5090/32

Biology 5090/32May/June 2011

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

3
questions
40
marks
75
minutes

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

Q111MExperimental ContextsUse of Techniques, Apparatus and MaterialsObservations and MeasurementsAnalysis, Conclusions and EvaluationFree sample

Read through the whole question before starting.

Do not taste the fruit sections provided.

(a)
(i)

Describe how you would carry out a test for reducing sugars using Benedict's solution and the results you would expect if reducing sugars were present.

3M
DifficultyMedium-Easy
Worked solution

Answer

  • Add an equal volume of Benedict's solution to the sample and heat the mixture (in a water-bath).
  • A positive result is a colour change from blue to green, yellow, orange or brick-red.
  • This shows that a reducing sugar is present.
Final answer

Heat equal volumes of the solution and Benedict's solution; blue changes to green/yellow/orange/red, showing a reducing sugar is present.

Detailed explanation

Walkthrough

Benedict's solution is the standard test for reducing sugars. The procedure has three creditable parts:

  1. Heating — Benedict's solution only reacts when the mixture is warmed, so 'heat with Benedict's solution' is a mark on its own. In practice you place the test-tube in a hot water-bath rather than heating it directly over a flame.
  2. Equal volumes — the mark scheme specifically wants 'equal volumes of each'. This makes the test a fair comparison between samples; if you used different amounts of Benedict's solution the intensity of the colour could not be compared between samples.
  3. The colour change — Benedict's solution is blue when it goes in. If a reducing sugar (such as glucose) is present, the blue colour changes through green and yellow to orange and finally brick-red, depending on how much sugar there is: more sugar gives a redder precipitate.

The conclusion is that the colour change is a positive result, showing a reducing sugar is present.

Key Takeaways

  • Benedict's solution tests for reducing sugars; it must be heated.
  • Blue → green → yellow → orange → brick-red; the redder the colour, the more sugar.
  • Use equal volumes of sample and reagent so results are comparable.

Common Mistakes

  • Forgetting to state that the mixture is heated — Benedict's does not react in the cold, so this loses a mark.
  • Writing just 'it changes colour' without naming the starting colour (blue) and the final colours — the mark scheme wants the actual change.
  • Saying 'sugar' instead of reducing sugar — Benedict's does not detect non-reducing sugars such as sucrose.
  • Omitting 'equal volumes', which the mark scheme credits as a separate point.

Things to Be Careful About

  • Give the full sequence of colours or at least the endpoints: blue to green/yellow/orange/red.
  • State the conclusion explicitly — the colour change alone is not the final mark; you must say a reducing sugar is present.
  • Do not confuse Benedict's solution (reducing sugars, blue to brick-red on heating) with iodine solution (starch, brown to blue-black) or biuret reagent (protein, blue to purple).
Techniques used
describe a food test procedurestate the colour change of Benedict's solution on heatinginterpret a positive result
(ii)

You are provided with a solution labelled S1.

Carry out the test you have described on a sample of S1 and record what you conclude about the solution.

______

1M
DifficultyEasy
Worked solution

Answer

S1 contains a reducing sugar (the Benedict's test was positive — the blue solution changed to orange/brick-red on heating).

Final answer

S1 — reducing sugar present.

Detailed explanation

Walkthrough

This part asks you to actually perform the test from (a)(i) on the unknown solution S1 and write what you conclude. In the examination your conclusion depends on the colour you observed, but the mark scheme shows the expected outcome: the test is positive, so S1 contains a reducing sugar. You would write the conclusion in the blank — 'reducing sugar present' — because the colour changed from blue to green/yellow/orange/red on heating.

Key Takeaways

  • A conclusion must follow from the observation: positive Benedict's test → reducing sugar present.

Common Mistakes

  • Writing an observation ('it went red') where a conclusion is asked for — the blank says 'what you conclude', so name the reducing sugar.
  • Saying 'glucose is present' — the test detects reducing sugars generally; you cannot identify which one.

Things to Be Careful About

  • The answer must be a conclusion, not a description of the colour change.
  • Use the exact phrase 'reducing sugar present' to mirror the mark scheme.
Techniques used
carry out a food test on an unknown sampledraw a conclusion from the observed colour change
(iii)

You are provided with some potato tissue covered in polythene.

  • Remove the polythene.
  • Cut the potato tissue into small pieces and place these in a clean test-tube.
  • Add some distilled water and shake the test-tube.

Carry out the test you described in (a)(i) on this mixture.
State your result and conclusion.

result = ______
conclusion = ______

1M
DifficultyMedium-Easy
Worked solution

Answer

result = the mixture stayed blue / showed only a slight green colour on heating with Benedict's solution

conclusion = little or no reducing sugar is present in the potato tissue

Final answer

Result: stays blue or only slight green change; conclusion: little or no reducing sugar present.

Detailed explanation

Walkthrough

Potato is mostly starch, which is a non-reducing polysaccharide, so Benedict's solution gives a negative (or only very weakly positive) result. The mark scheme accepts two outcomes:

  • the mixture stays blue (no reducing sugar), or
  • it changes only slightly to green (a trace of reducing sugar).

The conclusion follows: little or no reducing sugar is present. Note the question asks for the result and the conclusion separately — the result is the colour you see, the conclusion is what that colour means.

Key Takeaways

  • Benedict's solution detects reducing sugars only; starch is not a reducing sugar.
  • A negative Benedict's test stays blue; a trace gives a faint green.

Common Mistakes

  • Claiming a strong orange/brick-red result — potato stores starch, not reducing sugar, so a strongly positive result is wrong here.
  • Confusing this with the iodine test, which would be positive for starch (blue-black).
  • Giving the result without the conclusion, or vice versa — both blanks must be filled.

Things to Be Careful About

  • The mark scheme explicitly allows 'green for little' or 'blue/no change for none' — either observation with the matching conclusion scores.
  • Keep the wording 'little or no reducing sugar present'; do not overstate it as 'no sugar at all'.
Techniques used
carry out a food test on plant tissue extractstate the observation and the conclusion separately
(b)

You are provided with three dishes, each containing a similar piece of potato and a solution.
Each potato strip was cut exactly 5.0 cm5.0\text{ cm} in length before being placed in the solution at least an hour before the start of the examination.

Dish A – contains S1 solution.
Dish B – contains half S1 and half distilled water.
Dish C – contains distilled water.

  • Remove the potato strip from dish A.
  • Blot the strip carefully on a paper towel.
(i)

Accurately measure the longest length of this potato strip and record the length in Table 1.1.

  • Repeat the procedure with the potato strips in dishes B and C and record their lengths in Table 1.1.
DifficultyEasy
Worked solution

Answer

Measure the longest length of the blotted strip from dish A with a ruler, to the nearest millimetre (recorded to 1 decimal place in cm, e.g. 4.6 cm). Repeat for the strips in dishes B and C and record all three measured lengths in Table 1.1.

Final answer

Candidate-dependent measurement: the length of each blotted strip measured to the nearest mm and recorded in Table 1.1.

Detailed explanation

Walkthrough

This is a hands-on step. After blotting the strip (to remove surface liquid so the solution does not affect the reading), you measure its longest length with a ruler and record it in Table 1.1, then do the same for strips B and C. The initial length was exactly 5.0 cm, so any change you measure is due to water moving into or out of the potato cells by osmosis while the strips soaked.

Key Takeaways

  • Blot before measuring — surface liquid would add error.
  • Record lengths to a consistent precision (1 decimal place in cm, i.e. nearest mm).

Common Mistakes

  • Measuring a bent strip along a curve instead of its longest straight length.
  • Recording to inconsistent decimal places (4.6, then 4.55, then 5).
  • Forgetting to blot, which can change the apparent length and wet the ruler.

Things to Be Careful About

  • This step is not marked separately, but the values feed the marked table in (b)(ii) — an inaccurate measurement loses marks later.
  • Keep the unit (cm) as printed in the table header.
Techniques used
measure the length of a specimen accuratelyrecord measurements in a table with units
(ii)

Calculate the change in length between the initial and your measured length and complete Table 1.1.

Table 1.1

length of potato strip / cm\text{cm}
ABC
initial length5.05.05.0
measured length
change in length
2M
DifficultyMedium-Easy
Worked solution

Answer

Complete Table 1.1 by subtracting the initial length (5.0 cm) from each measured length, giving the change a sign:

change in length=measured length5.0\text{change in length} = \text{measured length} - 5.0
length of potato strip / cm\text{cm}
ABC
initial length5.05.05.0
measured length(candidate's reading, e.g. 4.6)(e.g. 5.0)(e.g. 5.4)
change in length0.4-0.40.00.0+0.4+0.4

(The measured lengths are the candidate's own readings; the changes must carry a + or − sign.)

Final answer

See working — all three measured lengths recorded and all three changes calculated with +/− signs.

Detailed explanation

Walkthrough

The table needs two things from you:

  1. Measured lengths — your own readings from (b)(i), recorded to 1 decimal place in cm.
  2. Change in length — measured length minus the initial 5.0 cm. The mark scheme insists on the sign: a strip that shrank gets a minus (−0.4), one that grew gets a plus (+0.4), and one that stayed the same is 0.0.

The sign matters because it shows the direction of water movement: minus means water left the potato by osmosis (into a more concentrated solution), plus means water entered the potato (from pure water). The mark scheme awards 1 mark for two correct columns and 2 marks for all three, including the signs.

Key Takeaways

  • change = final − initial, always with a sign.
  • Keep units in the header (length / cm), not in each cell.

Common Mistakes

  • Omitting the + or − sign on the change — the mark scheme explicitly requires it.
  • Calculating initial − measured, which reverses the sign.
  • Writing units in every cell when the header already carries '/ cm'.

Things to Be Careful About

  • The measured lengths are candidate-dependent; only the changes are checked for consistency with your readings (ecf applies).
  • Record all values to the same precision (1 decimal place).
Techniques used
complete a results table with measured and calculated valuescalculate change in length with a sign
(iii)

Describe and explain the changes in length.

4M
DifficultyMedium
Worked solution

Answer

  • A (S1 solution): the strip decreased in length — water moved out of the potato cells by osmosis (exosmosis) into the more concentrated S1 solution, whose water potential is lower than that of the potato cells.
  • B (half S1, half water): the length stayed nearly the same — the water potential inside and outside the cells was about equal, so water moved in and out at the same rate (water in = water out).
  • C (distilled water): the strip increased in length — water entered the potato cells by osmosis (endosmosis) because the distilled water has a higher water potential than the cell contents.
Final answer

A decreased (water lost by exosmosis to the more concentrated solution); B stayed nearly the same (water in = water out); C increased (water absorbed by endosmosis from pure water).

Detailed explanation

Walkthrough

This question links the data in Table 1.1 to the biology of osmosis. Osmosis is the movement of water from a region of higher water potential to a region of lower water potential through a partially permeable membrane — here, the cell membranes of the potato cells.

  • Dish A (S1): S1 contains a reducing sugar, so it is a concentrated solution with a lower water potential than the potato cell contents. Water leaves the cells — the mark scheme calls this exosmosis — and the vacuoles shrink, so the strip loses length. The cells become flaccid (slightly plasmolysed).
  • Dish B (half S1, half water): diluting S1 halves its solute concentration, bringing its water potential close to that of the potato cells. Water moves in and out at equal rates, so there is no net movement and the length barely changes. This dish acts like the 'equilibrium' condition.
  • Dish C (distilled water): pure water has the highest possible water potential (zero by convention), higher than the cell contents, so water enters the cells — endosmosis. The vacuoles swell, the cells become turgid, and the strip lengthens.

Each dish needs both a description (what happened to the length) and an explanation (why, in terms of water movement and water potential) — the mark scheme pairs them with '+'.

Key Takeaways

  • Water always moves by osmosis from higher to lower water potential through a partially permeable membrane.
  • Concentrated sugar solution → water leaves the potato (exosmosis, strip shrinks); pure water → water enters (endosmosis, strip swells); matched concentration → no net change.
  • Turgor: cells gaining water become turgid; cells losing water become flaccid.

Common Mistakes

  • Describing the change without explaining it — each mark pairs the observation with the reason.
  • Saying 'water moves from a dilute to a concentrated solution' without using the term osmosis or water potential — the precise term carries the mark.
  • Confusing osmosis (water) with diffusion (solute) or active transport — no energy is used here.
  • Saying the potato 'sucked up' water — water moves passively down the water potential gradient.
  • Mixing up which dish did what: A is the concentrated S1 (shrinks), C is pure water (swells).

Things to Be Careful About

  • The mark scheme uses the words exosmosis and endosmosis — include them.
  • Use 'water potential' rather than 'water concentration', which the scheme rejects.
  • For B, the key idea is that water in = water out, i.e. no net movement — not that nothing happens at all.
  • Give all three dishes; the maximum is 4 marks across the six listed points.
Techniques used
describe the direction of water movement in each solutionexplain the changes using osmosis and water potentialrelate solute concentration to the direction of osmosis

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