Biology 5090/32 — May/June 2016

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Experimental Contexts · Analysis, Conclusions and Evaluation · Observations and Measurements · Use of Techniques, Apparatus and Materials · Planning Experiments and Investigations · Microscopy and Biological Drawing

Q121MExperimental ContextsObservations and MeasurementsUse of Techniques, Apparatus and MaterialsPlanning Experiments and InvestigationsAnalysis, Conclusions and EvaluationFree sample

(a)

Dehydrogenase is an enzyme found in cells such as yeast.

Methylene blue is an indicator that can be used to show the activity of dehydrogenase.

In the presence of active dehydrogenase, methylene blue changes colour from blue to colourless.

You are required to carry out an experiment to investigate the effect of temperature on the activity of dehydrogenase in yeast cells.

You have been provided with two yeast suspensions, A and B. Yeast suspension B has been heated to boiling and then cooled.

Prepare a water bath by half filling a beaker with water and adjusting the temperature to $25\,^\circ\text{C}$ .
Label one test-tube A and one test-tube B.
Put $5\text{ cm}^3$ of yeast suspension A into the test-tube labelled A.
Put $5\text{ cm}^3$ of yeast suspension B into the test-tube labelled B.
Stand both test-tubes in the beaker of water kept at $25\,^\circ\text{C}$ . Leave the test-tubes in the water for 5 minutes.
After 5 minutes, add $1\text{ cm}^3$ of methylene blue solution to each test-tube, mix carefully and leave the test-tubes in the water.
Observe the colour of the contents of the test-tubes every minute for a period of 5 minutes, while keeping the water temperature at $25\,^\circ\text{C}$ .
Repeat this procedure using fresh yeast suspension and a beaker of water kept at $35\,^\circ\text{C}$ .

(i)

Record all your observations in Table 1.1.

Table 1.1

temperature of water / $^\circ\text{C}$	test-tube	observations at 1 min	observations at 2 mins	observations at 3 mins	observations at 4 mins	observations at 5 mins
25	A
25	B
35	A
35	B

DifficultyMedium-Easy

Worked solution

Answer

temperature of water / $^\circ\text{C}$	test-tube	observations at 1 min	observations at 2 mins	observations at 3 mins	observations at 4 mins	observations at 5 mins
25	A	blue	blue	slightly less blue	paler blue	almost colourless
25	B	blue	blue	blue	blue	blue
35	A	blue	paler blue	very pale / almost colourless	colourless	colourless
35	B	blue	blue	blue	blue	blue

All boxes completed; tube B shows no colour change at either temperature; tube A changes from blue towards colourless at both temperatures; the change is more rapid at $35\,^\circ\text{C}$ than at $25\,^\circ\text{C}$ .

Final answer

See working — completed Table 1.1 showing no colour change in B, gradual decolourisation in A, faster at 35 °C

Detailed explanation

Walkthrough

This is a candidate-dependent observation part: you write down what you actually saw. The mark scheme expects four things — every box filled in, no colour change in tube B, a colour change in tube A at both temperatures, and a faster change in A at $35\,^\circ\text{C}$ than at $25\,^\circ\text{C}$ . Methylene blue is blue when oxidised and turns colourless when it is reduced by active dehydrogenase in living yeast cells. Boiled yeast has dead cells with denatured enzymes, so it cannot reduce the dye and stays blue. Warmer temperature (within limits) speeds up enzyme activity, so the dye loses its colour sooner at $35\,^\circ\text{C}$ .

Key Takeaways

Methylene blue is a redox indicator: blue = oxidised, colourless = reduced by active dehydrogenase.
A table of observations must have every cell completed to earn the first mark.
Comparisons between tubes and temperatures are themselves marking points.

Common Mistakes

Leaving boxes blank because 'nothing happened' — write 'blue' or 'no change'; an empty box scores nothing.
Recording a colour change in tube B — boiled yeast is dead, so the scheme expects no change.
Writing vague terms like 'it changed' instead of naming the colours (blue → colourless).

Things to Be Careful About

Give exactly the pattern the scheme credits: no change in B, change in A at both temperatures, faster at $35\,^\circ\text{C}$ .
Your own readings must be internally consistent — if you wrote 'colourless' at 1 minute at $25\,^\circ\text{C}$ but slower at $35\,^\circ\text{C}$ , the comparison mark is lost.

Techniques used

record qualitative colour observations in a tablecompare the boiled and unboiled yeast suspensionscompare the rate of colour change at two temperatures

(ii)

Explain how you were able to check the temperature of the water in the beaker.

______

DifficultyEasy

Worked solution

Answer

Use a thermometer placed in the water of the beaker to read the temperature.

Final answer

Use a thermometer (or temperature probe) in the water

Detailed explanation

Walkthrough

To check the temperature of the water bath you need a measuring instrument in contact with the water. A thermometer (the scheme also accepts a temperature probe) stood in the beaker lets you read the water temperature directly.

Key Takeaways

Name the instrument, not just the action: 'thermometer' is the credited word.

Common Mistakes

Saying 'I looked at the thermometer' without stating that a thermometer was used to check the water.
Confusing the thermometer reading with controlling the temperature — this part only asks how you CHECKED it.

Things to Be Careful About

One mark only — one clear sentence is enough; do not pad.

Techniques used

identify apparatus used to measure temperature

(iii)

Explain how you kept the temperature of the water in the beaker at $25\,^\circ\text{C}$ and at $35\,^\circ\text{C}$ .

______

DifficultyEasy

Worked solution

Answer

Add small amounts of hot water (heat the beaker) if the temperature falls, or add cold water / ice if it rises, keeping the thermometer reading at $25\,^\circ\text{C}$ (and $35\,^\circ\text{C}$ ).

Final answer

Heat the water or add hot water if too cool; add cold water/ice if too warm

Detailed explanation

Walkthrough

Checking is not the same as keeping constant. Water cools by heat loss to the surroundings, so during the 5 minutes you must adjust it: warm it gently (Bunsen flame moved away between readings, or add hot water) when it drops below the set value, and add cold water or ice if it rises above. This part asks for any suitable method of adjustment — one mark.

Key Takeaways

Maintaining a water-bath temperature means actively correcting deviations in both directions.

Common Mistakes

Repeating the answer to (ii): 'I used a thermometer' does not explain how the temperature was KEPT constant.
Giving only one direction of correction when a full answer mentions both heating and cooling.

Things to Be Careful About

The scheme accepts 'suitable method of heating / add cold water / add ice' — any one workable method scores, but mentioning both directions reads best.

Techniques used

describe how a constant water-bath temperature is maintained

(iv)

Explain why the test-tubes were left in the beaker of water for 5 minutes before adding methylene blue solution.

______

DifficultyMedium-Easy

Worked solution

Answer

To allow the yeast suspension / enzyme / contents of the test-tubes to reach the temperature of the water bath (equilibration), so the reaction starts at the set temperature.

Final answer

To let the yeast/enzyme reach the water-bath temperature before the reaction starts

Detailed explanation

Walkthrough

If methylene blue were added straight away, the yeast would still be warming up, so the first minutes of the reaction would happen at temperatures below the intended one. Standing the tubes in the bath for 5 minutes lets the contents equilibrate — reach the same temperature as the water — so the timed reaction genuinely occurs at $25\,^\circ\text{C}$ or $35\,^\circ\text{C}$ . This makes the comparison between temperatures fair.

Key Takeaways

Pre-equilibration ensures the independent variable (temperature) applies to the whole measured period.
The credited idea is 'adjust to temperature' / equilibration.

Common Mistakes

Saying 'to give the yeast time to respire' — the purpose is thermal equilibration, not starting the reaction early.
Vague answers like 'for accuracy' without naming what adjusts to what.

Things to Be Careful About

Use the word 'temperature' explicitly: contents adjusting TO THE TEMPERATURE of the water is the mark.

Techniques used

explain the purpose of equilibration before starting timing

(v)

Suggest an explanation for your observations.

______

DifficultyMedium

Worked solution

Answer

In tube B the yeast was boiled, so the dehydrogenase enzymes are denatured / inactive (yeast dead), so methylene blue stays blue.
In tube A the enzymes are active, so they reduce methylene blue and it turns colourless.
At $35\,^\circ\text{C}$ the enzyme activity is greater than at $25\,^\circ\text{C}$ (closer to its optimum), so the colour changes faster.

Final answer

Enzymes denatured/inactive in B; active in A; higher activity at 35 °C so faster colour change

Detailed explanation

Walkthrough

Each observation needs its reason. Tube B never changes colour because boiling denatured the dehydrogenase — the heat altered the shape of the active sites, so the enzyme can no longer function (equivalently, the yeast cells are dead). Tube A contains living yeast with working enzymes, which reduce methylene blue to its colourless form. Between the two live-yeast temperatures, $35\,^\circ\text{C}$ gives a faster change than $25\,^\circ\text{C}$ because enzyme activity increases with temperature up to the optimum — molecules move faster and enzyme–substrate reactions occur more often. Three separate points, three marks.

Key Takeaways

Denaturation explains why boiled yeast fails; active enzymes explain the colour loss; temperature explains the rate difference.
Every 'suggest an explanation' point should pair observation with cause.

Common Mistakes

Saying the enzymes in B were 'killed' without linking to inactive/denatured enzymes — the scheme wants the enzyme made inactive.
Omitting one of the three points, especially the ORA comparison between $25\,^\circ\text{C}$ and $35\,^\circ\text{C}$ .
Claiming $35\,^\circ\text{C}$ denatures the enzyme — it does not; it increases activity.

Things to Be Careful About

Use 'denatured' or 'inactive' for tube B, not just 'dead yeast'.
State the direction of the temperature effect clearly: increased activity at $35\,^\circ\text{C}$ (or the reverse argument).

Techniques used

explain observations using enzyme denaturationexplain the effect of temperature on enzyme activity

(b)

Giving full experimental details, describe an experiment you could carry out to investigate the effect of pH on the activity of dehydrogenase, using yeast suspension and methylene blue solution.

DifficultyMedium-Hard

Worked solution

Answer

Prepare yeast suspensions at different pH values using buffers (e.g. pH 4, 6, 7, 8) — or add measured amounts of acid or alkali to make different pH values.
Add an equal, stated volume of each buffer solution to equal, stated volumes of yeast suspension (e.g. $5\text{ cm}^3$ yeast + $2\text{ cm}^3$ buffer in each tube).
Stand all tubes in a water bath kept at a stated constant temperature (e.g. $30\,^\circ\text{C}$ ) for 5 minutes.
Add the same stated volume ( $1\text{ cm}^3$ ) of methylene blue solution to each tube and mix.
Record the time taken for the blue colour to disappear (change to colourless) in each tube.
Repeat each pH and calculate mean times; compare times across pH values.

Final answer

See working — plan using buffers at ≥3 pH values, equal volumes, constant temperature, measuring time for decolourisation

Detailed explanation

Walkthrough

A planning question is marked against a checklist, and the scheme lists six creditable points for five marks, so hit every category:

Independent variable: pH — created with buffer solutions (or acid/alkali), covering at least three values across a sensible range.
Controlled variables: equal volumes of yeast, buffer and methylene blue, and a fixed temperature — name them specifically.
Dependent variable: the time taken for the blue colour to disappear, which measures enzyme activity.
Reliability: repeats and means.
Write it as a numbered method someone could follow, with actual volumes and temperatures stated rather than 'some' or 'suitable'.

Key Takeaways

Buffers are the standard way to hold solutions at set pH values.
Every quantity in a plan should be stated or described as equal/stated volume.
The dependent variable here is TIME for a colour change, not the colour itself.

Common Mistakes

Using only one or two pH values — the scheme requires a minimum of three.
Forgetting to state that volumes are equal, or omitting the controlled temperature.
Describing observing colour every minute instead of timing how long decolourisation takes.
Not adding the buffer to the yeast — the scheme credits '(buffer) added to yeast'.

Things to Be Careful About

Underline-worthy words: different pH values, equal/stated volumes, stated temperature, time taken for colour to disappear.
Five marks available from six listed points — include all six categories to be safe.

Techniques used

plan an investigation varying pHidentify variables to controlstate how results are measured

(c)

Yeast is used for the production of alcohol in the brewing industry.

In an investigation, the production of alcohol by yeast was recorded every 5 hours for 25 hours.

The results are shown in Table 1.2.

Table 1.2

time / hr	alcohol concentration / $\text{g per dm}^3$
0	0.0
5	3.7
10	5.6
15	6.8
20	7.5
25	7.8

(i)

Construct a line graph of the data in Table 1.2 on the grid below.

Draw a smooth curve through your points.

Fig. 1.1

DifficultyMedium

Worked solution

Answer

x-axis: time / hr, linear scale from 0 to at least 25; y-axis: alcohol concentration / g per dm $^3$ , linear scale from 0 to at least 8 — both axes fully labelled with units.
Scales chosen so more than half the grid is used in both directions.
All six points plotted correctly.
A single smooth curve drawn through all the points (rising steeply then levelling off).

Final answer

Line graph: time on x-axis, alcohol concentration on y-axis, both labelled with units, >half grid used, all points plotted, smooth curve

Detailed explanation

Walkthrough

Graph construction on Paper 3 earns marks for conventions, not artistic skill:

Put the independent variable (time) on the x-axis and the dependent variable (alcohol concentration) on the y-axis, and label BOTH axes fully including units ('time / hr', 'alcohol concentration / g per dm $^3$ ').
Choose linear scales starting at 0 that use more than half the grid in each direction — e.g. 2 cm per 5 hours horizontally and 2 cm per 1 g per dm $^3$ vertically.
Plot each of the six points precisely (within half a small square).
Join them with ONE smooth curve — not dot-to-dot straight segments — rising steeply at first and flattening after about 20 hours.

Key Takeaways

Axis labels must carry the slash-form unit exactly as the table prints it.
Scale choice is itself a mark: start at 0, fill more than half the grid.
'Smooth curve' means a flowing best-fit line through the points, not connecting them with rulers.

Common Mistakes

Swapping the axes or leaving off units.
Starting an axis at a non-zero value or choosing a scale that crams the plot into a corner.
Joining points with ruled straight lines when a smooth curve is demanded.
Extrapolating the curve beyond 25 hours — never draw beyond your data.

Things to Be Careful About

Plot points to within half a small square; a misplaced point costs the plotting mark.
Keep the curve thin and continuous, in sharp pencil.

Techniques used

construct a line graph with labelled axes and linear scalesplot points accuratelydraw a smooth curve through points

(ii)

Use your graph to find the concentration of alcohol after 12 hours.

concentration = ______

DifficultyMedium-Easy

Worked solution

Working

Read up from 12 hours on the x-axis to the curve, then across to the y-axis.

\text{concentration at 12 hr} \approx 6.2\text{ g per dm}^3

Answer

concentration = $6.2$ g per dm $^3$

Final answer

6.2 g per dm³ (any value consistent with the candidate's graph)

Detailed explanation

Walkthrough

12 hours lies between the plotted points at 10 and 15 hours, so this is interpolation: draw a vertical construction line up from 12 on the x-axis until it meets your curve, then a horizontal line across to the concentration axis. On a well-drawn smooth curve this lands at about 6.2 g per dm $^3$ . The mark scheme accepts any answer consistent with YOUR graph, so an honest read-off from a slightly different curve still scores. Two marks: the value and the unit.

Key Takeaways

Interpolation means reading between plotted points using construction lines.
Always quote the unit printed on the axis: g per dm $^3$ .

Common Mistakes

Reading from the wrong curve or misreading the scale division.
Omitting the unit — that alone loses one of the two marks.
Guessing a value inconsistent with their own plotted curve.

Things to Be Careful About

Draw light construction lines on the graph to show where the reading was taken — examiners look for them.
The accepted value follows your graph (ecf), but it must sit sensibly between 5.6 and 6.8.

Techniques used

read a value off a graph by interpolationstate the unit with the answer

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