Cambridge IGCSE Biology 0610: Biotechnology and genetic modification – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover topic 21, Biotechnology and genetic modification, of the Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028, sections 21.1–21.3. Questions without a label test Core content, which appears on every paper. Questions labelled (Extended) test Supplement content, which is examined only on Papers 2 and 4. Question 4 is a practical-style question on the washing powder investigation in 21.2.4.

Before you start, you may want the topic 21 study guide or the topic 21 revision notes. For the whole course, see the Cambridge IGCSE Biology hub and the printable 0610 checklist.

Questions

1. Define the term genetic modification. [2]

2. A baker mixes flour, water, sugar and yeast to make dough and leaves it in a warm room. The volume of the dough increases from 150 cm³ to 240 cm³ in 60 minutes.

(a) State the word equation for anaerobic respiration in yeast. [2] (b) Calculate the percentage increase in the volume of the dough. [2] (c) Explain why the dough increases in volume. [2]

3. A student crushed two 100 g samples of apple. Pectinase was added to one sample and water to the other. After 20 minutes both were filtered. The sample without pectinase gave 14.0 cm³ of juice; the sample with pectinase gave 23.8 cm³.

(a) Calculate the percentage increase in the volume of juice when pectinase was used. [2] (b) Explain why more juice was obtained with pectinase. [2] (c) State one other difference you would expect in the juice made with pectinase. [1]

4. A student investigated how temperature affects a biological washing powder. Squares of cloth stained with egg were placed in washing powder solution at different temperatures. The time for each stain to disappear was recorded.

Temperature / °C	20	30	40	50	60
Time for stain to disappear / min	18	11	6	9	not removed after 30

(a) Identify the independent variable. [1] (b) State two variables that should be kept constant. [2] (c) Calculate the rate of stain removal at 40 °C, using rate = 1 ÷ time. Give the unit. [2] (d) Explain the result at 60 °C. [2] (e) Describe a suitable control for this investigation. [1]

5. Outline three different ways in which crop plants have been genetically modified. [3]

6. (Extended) Discuss why bacteria are useful in genetic modification. [3]

7. (Extended) Explain how lactase is used to produce lactose-free milk, and why this milk is made. [3]

8. (Extended) A fungus is grown in a fermenter to produce mycoprotein.

(a) Explain why sterile air is bubbled into the fermenter. [2] (b) The cooling system failed and the temperature rose from 25 °C to 31 °C in 4 hours. (i) Calculate the mean rate of temperature rise. [2] (ii) Explain why the temperature rose and suggest how this would affect the yield of mycoprotein if it continued. [3] (c) Explain why the pH in the fermenter is monitored and controlled. [2]

9. (Extended) Outline how bacteria can be genetically modified to produce human insulin. [6]

10. (Extended) A farm grew conventional maize one year and insect-resistant GM maize the next year on the same fields.

	Conventional maize	GM maize
Yield / tonnes per hectare	8.0	9.2
Insecticide used / kg per hectare	2.5	0.6

(a) Calculate the percentage increase in yield. [2] (b) Calculate the percentage decrease in insecticide used. [2] (c) Suggest why less insecticide was used on the GM maize. [2] (d) Give two disadvantages of growing GM crops. [2]

11. Describe the role of anaerobic respiration in yeast in the production of ethanol for biofuels. [3]

Answers

1. Changing the genetic material of an organism [1]; by removing, changing or inserting individual genes [1]. [2] Examiner insight: The second mark needs the idea of individual genes being removed, changed or inserted; “changing DNA” alone earns only the first mark.

2. (a) glucose → alcohol + carbon dioxide: glucose on the left [1]; both alcohol and carbon dioxide on the right [1]. [2] (b) Increase = 240 − 150 = 90 cm³ [1]; 90 ÷ 150 × 100 = 60% [1]. [2] (c) Yeast respires anaerobically and releases carbon dioxide [1]; the gas forms bubbles trapped in the dough, so it rises [1]. [2] Examiner insight: In (b) a correct answer with no working usually earns both marks, but dividing by 240 instead of 150 loses the accuracy mark; in (c) naming alcohol as the cause of rising earns nothing.

3. (a) Increase = 23.8 − 14.0 = 9.8 cm³ [1]; 9.8 ÷ 14.0 × 100 = 70% [1]. [2] (b) Pectinase breaks down pectin [1]; in plant cell walls, so cells release more juice [1]. [2] (c) The juice is clearer (less cloudy) [1]. [1] Examiner insight: “Breaks down cell walls” alone is not enough for both marks in (b); you need the name of the substrate, pectin.

4. (a) Temperature [1]. [1] (b) Any two: volume of solution; concentration of washing powder; size of cloth square; type or amount of stain; type of cloth [1] [1]. [2] (c) 1 ÷ 6 = 0.17 (0.167) [1]; min⁻¹ (per minute) [1]. [2] (d) The enzymes (proteases) are denatured [1]; the active site changes shape so the protein stain can no longer bind and be digested [1]. [2] (e) Repeat at each temperature using boiled biological powder (or non-biological powder) [1]. [1] Examiner insight: In (d) “the enzyme is killed” earns no credit; “denatured” plus a reference to the active site changing shape is needed for both marks.

5. Any three: genes inserted to give resistance to herbicides [1]; genes inserted to give resistance to insect pests [1]; genes inserted to improve nutritional qualities [1]. [3] Examiner insight: Each mark needs a different purpose; giving two versions of the same idea (for example two different pests) scores only once.

6. Few ethical concerns over manipulating and growing bacteria [1]; bacteria contain plasmids [1]; a gene can be inserted into a plasmid and taken up by bacteria, so the bacteria make the gene’s product [1]. [3] Examiner insight: “Discuss” needs a reason attached to each feature; simply listing “plasmids” and “ethics” without explanation limits you to two marks.

7. Lactase breaks down lactose [1]; into glucose and galactose [1]; so people who cannot digest lactose can drink the milk [1]. [3] Examiner insight: Both products are needed for the second mark; “simple sugars” without naming them is usually not accepted.

8. (a) To provide oxygen for aerobic respiration [1]; which releases energy for growth of the fungus [1]. [2] (b)(i) Rise = 31 − 25 = 6 °C [1]; 6 ÷ 4 = 1.5 °C per hour [1]. [2] (b)(ii) Respiration by the fungus releases heat [1]; the heat was not removed [1]; enzymes would be denatured, so growth and yield would fall [1]. [3] (c) Enzymes work best at an optimum pH [1]; waste products such as carbon dioxide can change the pH, so it must be kept near the optimum for growth [1]. [2] Examiner insight: A rate without its unit (°C per hour) normally loses the final accuracy mark, and in (b)(ii) the idea that respiration produces heat is the key mark.

9. The human insulin gene is cut out using restriction enzymes [1], leaving sticky ends [1]. A bacterial plasmid is cut with the same restriction enzymes, giving complementary sticky ends [1]. The gene is inserted into the plasmid and joined using DNA ligase, forming a recombinant plasmid [1]. Recombinant plasmids are inserted into bacteria, which multiply [1]. The bacteria express the human gene and make human insulin [1]. [6] Examiner insight: Each named term (restriction enzymes, sticky ends, same enzymes, DNA ligase, recombinant plasmid, expression) carries a mark; a vague description such as “the gene is added to the bacteria” scores very little.

10. (a) 9.2 − 8.0 = 1.2 [1]; 1.2 ÷ 8.0 × 100 = 15% [1]. [2] (b) 2.5 − 0.6 = 1.9 [1]; 1.9 ÷ 2.5 × 100 = 76% [1]. [2] (c) The GM maize has a gene that gives resistance to insect pests [1]; so fewer pests damage the crop and less spraying is needed [1]. [2] (d) Any two: GM seed is expensive or must be bought each year; genes may transfer to wild relatives; reduced biodiversity; pests may become resistant; possible health concerns such as allergies [1] [1]. [2] Examiner insight: Allow error carried forward in a percentage if the subtraction is wrong but the method is correct; a decrease calculated as 0.6 ÷ 2.5 × 100 (24%) shows a method error and earns only the first mark if the subtraction is shown.

11. Yeast respires anaerobically, without oxygen [1]; it breaks down sugar (glucose) from plant material [1]; producing ethanol (alcohol), which is used as the fuel [1]. [3] Examiner insight: The answer must link respiration to the product: saying “yeast makes the fuel” without naming ethanol or alcohol loses the final mark.

Where marks are usually lost

• Naming alcohol instead of carbon dioxide as the reason dough rises.
• Saying pectinase breaks down cell membranes rather than pectin in cell walls.
• Writing “enzyme killed” instead of “denatured” and not mentioning the active site.
• Giving a rate or percentage without its unit, or dividing by the new value instead of the original.
• Missing a control in a washing powder investigation.
• Saying fermenters are heated instead of cooled.
• Leaving out “same restriction enzymes” or “DNA ligase” when outlining GM.
• Giving only advantages when a question asks you to discuss GM crops.

Next steps

• Recap with the topic 21 revision notes.
• Reread the explanations in the topic 21 study guide.
• See every topic on the Cambridge IGCSE Biology hub.
• Tick off topic 21 on the printable 0610 checklist.
• Try the free 0610 Core diagnostic or 0610 Extended diagnostic.
• Book a free trial class.

Official syllabus

Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028 (Version 3), published by Cambridge University Press & Assessment (Cambridge International).