Required practicals: A-level Biology AQA revision notes

RP1

Enzyme-controlled reaction rate

Enzyme rates, one factor at a time

Definitions

Independent variable: the factor changed, e.g. temperature (water bath) or pH (buffer solutions).
Dependent variable: time for the substrate to disappear, or the amount of product formed in a set time. Rate = $1/ t$ , or the gradient of a product–time graph.
Control variables: volume and concentration of enzyme and substrate, temperature (if not the IV), pH (buffer), and total volume.

Key results

Rate $= \frac{1}{time}$ (units s⁻¹), or the initial rate from the gradient of a tangent at $t = 0$ on a product–time graph.
Temperature coefficient $Q_{10} = \frac{rate at ( T + 10 )}{rate at T}$ ; about 2 below the optimum.

Method

Place the enzyme and substrate solutions in separate tubes in a water bath at the chosen temperature for about 5 minutes, so both start at that temperature.
Mix and start the timer at once.
Amylase and starch: every 30 s, transfer a drop of the mixture to iodine solution on a spotting tile; record the first time the iodine stays orange-brown (no starch left).
Repeat at least three times at each value of the independent variable and calculate a mean, discarding anomalies.
Set up a control with boiled (denatured) enzyme, to show that the change is caused by the enzyme.

In practiceStarch is fully broken down in 90 s at pH 7 and 150 s at pH 5. Compare the rates.

$rate = \frac{1}{t} : \frac{1}{90} = 0.0111, \frac{1}{150} = 0.0067 s^{- 1}$
$\frac{0.0111}{0.0067} = 1.7$ the rate at pH 7 is 1.7 times that at pH 5

Notes

Above the optimum, hydrogen and ionic bonds in the tertiary structure break, the active site changes shape, and fewer enzyme–substrate complexes form.
Sampling at 30 s intervals limits resolution: the true end point lies anywhere in the last interval. Shorter intervals or a colorimeter improve this.
Exam trap: 'the enzyme is killed'. Enzymes are proteins, not living things; they are denatured.

Measure how quickly substrate disappears or product appears, and change one factor at a time: temperature, pH, substrate concentration or enzyme concentration. Common systems are amylase breaking down starch (sampled into iodine), catalase releasing oxygen from hydrogen peroxide, and trypsin clearing a suspension of milk protein.

RP2

Mitosis in root tip squashes

Root tip squash: stages of mitosis and the mitotic index

Key results

Mitotic index $= \frac{number of cells in mitosis}{total number of cells counted}$ .
Time in a stage $= \frac{cells in that stage}{total cells} \times length of the cell cycle$ .
Stages: prophase (chromosomes condense, visible as threads), metaphase (chromosomes lined up on the equator), anaphase (chromatids pulled to opposite poles, V-shapes), telophase (two groups at the poles, nuclear envelopes reforming).

Method

Cut the terminal 5 mm of a growing root (e.g. garlic or onion), where the meristem is.
Warm in 1 mol dm⁻³ hydrochloric acid at about 60 °C for about 5 minutes: the acid breaks down the middle lamella between cells so they separate, and stops division.
Rinse in cold water, then place on a slide and add a stain such as toluidine blue (or acetic orcein), which binds to DNA and makes chromosomes visible.
Cover with a coverslip and press firmly (without twisting) to spread the cells into a single layer, so light passes through and cells do not overlap.
Examine at low power to find the meristem, then at high power (×400); count cells in each stage in a field of view, repeating in several fields.

In practiceOf 300 cells, 45 are in mitosis and 6 of those in anaphase. The cell cycle takes 24 hours. Find the mitotic index and the time spent in anaphase.

$mitotic index = \frac{45}{300} = 0.15$
$t_{anaphase} = \frac{6}{300} \times 24 \times 60 = 28.8 min$ proportion of cells × cycle length

Notes

Measure cells with an eyepiece graticule calibrated against a stage micrometer at the same magnification.
Safety: hydrochloric acid is an irritant (eye protection); cut away from your fingers on a tile.
Count only whole cells, and decide in advance how to treat cells on the edge of the field (e.g. count those touching two sides only).
Exam trap: counting interphase cells as 'not dividing' without including them in the total.

Cells divide by mitosis in the meristem just behind a root tip. Softening, staining and squashing a root tip spreads its cells into a single layer, so chromosomes are visible and the proportion of cells in each stage can be counted.

RP3

Water potential of plant tissue

Find the water potential of plant tissue

Key results

Percentage change in mass $= \frac{final - initial}{initial} \times 100$ ; percentage change allows pieces of different starting mass to be compared.
Read the concentration where the line crosses 0%; convert it to a water potential using a calibration table for sucrose solutions.

Method

Make a dilution series from a 1.0 mol dm⁻³ sucrose stock: for 20 cm³ of 0.4 mol dm⁻³, mix 8 cm³ of stock with 12 cm³ of distilled water. Prepare, for example, 0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³.
Cut cylinders of potato of the same diameter (cork borer) and length, so all have the same surface area.
Blot each dry, weigh it and place it in a labelled tube of one solution, fully covered.
Leave for the same time (e.g. 30 minutes or overnight) at the same temperature.
Remove, blot dry in the same way, and reweigh.
Calculate the percentage change in mass and plot it against concentration.

In practiceA cylinder weighs 3.20 g before and 3.52 g after soaking. Find the percentage change in mass and interpret it.

$\frac{3.52 - 3.20}{3.20} \times 100 = + 10%$
A gain: the solution had a higher (less negative) water potential than the tissue, so water entered by osmosis.

Notes

Blotting removes surface solution, which would add to the mass; blot every piece the same way.
Loss of water through evaporation while weighing, or uneven blotting, are random errors; using more concentrations near the intercept improves the estimate.
Exam trap: saying water moves 'from a high concentration'. Use water potential: water moves from higher (less negative) to lower (more negative) water potential.

Equal pieces of tissue are left in a dilution series of sucrose. Where the solution has a higher water potential than the tissue, the tissue gains water by osmosis and mass; where lower, it loses mass. The concentration at which the mass would not change has the same water potential as the tissue.

RP4

Membrane permeability

Leaky membranes: beetroot and pigment

Key results

Higher absorbance means more pigment has leaked out, so the membranes are more permeable.
With temperature: permeability rises gently, then sharply above about 40–50 °C.

Method

Cut equal-sized pieces of beetroot with a cork borer and scalpel.
Rinse them in water until no more colour comes out, to remove pigment released from cells cut open.
Place one piece in a fixed volume of water at each temperature (water baths) for a fixed time, e.g. 20 minutes.
Remove the beetroot and shake the tube to distribute the pigment.
Zero the colorimeter with distilled water, then measure the absorbance of each solution using a blue-green filter (red solutions absorb most strongly in the blue-green).
Repeat each temperature three times and calculate the mean.

In practiceRepeat absorbances: at 20 °C, 0.12, 0.10 and 0.14; at 60 °C, 0.85, 0.95 and 0.90. Find the means and compare.

$\overset{ˉ}{A}_{20} = \frac{0.12 + 0.10 + 0.14}{3} = 0.12$ range 0.10–0.14
$\overset{ˉ}{A}_{60} = \frac{0.85 + 0.95 + 0.90}{3} = 0.90$ range 0.85–0.95
The ranges do not overlap, so the higher permeability at 60 °C is unlikely to be chance variation.

Notes

Above that temperature, membrane proteins denature, leaving gaps; and phospholipids gain kinetic energy and become more fluid, so the bilayer is less tightly packed.
Ethanol (an organic solvent) dissolves phospholipids, so permeability rises with its concentration.
Exam trap: 'the membrane melts'. Describe denaturation of proteins and increased fluidity of the bilayer.

Beetroot cells store a red pigment (betalain) in their vacuoles. The pigment cannot cross intact membranes, so the amount that leaks into the surrounding water, measured with a colorimeter, indicates how damaged the tonoplast and cell-surface membrane are. Temperature or the concentration of a solvent such as ethanol is varied.

RP5

Dissection

Dissection of a gas exchange or mass transport system

Key results

Drawing rules: a sharp pencil; clear, continuous lines with no shading; drawn to fill at least half the space; label lines ruled, not crossing, and ending on the structure; a title and a magnification or scale.

Method

Wear gloves; place the specimen on a dissection board.
Use a sharp scalpel or scissors, cutting away from yourself and others, onto the board.
For a heart: identify the ventricles by their walls; insert a finger or probe into each vessel to trace where it leads; cut through each chamber to see the valves and tendons.
Make a labelled drawing of what you observe.
Dispose of the specimen as instructed, disinfect the tools and bench, and wash your hands.

In practiceA drawing of a heart valve is 90 mm long; the valve itself is 30 mm. Find the magnification of the drawing.

$magnification = \frac{drawing size}{actual size} = \frac{90}{30} = \times 3$ same units for both
Or draw a scale bar: a 30 mm bar on the drawing labelled 10 mm, which stays true if the drawing is enlarged or reduced.

Notes

Ethics: use specimens from a butcher or supplier (animals killed for food), treat them with respect, and avoid waste.
The left ventricle has a thicker muscular wall because it pumps blood at higher pressure around the whole body; the right ventricle pumps only to the lungs.
Exam trap: drawing what a textbook shows rather than what is in front of you; examiners reward an accurate drawing of the specimen.

Dissecting a heart, lungs or fish gills links structure to function: you see, for example, that the left ventricle wall is thicker than the right, or that each gill arch carries many filaments. The practical also assesses safe use of dissection tools and the conventions of a biological drawing.

RP6

Aseptic technique and antimicrobials

Aseptic technique and the effect of antimicrobial substances

Key results

Area of clear zone $= π r^{2}$ , with $r$ = half the measured diameter.
Compare the mean areas for each substance or concentration with the control.

Method

Disinfect the bench before and after; work near a lit Bunsen burner, whose updraught carries airborne microbes away.
Flame the neck of the culture bottle after opening and before closing it; sterilise the inoculating loop in the flame until red hot and let it cool.
Lift the Petri dish lid only slightly, at an angle, to add the culture; spread it evenly with a sterile spreader.
Place discs soaked in each substance (and a control disc soaked in sterile distilled water) with sterile forceps, evenly spaced.
Tape the lid with two strips (not sealed all round), label the base, and incubate upside down at 25 °C for 24–48 hours.
Measure the diameter of each clear zone and calculate its area.

In practiceA clear zone has a diameter of 14 mm. Find its area.

$r = \frac{14}{2} = 7 mm$ halve the diameter
$A = π r^{2} = π \times 7^{2} = 154 mm^{2}$

Notes

25 °C, not 37 °C: lower temperatures reduce the chance of growing human pathogens.
Not sealed all round, so oxygen can enter and anaerobic bacteria (which include dangerous species) are not encouraged; upside down, so condensation does not drip onto the agar.
The zone size also depends on how well the substance diffuses through agar, not only on how strongly it acts, which limits comparisons between different substances.
Exam trap: 'to kill all bacteria' as the purpose of aseptic technique. It prevents contamination of the culture by unwanted microorganisms and of the worker by the culture.

Bacteria are spread evenly over agar; paper discs soaked in antimicrobial substances (antibiotics, antiseptics, plant extracts) are placed on the surface. Where the substance diffuses out and kills or inhibits the bacteria, a clear zone forms. The larger the clear zone, the more effective the substance at that concentration. Aseptic technique stops contamination from the surroundings and protects the worker.

RP7

Chromatography of leaf pigments

Chromatography of photosynthetic pigments

Key results

$R_{f} = \frac{distance moved by pigment}{distance moved by solvent front}$ ; compare with known values for the same solvent.
Carotene, the least polar pigment, travels furthest; the order of the others depends on the solvent and the paper or plate.

Method

Grind leaves with a little sand and propanone (or another solvent) with a pestle and mortar to break the cells and dissolve the pigments.
Draw a pencil line about 1.5 cm from the bottom of the chromatography paper or TLC plate.
Spot the extract on the line with a capillary tube, letting each spot dry before adding the next, to build a small concentrated spot.
Stand the strip in solvent below the line in a covered tube; remove it before the solvent reaches the top and mark the solvent front at once.
Measure the distance from the line to the centre of each pigment spot and to the solvent front.

In practiceA pigment spot is 2.8 cm from the start line; the solvent front is 7.0 cm. Find its Rf.

$R_{f} = \frac{2.8}{7.0} = 0.40$ compare with known values for the same solvent
A spot with no match in the reference values may be a breakdown product of chlorophyll, such as phaeophytin; run fresh extract to check.

Notes

Shade-tolerant plants often have a higher proportion of chlorophyll b and accessory pigments, which absorb the wavelengths that pass through the canopy above.
Propanone is flammable and an irritant; keep it away from flames and work in a ventilated room.
Exam trap: drawing the start line in ink (it runs), or letting the solvent cover the spots (they dissolve into it).

Leaves contain several pigments: chlorophyll a, chlorophyll b, carotene and xanthophylls. Chromatography separates them because they differ in solubility in the solvent and in attraction to the paper or TLC plate. Comparing extracts from, say, shade-tolerant and sun plants shows differences in the pigments they contain.

RP8

Dehydrogenase activity in chloroplasts

Dehydrogenase activity in isolated chloroplasts

Key results

Rate $= 1/ t$ for decolourisation, or the gradient of absorbance against time.
Light intensity $\propto \frac{1}{d ^{2}}$ , where $d$ is the distance from the lamp: halving $d$ quadruples the intensity.

Method

Grind leaves in ice-cold isolation solution: buffered (to keep the pH constant), with the same water potential as the chloroplasts (so they do not burst or shrink by osmosis), kept cold (to slow enzymes that would damage the chloroplasts).
Filter, then centrifuge to separate the chloroplasts; resuspend them in a little isolation solution and keep on ice.
Mix chloroplast suspension with DCPIP in a tube and place it at a set distance from a lamp; time how long the blue colour takes to disappear, or record absorbance with a colorimeter at intervals.
Set up controls: a tube wrapped in foil (no light) and a tube with isolation solution instead of chloroplasts.
Repeat at different distances from the lamp.

In practiceDCPIP decolourises in 120 s with the lamp 15 cm away and 300 s at 30 cm. Compare light intensities and rates.

$\frac{I _{15}}{I _{30}} = (\frac{30}{15})^{2} = 4$ intensity ∝ 1/d²
$rates : \frac{1}{120} = 0.0083, \frac{1}{300} = 0.0033 s^{- 1}$ ratio 2.5
Four times the light gives only 2.5 times the rate: another factor is beginning to limit.

Notes

The dark control stays blue (no light energy, no electrons); the control without chloroplasts stays blue (no electron source). Together they show the change needs both light and chloroplasts.
A lamp heats the tube as well as lighting it; a heat filter (a beaker of water) between lamp and tube keeps the temperature constant.
Exam trap: saying DCPIP is oxidised. It accepts electrons, so it is reduced.

In the light-dependent reaction, electrons from chlorophyll are passed along an electron transport chain and normally reduce NADP. A blue dye, DCPIP, can take those electrons instead: it is reduced and turns colourless. The faster the dye decolourises, the faster the dehydrogenase-driven electron transfer in the chloroplasts. Light intensity, temperature or a herbicide can be varied.

RP9

Respiration rate in single-celled organisms

How fast does yeast respire?

Key results

Volume of oxygen absorbed $= π r^{2} l$ , with $r$ the capillary radius and $l$ the distance moved.
Rate of oxygen uptake = volume ÷ (time × mass of yeast), e.g. in mm³ g⁻¹ min⁻¹.

Method

Respirometer: put a known mass of yeast culture in the experimental tube above soda lime (which absorbs carbon dioxide), and glass beads of the same volume in the control tube.
Let the tubes equilibrate in a water bath at the chosen temperature for about 10 minutes with the tap open.
Close the tap and record the distance the coloured liquid moves along the capillary in a set time.
Methylene blue: add equal volumes of indicator to yeast suspensions at each temperature and time how long the blue colour takes to disappear.

In practiceIn a respirometer the liquid moves 18 mm in 6 minutes along a capillary of radius 0.5 mm. Find the rate of oxygen uptake.

$V = π r^{2} l = π \times 0. 5^{2} \times 18 = 14.1 mm^{3}$
$rate = \frac{14.1}{6} = 2.4 mm^{3} min^{- 1}$ divide by mass for a rate per gram

Notes

Soda lime absorbs carbon dioxide, so the change in gas volume is due to oxygen taken up alone.
The control tube compensates for changes in temperature and atmospheric pressure, which would otherwise move the liquid too.
Soda lime is corrosive; keep it away from the yeast and skin.
Exam trap: forgetting that anaerobic respiration releases CO₂ without taking up O₂, so a respirometer with soda lime records no movement for it.

Yeast respires glucose. Its rate can be followed by the oxygen it takes up (in a respirometer), the carbon dioxide it gives out (gas syringe or bubbles), or by the time a redox indicator such as methylene blue takes to decolourise as it is reduced by the electrons released. Temperature, substrate or concentration of yeast can be varied.

RP10

Animal responses: choice chamber or maze

Animal responses: choice chambers and mazes

Definitions

Taxis: a directional movement towards (positive) or away from (negative) a stimulus.
Kinesis: a non-directional response in which the rate of movement or turning changes with the intensity of the stimulus.

Key results

Chi-squared test for a difference from an even distribution: $χ^{2} = \sum \frac{( O - E ) ^{2}}{E}$ , with degrees of freedom = number of categories − 1.
If $χ^{2}$ exceeds the critical value at $p = 0.05$ , reject the null hypothesis that there is no preference.

Method

Set up the chamber with silica gel (dry) under one side and damp cotton wool (humid) under the other, keeping light even across it.
Place a set number of animals (e.g. ten) in the centre and cover; leave for a fixed time (e.g. 10 minutes).
Count the animals on each side; repeat several times with fresh animals or after resetting.
Handle animals gently with a soft paintbrush, minimise the time in the chamber, and return them to their habitat afterwards.

In practiceOf 30 woodlice, 9 end on the light side and 21 on the dark side. Test for a preference (critical value 3.84).

$E = 15 each side$ null hypothesis: no preference
$χ^{2} = \frac{( 9 - 15 ) ^{2}}{15} + \frac{( 21 - 15 ) ^{2}}{15} = 2.4 + 2.4 = 4.8$
4.8 > 3.84 (1 degree of freedom, p = 0.05): reject the null hypothesis; the woodlice move away from light more than chance predicts.

Notes

Control: run the chamber with identical conditions on both sides to check there is no bias from the apparatus itself.
Only one factor should differ between the sides: covering one side to make it dark may also change its humidity.
Exam trap: concluding 'woodlice like damp conditions' rather than describing the response and its survival value (avoiding desiccation).

Small invertebrates such as woodlice or maggots are given a choice between two conditions (dry and damp, light and dark) in a choice chamber, or a series of turns in a maze. Counting where they are after a set time shows whether they move towards or away from a stimulus, and a statistical test decides whether the pattern could be due to chance.

RP11

Glucose concentration by colorimetry

Finding glucose concentration with a calibration curve

Key results

Absorbance falls as glucose concentration rises (less blue copper(II) left); transmission rises.
Only read values within the range of the standards; outside it the curve is an extrapolation.

Method

Make a dilution series of glucose (e.g. 0, 2, 4, 6, 8 and 10 mmol dm⁻³) from a stock solution with distilled water.
Add the same volume of Benedict's reagent, in excess, to the same volume of each solution and of the unknown.
Heat all tubes in a boiling water bath for the same time.
Remove the precipitate by filtering or centrifuging, leaving the blue solution.
Zero the colorimeter with distilled water; measure the absorbance (or transmission) of each solution with a red filter.
Plot absorbance against glucose concentration and read off the concentration of the unknown.

In practiceThe calibration line is $A = 0.80 - 0.06 c$ . A sample gives $A = 0.56$ . Find its glucose concentration.

$0.06 c = 0.80 - 0.56 = 0.24$
$c = 4.0 mmol dm^{- 3}$ inside the range of the standards

Notes

Benedict's reagent must be in excess, otherwise above some concentration all the copper(II) is used and the curve flattens.
Use the same cuvette orientation and wipe the clear sides; fingerprints and scratches scatter light.
Exam trap: comparing colours by eye. A colorimeter gives a quantitative, less subjective reading.

Benedict's reagent is reduced by glucose, forming an orange-red precipitate of copper(I) oxide and using up the blue copper(II) ions. After heating, the more glucose there was, the paler the blue solution left. A colorimeter measures that blue solution for a series of known concentrations (a calibration curve), and the unknown sample is read off the curve.

RP12

Species distribution

Where a species grows, and why

Definitions

Random sampling: lay out two tapes at right angles and use random-number coordinates to place quadrats, avoiding bias in where samples are taken.
Systematic sampling: place quadrats at regular intervals along a belt (continuous) or interrupted transect across a change in conditions.
Abundance: percentage cover (for plants hard to count), frequency (proportion of quadrats in which it occurs) or density (number per m²).

Key results

Spearman's rank correlation for a relationship between two variables: $r_{s} = 1 - \frac{6 \sum d ^{2}}{n ( n ^{2} - 1 )}$ .
Chi-squared for an association between categories; Student's t-test for a difference between two means.
Compare the statistic with the critical value at $p = 0.05$ for the sample size.

Method

Decide the question and the type of sampling: random to compare two areas, a transect for a gradient.
Use enough quadrats that the running mean of abundance levels off.
In each quadrat record the abundance of the species and measure the abiotic factor (light meter, soil moisture probe, pH probe) at the same time.
Apply a statistical test.

In practiceFive quadrats give $\sum d^{2} = 2$ for ranks of soil moisture against plant cover. Find Spearman's coefficient.

$r_{s} = 1 - \frac{6 \sum d ^{2}}{n ( n ^{2} - 1 )} = 1 - \frac{6 \times 2}{5 \times 24} = 0.90$
Compare with the critical value for n = 5 before concluding; with so few quadrats it is high.

Notes

A correlation does not show that the factor causes the change in abundance; other factors may vary along the same gradient.
Estimating percentage cover is subjective; using a quadrat divided into 100 squares, and the same person, reduces this.
Exam trap: placing quadrats 'randomly' by throwing them, which is biased.

Measure how the abundance of a plant (or a slow-moving animal) changes with an abiotic factor such as light, soil moisture or trampling. Random sampling compares two areas; systematic sampling along a transect follows a gradient. A statistical test then tests whether the abundance is related to the factor.