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OCR A Level Biology

Topic questions Cell biology Food tests

Food tests

6 questions · 80 marks · about 2 hours

Food tests questions

OCR H420 Food tests 6 questions

Biological Diversity June 2023, Question 18 10 marks

Figure for question 18
(a)

State one other method for immobilising enzymes.

[1]
Reveal mark scheme Mark scheme for question 18(a)
(b)

One advantage of using immobilised enzymes is that the enzyme can be reused many times. However, the process of immobilisation can reduce the activity of enzymes.

With reference to Fig. 18.1, explain why the activity of immobilised enzymes might be lower than that of enzymes that are free in solution.

[2]
Reveal mark scheme Mark scheme for question 18(b)
(c)

Invertase is an enzyme that catalyses the following reaction:

Question 18, shown as an image

Invertase is present in cells of baker’s yeast, a type of fungus.

It is possible to immobilise invertase and it is also possible to immobilise yeast cells.

(c)(i)

Suggest one reason why using immobilised yeast might be more expensive than using immobilised invertase.

[1]
Reveal mark scheme Mark scheme for question 18(c)(i)
(c)(ii)
Figure for question 18 Figure for question 18
[6]
Reveal mark scheme Mark scheme for question 18(c)(ii)

Biological Diversity June 2022, Question 21 16 marks

Microorganisms can be used to produce a variety of food products.

(a)

Microorganisms have simple nutrient requirements, which helps to reduce production costs.

List two other advantages of using microorganisms in food production.

[2]
Reveal mark scheme Mark scheme for question 21(a)
(b)

Yoghurt is a food produced from milk using microorganisms.

Yoghurt production involves two bacteria: Lactobacillus delbrueckii and Streptococcus thermophilus.

(b)(i)

The bacteria convert the lactose present in milk into lactic acid.

Lactic acid is an important contributor to the flavour of yoghurt. Lactic acid also helps to give yoghurt a longer shelf life than milk.

Suggest how lactic acid helps to extend the shelf life of yoghurt.

[2]
Reveal mark scheme Mark scheme for question 21(b)(i)
(b)(ii)

Both bacteria also break down some of the protein casein, which is present in milk.

Name the product of protein breakdown and describe the type of reaction that takes place.

[2]
Reveal mark scheme Mark scheme for question 21(b)(ii)
(c)

The microorganism that is used to produce mycoprotein is a fungus.

The diagram is of a fermenter used for mycoprotein production.

Figure for question 21
(c)(i)

Use the diagram to name the type of fermentation process used for mycoprotein production.

Justify your answer.

[1]
Reveal mark scheme Mark scheme for question 21(c)(i)
(c)(ii)

Suggest and explain why a cooling system is necessary.

[2]
Reveal mark scheme Mark scheme for question 21(c)(ii)
(c)(iii)

The air inlet provides the fungus with oxygen for respiration, and ammonia.

Suggest and explain why the fungus is provided with ammonia.

[2]
Reveal mark scheme Mark scheme for question 21(c)(iii)
(d)

Mycoprotein foods contain protein.

The presence of protein in a solution can be detected using biuret reagent.

In the presence of protein, biuret reagent turns from blue to violet.

(d)(i)

A colorimeter can be used, along with biuret reagent, to determine the concentration of protein in a solution.

Outline how to use a colorimeter to determine the concentration of protein in a solution.

[4]
Reveal mark scheme Mark scheme for question 21(d)(i)
(d)(ii)

State one alternative method for determining the concentration of protein in a solution.

[1]
Reveal mark scheme Mark scheme for question 21(d)(ii)

Biological Processes November 2020, Question 16 16 marks

(a)

Fig. 16.1 shows the structure of ATP.

Figure for question 16
(a)(i)

Name the circled component in Fig. 16.1.

[1]
Reveal mark scheme Mark scheme for question 16(a)(i)
(a)(ii)

Name the type of reaction that occurs when ATP is converted to ADP.

[1]
Reveal mark scheme Mark scheme for question 16(a)(ii)
(a)(iii)

A teacher told his students that the human body makes the equivalent of its own mass in ATP every day.

Explain why, at the end of the day, only a small proportion of the students’ mass was ATP.

[2]
Reveal mark scheme Mark scheme for question 16(a)(iii)
(b)

A student investigated the uptake of two different substances in cultured cells.

The rate of uptake was measured at different concentrations of each substance in the medium surrounding the cells.

The results are shown in Fig. 16.2.

Figure for question 16
Question 16, shown as an image
(b)(i)

The student concluded that one substance entered the cells by simple diffusion and the other by active transport.

Evaluate the student’s conclusion.

[4]
Reveal mark scheme Mark scheme for question 16(b)(i)
(b)(ii)

The student then added 2,4-dinitrophenol (DNP) to the cells. DNP inhibits respiration.

Predict the effect of DNP on the uptake of each substance and explain your prediction.

Substance A

Substance B

[4]
Reveal mark scheme Mark scheme for question 16(b)(ii)
(c)

The student used a colorimeter to measure the concentration of each substance in the liquid surrounding the cells.

The colorimeter had an analogue display. The reading was indicated by a needle moving across a scale. The smallest divisions on the scale corresponded to 0.1 absorbance unit.

After the investigation the student suggested some improvements.

Draw a line between each of the improvements to the corresponding justification.

Figure for question 16

to 3 decimal places.

Figure for question 16

before use.

Figure for question 16

For each concentration, repeat the measurement of the rate of reaction

Figure for question 16

Ask students in several schools to

Figure for question 16
[4]
Reveal mark scheme Mark scheme for question 16(c)

Biological Processes June 2019, Question 17 12 marks

Gibberellin causes stem elongation in plants.

Fig. 17.1 shows the effect of gibberellin on cabbage plants.

Figure for question 17

Gibberellin causes an increase in the distance between the leaves on the stem, which is known as the internodal length.

(a)

Explain why gibberellin is classed as a plant hormone.

[3]
Reveal mark scheme Mark scheme for question 17(a)
(b)

A scientist carried out an investigation into the effect of gibberellin on cabbage plants.

The scientist applied a range of volumes of gibberellin and measured the rate of increase of internodal length over 30 days.

Table 17 shows the scientist’s results.

Figure for question 17
(b)(i)

Plot the results from Table 17 as a suitable graph.

[4]
Reveal mark scheme Mark scheme for question 17(b)(i)
(b)(ii)

Gibberellin causes an increase in internodal length.

State one other role of gibberellin in plants.

[1]
Reveal mark scheme Mark scheme for question 17(b)(ii)
(c)

A student carried out chemical tests on cabbage leaves to investigate which molecules were present.

The student’s method was as follows:

Figure for question 17
(c)(i)

Name a type of food molecule that the student will not be able to detect using these chemical tests.

[1]
Reveal mark scheme Mark scheme for question 17(c)(i)
(c)(ii)

The table below is a summary of some of the student’s findings.

Complete the table by writing in the missing observations and conclusions.

TubeObservationConclusion
1.........................................Protein present
2Yellow colour............................................................................
3Pale brown colour............................................................................
4.........................................Fat present
5.........................................Glucose concentration small (15 mg dm−3)
[2]
Reveal mark scheme Mark scheme for question 17(c)(ii)
(c)(iii)

The student then used a colorimeter to measure the absorbance of the contents of Tube 2.

Explain how the use of a colorimeter could improve the student’s conclusion.

[1]
Reveal mark scheme Mark scheme for question 17(c)(iii)

Unified Biology June 2018, Question 2 10 marks

ATP can be produced in various ways. Each stage of respiration contributes to the production of ATP.

(a)

Describe the production of ATP by substrate-level phosphorylation in different stages of respiration with reference to the number of ATP molecules produced.

[4]
Reveal mark scheme Mark scheme for question 2(a)
(b)

Glucose and other carbohydrates are present in respiring cells. The concentrations of carbohydrate molecules vary between tissues.

A student conducted tests on three tissues, A, B and C. Table 2 shows the results of these

Figure for question 2

Two of the tissues were known to be phloem tissue and liver tissue.

Use the evidence in Table 2 to identify which tissue, A, B or C, is phloem and which tissue is liver. Explain your answer.

Tissue........ must be phloem because........

Tissue........ must be liver because........

[3]
Reveal mark scheme Mark scheme for question 2(b)
(c)
Figure for question 2
(c)(i)

Using the information in Fig. 2, calculate the percentage of carbon atoms in the fatty acid that are able to enter the Krebs cycle.

[1]
Reveal mark scheme Mark scheme for question 2(c)(i)
(c)(ii)

The percentage of carbon atoms that a reaction makes available for use in the Krebs cycle can be described as the efficiency of the reaction.

Calculate the efficiency of the link reaction. Using your answer to part (i), state whether the link reaction is more, less or equally efficient when compared to the reactions described in Fig. 2.

Show your working.

Link reaction is........ efficient

[1]
Reveal mark scheme Mark scheme for question 2(c)(ii)
(c)(iii)

Fig. 2 shows the role of coenzyme A in beta oxidation.

Suggest a role for coenzymes other than coenzyme A in beta oxidation.

[1]
Reveal mark scheme Mark scheme for question 2(c)(iii)

Biological Diversity June 2017, Question 20 16 marks

Also involves Respiratory substrates and RQ

Fig. 20 shows the disaccharide lactose, which is found in milk.

Figure for question 20
(a)

Another disaccharide is maltose. Maltose and lactose both contain the same number of atoms of each element, C, H and O.

(a)(i)

State two other structural similarities between lactose and maltose.

[2]
Reveal mark scheme Mark scheme for question 20(a)(i)
(a)(ii)

Complete the table below to identify differences between the structures of lactose and maltose. The first one has been completed for you.

Figure for question 20
[3]
Reveal mark scheme Mark scheme for question 20(a)(ii)
(b)

One of the monomers of lactose is galactose.

The bacterium E. coli usually uses glucose as a respiratory substrate.

Under certain circumstances, E. coli is able to use galactose as a respiratory substrate by breaking down lactose into glucose and galactose and then using both glucose and galactose as respiratory substrates.

(b)(i)

Explain how the structure of galactose allows it to be used as a respiratory substrate.

[3]
Reveal mark scheme Mark scheme for question 20(b)(i)
(b)(ii)

E. coli usually grows in conditions where the extracellular concentration of lactose is low. In such conditions lactose does not easily cross the bacterial cell surface membrane.

Suggest and explain why lactose is unable to cross membranes.

[2]
Reveal mark scheme Mark scheme for question 20(b)(ii)
(b)(iii)

In order for lactose to enter the cytoplasm of E. coli a protein is required.

The E. coli living in the digestive system of young mammals are more likely to contain this protein than E. coli living in the digestive system of old mammals.

Suggest an explanation for this observation.

[2]
Reveal mark scheme Mark scheme for question 20(b)(iii)
(c)

Lactose is a reducing sugar.

Benedict’s reagent can be used to detect the presence of lactose in a solution.

A colorimeter can be used to measure the concentration of lactose.

The colorimeter first needs to be calibrated.

Describe how a method that uses Benedict’s reagent and a colorimeter could be calibrated to measure the concentration of lactose in an unknown sample.

[4]
Reveal mark scheme Mark scheme for question 20(c)

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