Digestion and enzymes
By the end of this chapter you will be able to
- Name the organs of the digestive system and say what each one does
- Describe how enzymes work using the lock and key model
- Name the enzyme that breaks down starch, protein and fat, and its products
- Explain the role of bile, and why enzymes are denatured at high temperatures
- Carry out and interpret the enzyme practical, and calculate a rate from a time
- [H] Interpret graphs of rate against concentration
You should already know
- Cells and their parts, including the cell membrane (Chapter 2)
- Diffusion and surface area (Chapter 3)
Where this leads
- Absorption in the small intestine (Chapter 3)
- Respiration and enzymes in cells (Chapter 10)
- Enzymes in photosynthesis (Chapter 9)
GCSE Biology. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Name the organs of the digestive system and their functions, Explain how large food molecules are broken down into small soluble molecules, Describe enzymes as biological catalysts with specific active sites, Name the digestive enzymes, where they act, and the role of bile, Investigate and explain how temperature and pH change enzyme activity, [H] Explain how enzyme and substrate concentration change the rate of reaction.
Exercise P Do you remember?
Method. Divide 1000 by 50.
- Divide. 1000 / 50 = 20.1000 / 50 = 20
- Write the answer. 20
Check. 20 x 50 = 1000. ✓
Method. Recall the boundary of the cell.
- Recall. The cell membrane controls what enters and leaves the cell.enters and leaves
- Write the answer. cell membrane
Check. The membrane is partially permeable. ✓
Method. Recall the process for water.
- Recall. The movement of water across a partially permeable membrane is osmosis.osmosis
- Write the answer. osmosis
Check. Osmosis is the movement of water only. ✓
Method. Add, then divide by 3.
- Add. 10 + 20 + 30 = 60.60
- Divide. 60 / 3 = 20.20
- Write the answer. 20
Check. 20 x 3 = 60. ✓
Method. Recall the waste gas of aerobic respiration.
- Recall. Aerobic respiration releases carbon dioxide as a waste product.carbon dioxide
- Write the answer. carbon dioxide
Check. Oxygen is used up, not released. ✓
The digestive system
Food passes through the alimentary canal from the mouth to the anus. Food is broken into smaller pieces by mechanical digestion, such as chewing with the teeth and churning in the stomach. It is broken down into small, soluble molecules by chemical digestion, using enzymes. Mechanical digestion increases the surface area for the enzymes to work on.
Glands add enzymes and bile to the gut. The liver makes bile, the gall bladder stores it, and the pancreas makes enzymes and sends them to the small intestine.
Key ideas
- Mechanical digestion breaks food into smaller pieces, increasing surface area.
- Chemical digestion uses enzymes to make small soluble molecules.
- The liver makes bile; the gall bladder stores it; the pancreas makes enzymes.
- Absorption of small molecules happens mainly in the small intestine.
Method. Recall where food is absorbed.
- Recall. Most digested food is absorbed through the walls of the small intestine.small intestine
- Name it. So the organ is the small intestine.small intestine
- Write the answer. small intestine
Check. The large intestine absorbs water, not most food. ✓
Method. Recall the gland that makes bile.
- Recall. Bile is made in the liver.made in the liver
- Name it. So the organ is the liver.liver
- Write the answer. liver
Check. The gall bladder stores the bile. ✓
Method. Think about the physical breaking of food.
- Think. Mechanical digestion breaks food into smaller pieces without enzymes.without enzymes
- Name one. Chewing with the teeth is an example.chewing
- Write the answer. chewing with the teeth
Check. Churning in the stomach is another example. ✓
Watch out
Watch out: mechanical and chemical digestion happen together. Chewing does not make molecules smaller; enzymes do.
Exam tip
Exam tip. Name the organ and what it does. “The small intestine” is the organ; “absorbs digested food” is its job.
Try it (with your teacher). With your teacher, use a food diagram or a labelled model of the digestive system. Trace the path of a piece of bread from the mouth to the anus, and name each organ.
Exercise 5A The digestive system
Method. Recall the gland that makes bile.
- Recall. Bile is made in the liver.made in the liver
- Write the answer. liver
Check. The liver also does many other jobs. ✓
Method. Recall where bile is kept.
- Recall. Bile is stored in the gall bladder until it is needed.stored
- Write the answer. gall bladder
Check. The gall bladder is a small sac below the liver. ✓
Method. Recall the gland that releases enzymes into the small intestine.
- Recall. The pancreas makes enzymes and sends them to the small intestine.sends them to the small intestine
- Write the answer. pancreas
Check. The salivary glands also make amylase. ✓
Method. Recall the organ with strong acid.
- Recall. The stomach churns food and adds hydrochloric acid.hydrochloric acid
- Write the answer. stomach
Check. The acid helps the enzyme pepsin work. ✓
Method. Recall the organ that absorbs water.
- Recall. Water is absorbed in the large intestine.absorbed in the large intestine
- Write the answer. large intestine
Check. Most digested food is absorbed earlier. ✓
Method. Recall the term.
- Recall. Breaking food into smaller pieces by chewing and churning is mechanical digestion.smaller pieces
- Write the answer. mechanical digestion
Check. Chemical digestion uses enzymes. ✓
Method. Recall the enzyme-based process.
- Recall. Enzymes break large molecules into small soluble ones in chemical digestion.small soluble
- Write the answer. chemical digestion
Check. Mechanical digestion does not change the molecules. ✓
Method. Recall the first place starch is digested.
- Recall. Saliva in the mouth contains amylase, which starts to digest starch.contains amylase
- Name it. So the first organ is the mouth.mouth
- Write the answer. mouth
Check. Digestion of starch is completed in the small intestine. ✓
Method. Recall the part of the alimentary canal after the mouth.
- Recall. Food passes from the mouth down the oesophagus to the stomach.down the oesophagus
- Write the answer. oesophagus
Check. The oesophagus carries food by muscle movements. ✓
Method. Think about where enzymes act.
- Think. Enzymes act on the surface of food molecules.act on the surface
- Conclude. More surface means more enzyme can act at once, so digestion is faster.faster
- Write the answer. Accept any sentence that says enzymes can act on more of the food at once, so digestion is faster.
Check. Mechanical digestion speeds up chemical digestion. ✓
Method. Check which food the stomach mainly digests.
- Check. The stomach uses pepsin, which mainly digests protein.digests protein
- Conclude. Fat is digested mainly by lipase in the small intestine, so the statement is not correct.not correct
- Write the answer. Accept any answer that says the stomach mainly digests protein, not fat, so the statement is not correct.
Check. Fat is digested in the small intestine. ✓
Method. Think about the liver’s role.
- Think. The liver makes bile, which helps digest fat.helps digest fat
- Conclude. Bile is stored and released into the small intestine.released
- Write the answer. Accept any sentence that says the liver makes bile, which is stored in the gall bladder and released into the small intestine.
Check. Bile is not an enzyme. ✓
Method. Give each type a job.
- Mechanical. Mechanical digestion breaks food into smaller pieces, giving more surface area.more surface area
- Chemical. Chemical digestion uses enzymes to make small, soluble molecules.small, soluble
- Conclude. The two work together, so digestion is faster and more complete.work together
- Write the answer. Accept any answer that says mechanical digestion makes more surface area, and chemical digestion uses enzymes to make small soluble molecules.
Check. Each type of digestion supports the other. ✓
Enzymes as biological catalysts
Enzymes are proteins that act as biological catalysts. A catalyst speeds up a reaction without being used up, so the same enzyme molecules can be used again and again. Each enzyme has an active site with a shape that fits only one substrate. This is the lock and key model: the substrate (the key) fits into the active site (the lock).
If the enzyme is heated too much, or the pH is far from its best, the active site changes shape. The enzyme is then denatured: the substrate no longer fits, and this change is permanent.
Key ideas
- Enzymes are proteins that act as biological catalysts.
- Each enzyme has an active site that fits only one substrate.
- Enzymes are not used up in the reaction.
- Denaturing changes the active site shape, so the substrate no longer fits.
Method. Recall the lock and key model.
- Recall. The substrate fits into a specific part of the enzyme.specific part
- Name it. This part is called the active site.active site
- Write the answer. active site
Check. A different enzyme has a different active site. ✓
Method. Think about the active site shape.
- Think. The active site has one particular shape.one shape
- Conclude. So it fits only one kind of substrate.only one kind
- Write the answer. Its active site has one shape, so it fits only one kind of substrate.
Check. Specific means one enzyme for one substrate. ✓
Method. Recall the effect of heat.
- Recall. High heat changes the shape of the active site.changes the shape
- Name it. The enzyme is denatured.denatured
- Write the answer. denatured
Check. Denaturing is permanent. ✓
Watch out
Watch out: an enzyme is not used up in the reaction. Do not say that the enzyme is “eaten” by the substrate.
Exam tip
Exam tip. When you explain denaturing, say that the active site changes shape, so the substrate no longer fits.
Try it (with your teacher). With your teacher, model the lock and key idea using two different shaped objects, such as a key and a coin. Show which one fits the lock and which does not. Explain why only one fits.
Exercise 5B Enzymes as biological catalysts
Method. Recall the role of enzymes.
- Recall. Enzymes speed up reactions in living things, so they are biological catalysts.biological catalysts
- Write the answer. biological catalysts
Check. A catalyst is not used up. ✓
Method. Recall the lock and key model.
- Recall. The substrate fits into the active site.fits into
- Write the answer. active site
Check. The active site is a specific shape. ✓
Method. Recall the model.
- Recall. In the lock and key model, the substrate is the key.the key
- Write the answer. the key
Check. The enzyme is the lock. ✓
Method. Recall what a catalyst does.
- Recall. A catalyst speeds up a reaction and is not used up.not used up
- Write the answer. So the answer is no.no
- Write the answer. no
Check. The same enzyme can be used again. ✓
Method. Recall the word for the change.
- Recall. When the active site changes shape permanently, the enzyme is denatured.changes shape permanently
- Write the answer. denatured
Check. Denaturing is permanent. ✓
Method. Recall the factors that denature.
- Recall. High temperature or a pH far from the best value can denature an enzyme.high temperature
- Choose one. Either factor is a correct answer.either
- Write the answer. Accept any one of: temperature, pH.
Check. Low temperature slows an enzyme but does not denature it. ✓
Method. Recall what specific means.
- Recall. An enzyme has one active site shape, so it fits only one substrate.one substrate
- Write the answer. So yes.yes
- Write the answer. yes
Check. Amylase and protease are different enzymes. ✓
Method. Recall the substrate of amylase.
- Recall. Amylase breaks down starch.breaks down starch
- Write the answer. starch
Check. Protease acts on protein. ✓
Method. Think about the particles.
- Think. At low temperature the particles move more slowly.move more slowly
- Conclude. So they collide less often, and the rate is lower.collide less often
- Write the answer. Accept any sentence that says the enzyme and substrate molecules have less kinetic energy, so they collide less often.
Check. A low temperature does not denature the enzyme. ✓
Method. Think about denaturing.
- Think. High heat denatures the enzyme, changing its active site.changing its active site
- Conclude. The change is permanent, so cooling does not restore the active site.permanent
- Write the answer. So the enzyme does not work at the same rate.no
- Write the answer. no
Check. Denatured enzymes cannot be restored by cooling. ✓
Method. Use the lock and key model.
- Think. Only the right substrate fits the active site.right substrate
- Conclude. So each enzyme acts on only one kind of substrate.one kind
- Write the answer. Accept any sentence that says the active site has a shape that only fits one substrate, like a key in one lock.
Check. The shape is the key to the specificity. ✓
Method. Link the shape to the fit.
- Change. High heat breaks the bonds that hold the enzyme in its shape.breaks the bonds
- Consequence. The active site is now a different shape, so the substrate does not fit.does not fit
- Conclude. Cooling does not restore the shape, so the enzyme is permanently denatured.permanently
- Write the answer. Accept any answer that says the active site has changed shape permanently, so the substrate no longer fits.
Check. The shape is what gives the enzyme its function. ✓
Digestive enzymes and bile
Three main groups of digestive enzymes are carbohydrases (such as amylase, which breaks starch down into sugars), proteases (which break proteins into amino acids) and lipases (which break lipids into fatty acids and glycerol).
Bile is made in the liver and released into the small intestine. It neutralises the acid from the stomach, so the small intestine is alkaline, which suits the enzymes there. Bile also emulsifies fat: it breaks large droplets into many smaller ones. This gives lipase a larger surface area to work on.
| Enzyme | Substrate | Products | Where it is made or acts | Note |
|---|---|---|---|---|
| amylase | starch | maltose | salivary glands, pancreas | small intestine (maltase) breaks maltose to glucose |
| protease (pepsin) | protein | amino acids | stomach | works best in acid |
| protease (trypsin) | protein | amino acids | pancreas, released into the small intestine | works best in alkali |
| lipase | lipid (fat) | fatty acids and glycerol | pancreas, released into the small intestine | bile helps by emulsifying the fat first |
| Bile | What it does |
|---|---|
| made in the liver, stored in the gall bladder | released into the small intestine |
| neutralises stomach acid | gives the alkaline conditions that the small intestine enzymes need |
| emulsifies fat | breaks large fat droplets into smaller ones, giving a larger surface area for lipase |
Key ideas
- Amylase: starch to maltose, then maltase to glucose.
- Protease: protein to amino acids.
- Lipase: lipid (fat) to fatty acids and glycerol.
- Bile neutralises stomach acid and emulsifies fat; it contains no enzymes.
Method. Recall the protein-digesting enzymes.
- Recall. Proteases break proteins into amino acids.proteases
- Name one. So the enzyme is a protease, such as pepsin.protease
- Write the answer. protease
Check. Pepsin is a protease made in the stomach. ✓
Method. Recall what lipase produces.
- Recall. Lipase breaks lipids into fatty acids and glycerol.fatty acids and glycerol
- Name them. So the products are fatty acids and glycerol.fatty acids and glycerol
- Write the answer. fatty acids and glycerol
Check. Lipase is made by the pancreas. ✓
Method. Think about what happens in the small intestine.
- Think. The stomach acid must be neutralised, and fat must be broken into droplets.neutralised, and fat
- Conclude. Bile does both of these jobs.both of these
- Write the answer. Accept any sentence that says bile neutralises the acid from the stomach and emulsifies fat.
Check. Bile contains no enzyme. ✓
Watch out
Watch out: bile does not contain enzymes. It helps the enzymes by neutralising acid and emulsifying fat.
Exam tip
Exam tip. For a digestive enzyme question, name the enzyme, its substrate, and its products.
Try it (with your teacher). With your teacher, look at a food label and list the main nutrients. For each one, name the enzyme that digests it and the products it makes.
Exercise 5C Digestive enzymes and bile
Method. Recall the carbohydrase for starch.
- Recall. Amylase breaks down starch into sugars.breaks down starch
- Write the answer. amylase
Check. Amylase is made in the salivary glands and the pancreas. ✓
Method. Recall the product of amylase.
- Recall. Amylase breaks starch into maltose, a sugar.maltose
- Write the answer. maltose
Check. Maltose is then broken into glucose by maltase. ✓
Method. Recall the enzyme for fats.
- Recall. Lipase breaks lipids into fatty acids and glycerol.fatty acids and glycerol
- Write the answer. lipase
Check. Lipase is made in the pancreas. ✓
Method. Recall the products of protease.
- Recall. Proteases break proteins into amino acids.amino acids
- Write the answer. amino acids
Check. Amino acids are small enough to be absorbed. ✓
Method. Recall where bile is kept.
- Recall. Bile is stored in the gall bladder.stored in the gall bladder
- Write the answer. gall bladder
Check. Bile is made in the liver. ✓
Method. Recall the role of bile.
- Recall. Bile neutralises the acid from the stomach.neutralises the acid
- Write the answer. neutralises
Check. Bile also emulsifies fat. ✓
Method. Recall how bile helps fat digestion.
- Recall. Bile breaks large fat droplets into smaller ones.smaller ones
- Name it. This is called emulsifying.emulsifies
- Write the answer. emulsifies
Check. Smaller droplets give lipase a larger surface area. ✓
Method. Recall the stomach enzyme.
- Recall. Pepsin is a protease made in the stomach.protease made in the stomach
- Write the answer. pepsin
Check. Pepsin works best in acid. ✓
Method. Recall the enzyme that finishes starch digestion.
- Recall. Maltase breaks maltose into two glucose molecules.two glucose molecules
- Write the answer. maltase
Check. Maltase is in the small intestine wall. ✓
Method. Think about the surface area.
- Think. Smaller droplets have more total surface.more total surface
- Conclude. Lipase can act on more of the fat at once.act on more
- Write the answer. Accept any sentence that says the smaller fat droplets give lipase a larger surface area to act on.
Check. Emulsification is a physical change. ✓
Method. Check what bile is.
- Check. Bile contains no enzyme; it is a substance that emulsifies fat.no enzyme
- Conclude. So the statement is not correct, and bile only helps lipase to work.not correct
- Write the answer. Accept any answer that says bile is not an enzyme; it only emulsifies fat so lipase can act.
Check. Lipase is the enzyme that digests fat. ✓
Method. Recall the stomach enzyme and its products.
- Enzyme. The stomach enzyme for protein is pepsin.pepsin
- Products. Pepsin breaks protein into amino acids.amino acids
- Write the answer. Accept pepsin and amino acids.
Check. Pepsin is a protease. ✓
Method. Think about the conditions in each organ.
- Stomach. The stomach is acidic, which suits pepsin but not most pancreatic enzymes.acidic
- Small intestine. The small intestine is alkaline, which suits the pancreatic enzymes.alkaline
- Conclude. Bile neutralises the acid, so the pancreatic enzymes can work in the small intestine.can work
- Write the answer. Accept any answer that says the small intestine has the alkaline conditions these enzymes need, which bile helps to create.
Check. Each enzyme works best in its own conditions. ✓
Required practical: enzymes and temperature (or pH)
In this practical the independent variable is the temperature (or the pH), and the dependent variable is the time for the starch to be digested. The rate of reaction is found from the time: rate = 1000 / time (in this data, a convenient scale). The control variables are the volume and concentration of the amylase and starch, the iodine test, and the time between drops.
Starch gives a blue-black colour with iodine. When all the starch is digested, the iodine stays orange-brown. Only the teacher heats the water bath; pupils must not put hot glass or water near the drop tiles.
| Temperature (C) | Time to clear (s) | Rate = 1000 / time |
|---|---|---|
| 20 | 250 | 4 |
| 30 | 125 | 8 |
| 40 | 50 | 20 |
| 50 | 100 | 10 |
| 60 | 250 | 4 |
Key ideas
- Independent variable: temperature (or pH). Dependent variable: time for the starch to clear.
- Rate of reaction = 1000 / time, so a shorter time means a higher rate.
- Iodine turns starch blue-black; the colour disappears when the starch is digested.
- Enzyme rates rise with temperature, then fall sharply when the enzyme denatures.
Method. Divide 1000 by the time.
- Write the formula. Rate = 1000 / time.1000 / 50
- Calculate. 1000 divided by 50 is 20.1000 / 50 = 20
- Write the answer. 20
Check. 20 x 50 = 1000. ✓
Method. Find the highest point on the graph.
- Find. The highest point is at 40 C, with a rate of 20.highest point
- Read. So the highest rate is at 40 C.40
- Write the answer. 40
Check. The rate at 40 C is higher than at 30 C or 50 C. ✓
Method. Think about what happens to the enzyme.
- Think. Above the best temperature the enzyme starts to denature.starts to denature
- Conclude. Its active site changes shape, so fewer substrate molecules fit.fewer fit
- Write the answer. Above the best temperature the enzyme denatures, so its active site changes shape and fewer substrate molecules fit.
Check. Denaturing is permanent. ✓
Watch out
Watch out: a rate is calculated from the time, so a short time gives a high rate. Do not divide the time by 1000 by mistake.
Exam tip
Safety. Iodine solution can stain skin and clothes and is harmful if swallowed. Wear eye protection and an apron, and wipe up spills at once. The teacher controls the water bath and the heating.
Try it (with your teacher). With your teacher, use amylase and starch at two temperatures, such as room temperature and a warm water bath. Test every 30 seconds with iodine on a spotting tile until the colour stops. Calculate the rate of reaction for each temperature.
Exercise 5D Required practical: enzymes and temperature (or pH)
Method. Find what is measured.
- Find it. The time for the blue-black colour to disappear is measured.time for the colour to disappear
- Write the answer. time for the starch to clear
Check. Temperature is the independent variable. ✓
Method. Find what is changed on purpose.
- Find it. The temperature of the water bath is changed on purpose.changed on purpose
- Write the answer. temperature
Check. The time is measured. ✓
Method. Recall the iodine test.
- Recall. Starch with iodine turns blue-black.blue-black
- Write the answer. blue-black
Check. When the starch is digested, the colour stops. ✓
Method. Divide 1000 by the time.
- Divide. 1000 / 125 = 8.1000 / 125 = 8
- Write the answer. 8
Check. 8 x 125 = 1000. ✓
Method. Divide 1000 by the time.
- Divide. 1000 / 250 = 4.1000 / 250 = 4
- Write the answer. 4
Check. 4 x 250 = 1000. ✓
Method. Read the table.
- Read. The time of 100 seconds is at 50 C.50 C
- Write the answer. 50
Check. The rate at 50 C is 10. ✓
Method. Compare the two rates.
- Read. The rate at 20 C is 4, and the rate at 30 C is 8.4 and 8
- Decide. The higher rate is at 30 C.30
- Write the answer. 30
Check. The rate rises as the temperature rises up to the optimum. ✓
Method. Think about the control variables.
- Think. The time between drops is a control variable.control variable
- Conclude. If it changes, the results cannot be compared fairly.cannot be compared
- Write the answer. Accept any sentence that says the test must be fair, so only the temperature changes the result.
Check. Keep the same volumes and concentrations too. ✓
Method. Think about the endpoint.
- Think. The end is when no blue-black colour forms with iodine.no blue-black
- Conclude. That endpoint is used in every test, so the rates can be compared.compared
- Write the answer. Accept any sentence that says the endpoint is when the colour no longer changes, so each test ends at the same stage.
Check. A clear endpoint makes the results comparable. ✓
Method. Read the graph from left to right.
- Read. From 20 C to 40 C the rate rises from 4 to 20.4 to 20
- Explain. Particles move faster and collide more often, so the rate rises.collide more often
- Write the answer. Accept any sentence that says the rate rises as the temperature rises, because particles move faster and collide more often.
Check. The rate rises most steeply at the optimum. ✓
Method. Look at the method and the results.
- Weakness. Only five temperatures were tested, spaced 10 C apart, so the best temperature may be missed.10 C apart
- Improvement. Testing more temperatures near the best value, and repeating each test, makes the result more reliable.more reliable
- Write the answer. Accept any sensible weakness, such as a wide spacing of temperatures, and an improvement such as testing more temperatures or repeating each one.
Check. A weakness and a specific improvement are both needed. ✓
Method. Think about why two results can match.
- Two causes. A low rate at 20 C is due to slow particles; a low rate at 60 C is due to denaturing.two causes
- Conclude. Two causes can give the same number, so the rate alone does not prove denaturing.does not prove
- Write the answer. Accept any answer that says two different reasons can give the same rate, so the rate alone does not prove denaturing.
Check. Compare the reasons, not only the numbers. ✓
Enzyme and substrate concentration [H]
[H] Higher tier only. This section is higher-tier content. The rate of reaction depends on the substrate concentration, until the enzymes are all working at full capacity.
As the substrate concentration increases, there are more collisions between substrate and enzyme, so the rate increases. At a high substrate concentration, all the active sites are full, so adding more substrate does not increase the rate. The rate then levels off. Increasing the enzyme concentration at a fixed substrate concentration increases the rate, because there are more active sites.
Key ideas
- [H] A higher substrate concentration gives more collisions, so the rate increases.
- [H] At high substrate concentration, the rate levels off because all the active sites are full.
- [H] Increasing the enzyme concentration increases the rate, while the substrate is not limiting.
- [H] Enzymes work at a fixed maximum rate for a given amount of enzyme.
Method. Find 3 on the x-axis and read up.
- Find 3. Find 3 on the substrate axis.3
- Read. The point at 3 is at a rate of 24.24
- Write the answer. 24
Check. The rate at 4 and 5 is about 25. ✓
Method. Look for the first point where the line becomes flat.
- Look. The rate is 24 at 3 and 25 at 4, so it nearly stops rising at 4.25 at 4
- Read. The rate is the same, 25, at 4 and at 5, so it levels off at 4.levels off at 4
- Write the answer. 4
Check. The rate is 25 for both 4 and 5. ✓
Method. Think about the active sites.
- Think. At high substrate concentration, every active site is occupied.every active site
- Conclude. So more substrate cannot increase the rate; the enzyme is working at its maximum.maximum
- Write the answer. Every active site is occupied, so adding more substrate cannot increase the rate; the enzyme is working at its maximum.
Check. The limit is the number of enzymes, not the substrate. ✓
Watch out
Watch out: the rate levels off because the enzymes are full, not because the substrate has run out. Adding more enzyme would raise the level.
Exam tip
Exam tip. [H] When a graph levels off, say which factor limits the rate: here, the number of active sites.
Try it (with your teacher). Sketch the graph of rate against substrate concentration. Mark the point where it levels off, and say what limits the rate at that point.
Exercise 5E [H] Enzyme and substrate concentration
Method. Read the start of the graph.
- Read. The rate rises from 0 to 10 to 18 as the substrate concentration goes up.rises from 0
- Write the answer. increases
Check. The rate levels off later. ✓
Method. Read the point at 1.
- Read. The graph is at 10 when the substrate concentration is 1.10
- Write the answer. 10
Check. The rate at 0 is 0. ✓
Method. Read the point at 5.
- Read. At substrate concentration 5 the rate is 25.25
- Write the answer. 25
Check. The rate is the same as at 4. ✓
Method. Think about collisions.
- Think. More substrate molecules are present.more substrate
- Conclude. So there are more collisions with the active sites, and the rate rises.rate rises
- Write the answer. Accept any sentence that says more substrate means more collisions with active sites.
Check. Below the limit, substrate is the limiting factor. ✓
Method. Think about what is full.
- Think. At high substrate concentration, the active sites are all full.all full
- Name it. So the number of active sites limits the rate.the number of active sites
- Write the answer. the number of active sites
Check. More enzyme would raise the level. ✓
Method. Think about the active sites.
- Think. Doubling the enzyme doubles the number of active sites.doubles the active sites
- Conclude. With more active sites free to use, the rate increases.rate increases
- Write the answer. So yes.yes
- Write the answer. yes
Check. Enzyme concentration is a limiting factor at high substrate. ✓
Method. Find the steepest part of the line.
- Look. The rise from 0 to 1 is 10, the largest single rise in the graph.0 to 1 is 10
- Conclude. So the steepest part is between 0 and 1.steepest
- Write the answer. Accept 0 and 1.
Check. The rise from 1 to 2 is 8, which is smaller. ✓
Method. Subtract the two readings.
- Read. The rate is 10 at 1, and 24 at 3.10 and 24
- Subtract. The change is 24 - 10 = 14.24 - 10 = 14
- Write the answer. 14
Check. 10 + 14 = 24. ✓
Method. Link the graph to the active sites.
- Think. Beyond 4 the rate hardly changes, so the enzymes are already working at full capacity.full capacity
- Conclude. All active sites are full, so the extra substrate cannot be used.cannot be used
- Write the answer. Accept any sentence that says the enzyme is working at its maximum, because all active sites are full.
Check. The limit is set by the enzyme, not the substrate. ✓
Method. Link the number of active sites to the maximum.
- Enzyme. Doubling the enzyme doubles the number of active sites.doubles the active sites
- Maximum. The maximum rate depends on the number of active sites, so it rises.maximum rate rises
- Write the answer. Accept any answer that says doubling the enzyme doubles the number of active sites, so the maximum rate rises.
Check. The maximum rate is set by the enzyme concentration. ✓
Chapter summary
| Skill | Method |
|---|---|
| Organs | Mouth, oesophagus, stomach, small intestine, large intestine; liver, gall bladder and pancreas as glands. |
| Mechanical and chemical digestion | Mechanical breaks food into pieces; chemical uses enzymes to make small soluble molecules. |
| Enzymes | Biological catalysts with an active site that fits one substrate (lock and key); not used up. |
| Denaturing | High temperature or extreme pH changes the active site shape permanently. |
| Digestive enzymes | Amylase: starch to maltose; protease: protein to amino acids; lipase: fat to fatty acids and glycerol. |
| Bile | Made in the liver, stored in the gall bladder; neutralises stomach acid and emulsifies fat. |
| Rate | Rate = 1000 / time in the amylase practical; a shorter time means a faster rate. |
| [H] Substrate concentration | Rate rises, then levels off when all active sites are full. |
Can you…?
- I can name the organs of the digestive system and say what each does
- I can explain enzymes as biological catalysts using the lock and key model
- I can name the enzymes for starch, protein and fat, and their products
- I can explain the role of bile, and why enzymes denature
- I can carry out the enzyme practical, calculate a rate and interpret a graph
- [H] I can explain how substrate concentration changes the rate
Key terms
- mechanical digestion
- breaking food into smaller pieces without chemicals, such as chewing
- chemical digestion
- breaking large molecules into small soluble ones using enzymes
- enzyme
- a protein that acts as a biological catalyst
- active site
- the part of an enzyme where the substrate fits
- substrate
- the molecule that an enzyme acts on
- denatured
- an enzyme whose active site has changed shape permanently
- bile
- a liquid made in the liver that neutralises acid and emulsifies fat
- emulsify
- to break large fat droplets into smaller ones
- rate of reaction
- how fast a reaction happens; in this practical, 1000 / time
- [H] limiting factor
- a factor that stops the rate increasing further, such as the number of active sites
Exercise 5Z Check-up
Method. Recall the gland that makes bile.
- Recall. Bile is made in the liver.made in the liver
- Write the answer. liver
Check. The gall bladder stores it. ✓
Method. Recall the carbohydrase.
- Recall. Amylase breaks down starch.breaks down starch
- Write the answer. amylase
Check. Amylase is in saliva and the pancreas. ✓
Method. Recall the products.
- Recall. Lipase breaks fats into fatty acids and glycerol.fatty acids and glycerol
- Write the answer. fatty acids and glycerol
Check. Fat is digested in the small intestine. ✓
Method. Recall the lock and key model.
- Recall. The substrate fits into the active site.fits into the active site
- Write the answer. active site
Check. Each enzyme has a specific active site. ✓
Method. Recall the process.
- Recall. High temperature changes the active site shape permanently; this is denaturing.denaturing
- Write the answer. denaturing
Check. Denatured enzymes cannot be restored by cooling. ✓
Method. Divide 1000 by the time.
- Divide. 1000 / 50 = 20.1000 / 50 = 20
- Write the answer. 20
Check. 20 x 50 = 1000. ✓
Method. Recall the role of bile.
- Recall. Bile neutralises the acid from the stomach.neutralises the acid
- Name it. So the answer is that bile neutralises stomach acid.bile neutralises stomach acid
- Write the answer. bile neutralises stomach acid
Check. Bile is made by the liver. ✓
Method. Recall the stomach enzyme.
- Recall. Pepsin is a protease made in the stomach.protease made in the stomach
- Write the answer. pepsin
Check. Pepsin works in acid. ✓
Method. Think about the particles.
- Think. Higher temperature makes particles move faster.move faster
- Conclude. So the enzyme and substrate collide more often, and the rate rises.collide more often
- Write the answer. Accept any sentence that says particles move faster and collide more often.
Check. Above the optimum the enzyme denatures. ✓
Method. Recall the physical processes.
- Recall. Chewing and churning break food into smaller pieces without enzymes.without enzymes
- Write the answer. Accept chewing, or churning in the stomach.
Check. These are mechanical, not chemical, processes. ✓
Total 22 marks. Allow about 22 minutes.
Exercise 5X Exam-style questions
Method. Recall the lipid enzyme.
- Recall. Lipase breaks down lipids.breaks down lipids
- Write the answer. lipase
Check. Protease breaks down protein. ✓
Method. Recall the products.
- Recall. Proteins are broken down into amino acids.amino acids
- Write the answer. amino acids
Check. Amino acids are soluble and can be absorbed. ✓
Method. Recall the storage organ.
- Recall. Bile is stored in the gall bladder.stored in the gall bladder
- Write the answer. gall bladder
Check. The liver makes bile. ✓
Method. Describe the fit, then the specificity.
- Fit. The substrate fits into the active site of the enzyme.fits into the active site
- Specific. Only a substrate with the right shape fits, so each enzyme is specific.right shape
- Write the answer. Accept any answer that says the substrate fits into the active site, and the enzyme only works with one shape of substrate.
Check. The key and the lock must match. ✓
Method. Give two jobs of bile.
- Job 1. Bile neutralises the acid from the stomach.neutralises
- Job 2. Bile emulsifies fat, breaking it into smaller droplets.emulsifies
- Write the answer. Accept any two of: neutralises stomach acid; emulsifies fat; gives alkaline conditions for enzymes in the small intestine.
Check. Bile contains no enzyme. ✓
Method. Divide, then find the other temperature with the same time.
- Calculate. 1000 / 250 = 4.1000 / 250 = 4
- Look up. The other temperature with a time of 250 seconds is 60 C.60 C
- Write the answer. Accept 4, and 60 C.
Check. Both temperatures give the same rate. ✓
Method. Think about the enzyme at high temperature.
- Denature. At 80 C the enzyme denatures; its active site changes shape.denatures
- Fit. The substrate no longer fits, so the reaction slows or stops.no longer fits
- Conclude. So a higher temperature does not always give a faster rate.not always faster
- Write the answer. Accept any answer that says the enzyme denatures at 80 C, so its active site no longer fits the substrate.
Check. The optimum is a balance of both effects. ✓
Method. Follow the starch through the gut.
- Mouth. Amylase in saliva starts to break starch down into maltose.saliva
- Small intestine. Amylase from the pancreas continues the digestion of the starch.pancreas
- Final step. Maltase breaks maltose into glucose, which can be absorbed.glucose
- Write the answer. Accept any three of: amylase in the mouth; amylase from the pancreas in the small intestine; maltase breaks maltose into glucose.
Check. Glucose is then absorbed into the blood. ✓
Method. Calculate the rates, then interpret.
- Rates. The rates are 4, 8, 20 and 10 for these temperatures.4, 8, 20, 10
- Highest. The highest rate is 20 at 40 C.20 at 40 C
- Explain. At 40 C the enzyme works best; the rate falls at higher temperatures as it denatures.optimum
- Write the answer. Accept 40 C, and an answer that says it is the optimum, where the enzyme works best.
Check. Calculate each rate before comparing. ✓
Method. Explain the plateau, then the change needed.
- Plateau. At substrate concentration 4 all the active sites are full.all the active sites
- Limit. Adding more substrate cannot increase the rate, because the enzymes are working at their maximum.maximum
- Raise. More enzyme gives more active sites, so the rate at the same substrate concentration would be higher.more active sites
- Write the answer. Accept any answer that says all active sites are full, and that more enzyme would raise the rate.
Check. The number of enzymes sets the maximum. ✓