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GCSE · Biology

B2a The Digestive System and Enzymes

AQA 8464 · Calculator allowed · about 95 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Breaking Down a Meal

Original text written for Revision Library.

Every meal contains large molecules, such as starch, proteins and fats, that are far too big to pass through the wall of the gut and into the blood. Before the body can use the energy and nutrients in food, the digestive system must break these large, insoluble molecules down into small, soluble ones. This happens through mechanical digestion, which physically breaks food into smaller pieces, and chemical digestion, in which enzymes catalyse (speed up) the breakdown of large molecules into smaller ones. Different digestive enzymes are produced in different parts of the gut, and each works fastest at its own optimum temperature and pH. Bile, produced by the liver and stored in the gall bladder, is not itself an enzyme but plays an important supporting role by neutralising stomach acid and by emulsifying fats, so that lipase can act on them more quickly. This worksheet covers the organs of the digestive system, the action of digestive enzymes, the effect of temperature and pH on enzyme activity, and the required practicals investigating food tests and enzyme activity.

The table below lists four organs involved in digestion. Complete the table by writing one main digestive function for each organ.

Organ	Main digestive function
Mouth	?
Stomach	?
Small intestine	?
Large intestine	?

(Total for Question 1 is 4 marks)

Enzymes are essential for digestion.

(a)State what is meant by a catalyst.(1)

(b)State the type of molecule that all enzymes are made of.(1)

(c)State why enzymes are needed to digest food.(1)

(Total for Question 2 is 3 marks)

Bile is produced and stored before it is released into the small intestine.

(a)Name the organ that produces bile.(1)

(b)Name the organ that stores bile before it is released.(1)

(c)State one function of bile in digestion.(1)

(Total for Question 3 is 3 marks)

Enzymes are described as being 'specific' to their substrate.

(a)Using the lock and key model, explain what is meant by an enzyme being 'specific'.(2)

(b)A student mixes protease enzyme with a starch solution. Use the lock and key model to explain why no reaction takes place.(2)

(Total for Question 4 is 4 marks)

The small intestine is lined with villi, which increase the efficiency of absorption of digested food molecules into the blood.

(a)State four features of villi that adapt them for efficient absorption.(4)

(Total for Question 5 is 4 marks)

Table 1 shows how the rate of reaction of amylase changes with pH, obtained by mixing amylase with starch solution at a fixed temperature using buffer solutions of different pH.

pH: 2, 4, 6, 7, 8, 10
Relative rate of reaction (arbitrary units): 0, 3, 8, 10, 6, 1

(a)Identify the optimum pH for amylase shown by the data in Table 1.(1)

(b)Describe the pattern shown by the data in Table 1 as pH increases from 2 to 10.(2)

(c)Explain, in terms of enzyme structure, why the rate of reaction is zero at pH 2.(2)

(Total for Question 6 is 5 marks)

Aisha investigates how pH affects the rate of starch digestion by amylase (required practical), using a water bath at 35 degC, amylase solution, starch solution, buffer solutions of different pH, a spotting tile and iodine solution.

(a)Describe a method Aisha could use to obtain reliable results for this investigation, and explain how she would use her results to determine the rate of reaction at each pH.(6)

(Total for Question 7 is 6 marks)

Table 2 shows the results of four food tests (iodine test, Benedict's test, Biuret test and emulsion test) carried out on four unknown food samples, W, X, Y and Z (required practical).

Sample	Iodine test	Benedict's test (after heating)	Biuret test	Emulsion test
W	turns blue-black	remains blue	remains blue	remains clear, no layer
X	remains browny-orange	forms a brick-red precipitate	remains blue	remains clear, no layer
Y	remains browny-orange	remains blue	turns purple/lilac	remains clear, no layer
Z	remains browny-orange	remains blue	remains blue	forms a cloudy white emulsion layer

(a)Identify which biological molecule is present in each of samples W, X, Y and Z.(4)

(b)State one safety precaution that should be taken when heating sample X with Benedict's solution, and explain why it is necessary.(2)

(Total for Question 8 is 6 marks)

Bile emulsifies large lipid droplets into many smaller droplets. A student models a large lipid droplet as a cube of side length 8 mm.

(a)Calculate the volume of the large cube-shaped droplet.(1)

(b)Bile breaks this large droplet into 8 smaller droplets of equal volume, also modelled as cubes. Calculate the side length of each smaller cube.(2)

(c)Calculate the total surface area of the 8 smaller cubes, and compare this with the surface area of the original large cube.(3)

(d)Use your answer to part (c) to explain why bile increases the rate of fat digestion by lipase.(2)

(Total for Question 9 is 8 marks)

A 1 m² section of small intestine wall lining, if it were completely flat and smooth, would have a surface area of 1 m^2. Folding of the wall into villi and microvilli increases the effective surface area of this section by a factor of 30.

(a)Calculate the effective surface area of this section of intestine wall after folding into villi and microvilli.(1)

(b)The total (folded) surface area of the entire human small intestine is approximately 250 m^2. Using the factor of 30 given above, estimate the surface area of the intestine wall if it were completely flat and unfolded. Give your answer to 2 significant figures.(2)

(Total for Question 10 is 3 marks)

Ryan investigates how temperature affects the rate of starch digestion by amylase (required practical), using the iodine spotting-tile method. Table 3 shows the time taken for the iodine test to stop giving a blue-black colour at each temperature.

Temperature (degC): 10, 20, 30, 40, 50, 60
Time for starch to be fully digested (s): 240, 120, 60, 40, 90, 300

(a)Describe the pattern shown by the data in Table 3 as temperature increases from 10 degC to 60 degC.(2)

(b)Calculate the rate of reaction at 30 degC. Give your answer in s^-1, to 3 significant figures.(2)

(c)Calculate the percentage decrease in the rate of reaction between 40 degC and 60 degC.(3)

(d)Explain, in terms of enzyme and substrate molecules, why the time taken increases sharply between 40 degC and 60 degC.(3)

(Total for Question 11 is 10 marks)

Lactase is an enzyme that catalyses the breakdown of lactose (a sugar found in milk) into glucose and galactose in the small intestine. People with lactose intolerance produce very little or no lactase.

(a)Predict what happens to undigested lactose in a lactose-intolerant person after they drink milk.(1)

(b)Tom, a food scientist, develops a milk-based drink with lactase enzyme already added to it before it is sold, to make it suitable for lactose-intolerant people. Suggest why adding lactase directly to the drink allows lactose-intolerant people to digest the lactose it contains.(2)

(c)Tom tests the new drink by measuring the concentration of lactose remaining over time after adding a fixed amount of lactase, at 37 degC. State two variables, other than temperature, that should be controlled between repeat tests to make the comparison valid.(2)

(Total for Question 12 is 5 marks)

A meal contains starch, protein and fat.

Explain how this meal is digested as it passes through the digestive system, referring to mechanical digestion, chemical digestion (naming enzymes and the products they produce) and the role of bile.

(Total for Question 13 is 6 marks)

Higher tier only. Enzymes are proteins with a specific tertiary structure.

(a)Explain, using ideas about bonds within the enzyme molecule, how a change in pH away from the optimum causes an enzyme to lose its catalytic function, and explain why this effect is usually permanent even if the pH is later returned to the optimum value.(4)

(Total for Question 14 is 4 marks)

Priya investigated the effect of temperature on the rate of starch digestion by amylase (as in Table 3, Question 11). She placed the test tube of iodine solution used for sampling into the same water bath as the tube containing the starch, buffer and amylase mixture.

(a)Evaluate this aspect of Priya's method, explaining one problem it could cause with her results.(2)

(b)Suggest one improvement to the method that would address this problem.(1)

(c)Priya repeated the investigation at each temperature only once. Explain why repeating each temperature reading and calculating a mean time would improve the investigation.(2)

(Total for Question 15 is 5 marks)

Higher tier only. Figure 1 shows how the rate of reaction of two protease enzymes, pepsin and trypsin, changes with pH. Pepsin's rate rises from 0 at pH 1 to a peak of 12 (arbitrary units) at pH 2, then falls back to 0 by pH 5. Trypsin's rate is 0 at pH 5, rises to a peak of 15 at pH 8, then falls to 2 by pH 10. At pH 7, trypsin's rate of reaction is 12 units and pepsin's rate of reaction is 0 units.

(a)State the optimum pH for pepsin and the optimum pH for trypsin.(2)

(b)Pepsin is produced in the stomach and trypsin is produced by the pancreas, acting in the small intestine. Using the optimum pH values from part (a), explain why it is an advantage that pepsin and trypsin have different optimum pH values.(3)

(c)Calculate the percentage of trypsin's maximum rate of reaction that its rate at pH 7 represents.(2)

(Total for Question 16 is 7 marks)

Mark scheme · B2a The Digestive System and Enzymes

Question 1

B1 mouth: mechanical digestion by chewing (mastication) and mixing food with saliva, which contains amylase to begin starch digestion, oe
B1 stomach: churns food (mechanical digestion) and produces protease (pepsin) and hydrochloric acid, which gives the acidic pH needed for pepsin and helps kill bacteria, oe
B1 small intestine: site where chemical digestion is completed (amylase, protease and lipase act here) and where the small, soluble products of digestion are absorbed into the blood through the villi, oe
B1 large intestine: absorbs water (and some mineral ions) from the undigested remains of food, forming faeces, oe
Answer: Mouth: mechanical digestion and amylase in saliva begins starch digestion. Stomach: churns food and produces pepsin and hydrochloric acid. Small intestine: completes chemical digestion and absorbs digested food into the blood. Large intestine: absorbs water, forming faeces.

Question 2

(a) B1 a substance that speeds up a reaction without being changed or used up itself, oe
(a) Answer: A substance that speeds up a reaction without being changed or used up itself.
(b) B1 protein
(b) Answer: Protein.
(c) B1 to speed up (catalyse) the breakdown of large, insoluble food molecules into small, soluble molecules fast enough to be useful at body temperature, oe
(c) Answer: They speed up the breakdown of large, insoluble food molecules into small, soluble ones fast enough for the body to use them.

Question 3

(a) B1 liver
(a) Answer: Liver.
(b) B1 gall bladder
(b) Answer: Gall bladder.
(c) B1 any one of: neutralises the acid from the stomach, producing an alkaline pH suitable for enzymes in the small intestine; OR emulsifies fats/lipids into smaller droplets, increasing the surface area for lipase to act on, oe
(c) Answer: E.g. neutralises stomach acid to give an alkaline pH suited to the small intestine's enzymes; or emulsifies fats, increasing the surface area for lipase.

Question 4

(a) B1 the enzyme's active site has a shape that is complementary to (fits) only one particular substrate, oe
(a) B1 the substrate binds to (fits into) the active site, forming an enzyme-substrate complex, allowing the reaction to take place, oe
(a) Answer: An enzyme's active site has a shape complementary to only one substrate; that substrate fits into (binds to) the active site, forming an enzyme-substrate complex.
(b) B1 protease's active site shape is complementary to protein molecules, not to starch, oe
(b) B1 the starch molecule's shape does not fit the protease's active site, so no enzyme-substrate complex can form and the reaction cannot take place, oe
(b) Answer: Protease's active site is complementary to protein, not starch, so starch does not fit the active site; no enzyme-substrate complex forms and no reaction occurs.

Question 5

(a) B1 any four from: a large surface area, due to their finger-like shape and folding, increasing the rate of diffusion
(a) B1 a single layer of epithelial cells, giving a short diffusion distance/thin wall
(a) B1 a good (dense) blood supply/network of capillaries, which maintains a steep concentration gradient by carrying absorbed molecules away quickly
(a) B1 microvilli on the surface of the epithelial cells, further increasing the surface area
(a) B1 a lacteal, for the absorption of fatty acids and glycerol into the lymphatic system, oe
(a) Answer: Any four from: large surface area (finger-like shape/folding); single layer of cells (short diffusion distance); good blood supply (capillary network) maintaining a steep concentration gradient; microvilli further increasing surface area; a lacteal for absorbing fatty acids and glycerol.

Question 6

(a) B1 pH 7 cao
(a) Answer: pH 7.
(b) B1 the rate of reaction increases as pH increases from 2 up to a peak at pH 7, oe
(b) B1 the rate of reaction then decreases as pH increases further, from pH 7 to pH 10, oe
(b) Answer: The rate increases from pH 2 to a peak at pH 7, then decreases from pH 7 to pH 10.
(c) B1 a pH this far from the optimum breaks the bonds holding the enzyme's tertiary structure together, changing the shape of its active site (the enzyme is denatured), oe
(c) B1 the substrate (starch) can no longer bind to/fit the active site, so no enzyme-substrate complexes form and the rate of reaction is zero, oe
(c) Answer: At pH 2 the enzyme is denatured (its active site shape is changed), so starch can no longer bind to the active site and the rate of reaction is zero.

Question 7

(a) B1 set up a water bath at a constant temperature (e.g. 35 degC) to control this variable
(a) B1 mix a set volume of starch solution with a set volume of amylase solution and a chosen pH buffer solution, and start a timer immediately
(a) B1 at regular time intervals (e.g. every 30 s), remove a drop of the reaction mixture and add it to a fresh well of iodine solution on the spotting tile
(a) B1 continue sampling until the iodine solution no longer changes from browny-orange to blue-black, and record this time as the end-point
(a) B1 repeat at each pH, keeping temperature, volumes and concentrations the same, and repeat each pH at least twice more to calculate a mean time and improve reliability
(a) B1 calculate the rate of reaction at each pH using rate = 1 / time, and use these values to identify the pH giving the fastest rate (the optimum pH), oe
(a) Answer: Mix starch, amylase and a buffer at a controlled temperature; sample onto iodine at regular intervals until the colour no longer changes, recording the end-point time; repeat at each pH (repeating each pH for a mean); calculate rate = 1/time at each pH to find the optimum.

Question 8

(a) B1 W: starch
(a) B1 X: reducing sugar (e.g. glucose)
(a) B1 Y: protein
(a) B1 Z: lipid (fat/oil)
(a) Answer: W: starch. X: reducing sugar. Y: protein. Z: lipid.
(b) B1 precaution: heat the test tube in a water bath rather than directly over a Bunsen flame, oe
(b) B1 reason: this heats the mixture gently and evenly, reducing the risk of the boiling liquid spitting out of the tube or the glass cracking, oe
(b) Answer: Heat in a water bath, not directly over a flame, because this heats the mixture gently and evenly, reducing the risk of the liquid spitting or the tube cracking.

Question 9

(a) B1 512 mm³ cao
(a) Answer: 512 mm^3.
(b) M1 volume of each small cube = 512 / 8 = 64 mm³, ft from (a), oe
(b) A1 4 mm cao (side = cube root of 64)
(b) Answer: 4 mm.
(c) M1 surface area of large cube = 6 * 8² = 384 mm², oe
(c) M1 surface area of one small cube = 6 * 4² = 96 mm², total for 8 cubes = 8 * 96 = 768 mm², ft from (b), oe
(c) A1 comparison: 768 mm² is double (2 times) the original 384 mm² cao (ft from their two surface areas)
(c) Answer: Total surface area of the 8 small cubes = 768 mm², which is double the original large cube's surface area of 384 mm^2.
(d) B1 breaking the large droplet into many smaller droplets increases the total surface area of the lipid, without changing its total volume/mass, oe
(d) B1 this gives lipase a much greater surface area to act on (more lipid molecules exposed at a surface), increasing the rate of fat digestion, oe
(d) Answer: Emulsification increases the total surface area of the lipid (without changing its total volume), giving lipase a greater surface area to act on and so increasing the rate of digestion.

Question 10

(a) B1 30 m² cao
(a) Answer: 30 m^2.
(b) M1 250 / 30, oe
(b) A1 8.3 m² awrt cao (2 s.f.)
(b) Answer: 8.3 m^2.

Question 11

(a) B1 the time taken decreases (the rate of reaction increases) as temperature rises from 10 degC to a minimum time (fastest rate) at 40 degC, oe
(a) B1 the time taken then increases sharply (the rate of reaction decreases) as temperature rises further, from 40 degC to 60 degC, oe
(a) Answer: Time decreases (rate increases) from 10 degC to a fastest rate at 40 degC, then time increases sharply (rate decreases) from 40 degC to 60 degC.
(b) M1 rate = 1 / time = 1 / 60, oe
(b) A1 0.0167 s^-1 awrt cao
(b) Answer: 0.0167 s^-1.
(c) M1 rate at 40 degC = 1 / 40 = 0.025 s^-1 and rate at 60 degC = 1 / 300 = 0.00333 s^-1, oe
(c) M1 percentage decrease = (0.025 - 0.00333) / 0.025 * 100, oe
(c) A1 86.7% awrt cso (allow 86 to 87%)
(c) Answer: 86.7% decrease.
(d) B1 above the optimum temperature (40 degC), the amylase enzyme starts to denature, oe
(d) B1 the increase in temperature breaks the bonds holding the enzyme's tertiary structure together, changing (destroying) the shape of its active site, oe
(d) B1 the substrate (starch) can no longer bind to/fit into the active site, so fewer enzyme-substrate complexes form and the rate of reaction decreases (time taken increases), oe
(d) Answer: Above 40 degC the amylase denatures: the extra heat breaks bonds holding its tertiary structure together, changing the active site's shape so starch can no longer bind, reducing the rate of reaction.

Question 12

(a) B1 the lactose is not broken down/absorbed in the small intestine and passes into the large intestine, where gut bacteria act on it, causing digestive discomfort (e.g. bloating, wind, diarrhoea), oe
(a) Answer: It is not digested or absorbed, and passes into the large intestine, where gut bacteria act on it, causing digestive discomfort.
(b) B1 the lactase added to the drink can catalyse the breakdown of the lactose in the drink itself, independently of the person's own (lacking) lactase, oe
(b) B1 so the lactose is already broken down into glucose and galactose before/as the drink is consumed, which the person can absorb normally without needing their own lactase, oe
(b) Answer: The added lactase breaks down the lactose in the drink itself, so it is already converted to glucose and galactose, which the person can absorb without needing their own lactase.
(c) B1 any two from: volume of milk/drink used; concentration/amount of lactase enzyme added; starting concentration of lactose in the drink; pH of the drink, oe
(c) Answer: Any two from: volume of drink used; concentration of lactase added; starting lactose concentration; pH of the drink.

Question 13

Level 3 (5-6): A detailed and logically structured explanation that clearly links mechanical digestion, the action of amylase, protease and lipase (with correct substrates and products), and the role of bile in emulsifying fat and neutralising stomach acid, to show how the meal is broken down into small, soluble molecules.
Level 2 (3-4): The answer describes some relevant features of both mechanical and chemical digestion, naming at least two digestive enzymes and referring to bile, but the explanation of how these processes work together may be incomplete or not fully logically ordered.
Level 1 (1-2): The answer includes simple, isolated statements about digestion, for example naming that food is broken down or naming one enzyme, with little or no reference to mechanical digestion, specific enzyme products, or the role of bile, and the answer is not well linked.
Level 0 (0): No relevant content.
Indicative content:
- Mechanical digestion (chewing in the mouth and churning in the stomach) breaks food into smaller pieces, increasing the surface area for enzymes to act on.
- Amylase (produced by the salivary glands and pancreas, acting in the mouth and small intestine) catalyses the breakdown of starch into sugars such as maltose and glucose.
- Protease (e.g. pepsin, produced in the stomach; also produced by the pancreas and small intestine) catalyses the breakdown of protein into amino acids.
- Lipase (produced by the pancreas, acting in the small intestine) catalyses the breakdown of lipids into fatty acids and glycerol.
- The stomach produces hydrochloric acid, which gives the acidic pH needed for pepsin to work and helps kill harmful bacteria in food.
- Bile is produced in the liver and stored in the gall bladder before being released into the small intestine.
- Bile neutralises the acidic mixture arriving from the stomach, producing an alkaline pH suited to the enzymes of the small intestine.
- Bile emulsifies lipids into smaller droplets, increasing the surface area available for lipase to act on and speeding up fat digestion.
- Each digestive enzyme is specific because the shape of its active site is complementary only to its particular substrate (the lock and key model).
- The small, soluble products of digestion (glucose, amino acids, fatty acids and glycerol) are absorbed into the blood across the villi in the small intestine.

Question 14

(a) B1 a change in pH away from the optimum breaks the bonds (e.g. hydrogen bonds) holding the enzyme's tertiary structure (its 3D shape) together, oe
(a) B1 this changes the shape of the active site so that it is no longer complementary to the substrate, oe
(a) B1 the enzyme is described as denatured and can no longer catalyse the reaction (the rate of reaction becomes zero), oe
(a) B1 the change in shape is usually permanent because the original bonds do not reform in the same arrangement once broken, so the enzyme cannot return to its original (functional) shape, oe
(a) Answer: A pH change breaks bonds holding the enzyme's tertiary structure together, changing the active site's shape so the substrate no longer fits (the enzyme is denatured). This is usually permanent because the original bonds do not reform in the same arrangement once broken.

Question 15

(a) B1 the iodine solution would also be heated to the test temperature, rather than staying at room temperature, oe
(a) B1 this introduces an extra, uncontrolled variable, which could affect the iodine's colour-change reaction and make the results at different temperatures an unfair comparison, oe
(a) Answer: Heating the iodine along with the reaction mixture means the iodine is also at the test temperature, introducing an extra uncontrolled variable that could make comparisons between temperatures unfair.
(b) B1 keep the iodine solution in a separate spotting tile at room temperature, only removing samples of the reacting mixture to test with the iodine, so the iodine itself is never heated, oe
(b) Answer: Keep the iodine at room temperature in a separate spotting tile, only removing small samples of the reaction mixture to test against it.
(c) B1 repeating reduces the effect of any anomalous results/random error on the data, giving a more precise value for the time at each temperature, oe
(c) B1 this increases confidence that any pattern seen between temperature and rate is a real effect of temperature, rather than due to chance variation, oe
(c) Answer: Repeats reduce the effect of anomalies/random error, giving a more precise mean time, and increase confidence that the pattern seen is a real effect of temperature rather than chance.

Question 16

(a) B1 pepsin: pH 2
(a) B1 trypsin: pH 8
(a) Answer: Pepsin: pH 2. Trypsin: pH 8.
(b) B1 the stomach is strongly acidic (contains hydrochloric acid), so pepsin having a low, acidic optimum pH (pH 2) means it works fastest in this environment, oe
(b) B1 the small intestine is alkaline (bile neutralises the acid from the stomach), so trypsin having a higher, alkaline optimum pH (pH 8) means it works fastest there, oe
(b) B1 if the enzymes had the 'wrong' optimum pH for their location they would work more slowly (or be denatured), so having different optima matched to each organ maximises the rate of protein digestion overall, oe
(b) Answer: Pepsin's low optimum pH suits the acidic stomach; trypsin's higher optimum pH suits the alkaline small intestine. Each enzyme therefore works fastest in its own organ, maximising the overall rate of protein digestion.
(c) M1 12 / 15, ft from the maximum rate identified for trypsin, oe
(c) A1 80% cao
(c) Answer: 80%.

The Digestive System and Enzymes - Worksheets, Questions and Revision