A student wrapped a strip of exposed photographic film around a boiling tube. The shiny film is coated with a dark layer of gelatine, which is a protein. Tu...
Question 1 Report
A student wrapped a strip of exposed photographic film around a boiling tube. The shiny film is coated with a dark layer of gelatine, which is a protein. Tube A held a solution of the protein-digesting enzyme trypsin at pH 8. Tube B (a control) held trypsin that had been made acidic to pH 3. After 8 minutes the strips were taken out. Where the gelatine had been digested away, the film became clear and transparent. Fig. 1.1 shows the two strips held against a ruler.
Table 1.1
| Tube | Conditions | Length of cleared film / mm |
|---|---|---|
| A | trypsin at pH 8 | ? |
| B | trypsin at pH 3 | ? |
(a) Name the substance in the film that trypsin digests. [1]
(b) Measure the length of the cleared, transparent part of the film in tube A, in mm, using Fig. 1.1. [1]
(c) Measure the length of cleared film in the control tube B. [1]
(d) Complete Table 1.1 by recording the two cleared lengths you measured. [2]
(e) Explain why the film in tube A became transparent. [2]
(f) Explain the result seen in tube B. [2]
(g) Describe how the student could measure the rate at which the film clears. [2]
(h) Calculate the rate of clearing in tube A, in mm per minute, given that it took 8 minutes. Show your working. [2]
(i) State the independent variable in this investigation. [1]
(j) State two variables that must be kept constant. [2]
(k) Suggest two ways to improve the accuracy of the length measurement. [2]
(l) Suggest why photographic film is a convenient material for showing protein digestion. [2]
Answer Details
This question tests protein digestion by an enzyme and the effect of pH on enzyme activity. Trypsin digests the gelatine (protein) coating of the film; where it is digested the film clears, so the cleared length is a visible measure of how much digestion occurred.
(a) Trypsin digests the gelatine (a protein) coating the film. [1]
(b) Measuring the cleared, transparent part of the film in tube A against Fig. 1.1 gives about 28 mm (accept 25 to 31 mm). [1]
(c) In the control tube B the film has not cleared, so the cleared length is 0 mm. [1]
(d) Recording the two measured lengths completes the table:
| Tube | Conditions | Length of cleared film / mm |
|---|---|---|
| A | trypsin at pH 8 | 28 |
| B | trypsin at pH 3 | 0 |
Tube A the value from part (b), tube B 0 mm; each correct [1]. [2]
(e) At pH 8, close to trypsin's optimum, the enzyme digests the gelatine protein [1], removing the dark layer so the clear transparent film shows through [1]. [2]
(f) At pH 3 the acidic conditions denature the trypsin (it is far from its optimum pH) [1], so it cannot digest the gelatine and the film stays dark, giving no clearing [1]. [2]
(g) To measure the rate of clearing, measure the cleared length at fixed time intervals (or measure the final cleared length and the time taken) [1], then calculate rate as the length cleared divided by the time [1]. [2]
(h) Rate of clearing in tube A over 8 minutes:
\[ \frac{28}{8} = 3.5 \ \text{mm per minute} \]
Working [1], answer \( 3.5 \) mm per minute (accept the value from (b) divided by 8) [1]. [2]
(i) The variable deliberately changed is the pH of the trypsin solution. [1]
(j) Any two controlled variables: temperature; concentration of trypsin; type/size of film strip; time left before measuring. [2]
(k) Any two ways to improve measurement accuracy: view the ruler at eye level to avoid parallax; use a ruler with clear millimetre divisions; measure more than once and take a mean; lay the film flat so its full length is measured. [2]
(l) Photographic film is convenient because the colour change from dark to clear is easy to see, giving an obvious end result [1], and the cleared length can be measured directly with a ruler to give a numerical result [1]. [2]
Exam tip: the control at pH 3 shows the clearing needs active enzyme, and comparing a length gives a measurable quantity from a colour change.
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