Biology · Simulation and virtual lab Free
Enzymes: how temperature, pH and concentration change the rate
Mix starch with amylase in a water bath, test a drop with iodine every 30 seconds and find how fast the starch disappears at different temperatures and pH.
- You will learn to
- Follow the digestion of starch by amylase with the iodine test on a spotting tile.
- Find the optimum temperature and pH of amylase from graphs of rate.
- Explain the results with the lock-and-key model and denaturation; use controls and an inhibitor.
- You will measure
- Time to the end point (s) at different temperatures, pH values and concentrations; then rate = 1000 ÷ t.
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: starch solution, amylase (or saliva), buffer solutions, iodine solution, spotting tile, test tubes, water baths, thermometer, dropping pipettes, stopwatch.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To investigate the effect of temperature, pH and enzyme concentration on the rate of starch digestion by amylase, and to explain the results with the lock-and-key model.
What to do
- Set the water bath to 35 °C, pH 7, 1 % amylase, nothing extra. Tap Mix and start the stopwatch.
- Watch the spotting tile: a drop is tested with iodine every 30 s. The end point is the first well that stays brown. Tap Record the result.
- Tap Clean tile, new tubes. Repeat at 5, 15, 25, 45, 55 and 65 °C, keeping pH 7 and 1 % amylase. Use clock × 20 to save time.
- Look at the graph of rate against temperature. Find the optimum temperature.
- Set 35 °C and change only the pH (4, 5, 6, 7, 8, 9). Choose “Graph: pH” and find the optimum pH. Then try 0.5 %, 1 % and 2 % amylase.
- S4: repeat at 35 °C, pH 7 with copper(II) sulfate, then with boiled amylase. Compare the end points with the normal tube.
Results
Questions
Why does iodine turn blue-black at the start but stay brown at the end?
Iodine turns blue-black with starch. Amylase breaks starch into maltose (a sugar), which does not react with iodine, so when no starch is left the iodine keeps its own brown colour.
Why is the rate calculated as 1000 ÷ time?
A shorter time to the end point means a faster reaction, so the rate is inversely proportional to the time. Multiplying by 1000 just gives convenient numbers.
Explain why the rate increases from 5 °C to about 40 °C.
The enzyme and starch molecules gain kinetic energy and move faster, so they collide more often and more enzyme–substrate complexes form per second.
Explain why the rate falls above about 45 °C.
Bonds holding the enzyme in shape break, the active site changes shape and the starch no longer fits: the enzyme is denatured. This cannot be reversed by cooling.
Why was the boiled amylase tube included?
It is a control: it shows that the starch is digested by the active enzyme and not by the water, the buffer or the heat.
Why do both tubes stay in the water bath for 5 minutes before mixing?
So that the starch and the enzyme are both at the temperature of the bath when the reaction starts; otherwise the temperature would change during the reaction.
For teachers: expected results and how the model works
Model: starch left = e^(−kt), end point when below 5 % (the first brown well; drops every 30 s, so times are multiples of 30 s). At pH 7 and 1 % amylase the end points are 5 °C: none in 10 min; 15 °C: 390 s; 25 °C: 210 s; 35 °C: 120 s; 40 °C: 90 s; 45 °C: 120 s; 50 °C: 210 s; 55 °C and above: none (denatured). Optimum ≈ 40 °C. pH at 35 °C: 5: 510 s; 6: 150 s; 7: 120 s; 8: 210 s; 4 and 9: none (optimum ≈ pH 7). Concentration at 35 °C: 0.5 %: 210 s; 1 %: 120 s; 2 %: 60 s. Copper(II) sulfate: 330 s (non-competitive inhibitor, 70 % less activity). Boiled amylase: no end point. The close-up shows the active site losing its shape above 45 °C and at extreme pH.
Exam practice Exam set
National-exam style questions on this simulation (16 marks). Try each one on paper first, then open the marking scheme.
A student found these end points for amylase and starch: 10 °C, 420 s; 20 °C, 210 s; 30 °C, 120 s; 40 °C, 90 s; 50 °C, 240 s; 60 °C, no end point. (a) Calculate the rate (1000 ÷ t) at each temperature. (b) Plot rate against temperature and state the optimum. (c) Explain the result at 60 °C. (7 marks)
Marking scheme: (a) 2.4, 4.8, 8.3, 11.1, 4.2, 0 (2 for all correct, 1 for four). (b) Axes labelled with units (1), points plotted and joined (1), optimum about 40 °C (1). (c) Amylase is denatured: the active site changes shape and starch no longer fits, so no starch is digested (2).
Use the lock-and-key model to explain why amylase digests starch but not protein. (3 marks)
Marking scheme: The active site has a specific shape (1) that is complementary to the starch molecule only (1); protein has a different shape and cannot fit the active site, so no enzyme–substrate complex forms (1).
State three variables that must be kept constant when investigating the effect of pH on amylase activity. (3 marks)
Marking scheme: Temperature (1); volume and concentration of the starch (1); volume and concentration of amylase (1) (also: time interval between drops, volume of iodine).
Copper(II) ions slow down amylase even when there is plenty of starch. Explain how a non-competitive inhibitor works. (3 marks)
Marking scheme: It binds to the enzyme at a site other than the active site (1); this changes the shape of the active site (1), so fewer enzyme–substrate complexes can form and the rate falls; adding more substrate does not overcome it (1).
Donat Sciences and MathsBiology · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Enzymes: how temperature, pH and concentration change the rate
Aim: To investigate the effect of temperature, pH and enzyme concentration on the rate of starch digestion by amylase, and to explain the results with the lock-and-key model.
Apparatus (real laboratory): starch solution, amylase (or saliva), buffer solutions, iodine solution, spotting tile, test tubes, water baths, thermometer, dropping pipettes, stopwatch.
What to do (with the simulation on www.donatstem.co.rw)
- Set the water bath to 35 °C, pH 7, 1 % amylase, nothing extra. Tap Mix and start the stopwatch.
- Watch the spotting tile: a drop is tested with iodine every 30 s. The end point is the first well that stays brown. Tap Record the result.
- Tap Clean tile, new tubes. Repeat at 5, 15, 25, 45, 55 and 65 °C, keeping pH 7 and 1 % amylase. Use clock × 20 to save time.
- Look at the graph of rate against temperature. Find the optimum temperature.
- Set 35 °C and change only the pH (4, 5, 6, 7, 8, 9). Choose “Graph: pH” and find the optimum pH. Then try 0.5 %, 1 % and 2 % amylase.
- S4: repeat at 35 °C, pH 7 with copper(II) sulfate, then with boiled amylase. Compare the end points with the normal tube.
Results
| # | T (°C) | pH | Amylase (%) | Extra | Time to end point (s) | Rate = 1000 ÷ t |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
| 9 | | | | | | |
| 10 | | | | | | |
| 11 | | | | | | |
| 12 | | | | | | |
Graph
Plot the rate (1000 ÷ t) on the vertical axis against temperature (°C). Join the points with a smooth curve and mark the optimum.
Questions
- Why does iodine turn blue-black at the start but stay brown at the end? [2]
- Why is the rate calculated as 1000 ÷ time? [2]
- Explain why the rate increases from 5 °C to about 40 °C. [2]
- Explain why the rate falls above about 45 °C. [2]
- Why was the boiled amylase tube included? [2]
- Why do both tubes stay in the water bath for 5 minutes before mixing? [2]
- A student found these end points for amylase and starch: 10 °C, 420 s; 20 °C, 210 s; 30 °C, 120 s; 40 °C, 90 s; 50 °C, 240 s; 60 °C, no end point. (a) Calculate the rate (1000 ÷ t) at each temperature. (b) Plot rate against temperature and state the optimum. (c) Explain the result at 60 °C. [7]
- Use the lock-and-key model to explain why amylase digests starch but not protein. [3]
- State three variables that must be kept constant when investigating the effect of pH on amylase activity. [3]
- Copper(II) ions slow down amylase even when there is plenty of starch. Explain how a non-competitive inhibitor works. [3]
Donat Sciences and MathsBiology · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Enzymes: how temperature, pH and concentration change the rate
Expected results
Model: starch left = e^(−kt), end point when below 5 % (the first brown well; drops every 30 s, so times are multiples of 30 s). At pH 7 and 1 % amylase the end points are 5 °C: none in 10 min; 15 °C: 390 s; 25 °C: 210 s; 35 °C: 120 s; 40 °C: 90 s; 45 °C: 120 s; 50 °C: 210 s; 55 °C and above: none (denatured). Optimum ≈ 40 °C. pH at 35 °C: 5: 510 s; 6: 150 s; 7: 120 s; 8: 210 s; 4 and 9: none (optimum ≈ pH 7). Concentration at 35 °C: 0.5 %: 210 s; 1 %: 120 s; 2 %: 60 s. Copper(II) sulfate: 330 s (non-competitive inhibitor, 70 % less activity). Boiled amylase: no end point. The close-up shows the active site losing its shape above 45 °C and at extreme pH.
Answers
- Why does iodine turn blue-black at the start but stay brown at the end? [2]
Iodine turns blue-black with starch. Amylase breaks starch into maltose (a sugar), which does not react with iodine, so when no starch is left the iodine keeps its own brown colour.
- Why is the rate calculated as 1000 ÷ time? [2]
A shorter time to the end point means a faster reaction, so the rate is inversely proportional to the time. Multiplying by 1000 just gives convenient numbers.
- Explain why the rate increases from 5 °C to about 40 °C. [2]
The enzyme and starch molecules gain kinetic energy and move faster, so they collide more often and more enzyme–substrate complexes form per second.
- Explain why the rate falls above about 45 °C. [2]
Bonds holding the enzyme in shape break, the active site changes shape and the starch no longer fits: the enzyme is denatured. This cannot be reversed by cooling.
- Why was the boiled amylase tube included? [2]
It is a control: it shows that the starch is digested by the active enzyme and not by the water, the buffer or the heat.
- Why do both tubes stay in the water bath for 5 minutes before mixing? [2]
So that the starch and the enzyme are both at the temperature of the bath when the reaction starts; otherwise the temperature would change during the reaction.
- A student found these end points for amylase and starch: 10 °C, 420 s; 20 °C, 210 s; 30 °C, 120 s; 40 °C, 90 s; 50 °C, 240 s; 60 °C, no end point. (a) Calculate the rate (1000 ÷ t) at each temperature. (b) Plot rate against temperature and state the optimum. (c) Explain the result at 60 °C. [7]
(a) 2.4, 4.8, 8.3, 11.1, 4.2, 0 (2 for all correct, 1 for four). (b) Axes labelled with units (1), points plotted and joined (1), optimum about 40 °C (1). (c) Amylase is denatured: the active site changes shape and starch no longer fits, so no starch is digested (2).
- Use the lock-and-key model to explain why amylase digests starch but not protein. [3]
The active site has a specific shape (1) that is complementary to the starch molecule only (1); protein has a different shape and cannot fit the active site, so no enzyme–substrate complex forms (1).
- State three variables that must be kept constant when investigating the effect of pH on amylase activity. [3]
Temperature (1); volume and concentration of the starch (1); volume and concentration of amylase (1) (also: time interval between drops, volume of iodine).
- Copper(II) ions slow down amylase even when there is plenty of starch. Explain how a non-competitive inhibitor works. [3]
It binds to the enzyme at a site other than the active site (1); this changes the shape of the active site (1), so fewer enzyme–substrate complexes can form and the rate falls; adding more substrate does not overcome it (1).