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Welcome to GCSE Edexcel Science revision.

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Unit S B 1: Key concepts in biology.

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Magnification describes how many times larger an image is than the real object.

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It has no unit.

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Write it as, for example, times four hundred.

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Image size equals magnification times actual object size.

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Rearrange this to magnification equals image size divided by actual size, or actual size equals image size divided by magnification.

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For a light microscope, total magnification equals eyepiece magnification times objective magnification.

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A times ten eyepiece with a times forty objective gives times four hundred.

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Use the same units for image and actual size before calculating.

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One millimetre equals one thousand micrometres.

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One micrometre equals one thousand nanometres.

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An image twenty millimetres long of a cell fifty micrometres long has magnification twenty thousand divided by fifty, which equals times four hundred.

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Resolution is the smallest distance between two points that can still be seen as separate points.

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Higher resolution shows finer detail.

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Making an image bigger (increasing magnification) does not necessarily improve its resolution.

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Electron microscopes have greater magnification and resolving power than light microscopes.

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They reveal smaller cell structures, including fine internal details.

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Microscopy practical: place a thin specimen on a slide, add an appropriate stain if needed, and lower a coverslip gently to reduce trapped air.

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Handle slides and stains safely.

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Start with the lowest-power objective.

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Use coarse focus to find the specimen, then fine focus to sharpen it.

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At high power, use fine focus and keep the lens clear of the slide.

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Make a large, clear biological drawing with single lines, no shading and labelled structures.

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Include a scale or magnification; calculate sizes using matching units.

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Animal and plant cells are eukaryotic: their D N A is enclosed in a nucleus.

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Both contain cytoplasm, a cell membrane, ribosomes and mitochondria.

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The nucleus contains genetic material and controls the cell’s activities.

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The cytoplasm is where many chemical reactions occur.

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The cell membrane controls which substances move into and out of the cell.

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Ribosomes are where proteins are made: this is called protein synthesis.

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Mitochondria are the site of most reactions of aerobic respiration, which transfers energy for cell processes.

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A typical photosynthetic plant cell also contains chloroplasts, a permanent vacuole and a cell wall.

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Some plant cells, such as root cells, do not contain chloroplasts.

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Cell structures: simplified overview.

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The surrounding notes explain each structure’s function.

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Chloroplasts contain chlorophyll, which absorbs light for photosynthesis.

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The vacuole contains cell sap and helps maintain pressure inside the cell.

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The plant cell wall is made of cellulose and supports the cell.

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It lies outside the cell membrane; plant cells have both a wall and a membrane.

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Specialised cells have structures suited to their functions.

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Explain an adaptation by linking a structure to the job it helps the cell perform.

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A sperm cell has a tail for movement, many mitochondria to supply energy, and an acrosome containing enzymes that help it penetrate the egg’s outer layers.

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An egg cell contains plenty of cytoplasm with nutrients for early development.

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Its outer layer changes after fertilisation to help prevent more sperm entering.

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Cells lining the small intestine have microvilli: tiny folds that provide a large surface area for absorbing digested nutrients.

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Ciliated epithelial cells have hair-like cilia that move substances.

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In airways, cilia move mucus containing trapped particles towards the throat.

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Bacteria are prokaryotic cells.

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They have cytoplasm, a cell membrane, ribosomes and a cell wall, but no nucleus or mitochondria.

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Bacterial genetic material includes a large circular loop of D N A in the cytoplasm; some bacteria also contain small D N A rings called plasmids.

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Some bacteria have a flagellum for movement and a slime capsule for protection.

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Their cell walls are not made of cellulose.

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Enzymes are biological catalysts, usually proteins.

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They speed up reactions without being used up, including reactions that build molecules (synthesis) and break them down.

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A substrate is the substance an enzyme acts on.

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It fits into the enzyme’s active site because their shapes match (are complementary).

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This makes the enzyme specific: it usually catalyses one particular reaction.

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The active site is complementary to its substrate; denaturation changes that fit.

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An enzyme, substrate complex forms, the reaction produces products, and the products leave.

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The enzyme can then catalyse another reaction.

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Proteases break proteins down into amino acids.

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Carbohydrases break carbohydrates down into sugars; amylase breaks starch down into maltose, rather than directly producing glucose.

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Lipases break lipids down into fatty acids and glycerol.

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Synthesis reactions join smaller molecules to make larger biological molecules.

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Increasing temperature initially increases reaction rate because particles have more kinetic energy and collide more often.

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The optimum temperature gives the fastest rate for that enzyme under those conditions.

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At high temperatures an enzyme may denature: its active site changes shape so the substrate no longer fits.

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Low temperature usually slows activity rather than denaturing the enzyme.

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Each enzyme has an optimum P H.

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A P H far from the optimum can change the active site and reduce activity or denature the enzyme.

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Increasing substrate concentration raises the rate until all active sites are occupied.

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Beyond this point, enzyme availability limits the rate.

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Reaction rate equals amount of product formed divided by time, or amount of substrate used divided by time.

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For a fixed endpoint, relative rate can be estimated as one divided by time.

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With time in seconds, its unit is per second.

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Enzyme practical: investigate P H using amylase and starch.

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Use buffer solutions to change P H and sample at regular intervals onto iodine on a spotting tile.

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Iodine turns blue-black if starch remains.

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When a sample stays orange-brown, the starch has been broken down.

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Record the time to reach this endpoint.

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For a fair P H investigation, keep temperature, enzyme and starch concentrations, volumes and sampling intervals constant.

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Repeat measurements, calculate a mean and compare rates; use eye protection and follow safe handling guidance.

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Diffusion is the overall (net) movement of particles from a region of higher concentration to a region of lower concentration.

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This is movement down a concentration gradient.

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It happens because particles move randomly.

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Particles move randomly in both directions, but more move from high to low concentration than the other way round.

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Once the concentrations are equal (at equilibrium), particles still move, but there is no overall movement in either direction.

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Diffusion is faster with a steeper concentration gradient, higher temperature, larger surface area or shorter diffusion distance.

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It does not require energy from respiration.

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Osmosis is the net movement of water through a partially permeable membrane from a more dilute solution (higher water concentration) to a more concentrated solution (lower water concentration).

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Small aqua dots represent water; large navy dots represent solute.

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Water moves both ways, with a net movement to the right.

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A partially permeable membrane lets some particles through but not others.

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In osmosis, water crosses the membrane; it is not the net movement of solute.

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In a dilute solution, water enters a plant cell by osmosis and it becomes firm (turgid).

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The cell wall stops it bursting.

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In a concentrated solution, water leaves and the cell becomes limp (flaccid).

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If it loses enough water, the cell membrane pulls away from the wall: this is plasmolysis.

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An animal cell may swell and burst when too much water enters, or shrink when water leaves.

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It has no cell wall to resist swelling.

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Active transport moves substances against their concentration gradient, from lower to higher concentration, using carrier proteins in the cell membrane.

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Active transport requires energy transferred by respiration.

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For example, root hair cells can absorb mineral ions from dilute soil solutions against a concentration gradient.

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Cut potato cylinders of equal size.

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Measure their initial masses, place them in a range of sucrose concentrations, and leave them for the same amount of time.

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Remove the cylinders, gently blot off surface solution in the same way, and measure final mass.

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Keep temperature, solution volume and potato size/type controlled; take care with cutting equipment.

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Percentage change in mass equals final mass minus initial mass, divided by initial mass, times one hundred.

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Subtract the masses first.

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A positive result means mass gain.

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A negative result means mass loss.

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A potato increasing from two point zero grams to two point three grams has a percentage mass change of plus fifteen percent.

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Water entered its cells by osmosis.

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A decrease from two point zero grams to one point eight grams gives minus ten percent.

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Repeat each concentration and calculate a mean.

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Plot percentage mass change against sucrose concentration; where the graph crosses 0 percent estimates the concentration giving no net water movement.

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Prepare a food solution by crushing a sample with distilled water and filtering if necessary.

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Use separate portions for each test so reagents do not interfere.

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Iodine solution changes from orange-brown to blue-black when starch is present.

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A negative result stays orange-brown.

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Use separate samples for the four tests.

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For reducing sugars, add Benedict’s solution and heat in a hot water bath.

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A positive result forms a green, yellow, orange or brick-red precipitate; a negative result remains blue.

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For protein, add the alkaline Biuret reagents: potassium hydroxide, then a small amount of copper sulfate solution.

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A lilac or purple colour is a positive result, showing the peptide bonds that join amino acids in proteins.

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A negative result stays blue.

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For lipids, shake the sample with ethanol, then add water.

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A cloudy white emulsion indicates lipid; a clear mixture is a negative result.

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Ethanol is flammable: keep it away from flames.

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Use known positive samples and distilled-water negative controls, equal sample volumes, clean apparatus and consistent heating.

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Wear eye protection; follow the teacher’s risk assessment for the alkaline reagents and hot water.

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Burn a weighed food sample beneath a boiling tube containing a known mass of water.

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Measure the water’s temperature before and after heating; reweigh the food to find the mass actually burned.

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Energy transferred to water equals mass of water times specific heat capacity times temperature rise.

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If mass is in grams and water’s heat capacity is 4.2 joules per gram per degrees Celsius, the result is joules.

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Divide the energy gained by the water by the mass of food burned to estimate energy per gram.

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For 20 grams water heated by 15 degrees Celsius using 0.5 grams food, energy equals 1260 joules and energy per gram equals 2520 joules per gram.

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The estimate is usually too low: energy heats the air and apparatus, combustion may be incomplete, and water may evaporate.

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A lid, insulation and a draught shield reduce some losses; repeat measurements and compare foods fairly.

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That completes Key concepts in biology.

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Revisit the notes and test yourself on the revision website.
