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Biology 0610/31 October/November 2019: question paper and mark scheme

Cambridge IGCSE Biology (0610), Paper 3 Theory (Core), October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

0610_w19_qp_310610_w19_ms_310610/31/O/N/19

Paper
Paper 3 Theory (Core)
Total marks
80
Time
1 hour 15 minutes
Questions
9
Code on the paper
0610/31/O/N/19
0610_w19_ms_31.pdf

0610/31 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 111 marks

Question

Fig. 1.1 is a diagram of a molecule of fat. Fig. 1.1 (a) (i) List the chemical elements present in fat. …… [1] (ii) State the smaller units that fats are made from. …… [2] (b) Fats are an important part of a balanced diet. State the name of three other components of a balanced diet. 1 …… 2 …… 3 …… [3] (c) Marmots and lynx are mammals that can live in a variety of environments. The percentage of fat in the bodies of these two species was measured. Measurements were taken from marmots and lynx living in Alaska and in Virginia. Alaska is a cold environment and Virginia is a warm environment. The results are shown in Table 1.1. Table 1.1 percentage of fat in the body difference in the species percentage of in Alaska in Virginia fat in the body marmot 36 5 31 lynx 15 11 (i) Complete Table 1.1 by calculating the difference in the percentage of fat in the body for the lynx. Write your answer in Table 1.1. [1] (ii) Describe the results shown in Table 1.1. …… [2] (iii) Explain the difference in the percentage of fat in the body between the mammals living in Alaska and Virginia. …… [2] [Total: 11]

Mark scheme

1(a)(i) carbon, hydrogen, oxygen ; 1 1(a)(ii) fatty acids ; 2 glycerol ; 1(b) any three from: 3 carbohydrates ; proteins ; (named)vitamins ; water ; (named) minerals ; fibre ; 1(c)(i) 4; 1 1(c)(ii) marmots and lynx have a higher percentage of body fat 2 in Alaska than in Virginia /AW ; ora marmots have a greater difference in percentage body fat (than lynx) ; ora marmots have the higher percentage body fat than lynx in Alaska / lynx have a higher percentage body fat than marmots in Virginia ; ora 1(c)(iii) (Alaska is colder than Virginia) 2 so (more) fat is needed : ora for insulation ; fat helps to, maintain body temperature / keep (mammal) warm / AW ; (lynx and marmots) are mammals so they regulate their body temperature ;

Question 28 marks

Question

(a) Complete the definition of the term adaptive feature by filling in the gaps with the correct words. An adaptive feature is an …… feature that helps an organism to …… and reproduce in its …… . [3] (b) The flagellum is one of the adaptive features of a sperm. A sample of sperm was taken and the length of each flagellum was recorded. Fig. 2.1 shows a graph of the results. 22 20 18 16 14 frequency 12 10 8 6 4 2 0 3.6 – 3.7 3.8 – 3.9 4.0 – 4.1 4.2 – 4.3 4.4 – 4.5 4.6 – 4.7 4.8 – 4.9 5.0 – 5.1 flagellum length / μm Fig. 2.1 (i) State the most frequent range for flagellum length. …… µm [1] (ii) State the frequency of sperm with flagellum length between 4.8 µm to 4.9 µm. …… [1] (iii) State the type of variation shown by flagellum length. …… [1] (c) Some scientists have suggested that the longer the flagellum the more likely the sperm is to fertilise the egg cell. Suggest a reason why. …… [1] (d) State the name of the organ that produces sperm. …… [1] [Total: 8]

Mark scheme

2(a) inherited / hereditary ; 3 survive ; environment / habitat ; 2(b)(i) 4.4–4.5 (μm) ; 1 2(b)(ii) 9; 1 2(b)(iii) continuous / phenotypic ; 1 2(c) to swim, faster / further ; 1 2(d) testis ; 1

Question 36 marks

Question

(a) A student investigated the conditions required for germination. Seeds were placed on cotton wool in Petri dishes and exposed to different conditions. The conditions used are shown in Table 3.1. Table 3.1 temperature condition of Petri dish access to light / °C cotton wool A 20 damp yes B 3 damp yes C 20 dry yes D 20 damp no Seeds in two of the Petri dishes did not germinate. Predict in which Petri dishes the seeds did not germinate. Give reasons for your answer. Petri dishes …… reasons …… [3] (b) In another investigation, the germination ratio of the seeds was calculated. 60 cress seeds were used in the investigation. 20 seeds germinated and 40 seeds did not germinate. Calculate the ratio of the seeds that germinated to the seeds that did not germinate. ratio …… : …… [1] (c) Plants need mineral ions for healthy growth. State why a plant needs magnesium ions and nitrate ions. magnesium ions …… nitrate ions …… [2] [Total: 6]

Mark scheme

3(a)(i) B and C ; 3 (B) is too cold / not optimum temperature / needs warmth ; (C) is dry / lacks water / needs to be damp ; 3(b) 1:2; 1 3(c) chlorophyll (production) ; 2 (named), amino acids / proteins ;

Question 411 marks

Question

(a) Fig. 4.1 is a graph showing the changes to the thickness of the lining of the uterus during the menstrual cycle. thickness of uterus lining 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 day of menstrual cycle Fig. 4.1 (i) Describe the changes to the thickness of the lining of the uterus during the menstrual cycle as shown in Fig. 4.1. …… [3] (ii) State the days, shown on Fig. 4.1, on which the lining of the uterus is broken down and lost. …… [1] (iii) Draw an X on Fig. 4.1 to show when ovulation occurs. [1] (b) The menstrual cycle is controlled by hormones. (i) Complete the sentence to define the term hormone. A …… substance produced by a …… , carried by the …… , which alters the activity of one or more specific target organs. [3] (ii) Adrenaline is a hormone involved in ‘fight or flight’ situations. Describe two effects of adrenaline on the body. 1 …… 2 …… [2] (iii) State the name of the organ that produces adrenaline. …… [1] [Total: 11]

Mark scheme

4(a)(i) decreases then increases and levels off ; 3 lowest thickness (reached) at day 6 / decreases from day 0/1 to day 6 ; increases in thickness from day 6 to day 15 ; constant thickness between days 15 and 28 / 0 / end ; 4(a)(ii) day 0 / 1, to day 6 ; 1 4(a)(iii) X drawn on the graph at approximately day 14 ; 1 A ± 2 days 4(b)(i) chemical ; 3 gland ; blood / plasma ; 4(b)(ii) increased rate of breathing ; 2 increased, pulse / heart, rate ; widening of pupils ; AVP ;; e.g. more glucose in the blood / more alert 4(b)(iii) adrenal (gland) ; 1

Question 510 marks

Question

(a) Aerobic and anaerobic respiration both release energy. Describe the other similarities and differences between aerobic and anaerobic respiration in muscles. …… [4] (b) Place ticks (3) in the boxes to show two uses of the energy released by respiration in humans. active transport diffusion osmosis protein synthesis transpiration [2] (c) Anaerobic respiration in yeast produces alcohol. The boxes on the right show some sentence endings. Draw lines from the word alcohol to make three correct sentences. abuse decreases instances of crime. can be addictive. causes lung cancer. Alcohol increases levels of self-control. increases reaction times. is a depressant. [3] (d) State the name of an organ damaged by long-term alcohol abuse. …… [1] [Total: 10]

Mark scheme

5(a) both use glucose ; 4 both occur in cells ; aerobic respiration uses oxygen ; ora aerobic respiration releases more energy (than anaerobic respiration) ; ora anaerobic respiration produces lactic acid ; ora aerobic respiration produces carbon dioxide ; ora 5(b) active transport ticked ; 2 protein synthesis ticked ; 5(c) Alcohol linked to: 3 can be addictive (box 2) ; increases reaction times (box 5) ; is a depressant (box 6) ; 5(d) liver / brain / pancreas / heart / stomach ; 1

Question 615 marks

Question

(a) (i) State the names of three structures that are found in a mammalian heart. 1 …… 2 …… 3 …… [3] (ii) Describe how the structure of a vein differs from the structure of an artery. …… [3] (iii) State the function of arteries in the human circulatory system. …… [1] (b) Fig. 6.1 is a photomicrograph of blood. Fig. 6.1 (i) Label one red blood cell on Fig. 6.1 with a label line and the letter X. [1] (ii) State the name of one other type of blood cell in Fig. 6.1. …… [1] (iii) State the names of two other components of blood. 1 …… 2 …… [2] (c) Coronary heart disease (CHD) is a disease of the circulatory system. (i) State three risk factors for developing CHD. 1 …… 2 …… 3 …… [3] (ii) State the name of the blood vessel that becomes blocked in CHD. …… [1] [Total: 15]

Mark scheme

6(a)(i) any three from: 3 capillaries / coronary artery / coronary vein / pulmonary artery / aorta / pulmonary vein / vena cava / left ventricle / right ventricle / left atrium / right atrium / septum / (named) valve / AVP ;;; 6(a)(ii) presence of valves ; 3 thinner wall ; thinner muscle layer / AW ; thinner elastic layer / AW ; wider lumen /AW ; AVP ; 6(a)(iii) carries blood, away from the heart ; 1 6(b)(i) red blood cell labelled with line and X ; 1 6(b)(ii) (named) white (blood cell) ; 1 6(b)(iii) plasma ; 2 platelets ; 6(c)(i) high, fat / cholesterol / salt, diet ; 3 smoking ; genetic predisposition ; age ; stress ; sex ; obesity ; lack of exercise / sedentary lifestyle / AW ; AVP ; high blood pressure / alcohol consumption 6(c)(ii) coronary artery ; 1

Question 76 marks

Question

Fig. 7.1 is a photograph of a large-scale monoculture of soybeans which are a crop plant. Fig. 7.1 (a) Describe the disadvantages of large-scale monocultures. …… [3] (b) Some crop plants have been selectively bred to be drought resistant. The sentences show stages in the process of selective breeding. They are not in the correct order. 1 A farmer identifies crop plants that survive in drought conditions. 2 Offspring that survive drought conditions are selected and bred again. 3 The drought resistant plants are bred together and seeds collected. 4 The farmer repeats the process over many generations. 5 The seeds are germinated and grown in drought conditions. Write the statement numbers in the boxes to show the correct order of the stages in selective breeding. [3] [Total: 6]

Mark scheme

7(a) lack of biodiversity / fewer (wild) species / AW ; 3 extinction of species ; loss of, habitats / resources (for species) / deforestation ; (crop) disease easily spread ; (crops) more susceptible to pests ; depletes soil nutrients ; ref. to pollution ; e.g. herbicides / pesticides / fungicides / fertilisers AVP ;; e.g. flooding / droughts 7(b) 1 (first) ; 3 (then) 3, 5, 2 ; 4 (at the end) ;

Question 88 marks

Question

(a) The government of a country introduced a law called the Endangered Species Act. It was hoped that the Act would help to conserve species that were at risk from extinction. Table 8.1 shows the numbers of birds from different species before and after the Act was introduced. Table 8.1 number of birds species before the Act after the Act bald eagle 416 9789 Kirtland’s warbler 210 1415 nene goose 400 1275 peregrine falcon 324 1700 whooping crane 54 513 (i) State which species was the most at risk from extinction in Table 8.1. …… [1] (ii) Calculate the percentage increase in the number of Kirtland’s warblers. Give your answer to the nearest whole number. …… % [3] (b) (i) List three reasons why species become endangered or extinct. 1 …… 2 …… 3 …… [3] (ii) Describe one method of conserving endangered plant species. …… [1] [Total: 8]

Mark scheme

8(a)(i) whooping crane ; 1 8(a)(ii) 574 ;;; 3 one mark for correct numbers from table one mark for correct calculation one mark for correct rounding to a whole number 8(b)(i) hunting / collecting / poaching ; 3 introduced species / competition (for named resource) / AW ; food, chain / web, disrupted or over predation or lack of food : (named) pollution ; disease ; loss / change, of habitat or natural disasters ; global warming / climate change ; 8(b)(ii) growing in a glasshouse / AW ; 1 seed banks ; legislation ; nature reserves / protected areas ; botanical gardens ; AVP ;

Question 95 marks

Question

Sewage contains water and other substances. Sewage should be treated before it goes into a river. (a) Describe two reasons why sewage should be treated before it goes into a river. 1 …… 2 …… [2] (b) Treatment of sewage has several stages. One of the stages is filtration. State the name of one other stage in the treatment of sewage. …… [1] (c) Fig. 9.1 shows a trickling filter in a sewage treatment plant. gravel Fig. 9.1 The untreated sewage trickles through gravel. There are organisms on the surface of the gravel. (i) State the name of the type of organism on the surface of the gravel. …… [1] (ii) Describe the function of these organisms. …… [1] [Total: 5]

Mark scheme

9(a) kill / remove, pathogens / microorganisms ; 2 to make it safe to, use / drink ; prevent disease ; to remove, toxic waste / nitrate pollution ; it can harm species in, river / sea / oceans ; AVP ; 9(b) screening ; 1 primary treatment ; secondary treatment ; separation ; aeration ; settlement / sedimentation ; anaerobic / sludge, digestion ; disinfection / chlorination / UV light ; AVP ; 9(c)(i) microorganism / AVP ; 1 9(c)(ii) decompose / digest / breakdown, (sewage) ; 1

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Where can I find the 0610/31 October/November 2019 mark scheme?

On this page, free: the mark scheme for IGCSE Biology 0610/31, October/November 2019 (0610_w19_ms_31), and the question paper (0610_w19_qp_31), to read or download, with each question's marking points written out underneath.

How many marks is 0610/31 out of?

Biology 0610/31 for October/November 2019 is out of 80 marks, with 1 hour 15 minutes to answer it, across 9 questions.

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Yes. Upload your answers to 0610/31/O/N/19 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.