Biology 9700/31 — October/November 2021

Background Concept

A serial dilution is a stepwise dilution in which the same dilution factor is applied at each step. A 1:1 (halving) serial dilution gives a concentration that is the starting concentration divided by $2^n$ after $n$ steps. To keep the dilution running, the volume transferred into each new beaker must equal the volume of diluent added, so that the total in each beaker is constant and a known sub-volume can be carried forward to the next step.

Understanding the Question

Fig. 1.1 shows Beaker 1 (already containing $20\ \text{cm}^3$ of 1.0% P and $0\ \text{cm}^3$ of W, with an arrow showing '10 cm³ to use') and an empty Beaker 2 receiving a curved transfer arrow from Beaker 1. You must label Beaker 2 with the transferred volume, the volume of W added and the resulting concentration, then draw three more beakers (3, 4 and 5) to take the dilution through four halving steps. Each finished beaker must provide $10\ \text{cm}^3$ for the test.

Approach

Apply the halving rule: each beaker receives $10\ \text{cm}^3$ of the previous concentration plus $10\ \text{cm}^3$ of W, giving $20\ \text{cm}^3$ total at half the previous concentration. After mixing, $10\ \text{cm}^3$ is taken 'to use' and $10\ \text{cm}^3$ is carried forward to the next beaker.

The concentrations therefore follow:

Step-by-Step Reasoning

• Beaker 1 is the starting solution of 1.0% P. A 'to use' arrow is already drawn.
• Beaker 2: take $10\ \text{cm}^3$ from Beaker 1, add $10\ \text{cm}^3$ of W. Total $20\ \text{cm}^3$. Concentration halves: $1.0\% \div 2 = 0.5\%$. Take $10\ \text{cm}^3$ to use and $10\ \text{cm}^3$ forwards.
• Beaker 3: $0.5\% \div 2 = 0.25\%$. Same pattern of arrows.
• Beaker 4: $0.25\% \div 2 = 0.125\%$.
• Beaker 5: $0.125\% \div 2 = 0.0625\%$.

Each new beaker therefore needs the same three arrows: $10\ \text{cm}^3$ transferred in (with the relevant concentration), $10\ \text{cm}^3$ of W added, and $10\ \text{cm}^3$ taken to use. The concentration under each beaker must include the % sign.

Key Takeaways

• In a halving serial dilution the concentration after $n$ steps = starting concentration $\div\ 2^n$.
• Keep the volume transferred equal to the volume of diluent added so that both a sample AND a forward-transfer volume are available at every step.

Common Mistakes

• Forgetting to halve each time (writing 0.5%, 0.4%, 0.3%, 0.2% is not a true serial dilution).
• Labelling only the concentration and not the transferred volume / water volume.
• Drawing the 'to use' arrow only on Beaker 1 — every beaker must carry one because each concentration is tested later.

Things to Be Careful About

• Mix thoroughly with a glass rod between transfers so that carry-over of the wrong concentration does not occur.
• The % sign on every concentration is required by the mark scheme.

Background Concept

The Biuret test detects peptide bonds. Copper(II) sulfate in alkaline solution (provided here as C + K) forms a violet/purple complex with peptide bonds; the more peptide bonds, the deeper the colour. A 0.0% control gives a pale blue colour from the Cu²⁺ ions alone (no peptide bonds present).

Understanding the Question

You have added KOH and CuSO₄ to each standard, mixed, left for 1 minute and observed the colour against a white background. You must record the colour observed in each tube using the symbols given in Table 1.2.

Approach

• Make a proper results table — the mark scheme requires column headings for both the independent variable (protein concentration, with units) and the dependent variable (symbol).
• Order the rows from lowest to highest concentration so the colour trend is visible.
• Match each observation to one of the symbols from Table 1.2; the same symbol can appear in more than one row.

Step-by-Step Reasoning

• Independent variable heading: 'percentage protein concentration / %'. Do not abbreviate to '% protein' or 'conc' — the / % unit is required.
• Dependent variable heading: 'symbol for colour' or simply 'symbol'.
• Six rows: 0.0% (distilled water control), 0.0625%, 0.125%, 0.25%, 0.5%, 1.0%.
• Expected trend: blue at 0.0% (no peptide bonds) → deepening purple as peptide bond concentration rises. Representative symbols are + through ++++++ as shown in the table.

Key Takeaways

• Biuret is qualitative — results are recorded as colour descriptions or a coded scale, not as numerical absorbances.
• Always include a 0.0% control so the negative result is recorded.
• Use a white background to judge subtle differences reliably.

Common Mistakes

• Missing units from the concentration heading (must include / %).
• Putting the rows in random order so the trend is hidden.
• Writing 'colour' instead of the agreed symbol (the mark scheme expects symbols from Table 1.2).

Things to Be Careful About

• A gentle shake between additions ensures Cu(OH)₂ is fully dispersed and reacts uniformly.
• Read the colour after the full 1-minute wait — colours deepen as the reaction proceeds.

Background Concept

The independent variable is what the experimenter deliberately changes; everything else is held constant. The dependent variable is what is measured as a result.

Understanding the Question

The question simply asks what is being varied across the six tubes in the investigation just carried out.

Approach

Scan the method: volumes ($1\ \text{cm}^3$), reagents (K then C), and timing (1 min) are identical for every tube. The only quantity that differs is the protein concentration of the standard added, so that is the independent variable.

Step-by-Step Reasoning

• Independent variable → what is changed → protein concentration (0.0% to 1.0%).
• Dependent variable → what is measured → colour produced (recorded as a symbol from Table 1.2).
• Controlled variables → $1\ \text{cm}^3$ volume, same reagents, same mixing, same 1-minute wait.

Key Takeaways

• Identifying variables is the first step before drawing any conclusion.
• The independent variable goes on the x-axis if a graph is drawn; the dependent goes on the y-axis.

Common Mistakes

• Saying 'volume' — volume is kept constant.
• Saying 'colour' — colour is the dependent variable, not the independent.

Things to Be Careful About

• State the variable in biological terms ('protein concentration'), not in technique terms ('amount of P added').

Background Concept

The same Biuret reaction is applied to the unknown sample. The colour obtained is matched to the colour of the standards prepared in (a)(ii) to estimate the protein concentration of U.

Understanding the Question

You test the mock urine U exactly as you tested the standards and write the symbol that best matches the colour observed.

Approach

• Add $1\ \text{cm}^3$ of K, then $1\ \text{cm}^3$ of C, to the tube labelled U.
• Mix gently, leave 1 min, then hold against a white background.
• Choose one symbol from Table 1.2 that best matches the colour.

Step-by-Step Reasoning

• 'No condition' samples give + (no colour change).
• A very pale result would be ++.
• A clear blue/purple = +++.
• A pale purple = ++++.
• A definite purple = +++++.
• A dark purple = ++++++.

The representative example ++++ is between the standards 0.125% (+++) and 0.5% (+++++), indicating an intermediate protein concentration.

Key Takeaways

• The same symbol scale is used for both standards and unknowns so that they can be compared directly.

Common Mistakes

• Writing a colour description rather than a symbol.
• Using a symbol not in Table 1.2.

Things to Be Careful About

• Mix gently so the Cu(OH)₂ does not form clumps that give a misleadingly dark patch.
• Read the colour after exactly the same 1-minute wait used for the standards.

Background Concept

A halving serial dilution is geometric, not linear, in concentration: each step multiplies the remaining concentration by 1/2. On a linear distance scale the tick marks therefore bunch up towards the 0.0% end, each one halving the previous distance from 0.0%.

Understanding the Question

Fig. 1.2 shows a straight line with 1.0% at the left and 0.0% at the right. You must mark on it the four intermediate concentrations used in the dilution (0.5%, 0.25%, 0.125%, 0.0625%).

Approach

For a halving series on a linear scale, the distance from the 1.0% end is given by

where $n$ is the number of dilutions from 1.0%:

• $n = 1$ → position $1 - 1/2 = 1/2$ (0.5%)
• $n = 2$ → position $1 - 1/4 = 3/4$ (0.25%)
• $n = 3$ → position $1 - 1/8 = 7/8$ (0.125%)
• $n = 4$ → position $1 - 1/16 = 15/16$ (0.0625%)

Step-by-Step Reasoning

• Place a short vertical tick at the midpoint of the line, label 0.5%.
• Place the next tick three-quarters of the way along, label 0.25%.
• Place the next tick seven-eighths of the way along, label 0.125%.
• Place the final tick fifteen-sixteenths of the way along, very close to the 0.0% end, label 0.0625%.

Key Takeaways

• Serial dilution scales are geometric: each mark halves the previous distance from the 0.0% end.

Common Mistakes

• Spacing the tick marks equally (linear) instead of geometrically — this is wrong because each step halves concentration, not subtracts a fixed amount.

Things to Be Careful About

• Label each tick with both the number and the % sign.
• Draw the ticks with a ruler so they are clear and uniform.

Background Concept

The completed scale in Fig. 1.2 is effectively a colour key for the protein concentrations used. An unknown sample can be read off this scale by matching its Biuret colour to the closest standard colour on the scale.

Understanding the Question

You must use the colour symbol recorded for U in (a)(iv) and place an arrow on the scale at the position corresponding to that colour intensity.

Approach

• Locate the symbol recorded for U in (a)(iv).
• Find the position on the Fig. 1.2 scale that corresponds to that symbol (using the colour order already established when filling in the standards in (a)(ii)).
• Draw a short downward arrow ↓ at that position and label it U underneath.

Step-by-Step Reasoning

For the representative example:

• U = ++++ (pale purple) matches the colour of the 0.25% standard.
• The 0.25% position is already marked on the scale at the three-quarters point.
• Place the arrow ↓ directly above the 0.25% tick.

If the student had recorded U = +++ (blue), the arrow would go at the 0.125% position; for U = +++++ (purple) it would go at 0.5%; and so on.

Key Takeaways

• A standard colour scale lets you turn a colour observation into a numerical estimate of concentration.

Common Mistakes

• Drawing the arrow the wrong way (must be ↓, into the scale).
• Forgetting the U label underneath the arrow.

Things to Be Careful About

• The estimate is only as good as the matching between the unknown colour and the standard colours — small differences in shade can shift the position by one dilution step.

Background Concept

Table 1.3 gives reference ranges for total urinary protein per 24 h for four clinical categories. The 1.0% protein solution represents $1000\ \text{mg}$ per $24\ \text{h}$, so a concentration of $x\%$ corresponds to $x \times 10\ \text{mg}\ 24\ \text{h}^{-1}$.

Understanding the Question

You must convert the % concentration estimated for U into a 24 h mass of protein using the 1.0% = $1000\ \text{mg}$ reference, then state which clinical condition in Table 1.3 matches.

Approach

• Convert the % concentration of U to mg per 24 h by multiplying by 10 (because 1% = $1000\ \text{mg}$).
• Locate the resulting value in the second column of Table 1.3.
• State the condition in the first column of the matching row.

Step-by-Step Reasoning

Conversion table:

For U at 0.25%:

This lies in the row 'kidney tubular disease (200–500 mg $24\ \text{h}^{-1}$)'.

Key Takeaways

• A simple proportion ($x\% \to 10x\ \text{mg}\ 24\ \text{h}^{-1}$) bridges the mock concentration and the clinical reference table.

Common Mistakes

• Forgetting to multiply by 10 and reporting the % as the mass.
• Reading off the wrong row of Table 1.3 — the boundaries 150, 200 and 500 are common trap points.

Things to Be Careful About

• The 1.0% = $1000\ \text{mg}$ relationship is given in the question stem — make the conversion explicit, do not skip it.

Background Concept

Repeating a measurement and looking for agreement is the simplest way to assess whether an observation is reliable. A single colour match against a scale is subjective and easily influenced by lighting, viewing angle and reagent ageing.

Understanding the Question

You are told your single result for U may be anomalous and you must say how to check it.

Approach

Apply the principle of replication: carry out the same procedure on more than one sample of U and check whether the same symbol is obtained.

Step-by-Step Reasoning

• Take a fresh portion of U and repeat steps 5–8.
• Carry out at least three repeats in total (three independently mixed tubes).
• If all three (or the majority) give the same symbol, the original result is confirmed.
• If they differ, the most likely symbol is the mode of the results, and you would mark the estimate as 'not reliable'.

Key Takeaways

• Replication distinguishes a one-off error from the true value.

Common Mistakes

• Saying 'repeat it once more' — one repeat is not enough to call a result reliable.
• Saying 'use a different method' — the question asks how to confirm this method's result, not to replace it.

Things to Be Careful About

• Each repeat should use a fresh, independently mixed tube so that an error in one tube does not propagate.

Background Concept

Benedict's reagent detects reducing sugars such as glucose. Cu²⁺ is reduced to Cu⁺ which precipitates as red copper(I) oxide; the higher the concentration of reducing sugar, the greater the colour change from blue → green → yellow → orange → brick-red. The intensity of the colour can therefore be used semi-quantitatively by comparison with standard solutions of known concentration.

Understanding the Question

The previous parts used a Biuret standard scale to estimate the protein concentration of U. The question asks how to apply the same idea to estimate the concentration of glucose in a urine sample known to contain glucose.

Approach

• Prepare a series of glucose standards of known concentration.
• Treat the unknown sample in exactly the same way as the standards.
• Match the resulting colour to the closest standard and read off the concentration.

Step-by-Step Reasoning

• Standards: at least three known concentrations of glucose (the more standards, the finer the resolution). These could be prepared by serial dilution of a stock 1% glucose solution.
• Benedict's test on standards: add a fixed volume of Benedict's reagent to each standard, heat in a boiling water bath for ~2 min, allow to cool, and record the colour obtained against a white background.
• Benedict's test on the unknown: treat an identical volume of the patient's urine in the same way.
• Compare: visually match the colour of the unknown to the colour of the standard that most closely resembles it; report that standard's concentration as the estimate. A more rigorous version would measure absorbance in a colorimeter.

Key Takeaways

• Semi-quantitative tests work by comparing an unknown against a series of standards.
• The same logic (standards → unknown → match) applies regardless of which reagent is used.

Common Mistakes

• Saying 'use a glucose test strip and read the value' — this is a valid practical method but the mark scheme rewards the multi-standard comparison method that mirrors the protein estimation just done.
• Forgetting that Benedict's must be heated — many students omit the boiling step.
• Suggesting only one standard concentration — at least three are needed to span the likely range.

Things to Be Careful About

• Use the same volume of reagent, the same heating time and the same temperature for every tube so that any colour difference is due to glucose alone.

Background Concept

A bar chart is used for discrete categorical data (different antibiotics, in this case) where the bars do not touch. The y-axis must start at zero, use a sensible scale that fills at least half the grid, and be labelled with the quantity and its units. The x-axis carries the categories; each bar has the same width with even gaps between bars.

Understanding the Question

Plot the data in Table 1.4 as a bar chart on the supplied grid.

Approach

• Decide the y-axis scale: the largest value is 59.0%, so a scale of 10% per 2 cm (i.e. 1% per 2 mm) reaches 60% at 12 cm — easily fitting on a 20 cm grid and using more than half of it.
• Order the antibiotics on the x-axis as listed in the table so the chart matches the data table.
• Draw each bar with a sharp pencil and ruler, using straight lines that meet cleanly.

Step-by-Step Reasoning

• Scale: 10% = 2 cm → 1% = 2 mm. The grid accommodates up to 100% — more than enough.
• Bar widths: divide the 16 cm plot width into 6 equal bar-and-gap units; bars of ~2 cm wide with ~0.7 cm gaps is typical.
• Bar heights:
  • 23.5% → 4.7 cm
  • 31.5% → 6.3 cm
  • 0.0% → 0 cm (no bar visible, but still draw the axis line at 0)
  • 0.5% → 0.1 cm (very thin sliver — still draw a hairline so the category is represented)
  • 59.0% → 11.8 cm
  • 2.0% → 0.4 cm

Key Takeaways

• Bar charts require evenly spaced bars of even width drawn with a ruler.
• The y-axis must start at 0 for fair visual comparison.

Common Mistakes

• Inverted axes (categories on y, percentage on x).
• Scale that is awkward (e.g. 1 cm = 7%) or that does not use at least half the grid.
• Bars not drawn with a ruler or with gaps that are not equal.
• Missing the 0.0% bar for imipenem — it is still a category and must be shown as a flat line at 0.

Things to Be Careful About

• Use a sharp pencil for the bar chart — the question reminds you to.
• Plotting bars accurately is what earns the plotting mark; rough freehand bars lose marks.

Background Concept

Antibiotic resistance evolves by natural selection. When a bacterial population is exposed to an antibiotic, susceptible cells die while any rare mutants that happen to carry a resistance allele survive and reproduce, passing the allele to their offspring. Over many generations the proportion of resistant cells in the population therefore rises. The longer a population is exposed, the more rounds of selection occur and the higher the resistance frequency becomes.

Understanding the Question

The data show that 59.0% of E. coli are resistant to ampicillin but only 0.0% to imipenem. Ampicillin was introduced in 1961; imipenem in 1985. Suggest why ampicillin resistance is higher.

Approach

• Connect the difference in deployment time (24 years) to the number of generations of selection that each antibiotic has exerted on E. coli populations.
• Mention mutation as the source of new resistance alleles.
• Mention selection pressure as the mechanism that increases their frequency.

Step-by-Step Reasoning

• Time: ampicillin has had ~24 more years of use. Bacterial generations are short (minutes to hours), so 24 years represents an enormous number of cell divisions and opportunities for new mutations to arise.
• Mutation: random mutations in genes encoding penicillin-binding proteins, β-lactamases or porins can confer ampicillin resistance; the longer the period of use, the more likely such a mutation has occurred and spread.
• Selection: each time ampicillin is used, susceptible E. coli are killed; resistant mutants survive and reproduce, so their proportion in the population increases over time.
• Imipenem: introduced much later and used less widely, so selection pressure has been lower and there has been less time for resistance mutations to arise and spread.

Key Takeaways

• Antibiotic resistance is an evolutionary process driven by mutation and selection.
• The longer an antibiotic has been in widespread use, the higher the typical resistance frequency.

Common Mistakes

• Saying 'bacteria become resistant because they get used to the drug' — this is the lay misconception and earns no credit.
• Failing to mention mutation; without it there is no source of new resistance alleles.
• Failing to mention selection; without it the alleles would not increase in frequency.

Things to Be Careful About

• The question is about E. coli isolated from patients with UTIs specifically — these bacteria have been exposed to ampicillin in real clinical practice for decades, providing the selection pressure.