf (allowing all cells optimal access)

9.1.4 Identify modifications of roots, stems and leaves for different functions: bulbs, stem tubers, storage roots and tendrilsA storage organ is a part of a plant specifically modified to store energy (e.g. carbohydrates) or waterThey are usually found underground (better protection from herbivores) and may result from modifications to roots, stems or leaves:Storage roots:Modified roots that store water or food (e.g. carrots)Stem tubers:Horizontal underground stems that store carbohydrates (e.g. potato)Bulbs:Modified leaf bases (may be found as underground vertical shoots) that contain layers called scales (e.g. onion)Some plants (called succulents) have modified leaves or stems (thickened, fleshy and wax-covered) to enable water storage (e.g. cacti)Other plants (e.g. vines) have modifications to their leaf or stem to enable climbing support and attachment - these are calledtendrils

Modifications to Plant Structure

9.1.5 State that dicotyledonous plants have apical and lateral meristemsA meristem is a tissue in a plant consisting of undifferentiated cells (meristematic tissue) and are found in zones where growth can take placeMeristematic cells are analogous to stem cells in animals, however have specific regions of growth and development (unlike stem cells)Dicotyledonous plants have apical and lateral meristems

9.1.6 Compare growth due to apical and lateral meristems in dicotyledonous plantsSimilarities:Both are composed of totipotent cells (able to divide and differentiate)Both are found in dicotyledonous plants

Differences:

9.1.7 Explain the role of auxin in phototropism as an example of the control of plant growthPhototropism is the growing or turning of an organism in response to a unidirectional light sourceAuxins (e.g. IAA) are plant hormones that are produced by the tip of a shoot and mediate phototropismAuxin makes cells enlarge or grow and, in the shoot, are eradicated by lightThe accumulation of auxin on the shaded side of a plant causes this side only to lengthen, resulting in the shoot bending towards the lightAuxin causes cell elongation by activating proton pumps that expel H+ions from the cytoplasm to the cell wallThe resultant decrease in pH within the cell wall causes cellulose fibres to loosen (by breaking the bonds that hold them together)This makes the cell wall flexible and capable of stretching when water influx promotes cell turgorAuxin can also alter gene expression to promote cell growth (via the upregulation of expansins)

The Role of Auxin in Phototropism

9.2- Transport in Angiosperms

9.2.1 Outline how the root system provides a large surface area for mineral ion and water uptake by means of branching and root hairsPlants take up water and essential minerals via their roots and thus need a maximal surface area in order to optimise this uptakeThe monocotyledon root has a fibrous, highly branching structure which increases surface area for maximal absorptionThe dicotyledon root has a main tap root which can penetrate deeply into the soil to access deeper reservoirs of water and minerals, as well as lateral branches to maximise surface areaThe root epidermis may have extensions called root hairs which further increase surface area for mineral and water absorptionThese root hairs have carrier proteins and ion pumps in their plasma membrance, and many mitochondria within the cytoplasm, to aid active transport

9.2.2 List the ways in which mineral ions in the soil move into the rootMinerals move into the root system via the following pathways:Diffusion:Movement of minerals along a concentration gradientMass Flow:Uptake of mineral ions by means of a hydrostatic pressure gradientWater being taken into roots via osmosis creates a negative hydrostatic pressure in the soilMinerals form hydrogen bonds with water molecules and are dragged to the root, concentrating them for absorptionFungal Hyphae:Absorb minerals from the soil and exchange with sugars from the plant (mutualism)

9.2.3 Explain the process of mineral ion absorption from the soil into roots by active transportMinerals that need to be taken up from the soil include K+, Na+, Ca2+, NH4+, PO43-and NO3-Fertile soil invariably contains negatively charged clay particles to which positively charged minerals may attachRoot cells contain proton pumps that actively pump H+ions into the surrounding soil, which displaces the positively charged minerals allowing for their absorption (the negatively charged minerals may bind to the H+ions and be reabsorbed with the proton)

Mineral Ion Absorption

This mode of absorption is calledindirectactive transport - it uses energy (and proton pumps) to establish an electrochemical gradient by which mineral ions may be absorbed via diffusionAlternatively, the root cells may absorb mineral ions viadirectactive transport - using protein pumps to actively translocate ions against their concentration gradient

9.2.4 State that terrestrial plants support themselves by means of thickened cellulose, cell turgor and lignified xylemThree ways by which terrestrial plants may support themselves are:Thickened cellulose:Thickening of the cell wall provides extra structural supportCell turgor:Increased hydrostatic pressure within the cell exerts pressure on the cell wall, making cells turgidLignified xylem:Xylem vessels run the length of the stem and branches, lignification of these vessels provides extra support

9.2.5 Define transpirationTranspiration is the loss of water vapour from the leaves and stems of plants

9.2.6Explain how water is carried by the transpiration stream, including the structure of xylem vessels, transpiration pull, cohesion, adhesion andevaporation

Some of the light energy absorbed by leaves changes into heat, converting water in the spongy mesophyll into vapourThis vapour diffuses out of the stomata and is evaporated, creating a negative pressure gradient in the leafNew water is drawn from the xylem (mass flow), which is replaced by water from the roots (enters from soil via osmosis)The flow of water through the xylem from the roots to the leaf is called the transpiration streamWater rises through xylem vessels because of two qualities:Cohesion:Water molecules are weakly attracted to each other via hydrogen bondsAdhesion:Water molecules form hydrogen bonds with the xylem cell wallThese properties create a suction effect (or transpiration pull) in the xylemThe xylem has a specialised structure to facilitate transpiration:The inner lining is composed of dead cells that have fused to create a continuous tubeThese cells lack a cell membrane, allowing water to enter the xylem freelyThe outer layer is perforated (contains pores), allowing water to move out of the xylem into the leavesThe outer cell wall contains annular lignin rings which strengthens the xylem against the tension created by the transpiration stream

9.2.7 State that guard cells can regulate transpiration by opening and closing stomataThe transpiration pull is generated by the negative hydrostatic pressure created by the evaporation of water vapor from the leafGuard cells line stomata and regulate transpiration by controlling how much water vapor can exit the leafWhen stomata are open the rate of transpiration will be higher than when they are closed

9.2.8 State that the plant hormone abscisic acid causes the closing of the stomataWhen a plant begins to wilt from water stress, dehydrated mesophyll cells release the plant hormone abscisic acid (ABA)Abscisic acid triggers the efflux of potassium from guard cells, decreasing the water pressure within these cells and making them flaccidThis causes the stomatal pore to close

Opening and Closing of Stomata by Abscisic Acid

9.2.9 Explain how the abiotic factors light, temperature, wind and humidity, affect the rate of transpiration in a typical terrestrial plantLight Increasing the intensity of light increases the rate of transpirationLight stimulates the opening of stomata (gas exchange required for photosynthesis to occur)Some of the light energy absorbed by leaves is converted into heat, which increases the rate of water evaporation

Temperature Increasing the temperature increases the rate of transpirationHigher temperatures cause an increase in water vaporisation in the spongy mesophyll and an increase in evaporation from the surface of the leafThis leads to an increase in the diffusion of water vapour out of the leaf (via the stomata) which increases the rate of transpiration

WindGreater air flow around the surface of the leafincreases the rate of transpirationWind removes water vapour (lower concentration of vapour on leaf surface), increasing the rate of diffusion from within the spongy mesophyll

HumidityIncreasing the humidity decreases the rate of transpirationHumidity is water vapour in the air, thus a high humidity means there is a high concentration of water vapour in the airThis reduces the rate of diffusion of water vapour from inside the leaf (concentration gradient is smaller resulting in less net flow)

9.2.10 Outline four adaptations of xerophytes that help to reduce transpirationXerophytes are plants that can tolerate dry conditions (such as deserts and high altitudes) due to a number of specialised adaptations:Reduced leaves:Reducing the surface area of the leaf will reduce the area for water loss and thus reduce transpirationRolled leaves:Rolling up leaves (lower epidermis inside) reduces exposure of stomata to air and thus reduces transpirationThick waxy cuticle:A thickened cuticle prevents water loss from the surface of the leaf and thus reduces transpirationStomata in pits:Having stomata in pits, surrounded by hairs, concentrates water vapour near the stomata, reducing the rate of transpirationLow growth:Plants located near the ground are less exposed to wind and may be shaded, reducing the rate of transpirationC4/ CAM physiolog