Seventeen lessons on molecules, cells, metabolism, genetics, evolution, physiology, ecology and the scientific investigation for the current IB Biology course.
A skills course on calculation, data interpretation and explanation across the unity and diversity, form and function, interaction and interdependence, and continuity and change themes. It does not replace the syllabus. Higher-level-only topics (origins of cells, viruses, cladistics, muscle and motility, chemical signalling, gene expression) are not covered.
Basic cell biology and arithmetic with percentages and ratios.
Course outline
Water and the molecules of life
Link the properties of water, carbohydrates and lipids to their structure and function.
Proteins and nucleic acids
Relate amino acid sequence to protein shape and DNA structure to its function.
Cells: structure, magnification and scale
Calculate actual size and magnification and compare prokaryotic and eukaryotic cells.
Membranes, osmosis and water potential
Predict water movement and calculate percentage change in mass.
Enzymes and metabolism: read the graph before explaining
Explain enzyme rate graphs using active site, substrate and conditions.
Cell respiration and the respiratory quotient
Distinguish aerobic and anaerobic respiration and use the respiratory quotient.
Photosynthesis: light, carbon and limiting factors
Explain the light-dependent and light-independent reactions and interpret limiting factor graphs.
DNA replication, transcription and translation
Count codons, predict amino acid sequences and describe replication.
Cell division, mutations and gene editing
Distinguish mitosis from meiosis, describe mutation effects and calculate a mitotic index.
Inheritance: crosses, pedigrees and the chi-squared test
Predict ratios from monohybrid crosses and test whether data fit an expected ratio.
Natural selection, speciation and diversity
Explain adaptation by natural selection and the formation of new species.
Gas exchange and transport
Link surface area, gradients and transport structures to function in animals and plants.
Neural signalling, hormones and homeostasis
Describe nerve impulses and negative feedback in the control of blood glucose and temperature.
Defence against disease
Explain barriers, the immune response, vaccination and antibiotic action.
Populations, communities and mark-recapture
Estimate population size and interpret growth curves and interactions.
Energy flow, nutrient cycles and climate change
Calculate efficiency of energy transfer and explain the greenhouse effect and its consequences.
Uncertainty, evaluation and the scientific investigation
Handle error bars, statistics and the structure of the investigation report.
Sources and curriculum note
Reviewed October 5, 2026 against the IB subject brief (first assessment 2025). The brief lists Paper 1 and Paper 2 as external assessment worth 80 percent and the internal assessment worth 20 percent, with a 3,000-word maximum for the scientific investigation report. Numerical values (resting potential, chi-squared critical value 3.84 at 0.05 with 1 degree of freedom, RQ values) are standard teaching values; check your current IB data booklet and statistics guidance. Check the official IB page for changes.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Water and the molecules of life
Learning goal: Link the properties of water, carbohydrates and lipids to their structure and function.
Water is a polar molecule: the oxygen atom has a slight negative charge and the hydrogen atoms a slight positive charge. This allows hydrogen bonds to form between water molecules. Hydrogen bonding explains most of the properties of water that matter in living things. Cohesion between water molecules supports transport of water up a plant stem and surface tension. Adhesion to the walls of xylem vessels helps water climb. A high specific heat capacity means large amounts of energy are needed to change its temperature, which makes aquatic habitats thermally stable.
Water is also an excellent solvent for polar and charged substances, which dissolve because water molecules cluster around them. Hydrophilic substances dissolve or mix easily with water, while hydrophobic substances, such as fats, do not. Because of this, the transport of fats in the blood requires them to be packaged with proteins in lipoproteins. Water is also a medium for metabolic reactions, and its thermal properties make evaporation an effective cooling mechanism in sweating and transpiration.
Carbohydrates are made of carbon, hydrogen and oxygen. Monosaccharides such as glucose are single sugar units. Two monosaccharides join by condensation, with loss of water, to form a disaccharide. Many glucose units form polysaccharides. Starch is a plant energy store and has branched and unbranched forms. Glycogen is the branched energy store in animals and fungi. Cellulose is a structural polysaccharide in plant cell walls whose chains are held together by hydrogen bonds into fibres of high tensile strength.
Lipids include triglycerides, phospholipids and steroids. A triglyceride has a glycerol and three fatty acids joined by ester bonds. Fats and oils are used for long-term energy storage, thermal insulation and protection of organs. A phospholipid has a hydrophilic phosphate head and two hydrophobic fatty acid tails, so in water it forms a bilayer with the heads outward. Compare energy storage: lipids store more energy per gram than carbohydrates, but carbohydrates release energy faster and are more soluble, which suits short-term supply.
Worked example
Name the bond that forms when two glucose molecules join, and the small molecule released.
Two monosaccharides join in a condensation reaction.
The bond formed is a glycosidic bond.
A molecule of water is released.
The product is maltose, a disaccharide.
Practice problem and solution
How many fatty acid molecules join to one glycerol molecule in a triglyceride? Enter the number.
A triglyceride has three fatty acids joined to one glycerol by ester bonds.
Mental model: Link properties to structure and structure to function.
Common trap: Saying hydrogen bonds are the covalent bonds inside a water molecule.
2. Proteins and nucleic acids
Learning goal: Relate amino acid sequence to protein shape and DNA structure to its function.
Proteins are polymers of amino acids joined by peptide bonds in a condensation reaction. There are twenty different amino acids in proteins, and their sequence is determined by the genetic code. The sequence of amino acids is the primary structure. Hydrogen bonds between backbone groups produce secondary structures such as alpha helices and beta pleated sheets. The overall three-dimensional folding of one polypeptide is the tertiary structure, held together by interactions between the side chains, including hydrogen bonds, ionic bonds and disulfide bridges.
Some proteins have quaternary structure, which is the association of more than one polypeptide, as in haemoglobin with four. The shape of a protein determines its function. Enzymes have active sites with a specific shape, antibodies have variable regions that bind antigens, and structural proteins such as collagen form strong fibres. Denaturation is the loss of the three-dimensional structure because of heat or a change in pH, which breaks the weak bonds, and the protein stops working. The primary structure is not changed by denaturation.
Nucleic acids are polymers of nucleotides. A nucleotide has a phosphate, a pentose sugar and a nitrogenous base. In DNA the sugar is deoxyribose and the bases are adenine, thymine, cytosine and guanine. DNA is a double helix of two antiparallel strands, held together by hydrogen bonds between complementary bases. Adenine pairs with thymine through two hydrogen bonds and cytosine pairs with guanine through three. The sugar-phosphate backbone is on the outside, and the order of bases along a strand carries the genetic information.
RNA differs from DNA: it is usually single-stranded, its sugar is ribose and it contains uracil in place of thymine. Messenger RNA carries a copy of a gene to the ribosome, transfer RNA brings amino acids and ribosomal RNA is part of the ribosome. Base-pairing rules let you predict the complementary strand: an A on the template strand gives a U in the messenger RNA. Because DNA is complementary, the percentage of adenine equals that of thymine, and the percentage of cytosine equals that of guanine. In questions on base percentages, use these rules to fill in the missing values.
Worked example
A DNA sample has 30 percent adenine. Find the percentage of cytosine.
Adenine pairs with thymine, so thymine is also 30 percent.
A plus T = 60 percent.
C plus G = 40 percent.
C equals G, so cytosine = 20 percent.
Practice problem and solution
In a DNA molecule, 22 percent of the bases are thymine. What percentage of the bases are guanine?
A = T = 22 percent. A + T = 44, so C + G = 56, and G = 28 percent.
Common trap: Saying that denaturation breaks peptide bonds.
3. Cells: structure, magnification and scale
Learning goal: Calculate actual size and magnification and compare prokaryotic and eukaryotic cells.
Magnification is the ratio of the size of the image to the actual size of the specimen: magnification = image size / actual size. The most common error is mixing units. Convert both measurements to the same unit before dividing, usually micrometres for cells. One millimetre is 1000 micrometres, and one micrometre is 1000 nanometres. When a scale bar is given on a micrograph, measure the bar and use its labelled length, not the whole image, to find the magnification.
To find the actual size from a drawing or micrograph, rearrange the equation: actual size = image size / magnification. If a cell appears 30 mm long in an image with magnification 1500, convert 30 mm to 30000 micrometres, then divide by 1500 to get 20 micrometres. Always give the unit and state the answer to the appropriate number of significant figures, matching the precision of the least precise measurement.
Prokaryotic cells have no nucleus; their DNA is a naked circular molecule in a region called the nucleoid, and they lack membrane-bound organelles. They have 70S ribosomes and a cell wall made of peptidoglycan in bacteria. Eukaryotic cells have a nucleus, linear chromosomes associated with histone proteins, 80S ribosomes in the cytoplasm and membrane-bound organelles such as mitochondria and the endoplasmic reticulum. Eukaryotic mitochondria and chloroplasts contain 70S ribosomes and their own circular DNA, which is evidence for the endosymbiotic origin of these organelles.
Organelles carry out specialised tasks in compartments. Mitochondria produce ATP by aerobic respiration, ribosomes make proteins, rough endoplasmic reticulum folds proteins for secretion and the Golgi apparatus modifies and packages them. Lysosomes contain digestive enzymes. Plant cells also have a cell wall, chloroplasts and a large central vacuole. Compartmentalisation matters because it allows incompatible reactions to occur in separate environments, concentrates substrates and enzymes, and lets conditions such as pH be controlled in each organelle.
Worked example
A cell measures 45 mm on a drawing with a magnification of 900. Find the actual length in micrometres.
Convert 45 mm to micrometres: 45 000.
Actual = image / magnification.
45 000 / 900.
Actual length = 50 micrometres.
Practice problem and solution
A scale bar on a micrograph is 20 mm long and is labelled 5 micrometres. Find the magnification.
Mental model: Always convert units before dividing.
Common trap: Measuring the whole image instead of the scale bar.
4. Membranes, osmosis and water potential
Learning goal: Predict water movement and calculate percentage change in mass.
The cell membrane is a phospholipid bilayer with embedded proteins, which is described by the fluid mosaic model. The hydrophobic core limits the passage of ions and large polar molecules, but small non-polar molecules such as oxygen cross freely. Cholesterol in animal membranes affects fluidity. Integral proteins act as channels, carriers, pumps, receptors and enzymes. Membrane proteins are chosen by the cell, so membranes are selectively permeable.
Passive transport needs no cell energy. Simple diffusion moves substances down a concentration gradient across the bilayer. Facilitated diffusion moves them through channel or carrier proteins, still down the gradient. Active transport uses energy from ATP to move substances against a gradient, through pump proteins, such as the sodium-potassium pump. Vesicles carry larger objects in endocytosis and exocytosis. When describing a transport process, state the direction relative to the gradient and whether energy and a protein are needed.
Osmosis is the net movement of water across a partially permeable membrane from a region of higher water potential to a region of lower water potential. Water potential is highest in pure water and falls as solute concentration rises. A cell placed in a solution with a lower solute concentration than its cytoplasm takes up water; in a solution with a higher solute concentration it loses water. Plant cells become turgid or plasmolysed, while animal cells swell and may burst, or shrink and crenate.
Osmosis experiments often use potato cylinders in different sugar solutions. Because the starting masses differ, compare the percentage change in mass: (final mass minus initial mass) divided by initial mass, times 100. Plot percentage change against concentration. The point where the line crosses zero gives the concentration at which there is no net water movement, so the solution is isotonic with the tissue. Use a graph, not the raw changes, to find this, and note the sources of error such as incomplete blotting of the cylinders.
Worked example
A potato cylinder has an initial mass of 2.40 g and a final mass of 2.04 g in a sugar solution. Find the percentage change in mass.
Change = 2.04 - 2.40 = -0.36 g.
Divide by initial mass: -0.36 / 2.40.
Result = -0.15.
Percentage change = -15 percent.
Practice problem and solution
A tissue sample increases in mass from 4.00 g to 4.60 g. Find the percentage change in mass.
(4.60 - 4.00) / 4.00 x 100 = 15 percent.
Mental model: Use percentage change when starting masses differ.
Common trap: Using the raw change in mass for comparisons.
5. Enzymes and metabolism: read the graph before explaining
Learning goal: Explain enzyme rate graphs using active site, substrate and conditions.
Enzymes are biological catalysts, almost all of them proteins. They lower the activation energy of a reaction by holding the substrate in the active site in a way that makes bond breaking or forming easier. The substrate binds to the active site, forming an enzyme-substrate complex, and the products are released, leaving the enzyme unchanged. The induced fit model says that the active site changes shape slightly as the substrate binds, improving the fit and the catalysis.
Several factors affect the rate. Raising the temperature increases the kinetic energy of molecules, which gives more frequent collisions and more successful ones, until the optimum. Above the optimum, the weak bonds that maintain the tertiary structure break, the active site changes shape and the enzyme denatures, so the rate falls rapidly. Each enzyme has an optimum pH, and extremes of pH denature the enzyme by changing the charges on amino acid side chains.
Substrate concentration increases the rate at first because more collisions occur, but the curve levels off at the maximum rate when all active sites are occupied, which is saturation. At that point adding more substrate does not raise the rate, and only increasing the enzyme concentration would. When reading a graph, state the trend and the data values, then give the explanation, then identify the limiting factor in each region.
Competitive inhibitors resemble the substrate and bind to the active site, so they can be overcome by raising the substrate concentration. Non-competitive inhibitors bind elsewhere, changing the shape of the active site, so increasing the substrate does not restore the maximum rate. In metabolic pathways, end-product inhibition is common: the final product acts as an inhibitor of an earlier enzyme in the pathway, preventing overproduction. To compare rates from experiments, calculate rate as the quantity of product per unit time, or the reciprocal of the time taken.
Worked example
An enzyme produces 12 cm^3 of oxygen in 40 s. Find the rate in cm^3 per second.
Rate = amount of product / time.
12 / 40.
Rate = 0.30.
The unit is cm^3 s-1.
Practice problem and solution
An enzyme reaction takes 25 s to complete. Find the rate as 1 / time in s-1.
Rate = 1 / 25 = 0.04 s-1.
Mental model: State the trend, explain, then name the limiting factor.
Common trap: Saying that an enzyme is used up in the reaction.
6. Cell respiration and the respiratory quotient
Learning goal: Distinguish aerobic and anaerobic respiration and use the respiratory quotient.
Cell respiration is the controlled release of energy from organic compounds to produce ATP. ATP is the immediate energy source for cell processes because it can release energy quickly in a single hydrolysis reaction. Respiration is not the same as breathing, which is gas exchange at the level of the organism. The overall equation for aerobic respiration of glucose is glucose plus oxygen giving carbon dioxide and water, with the release of energy that is used to make ATP.
Glycolysis occurs in the cytoplasm. Glucose is broken down to two pyruvate molecules, with a small net gain of ATP and the reduction of NAD. When oxygen is available, pyruvate enters the mitochondrion and is oxidised in the link reaction and the Krebs cycle, with carbon dioxide released. The electron transport chain in the inner mitochondrial membrane uses the energy from electrons to pump protons, and ATP is produced when protons return through ATP synthase. Oxygen is the final electron acceptor and forms water.
Anaerobic respiration occurs without oxygen. In humans, pyruvate is converted to lactate, which regenerates NAD so glycolysis can continue, giving a small yield of ATP. In yeast, pyruvate is converted to ethanol and carbon dioxide. Anaerobic respiration releases much less energy per glucose than aerobic respiration because the energy in the products is not extracted. Lactate must later be removed, which requires oxygen, and this is described as an oxygen debt.
The respiratory quotient is the ratio of the volume of carbon dioxide produced to the volume of oxygen consumed. For carbohydrate the RQ is about 1.0, for lipid about 0.7 and for protein about 0.8 to 0.9. A measured RQ between these values suggests a mixture of fuels. In a respirometer, soda lime absorbs the carbon dioxide, so the movement of the liquid shows the oxygen consumed. To find the RQ, repeat the measurement without the absorber and compare the volumes, then divide.
Worked example
In a respiring tissue, 40 cm^3 of carbon dioxide is released while 50 cm^3 of oxygen is consumed. Find the respiratory quotient.
RQ = CO2 produced / O2 consumed.
RQ = 40 / 50.
RQ = 0.8.
This suggests a fuel such as protein or a mixture.
Practice problem and solution
A seed uses 25 cm^3 of oxygen and releases 25 cm^3 of carbon dioxide. Find the respiratory quotient.
RQ = 25 / 25 = 1.0, which is the value for carbohydrate.
Mental model: Respiration releases energy to make ATP.
Common trap: Using the RQ equation upside down.
7. Photosynthesis: light, carbon and limiting factors
Learning goal: Explain the light-dependent and light-independent reactions and interpret limiting factor graphs.
Photosynthesis converts light energy into chemical energy stored in organic molecules. The overall equation is carbon dioxide plus water giving glucose plus oxygen in the presence of light and chlorophyll. It occurs in chloroplasts, which contain stacks of thylakoids, called grana, surrounded by the fluid stroma. The process has two linked stages, each in a different location, and the products of one stage are the inputs of the other.
In the light-dependent reactions on the thylakoid membranes, chlorophyll absorbs light, exciting electrons that pass along an electron transport chain. Photolysis splits water, releasing oxygen, protons and electrons. The energy released pumps protons into the thylakoid space, and ATP is produced as protons flow through ATP synthase, a process called photophosphorylation. Electrons end up on NADP, forming reduced NADP. The oxygen released comes from water, not from carbon dioxide, which was shown using isotope labelling.
In the light-independent reactions in the stroma, the Calvin cycle uses ATP and reduced NADP to fix carbon dioxide. The enzyme rubisco attaches carbon dioxide to ribulose bisphosphate. The product is reduced to triose phosphate, some of which is used to regenerate ribulose bisphosphate and the rest to make glucose, starch, cellulose and other compounds. The cycle does not directly need light but depends on the products of the light reactions, so it stops soon in darkness.
Pigments absorb mainly red and blue light and reflect green, so an absorption spectrum has peaks in red and blue, and the action spectrum, showing the rate of photosynthesis at each wavelength, resembles it. Limiting factors are light intensity, carbon dioxide concentration and temperature. On a graph, the rate rises with the factor and then levels off when another factor limits. To identify the limiting factor at a plateau, ask which condition has not been increased. A rise in temperature beyond the enzyme optimum lowers the rate through denaturation of enzymes such as rubisco.
Worked example
A leaf disc rate graph levels off as light intensity increases, but carbon dioxide concentration was low. Name the limiting factor in the plateau region.
The rate stops increasing with light.
Light is no longer limiting.
Carbon dioxide and temperature were not increased.
Carbon dioxide concentration is likely limiting.
Practice problem and solution
In which part of the chloroplast does the Calvin cycle take place? Enter 1 for the stroma or 2 for the thylakoid membrane.
The Calvin cycle takes place in the stroma.
Mental model: Link each stage to its location and products.
Common trap: Saying the oxygen comes from carbon dioxide.
DNA replication is semi-conservative: each new molecule has one original strand and one new strand. The enzyme helicase unwinds the double helix and breaks hydrogen bonds between the bases, and single-strand binding proteins keep the strands apart. DNA polymerase adds nucleotides to the 3 prime end of a growing strand, so synthesis always proceeds in the 5 prime to 3 prime direction. The leading strand is made continuously, while the lagging strand is made in short Okazaki fragments joined by DNA ligase, after primase has made RNA primers.
Transcription makes messenger RNA from the DNA template strand. RNA polymerase binds to a promoter, unwinds the DNA and adds RNA nucleotides complementary to the template strand, with uracil in place of thymine. In eukaryotes the primary transcript is processed: introns are removed by splicing and exons are joined. The mature messenger RNA moves from the nucleus to the cytoplasm through nuclear pores. The sense strand has the same base sequence as the mRNA, except that it has thymine instead of uracil.
The genetic code is read as codons, groups of three bases on the messenger RNA, each specifying an amino acid or a stop signal. The code is degenerate, because most amino acids have more than one codon, and nearly universal, because almost all organisms use the same code. A sequence of n bases that codes for a polypeptide therefore carries n divided by 3 codons, and a polypeptide of m amino acids needs at least 3m bases, plus a stop codon. Always count the stop codon if the question asks for the length of the coding region.
Translation occurs at ribosomes. A transfer RNA molecule has an anticodon complementary to the codon and carries the matching amino acid. The ribosome has sites where tRNA binds, a peptide bond forms between adjacent amino acids and the ribosome moves along the messenger RNA until it reaches a stop codon. Several ribosomes can translate one mRNA at once, forming a polysome. In prokaryotes, where there is no nucleus, translation can begin while the mRNA is still being transcribed. To work out a sequence, write the mRNA from the template, split it into triplets, then use the codon table in your booklet.
Worked example
A template DNA strand reads 3 prime TAC GGT 5 prime. Write the mRNA sequence.
Pair each base: T with A, A with U, C with G.
TAC gives AUG.
GGT gives CCA.
mRNA: 5 prime AUG CCA 3 prime.
Practice problem and solution
A polypeptide has 120 amino acids. What is the minimum number of bases in the coding region of the mRNA, including the stop codon? Enter the number.
120 codons x 3 = 360 bases, plus a stop codon of 3 bases: 363.
Mental model: Write the mRNA, split into triplets, then look up the code.
Common trap: Reading the template as if it were the mRNA.
9. Cell division, mutations and gene editing
Learning goal: Distinguish mitosis from meiosis, describe mutation effects and calculate a mitotic index.
Mitosis produces two genetically identical diploid daughter cells and is used for growth, tissue repair and asexual reproduction. The stages are prophase, when chromosomes condense and the nuclear envelope breaks down; metaphase, when chromosomes line up at the equator; anaphase, when sister chromatids separate; and telophase, when nuclear envelopes reform. Cytokinesis, the division of the cytoplasm, follows. Mitosis is part of the cell cycle, which also includes interphase, in which the DNA is replicated.
The mitotic index is the number of cells in mitosis divided by the total number of cells observed. It is used as an indicator of how fast tissue is dividing, for example in tumour tissue, which usually has a higher mitotic index. When counting, use several fields of view and count enough cells to reduce random error. State whether the result is a fraction or a percentage. Cells in interphase are the majority in most tissues, so the index is typically small.
A mutation is a change in the base sequence of DNA. A substitution changes one base and can be silent if the new codon codes for the same amino acid, a missense mutation if it changes the amino acid, or a nonsense mutation if it creates a stop codon. An insertion or deletion of bases that is not a multiple of three causes a frameshift, which changes every codon after the mutation and often gives a non-functional protein. Mutations in body cells are not passed to offspring, but those in gametes can be.
Gene editing techniques can make targeted changes in DNA. The CRISPR-Cas9 system uses a guide RNA to direct the Cas9 enzyme to a specific sequence, where it cuts both strands. The cell then repairs the break, and a change can be introduced in this process. Potential uses include correcting disease-causing mutations and improving crops, and there are ethical issues about safety, inheritable changes and fair access. When evaluating a use of gene editing, state the benefit, the risk, who is affected and whether the change can be inherited.
Worked example
In a root tip, 18 of 150 cells counted are in mitosis. Find the mitotic index as a percentage.
Mitotic index = cells in mitosis / total cells.
18 / 150 = 0.12.
Convert to a percentage.
Mitotic index = 12 percent.
Practice problem and solution
In a tissue sample, 9 of 60 cells are in mitosis. Find the mitotic index as a percentage.
9 / 60 = 0.15, which is 15 percent.
Mental model: Mutations are changes in sequence; their effect depends on where and what type.
Common trap: Counting cells in too few fields of view.
10. Inheritance: crosses, pedigrees and the chi-squared test
Learning goal: Predict ratios from monohybrid crosses and test whether data fit an expected ratio.
Alleles are different forms of the same gene. An individual is homozygous if both alleles are the same and heterozygous if they differ. The genotype is the combination of alleles, and the phenotype is the observable trait. A dominant allele shows its effect in the heterozygote, while a recessive allele shows only in the homozygote. Meiosis separates alleles into different gametes, so each gamete has one allele of each gene. In a cross between two heterozygotes (Aa x Aa), the expected genotype ratio is 1 AA, 2 Aa and 1 aa, and the phenotype ratio is 3 dominant to 1 recessive.
A Punnett grid sets out the gametes of each parent and combines them. Always list the possible gametes first, write the cross with symbols, and give the ratio as whole numbers. A test cross crosses an individual of unknown genotype with a homozygous recessive. If any offspring show the recessive phenotype, the parent must be heterozygous. For codominance, both alleles are expressed in the heterozygote, as in the blood group alleles. For sex-linked traits carried on the X chromosome, males have only one allele and are more likely to show a recessive trait.
A pedigree chart shows inheritance in a family. Squares are males, circles are females and shaded symbols are affected individuals. To decide if a trait is recessive, look for two unaffected parents with an affected child, which is possible only if the trait is recessive and both parents are carriers. For a dominant trait, an affected child must have at least one affected parent. Test each possible mode with each family, and state which fits all the data.
The chi-squared test compares observed counts with the counts expected from a hypothesis. The statistic is the sum over all categories of (observed minus expected) squared divided by expected. Degrees of freedom equal the number of categories minus one. For two categories, one degree of freedom, the critical value at the 0.05 significance level is 3.84. If the calculated value is below the critical value, the difference is not significant, and the data are consistent with the expected ratio. If it is above, reject the hypothesis. Use counts, never percentages, in the test.
Worked example
A cross gives 70 tall and 30 short plants (total 100). The expected ratio is 3 : 1. Calculate chi-squared.
Expected: 75 tall and 25 short.
Tall: (70 - 75)^2 / 75 = 25 / 75 = 0.333.
Short: (30 - 25)^2 / 25 = 25 / 25 = 1.0.
Chi-squared = 1.33, below 3.84, so the data fit 3 : 1.
Practice problem and solution
In a cross of two heterozygotes (Aa x Aa), what percentage of offspring is expected to be homozygous recessive?
The expected ratio is 1 AA : 2 Aa : 1 aa, so aa is 1 in 4, which is 25 percent.
Mental model: Use counts, not percentages, in chi-squared.
Common trap: Applying the test to percentages.
11. Natural selection, speciation and diversity
Learning goal: Explain adaptation by natural selection and the formation of new species.
Natural selection explains how populations change over generations. It rests on observations and inferences. Individuals in a population vary, and some of this variation is heritable. More offspring are produced than can survive, so there is competition for resources. Individuals with traits that give them a better chance of surviving and reproducing in their environment leave more offspring, so the alleles for those traits become more common in the population. Evolution is a change in allele frequency in a population over time, so the unit of evolution is the population, not the individual.
Variation arises from mutation, meiosis and sexual reproduction. Mutation produces new alleles, crossing over and independent assortment produce new combinations, and fertilisation combines alleles from two parents. Natural selection acts on phenotypes, but the response is a change in the frequency of the underlying alleles. A common mistake is to say that organisms evolve because they need to. Selection acts on variation that already exists, and individuals do not change their genes in response to the environment.
Evidence for evolution includes the fossil record, homologous structures such as the pentadactyl limb, comparisons of DNA and protein sequences and observed examples such as antibiotic resistance in bacteria. When bacteria are exposed to an antibiotic, resistant individuals survive and reproduce, and the proportion of resistant bacteria increases. Selection can be directional, shifting the mean of a trait, stabilising, favouring the average, or disruptive, favouring both extremes.
A species is a group of organisms that can interbreed and produce fertile offspring. Speciation occurs when populations become reproductively isolated. Geographic isolation, as on islands, separates populations that then diverge by different selection and genetic drift. Over time, differences accumulate until the populations can no longer interbreed. In sympatric speciation, isolation occurs without a geographic barrier, as with polyploidy in plants. Classification groups organisms in a hierarchy from domain to species, and the binomial system names each species with a genus and a species name, such as Homo sapiens.
Worked example
A population of beetles has 20 green and 80 brown individuals. Predators eat green beetles more often. Predict the change over generations.
Brown colour gives a survival advantage.
Brown beetles reproduce more.
The allele for brown becomes more frequent.
The proportion of brown beetles increases.
Practice problem and solution
A population of 200 has 50 individuals with a recessive phenotype. What fraction of the population is that? Give a decimal.
50 / 200 = 0.25.
Mental model: Evolution is a change in allele frequency.
Common trap: Saying individuals adapt by wanting to change.
12. Gas exchange and transport
Learning goal: Link surface area, gradients and transport structures to function in animals and plants.
Gas exchange surfaces share features: a large surface area, thin walls to give a short diffusion distance, moisture so gases can dissolve, and a maintained concentration gradient. In human lungs, alveoli provide a large area and are surrounded by capillaries, so blood removes oxygen and brings carbon dioxide, which keeps the gradients steep. Ventilation moves air in and out, which keeps oxygen high and carbon dioxide low in the alveoli. The same ideas apply to gills in fish, where water flows over the filaments in the opposite direction to blood, a countercurrent arrangement that maintains a gradient along the whole length.
Fick's law summarises the factors: the rate of diffusion is proportional to the surface area multiplied by the concentration difference and divided by the diffusion distance. Increasing area or gradient raises the rate, and increasing the distance lowers it. This explains why large organisms need transport systems, because the surface area to volume ratio falls as size increases. When comparing cell sizes, calculate the surface area and volume of simple shapes, then divide, to show how the ratio changes.
In mammals the circulatory system is a double circulation. The right side of the heart pumps blood to the lungs, and the left side pumps blood to the body at higher pressure. Valves prevent backflow. Arteries carry blood away from the heart with thick elastic walls, veins carry blood back with valves and capillaries have walls one cell thick for exchange. The heart beat is initiated by the sinoatrial node, which sets the pace, and a signal spreads through the walls of the atria and then the ventricles.
In plants, xylem vessels carry water and minerals upward from the roots, driven by transpiration, the evaporation of water through stomata. The cohesion of water molecules pulls the column up, producing tension in the xylem. Phloem transports sugars from sources, such as leaves, to sinks, such as roots and fruits, in a process called translocation, which needs energy. Stomata open to let carbon dioxide in, but at the cost of water loss, so plants close them in dry conditions. Compare xylem and phloem by direction, content, driving force and the cells involved.
Worked example
Cube A has sides 1 cm and cube B has sides 2 cm. Compare the surface area to volume ratios.
Cube A: surface area 6 cm^2, volume 1 cm^3, ratio 6.
Cube B: surface area 24 cm^2, volume 8 cm^3, ratio 3.
The larger cube has the smaller ratio.
Larger organisms need transport systems to supply their cells.
Practice problem and solution
A cube has sides of 3 cm. Find its surface area to volume ratio in cm-1.
Surface area = 6 x 9 = 54 cm^2. Volume = 27 cm^3. Ratio = 54 / 27 = 2.
Mental model: Large surface, thin walls, steep gradient.
Common trap: Confusing xylem with phloem.
13. Neural signalling, hormones and homeostasis
Learning goal: Describe nerve impulses and negative feedback in the control of blood glucose and temperature.
A neuron transmits electrical impulses. At rest, the inside of the neuron is negative relative to the outside, giving a resting potential of about minus 70 millivolts, maintained by sodium-potassium pumps and potassium channels. When a stimulus exceeds a threshold, sodium channels open and sodium ions enter, depolarising the membrane. Then potassium channels open, potassium leaves and the membrane repolarises. This all-or-nothing change is an action potential, and it moves along the axon as local currents trigger the next section.
In myelinated neurons, the action potential jumps between gaps in the myelin sheath, called nodes of Ranvier, which speeds up the impulse. At a synapse, the impulse triggers the release of neurotransmitter from vesicles by exocytosis, after calcium ions enter the presynaptic neuron. The neurotransmitter diffuses across the synaptic cleft and binds to receptors on the postsynaptic membrane, opening ion channels and starting a new impulse. The neurotransmitter is then broken down or taken back up, so the signal stops.
Homeostasis maintains the internal environment within narrow limits. It generally works by negative feedback: a receptor detects a change from the set point, a control centre processes the information and an effector acts to reverse the change. In blood glucose control, a rise after a meal is detected by cells in the pancreas, which release insulin, and cells take up glucose, with the liver storing it as glycogen. A fall leads to the release of glucagon, which causes the liver to break down glycogen into glucose.
In type 1 diabetes the body does not produce enough insulin, so the blood glucose stays high. In type 2 diabetes, cells respond poorly to insulin. In temperature control, the hypothalamus receives information from temperature receptors. If the body is too warm, sweating and vasodilation increase heat loss. If too cool, shivering and vasoconstriction conserve and generate heat. Hormones are chemical messengers carried in the blood, acting more slowly and for longer than nerve impulses. When describing a feedback loop, identify the stimulus, receptor, control centre, effector and response.
Worked example
Describe how the body responds to a rise in body temperature.
Receptors detect the rise and signal the hypothalamus.
Effectors respond.
Skin arterioles dilate and sweating increases.
Heat loss rises and temperature returns toward the set point.
Practice problem and solution
An impulse travels 0.30 m along an axon in 0.0025 s. Find its speed in m s-1.
Speed = distance / time = 0.30 / 0.0025 = 120 m s-1.
Mental model: Negative feedback reverses a change.
Common trap: Calling feedback a one-way chain without a set point.
14. Defence against disease
Learning goal: Explain barriers, the immune response, vaccination and antibiotic action.
Pathogens are organisms or viruses that cause disease. The body has barriers that prevent entry: the skin, mucous membranes, stomach acid and secretions such as tears that contain lysozyme. If a pathogen enters, the non-specific response begins. Phagocytes, a type of white blood cell, engulf pathogens by endocytosis and digest them with lysosome enzymes. Inflammation increases blood flow to the area, bringing more white blood cells. These responses act in the same way on any pathogen.
The specific immune response recognises antigens, molecules on the surface of the pathogen. Each lymphocyte has receptors for one specific antigen. When an antigen binds to the matching B lymphocyte, with help from a helper T cell, the B cell divides by clonal selection into plasma cells that secrete antibodies and memory cells. Antibodies bind to antigens, which can neutralise the pathogen, cause clumping or mark it for phagocytosis. Cytotoxic T cells destroy infected body cells.
Immunity follows from memory cells. On a second exposure to the same antigen, memory cells respond faster and produce more antibodies at a higher level, so the person usually has no symptoms. This is the basis of vaccination, which exposes the immune system to a harmless form of an antigen, such as a weakened or inactivated pathogen or a protein from it, without causing the disease. When enough of a population is immune, the spread of the disease is reduced, which indirectly protects those who cannot be vaccinated; this is called herd immunity.
Antibiotics are used against bacteria, not viruses. They work by targeting structures or processes in bacteria that human cells do not have, such as the bacterial cell wall or 70S ribosomes. They do not work on viruses, which depend on the host cell's processes. Overuse of antibiotics selects for resistant bacteria, as resistant individuals survive and reproduce. To limit this, complete the prescribed course, avoid use when it is not needed and use targeted treatment. In a graph of antibody concentration after two exposures, the second peak is higher and rises faster.
Worked example
Explain why a second exposure to an antigen produces a faster, larger antibody response.
Memory cells were formed in the first response.
Memory cells persist in the body.
On re-exposure they divide quickly into plasma cells.
More antibodies are made sooner.
Practice problem and solution
In a population of 400 people, 300 are immune after vaccination. What percentage is immune?
Common trap: Saying antibiotics treat viral infections.
15. Populations, communities and mark-recapture
Learning goal: Estimate population size and interpret growth curves and interactions.
A population is a group of organisms of the same species living in the same area. A community is all the populations in an area, and an ecosystem includes the community and its non-living environment. Ecologists cannot usually count every individual, so they sample. A quadrat is used for plants and slow-moving animals, with random placement in many locations so that the sample represents the area. Transects, lines across a gradient, are used to study how distribution changes with conditions.
The mark-recapture method estimates the size of mobile animal populations. Capture a sample, mark the individuals in a harmless way and release them. After they mix with the population, capture a second sample and count the marked individuals. The estimated population size is the number marked in the first sample multiplied by the total second sample, divided by the number of marked individuals in the second sample. The method assumes that marking does not harm the animals, that they mix randomly and that no migration, births or deaths significantly change the population.
Population growth in ideal conditions is exponential, as there are no limits, giving a J-shaped curve. In real conditions, limiting factors such as food, space and disease slow the growth, and the population levels off near the carrying capacity, giving an S-shaped, or sigmoid, curve. Density-dependent factors, such as competition and predation, have a stronger effect as the population is larger, while density-independent factors, such as a flood or a fire, act regardless of density.
Species interact in several ways. Competition is a negative interaction between organisms using the same resource. Predation benefits the predator and harms the prey. Herbivory is a plant being eaten. Mutualism benefits both species, as with nitrogen-fixing bacteria in root nodules. Parasitism benefits the parasite and harms the host. Each species has a niche, its role and the conditions it requires, and two species cannot occupy the same niche in the same place for long, which is the competitive exclusion principle. When describing an interaction, state who benefits and who is harmed.
Worked example
In a first sample, 40 beetles are marked and released. A second sample has 50 beetles, 10 of which are marked. Estimate the population size.
Population = (marked first x second sample) / marked in second.
Population = (40 x 50) / 10.
2000 / 10.
Estimated population = 200 beetles.
Practice problem and solution
A first sample marks 30 fish. A second sample of 60 fish contains 6 marked fish. Estimate the population size.
(30 x 60) / 6 = 300.
Mental model: The S-shaped curve shows a carrying capacity.
Common trap: Counting all marked animals from the first sample as recaptured.
16. Energy flow, nutrient cycles and climate change
Learning goal: Calculate efficiency of energy transfer and explain the greenhouse effect and its consequences.
Energy enters most ecosystems as sunlight, captured by producers in photosynthesis. It flows to consumers through food chains, and at each trophic level much of the energy is lost, mainly as heat from respiration, in waste and in parts that are not eaten. Only a fraction passes to the next level, so food chains rarely have more than four or five levels. Pyramids of energy show the energy available at each level and are always wide at the base, narrowing upward, which pyramids of numbers or biomass may not do.
The efficiency of transfer between levels is the energy at the higher level divided by the energy at the lower level, multiplied by 100 percent. If producers contain 10000 kJ and primary consumers 1000 kJ, the efficiency is 10 percent. Compare values with the same units and time period, such as kJ per square metre per year. Energy is not recycled in ecosystems, since it is lost as heat, but matter such as carbon and nitrogen is recycled through decomposers.
In the carbon cycle, carbon dioxide is taken up by photosynthesis and returned by respiration and combustion. Organic carbon in dead organisms can be stored in fossil fuels. Decomposition by fungi and bacteria releases carbon dioxide, but in waterlogged conditions decomposition is slow and peat forms. Methane, another greenhouse gas, is produced by some microorganisms in anaerobic conditions. Human activities such as burning fossil fuels and deforestation add carbon dioxide to the atmosphere faster than natural processes remove it.
The greenhouse effect is a natural process. Short-wave radiation from the Sun passes through the atmosphere and warms the Earth, which re-emits longer-wave infrared radiation. Greenhouse gases such as carbon dioxide, methane and water vapour absorb some of this infrared radiation and re-emit it in all directions, including back to the surface, so the Earth is warmer than it would otherwise be. The enhanced greenhouse effect results from higher concentrations of these gases from human activity. Consequences described in the IB course include changes in the distribution of species, effects on sea ice and sea level and changes in phenology. Always cite the data source when quoting figures.
Worked example
Producers contain 8000 kJ and primary consumers contain 800 kJ. Find the percentage efficiency of transfer.
Efficiency = energy at higher level / energy at lower level x 100.
800 / 8000 = 0.10.
0.10 x 100.
Efficiency = 10 percent.
Practice problem and solution
Primary consumers hold 2500 kJ and secondary consumers hold 250 kJ. Find the percentage efficiency of transfer.
250 / 2500 x 100 = 10 percent.
Mental model: Energy flows, matter cycles.
Common trap: Saying energy is recycled in ecosystems.
17. Uncertainty, evaluation and the scientific investigation
Learning goal: Handle error bars, statistics and the structure of the investigation report.
Every biological measurement has an uncertainty. The uncertainty of an instrument is usually half of the smallest division for analogue scales. Biological samples also vary, so repeats and larger samples reduce the effect of random variation. Take the mean of the repeats and the standard deviation as a measure of spread. A larger standard deviation shows greater variability in the data. Error bars on a graph, whether standard deviation or range, show the variation and allow the reader to judge whether differences are meaningful.
When comparing two means, look at the overlap of the error bars as a first guide. A statistical test gives a more reliable decision. A t-test compares the means of two samples and gives a probability that a difference this large would occur by chance. If the probability is less than 0.05, the difference is called significant at the 5 percent level. Do not say that a test proves the hypothesis. Say whether the data support it. Correlation does not show cause, so a relationship between two variables needs a controlled experiment to indicate causation.
The scientific investigation is the internal assessment in IB Biology. According to the IB subject brief, it is an open-ended task in which the student gathers and analyses data to answer their own research question, written up in a report of at most 3,000 words. Choose a question with a measurable independent variable and dependent variable, with controlled variables listed and a method that can be repeated. Check safety and ethics before starting, particularly for living organisms.
In the evaluation, name specific limitations and explain how each affected the data, for example incomplete blotting of potato cylinders or uneven temperature in a water bath. Separate random and systematic error. Suggest improvements that deal with each limitation, such as a larger sample or a thermostatic water bath. A strong conclusion refers to the data, compares with the hypothesis and with accepted scientific understanding, and says how far the result can be generalised. Avoid claiming more than the data can support.
Worked example
Four repeats of a reaction time give 12, 14, 13 and 15 s. Find the mean.
Add the values: 12 + 14 + 13 + 15 = 54.
Divide by four.
54 / 4.
Mean = 13.5 s.
Practice problem and solution
Five repeats give 10, 11, 9, 10 and 10 (units). Find the mean.
Sum = 50; 50 / 5 = 10.
Mental model: Use several repeats and report the spread.
Common trap: Claiming a correlation proves a cause.