Biology › Practicals › The Hill reaction: DCPIP and isolated chloroplasts
The Hill reaction: DCPIP and isolated chloroplasts
DCPIP is blue when it is oxidised and colourless when it is reduced. Put it in a suspension of isolated chloroplasts and it takes electrons from the electron transport chain in place of NADP, so in the light the blue fades and in the dark it stays. Timing that fade measures the rate of the light-dependent stage on its own, in a tube where no Calvin cycle is running.
The capability this develops
This guide is one investigation under capability P11: Measuring respiration and photosynthesis rates, which sets out the techniques the investigation belongs to, the calculations it carries and the errors that cost marks in all of them. The notes behind the biology are in Photosynthesis and primary productivity.
Where this sits on the specifications
This library maps a qualification to its own specification one numbered heading at a time, against a document whose sha256 is recorded, and publishes the result as a route map. These are the headings whose practical work this investigation serves, on the five qualifications mapped so far. Each links to the route map the row belongs to.
| Qualification | Heading |
|---|---|
| AQA 7401/7402 | 3.5.1 Photosynthesis |
| OCR H020/H420 | 5.2.1 Photosynthesis |
| Cambridge International 9700 | 13.2 Investigation of limiting factors |
| Pearson Edexcel Biology A (Salters-Nuffield) 8BN0/9BN0 | 5.7 The light-dependent reactions |
| Pearson Edexcel Biology B 8BI0/9BI0 | 5.6 i Photosynthetic pigments |
Pearson prints its core practical activities among the statements of the specification itself, so that list was read here with the rest of it: this investigation is Pearson Edexcel Biology A (Salters-Nuffield) 8BN0/9BN0 core practical 11.
Every heading in the table is A level content, so this investigation is not part of any of the AS courses.
The other qualifications above are matched by specification heading rather than by activity number. Their practical activity lists have not yet been mapped here number by number, so the lack of a number means only that, not that anything was checked and found not to match. Use your awarding organisation’s current specification and its practical handbook as the authority on which activities your course requires and how they are assessed.
The principle
In an intact chloroplast the electron that leaves photosystem II travels along the carriers of the thylakoid membrane and is finally handed, with a proton, to NADP. An artificial electron acceptor intercepts it before that happens. DCPIP is the usual one: it is a redox indicator, blue while oxidised and colourless once reduced, so the amount of it that has taken an electron can be read as a colour.
Robert Hill showed in 1937 that illuminated chloroplasts reduce such an acceptor and release oxygen while doing it, with no carbon dioxide supplied and no carbon being fixed. It was early evidence that the oxygen of photosynthesis comes from the photolysis of water rather than from carbon dioxide, and the reaction carries his name. Some specifications describe the same investigation as measuring the rate of dehydrogenase activity in extracts of chloroplasts.
Two consequences follow, and both are examined. Because DCPIP has replaced NADP, no reduced NADP leaves the membranes, so nothing drives the Calvin cycle and nothing in the tube fixes carbon. So the rate measured here is the rate of the light-dependent stage, and carbon dioxide concentration is not a variable in this experiment at all.
- Redox indicator
- A substance whose colour depends on whether it has been oxidised or reduced, used to follow electron transfer that would otherwise be invisible.
- Artificial electron acceptor
- A substance added to a preparation to take electrons from an electron transport chain in place of the acceptor that normally takes them.
- The Hill reaction
- The light-driven reduction of an added electron acceptor by illuminated chloroplasts, with oxygen released from water.
Apparatus
- Fresh dark-green leaves, spinach or lettuce, about 40 g once the midribs are cut out
- Isolation medium, ice-cold: sucrose at 0.4 mol dm⁻³ in a phosphate buffer at pH 7.0
- A blender, four layers of muslin, a funnel and two beakers, all chilled, and an ice bath to stand them in
- A centrifuge and balanced tubes
- DCPIP solution at 1 × 10⁻⁴ mol dm⁻³
- Graduated pipettes or syringes, 1 cm³ and 5 cm³
- Test tubes, a rack that holds them upright, and aluminium foil
- A water bath at 20 °C, deep enough to cover the liquid in the tubes
- A lamp of fixed power on a stand, and, unless the lamp is an LED that gives out little heat, a heat filter: a flat-sided flask of water
- A metre rule and a stopwatch
- Eye protection and a lab coat
Safety
Work to your centre’s own risk assessment. The blender is the sharpest thing on this bench and is unplugged while the jar is off. The centrifuge is run only with its tubes balanced and the lid shut, and is not opened until the rotor has stopped. Boiling the control tube is done in a water bath rather than over a flame, and the hot tube is lifted out with a holder because boiling water scalds. DCPIP stains skin and clothing: check its hazards on the CLEAPSS Hazcard or in your centre’s risk assessment, wear eye protection, and wash any splash off at once. A filament or halogen lamp gets hot enough to burn, so move it by its stand, and keep the lamp and its cable clear of the water bath.
Variables
Independent variable
Relative light intensity, set by the distance d from the lamp to the centre of the tube and measured with a metre rule. Six distances between 0.10 m and 0.60 m. Intensity falls as 1 ÷ d² for a point source, so the relative intensity of each tube is quoted as (0.60 ÷ d)², which makes the furthest tube 1.00 and gives a quantity with no units.
Dependent variable
The rate of DCPIP reduction, taken as 1 ÷ t, where t is the time in seconds from the suspension going into the tube until the tube cannot be told apart from the colour standard. The unit is s⁻¹. Rate, not time: a longer time is a slower reaction, so a graph of t against intensity falls where the rate is rising.
Control variables
- The amount of DCPIP in each tube. Every tube except the colour standard gets 1.0 cm³ of the same solution, which is 1.0 × 10⁻⁷ mol of dye. The reading is the time to reduce a fixed amount, so if the amount differs the times are not comparable.
- The volume and concentration of the chloroplast suspension. One preparation, 1.0 cm³ per tube, swirled immediately before each sample is drawn because the chloroplasts settle.
- Temperature. Held by the water bath, with the liquid in every tube below the water surface, and protected from the lamp by the heat filter.
- pH. Held by the buffer in the isolation medium. The carriers work over a narrow range and a drifting pH would change the rate for a reason nobody recorded.
- The endpoint. One colour standard, one person judging against it, for every tube.
- Background light. Blinds down, room lights off, and the bath screened from the next bench, because any light the metre rule did not measure is intensity that is not in the independent variable.
- The lamp, the bulb and its setting. Changing any of them changes the intensity at every distance at once.
- The age of the preparation. A chloroplast suspension loses activity as it stands, so the tubes are run in a randomised order and not from far to near.
Method
Isolating the chloroplasts, at 0 °C to 4 °C throughout
- Chill the isolation medium, the blender jar, the funnel and both beakers in an ice bath. Everything from here is done cold and quickly: breaking the cells releases hydrolytic enzymes that will digest the chloroplasts, and cold slows them down.
- Cut the midribs out of about 40 g of leaves, tear the laminae up and drop them into the chilled jar with 100 cm³ of ice-cold medium.
- Blend in three bursts of 5 s. Longer than that shears the thylakoid membranes, which are the part of the chloroplast this experiment needs intact.
- Filter through four layers of muslin into a chilled beaker, squeezing gently. The muslin holds back cell wall debris and whole pieces of leaf.
- Balance two centrifuge tubes with the filtrate and spin at about 200 g for 2 minutes. Discard the pellet, which is whole cells and wall fragments, and keep the supernatant.
- Spin the supernatant at about 1500 g for 10 minutes. The green pellet is the chloroplast fraction. Pour the supernatant away.
- Resuspend the pellet in 30 cm³ of ice-cold medium, drawing it gently in and out of a pipette rather than shaking it. Keep it on ice, in the dark, and use it within the hour.
Setting up the tubes
- Set the water bath to 20 °C. Stand the lamp at one end of the bench with the heat filter between it and the bath, and put the rack in the bath so that the liquid in the tubes will be below the water surface.
- Every tube finishes at 5.0 cm³, and every tube that carries dye carries the same amount of it: 1.0 cm³ of DCPIP solution, 3.0 cm³ of isolation medium and 1.0 cm³ of suspension. Tube 4, the colour standard, takes 4.0 cm³ of medium and 1.0 cm³ of suspension with no DCPIP, so it is the colour a tube reaches when all of its dye has been reduced. That colour is the green of the chloroplasts: the sample starts blue-green and finishes green, never colourless.
- Make up tube 2, the dark control, and tube 4, the colour standard, now. Wrap tube 2 in aluminium foil, over the top as well as round the sides. Both stand in the bath for the whole session.
- Boil about 3 cm³ of the suspension in a water bath for 2 minutes and cool it on ice, then make up tube 3, the boiled control, with it. That tube stands in the bath too.
- Make up one tube 1 for every reading you intend to take, which is three at each of six distances: 1.0 cm³ of DCPIP and 3.0 cm³ of medium in each, and no suspension yet. The suspension goes into tube 1 at the moment the stopwatch starts, because that is the moment the reaction begins.
Taking the readings
- Work through the six distances in an order chosen at random. Running them from far to near lets the preparation age in step with the independent variable, and the fall in rate that produces cannot afterwards be separated from the effect of the distance.
- Set the lamp at the first distance, measured with the metre rule from the lamp to the centre of tube 1.
- Swirl the suspension, add 1.0 cm³ of it to a fresh tube 1, start the stopwatch as it goes in, invert the tube once and stand it in the bath.
- Compare tube 1 with the colour standard where they stand, without lifting either out. A tube lifted out of the beam is a tube whose light intensity you have just changed, and one lifted out of the bath is one whose temperature you have changed with it.
- Stop the stopwatch the first time the two tubes cannot be told apart, and record the time in seconds.
- Repeat each distance three times with a fresh tube and take the mean. Run the three repeats for a distance together, in a row across the beam at the same distance from the lamp, started 20 s apart. Taken one after another, the eighteen readings need about 50 minutes of timing alone, which leaves no room inside the hour the preparation stays active.
- Run the first distance again at the end of the session. If its time has risen, the preparation lost activity while you worked, and you now know by how much rather than hoping it did not.
- Record the two controls at the end. Tubes 2 and 3 should both still be blue.
Sample data
The readings below are idealised. They were computed from a model of the relationship rather than measured, so that the analysis can be worked without scatter in the way. Real readings scatter, and a table invented to look real teaches you to read noise that nobody measured. Nothing here is a record of an experiment and none of it should be quoted as one.
| Distance d / m | Relative light intensity | Time to the endpoint t / s | Rate 1 ÷ t / 10⁻³ s⁻¹ |
|---|---|---|---|
| 0.60 | 1.00 | 500 | 2.00 |
| 0.40 | 2.25 | 222 | 4.50 |
| 0.30 | 4.00 | 125 | 8.00 |
| 0.20 | 9.00 | 58 | 17.2 |
| 0.15 | 16.0 | 51 | 19.6 |
| 0.10 | 36.0 | 50 | 20.0 |
Neither control reached the endpoint. Tube 2, in the dark, and tube 3, boiled, were both still blue after 600 s, and both are recorded that way rather than left out.
Take the third row. The relative light intensity is (0.60 ÷ 0.30)² = 2.00² = 4.00, which has no units because it is a ratio of one intensity to another. The rate is 1 ÷ 125 s = 0.00800 s⁻¹, which is 8.00 × 10⁻³ s⁻¹.
Every timed tube held 1.0 cm³ of DCPIP at 1 × 10⁻⁴ mol dm⁻³, so every one had the same amount of dye to reduce: n = c × V = 1 × 10⁻⁴ mol dm⁻³ × 1.0 × 10⁻³ dm³ = 1.0 × 10⁻⁷ mol. Diluted into 5.0 cm³ that is a final concentration of 2 × 10⁻⁵ mol dm⁻³.
In a real set of results each time would be the mean of three repeats, with the three readings and their range shown beside it. Judging the endpoint by eye carries an uncertainty of a few seconds at any time. Taken as ±3 s, that is ±0.6 % of 500 s but ±6 % of 50 s, so the points on the plateau are the least precise in the table. Plot each mean with a bar showing its range; where the bars of two points overlap, the data do not show that those two rates differ.
Reading the graph
Plot the rate on the vertical axis against the relative light intensity on the horizontal axis. The rate is the dependent variable and the intensity is the one you set, which is the way round the axes go.
The straight part. Divide each rate by its intensity: 2.00 ÷ 1.00, 4.50 ÷ 2.25 and 8.00 ÷ 4.00 all give 2.00, in units of 10⁻³ s⁻¹ per unit of relative intensity. A constant ratio is what directly proportional means, so over the first three points the line is straight and passes through the origin, and light intensity is the factor limiting the rate there.
The bend and the plateau. By 9.00 units the ratio has fallen to 1.91 and by 36.0 units to 0.56. Between 16.0 and 36.0 units the intensity more than doubles and the time to the endpoint changes by one second, from 51 s to 50 s. A colour match judged by eye is less certain than that, so within the uncertainty of the measurement the rate has stopped rising, and light is no longer the limiting factor.
What is limiting instead. Whatever was held constant: the amount of chloroplast suspension in the tube, which sets how many electron transport chains there are to work, and the temperature, which sets how fast each one can turn over. It is not carbon dioxide. None was supplied, and there is no Calvin cycle to use whatever dissolved in from the air. Carbon dioxide can limit the plateau in a whole-plant experiment, but not in this one.
What the plateau is not. It is a property of this suspension at this temperature, not a ceiling on photosynthesis. Put twice the volume of suspension in the tube and the plateau rises.
The origin. The lowest intensity measured is 1.00, not zero, so a line taken back to the origin is an extrapolation rather than a reading. The dark control supports it, but it sets a limit rather than measuring a value. A tube still blue after 600 s has a rate below 1 ÷ 600 s = 1.7 × 10⁻³ s⁻¹, not far under the 2.00 × 10⁻³ s⁻¹ measured at the lowest intensity. Still blue at the end of a session of about an hour, it puts the limit below 0.3 × 10⁻³ s⁻¹, which is much stronger evidence that the line passes through the origin.
Changing the colour of the light. The same tubes can be run at one fixed distance with a coloured filter, or a single-colour LED, in place of white light. Filters pass different amounts of light as well as different colours, so put a light meter where the tube stands and move the lamp until the meter reads the same for every colour. Plot the rate for each colour as a bar chart. Red and blue light give faster rates than green, which follows the absorption spectrum of chlorophyll. Methylene blue can replace DCPIP: it too is blue when oxidised and colourless when reduced.
Evaluation
Each of these is a real source of error in this investigation, and the middle column is the part an answer earns marks for: what it does to the result, and in which direction.
| Source of error | What it does to the result | What to do about it |
|---|---|---|
| Judging the endpoint by eye | A colour match is a judgement, so repeats scatter and two people get different times for the same tube. It does not push the result in a particular direction. | Read the tubes on a colorimeter at about 600 nm, through a red filter because DCPIP is blue and absorbs red, and take a fixed fall in absorbance as the endpoint. |
| Light from the room and the next bench | It adds an intensity the metre rule never measured, and it is a larger share of the total at the far tubes than at the near ones. If it is about the same at every distance, every rate on the straight part comes out too high by about the same amount, so the line keeps its gradient but is lifted, and taken back it cuts the rate axis above the origin. | Blinds down, room lights off, and a screen between the bath and anything else that is lit. |
| Assuming intensity falls as 1 ÷ d² | That holds for a point source in a vacuum. A lamp with a filament of real size is furthest from being a point source at the shortest distances, and a reflector behind the bulb makes the beam spread less than a point source would, so the highest relative intensities, at the foot of the table, are overstated and the top of the curve belongs further to the left than it is drawn. | Put a light meter where the tube stands and use its reading as the independent variable instead of a calculation from distance. |
| Heating by the lamp | Without the heat filter the near tubes run warmer than the far ones, and a warmer tube reacts faster for a reason that is not light. It exaggerates the rise, and it can be large enough to hide the plateau altogether. | Keep the heat filter in place and measure the bath temperature at the nearest and the furthest lamp position, so the claim that it was held constant is a reading rather than an assumption. |
| The preparation losing activity as it stands | Whatever is run last is slowed by the age of the suspension rather than by its treatment. Run far to near and the fall of rate with distance is masked; run near to far and it is exaggerated. | Randomise the order, keep the suspension on ice in the dark, and repeat the first distance at the end so the loss is a number rather than a hope. |
| Chloroplasts settling in the beaker | Later samples then carry more or fewer chloroplasts than earlier ones, which moves the rate up or down at random rather than in one direction. | Swirl the suspension immediately before every sample is drawn. |
| One tube per distance | There is no measure of spread, so there is no basis for saying that two points differ rather than that one reading was unlucky. | Three repeats at each distance, a mean, and the range quoted beside it. |
| A chloroplast-rich fraction, not pure chloroplasts | Much of what is in the tube is broken chloroplasts and other membrane. That does no harm here, because the reaction happens on the thylakoid membranes and a broken chloroplast still has them, but it means the rate is per cm³ of one preparation and not per chloroplast. | Say what the rate is per, and compare rates only within a single preparation on a single day. |
| A boiled control that decolourises | If tube 3 loses its blue, something other than a working electron transport chain is reducing the dye, and every reading in the set is measuring that as well. | Stop and find it before analysing anything. Check that the suspension was held in boiling water for the full 2 minutes. Leaf extract also carries reducing substances such as ascorbate, which reduce DCPIP with no enzyme involved. |
Questions
Question 1 (3 marks)
Explain why the DCPIP loses its blue colour in tube 1 but not in tube 2.
Check your answer
DCPIP is blue when oxidised and colourless when reduced;
in tube 1, light excites electrons in the chlorophyll of photosystem II, and these pass along the electron transport chain and are accepted by DCPIP, which is reduced;
in tube 2 no light reaches the chlorophyll, so no electrons are excited or released and the DCPIP stays oxidised.
Question 2 (3 marks)
A student says the plateau at high light intensity is caused by a shortage of carbon dioxide. Explain why that cannot be the reason in this experiment.
Check your answer
DCPIP has replaced NADP as the electron acceptor, so no reduced NADP is produced;
the Calvin cycle therefore does not run, so no carbon dioxide is being used, and none was supplied to the tube;
so the plateau must be set by something else that was held constant, such as the amount of chloroplast suspension or the temperature.
Question 3 (2 marks)
The tube at 0.30 m reached the endpoint in 125 s. Calculate the rate of the reaction and give its unit.
Check your answer
rate = 1 ÷ 125;
= 8.00 × 10⁻³ s⁻¹ (0.00800 s⁻¹).
Question 4 (2 marks)
Use the table to show that the rate is directly proportional to relative light intensity over the first three rows.
Check your answer
rate ÷ intensity is 2.00 ÷ 1.00 = 2.00, 4.50 ÷ 2.25 = 2.00 and 8.00 ÷ 4.00 = 2.00, in 10⁻³ s⁻¹ per unit;
the ratio is constant, so the rate is proportional to the intensity and a line through the points passes through the origin.
Question 5 (3 marks)
Every timed tube received 1.0 cm³ of DCPIP solution at 1 × 10⁻⁴ mol dm⁻³. Calculate the amount, in moles, of DCPIP each tube had to reduce, and explain why the same amount in every timed tube matters.
Check your answer
1.0 cm³ = 1.0 × 10⁻³ dm³, so n = c × V = 1 × 10⁻⁴ × 1.0 × 10⁻³;
= 1.0 × 10⁻⁷ mol;
the measurement is the time to reduce a fixed amount of dye, so if the tubes held different amounts a longer time could mean more dye rather than a slower reaction.
Question 6 (3 marks)
The room lights were left on for one set of readings. Explain the effect this has on the plotted line.
Check your answer
the room adds light that the distance measurement does not account for;
this added light is a larger proportion of the total at the greater distances, so the rates at low relative intensity are higher than they should be;
a steady background adds about the same to every rate on the straight part, so the line keeps its gradient but is lifted and, extrapolated back, meets the rate axis above the origin instead of passing through it.
Question 7 (3 marks)
Explain why the chloroplasts are isolated in an ice-cold buffered solution of sucrose rather than in water.
Check your answer
the sucrose makes the medium close to isotonic with the chloroplast contents, so water does not enter by osmosis and burst them;
the buffer holds the pH, so the proteins of the electron transport chain are not denatured by a pH change;
the low temperature slows the hydrolytic enzymes released when the cells are broken, which would otherwise digest the chloroplasts.
Coverage maps are available for AQA 7401/7402, OCR Biology A H020/H420, Cambridge International 9700, Pearson Edexcel Biology A (Salters-Nuffield) 8BN0/9BN0 and Pearson Edexcel Biology B 8BI0/9BI0. Use the relevant route map to check what InkBiology covers, and your awarding organisation’s current specification as the final authority on examinable content.