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IB Biology C3.2: Defence against disease worked solutions

Worked solutions to every practice question in C3.2: Defence against disease of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which statement defines a pathogen? (A) An organism or virus causing infection (B) A molecule that triggers antibodies (C) The symptoms of an infection (D) Any bacterium in the human body
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A pathogen is the agent that causes an infectious disease, and it can be a virus, bacterium, fungus or protist (A). B describes an antigen, C describes the disease and D is wrong because many bacteria in the body are harmless.

Final answer

A

Question A2

[1 mark]
Which feature belongs to the adaptive immune system and not to the innate system? (A) Phagocytes engulf foreign cells (B) Memory cells speed a second response (C) Mucus traps pathogens (D) Platelets seal a cut
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Only the adaptive system builds a memory, so a later response is faster (B). Phagocytes, mucus and platelets are all innate, non-specific defences (A, C, D).

Final answer

B

Question A3

[1 mark]
Which row correctly matches a barrier with its action? (A) Skin: plasma cells secrete antibody (B) Airway: mucus traps, cilia sweep out (C) Skin: phagocytes engulf on the surface (D) Gut lining: platelets make fibrin
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Sticky mucus traps pathogens and cilia sweep the mucus away (B). Antibody secretion, phagocytosis and clotting belong to the cellular defences inside the body, not to the barriers (A, C, D).

Final answer

B

Question A4

[1 mark]
Which sequence describes how a phagocyte destroys a bacterium? (A) Endocytosis, then lysosome digestion (B) Lysosome enzymes first, then endocytosis (C) Diffusion in, then the phagocyte divides (D) Antibody made, then digestion
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The phagocyte engulfs the bacterium by endocytosis into a vesicle, and lysosomes then release enzymes into the vesicle (A). B reverses the order, C has no diffusion or division, and D gives the phagocyte the job of a plasma cell.

Final answer

A

Question A5

[1 mark]
Platelets at a cut release clotting factors. Which sequence follows? (A) Prothrombin to thrombin; fibrinogen to fibrin (B) Fibrin to thrombin; prothrombin to fibrinogen (C) Thrombin to fibrin; fibrinogen to platelets (D) Fibrinogen to prothrombin; thrombin to fibrin
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The cascade activates the enzyme thrombin from inactive prothrombin, and thrombin turns soluble fibrinogen into insoluble fibrin (A). The other options reverse substrate and product.

Final answer

A

Question A6

[1 mark]
A person is infected by a bacterium that carries antigen X. Which statement about the antibodies made is correct? (A) They bind antigen X only (B) They bind any antigen (C) They are made by phagocytes (D) They are carbohydrates
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An antibody has binding sites that fit one antigen shape only (A). They are not general (B), they are proteins made by plasma cells (C, D).

Final answer

A

Question A7

[1 mark]
Which cell activates a B-cell that has bound an antigen? (A) Phagocyte (B) Plasma cell (C) Helper T-cell (D) Memory cell
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A helper T-cell activated by the same antigen makes contact with the B-cell and activates it (C). Plasma cells and memory cells are products of the activated B-cell, and the phagocyte is not the activator of the B-cell.

Final answer

C

Question A8

[1 mark]
Why is an antibiotic of no use against a viral infection such as flu? (A) Viruses are too small (B) A virus has no metabolism to block (C) Viruses mutate more than bacteria (D) Antibiotics stimulate antibody only
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Antibiotics block processes that occur in bacteria, but a virus has no metabolism and uses the host cell, so there is nothing to block (B). Size, mutation rate and stimulating antibodies are not the reason (A, C, D).

Final answer

B

Question A9

[1 mark]
Which disease is a zoonosis? (A) A cold passed on by droplets (B) Rabies from the bite of an infected dog (C) HIV passed on in a blood transfusion (D) Athlete's foot on a damp floor
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A zoonosis passes from another animal to humans, as rabies does from a dog bite (B). The other three pass from person to person.

Final answer

B

Question A10

[1 mark]
Why does herd immunity protect babies who are too young to be vaccinated? (A) Their B-cells respond to others' vaccines (B) Immune people rarely pass the pathogen on (C) The vaccine passes through the air (D) Vaccinated people kill it on their skin
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When enough of the population is immune, the pathogen cannot keep passing from person to person, so it seldom reaches the people who are not immune (B). Antibody is not transferred by the vaccine of others (A, C) and immune people do not destroy pathogens on their skin (D).

Final answer

B

Tutorial practice

Question B1

[5 marks]
Some people with haemophilia cannot make one of the clotting factors. (a) Outline how a clot forms at a cut in the skin of a person without haemophilia. [3] (b) Explain why a small cut is more dangerous for a person with haemophilia. [2]
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(a) Any three: platelets at the cut release clotting factors (1); a cascade converts inactive prothrombin to thrombin (1); thrombin converts soluble fibrinogen to insoluble fibrin (1); fibrin forms a mesh that traps red cells and platelets (1). (b) Without the factor the cascade fails and no clot forms (1), so bleeding continues and the open wound lets pathogens enter (1).

Final answer

(a) Platelets release clotting factors; prothrombin is activated to thrombin; thrombin converts fibrinogen to fibrin, which forms a mesh. (b) No clot, so blood loss continues and pathogens can enter.

Question B2

[6 marks]
A volunteer received a vaccine on day 0 and again on day 28. Antibody is in arbitrary units.\begin{array}{lccccccc}\hline \text{Day}&\text{0}&\text{7}&\text{14}&\text{21}&\text{28}&\text{35}&\text{42}\\ \hline \text{Antibody}&\text{0}&\text{3}&\text{20}&\text{12}&\text{6}&\text{150}&\text{110}\\ \hline\end{array} (a) State the day on which antibody concentration peaked after the first dose. [1] (b) Calculate how many times higher the peak after the second dose is than the peak after the first dose. [2] (c) Explain why the response to the second dose is faster and larger. [3]
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(a) Day 14 (1). (b) The peak after the first dose is 20 and after the second is 150 (1); 150/20=7.5 times (1). (c) The first dose produced memory cells that persist (1); on the second dose they are activated and divide quickly into many plasma cells (1); so more antibody appears sooner (3 units at 7 days after the first dose but 150 at 7 days after the second) (1).

Final answer

(a) Day 14. (b) 150/20=7.5 times. (c) Memory cells formed after the first dose; they divide quickly into many plasma cells, so antibody appears sooner and in larger amounts.

Question B3

[5 marks]
A person with untreated HIV infection later developed pneumonia, a common infection. (a) State two body fluids by which HIV can be passed from one person to another. [2] (b) Explain why this person could not control the pneumonia. [3]
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(a) Any two: blood (1); semen (1); vaginal fluid (1); breast milk (1). (b) HIV infects and destroys helper T-cells, so their number falls (1); fewer B-cells are activated, so fewer plasma cells form (1); little antibody is made, so the person cannot control the pneumonia, which is an opportunistic infection typical of AIDS (1).

Final answer

(a) Any two of blood, semen, vaginal fluid, breast milk. (b) HIV destroys helper T-cells; fewer B-cells are activated; little antibody is made, so infection is not controlled.

Question B4

[4 marks]
A vaccine against a bacterial disease contains a harmless fragment of a surface protein of the bacterium. (a) Explain how the vaccine produces immunity without causing the disease. [2] (b) Explain why a second dose given weeks later raises the protection. [2]
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(a) The fragment acts as an antigen that triggers B-cells and helper T-cells, so antibody is made and memory cells form (1); it is not a whole living bacterium, so it cannot cause the disease (1). (b) The second dose meets the memory cells made by the first dose (1), so more memory cells and plasma cells form and protection is greater (1).

Final answer

(a) The fragment is an antigen; it triggers antibody production and memory cells; there is no whole bacterium so no disease. (b) The second dose meets memory cells and gives a larger response.

Question B5

[6 marks]
A hospital tested bacteria from infected wounds.\begin{array}{lcc}\hline \text{Year}&\text{Samples tested}&\text{Resistant to antibiotic Z}\\ \hline \text{2015}&\text{400}&\text{8}\\ \hline \text{2024}&\text{400}&\text{72}\\ \hline\end{array} (a) Calculate the increase in the percentage of resistant samples, in percentage points. [2] (b) Explain how antibiotic use led to the rise. [3] (c) Suggest one way to slow the spread of resistance. [1]
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(a) 8/400\times100=2\% and 72/400\times100=18\% (1); 18-2=16 percentage points (1). (b) Some bacteria already carry a resistance gene from a random mutation (1); the antibiotic kills the susceptible bacteria and the resistant ones survive (1); the survivors divide and pass the gene on, so the resistant proportion rises (1). (c) Any one: use antibiotics only for bacterial infections; complete the full course; avoid unnecessary prescriptions (1).

Final answer

(a) 2 % to 18 %: 16 percentage points. (b) A resistance mutation occurs at random; the antibiotic kills the susceptible bacteria; the resistant ones survive, divide and pass on the gene. (c) Use antibiotics only for bacterial infections and finish the course.

Question B6

[6 marks]
A child was infected with a bacterium for the first time. (a) Outline how antibodies against the bacterium are produced. [4] (b) Explain why the child is unlikely to be infected by the same bacterium again. [2]
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(a) A helper T-cell activated by the antigen contacts a B-cell that has bound the same antigen (1); the activated B-cell divides by mitosis to form a clone (1); the clone includes plasma cells that secrete the specific antibody (1); some cells become memory cells (1). (b) Memory cells persist for a long time (1); on a second meeting with the antigen they respond quickly and strongly, so the bacterium is destroyed before disease develops (1).

Final answer

(a) Helper T-cell activates B-cell that binds the antigen; B-cell divides by mitosis to a clone; plasma cells secrete antibody; memory cells form. (b) Memory cells persist and give a fast, large response.

Question B7

[4 marks]
Three diseases are described. Q: a virus carried by wild birds and passed to people by mosquitoes that bite both. R: a bacterium in cattle; people fall ill after drinking milk that has not been heat-treated. S: a virus that passes only from person to person in droplets. (a) Deduce which two diseases are zoonoses. [1] (b) State how each zoonosis reaches people. [2] (c) Suggest one measure that would reduce infection by R. [1]
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(a) Q and R, because each passes from an animal to people (1). (b) Q is carried by mosquitoes that bite people (1); R is taken in with unpasteurised milk (1). (c) Heat-treat (pasteurise) the milk (1).

Final answer

(a) Q and R. (b) Q by mosquito bite; R by dairy products. (c) Pasteurise the milk.

Question B8

[5 marks]
A town has 20\,000 people, of whom 15\,000 are immune to a disease. In a second town only 40\% are immune. (a) Calculate the percentage of the first town that is immune. [1] (b) Explain how a high proportion of immune people reduces spread to those who are not immune. [2] (c) Suggest why an outbreak is more likely to spread in the second town. [2]
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(a) 15\,000/20\,000\times100=75\% (1). (b) An immune person cannot pass the pathogen on, so the chain of transmission is broken (1); an infected person rarely meets a susceptible person, so the outbreak dies out (1). (c) Six in ten people are not immune (1), so an infected person meets many susceptible people and the disease keeps spreading (1).

Final answer

(a) 75 %. (b) Each immune person breaks the chain, so an infected person is seldom in contact with a susceptible one. (c) Most people are susceptible, so each case infects others and the chain continues.

Question B9

[5 marks]
Weekly new cases of a respiratory disease in two regions.\begin{array}{lcc}\hline \text{}&\text{Region X}&\text{Region Y}\\ \hline \text{Week 1}&\text{1250}&\text{800}\\ \hline \text{Week 3}&\text{2000}&\text{840}\\ \hline \text{Total over the month}&\text{38\,000}&\text{24\,000}\\ \hline\end{array} (a) Calculate the percentage change in new cases in region X from week 1 to week 3. [2] (b) Calculate the percentage difference between the monthly totals of region X and region Y. [2] (c) Suggest why percentages are more informative than raw numbers when the regions differ in population. [1]
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(a) (2000-1250)/1250\times100=+60\% (1 for the method, 1 for the value with its sign). (b) The mean of the totals is (38\,000+24\,000)/2=31\,000 and the difference is 14\,000 (1); 14\,000/31\,000\times100\approx45\% (1). (c) A percentage scales the change to the size of the population, so a large and a small region can be compared fairly (1).

Final answer

(a) +60 %. (b) 45 %. (c) They allow regions of different size to be compared.

Advanced practice

Question C1

[8 marks]
New cases of a respiratory disease were recorded each day in a town. Rules requiring masks in public places began on day 5.\begin{array}{lcccccccc}\hline \text{Day}&\text{1}&\text{2}&\text{3}&\text{4}&\text{5}&\text{6}&\text{7}&\text{8}\\ \hline \text{New cases}&\text{4}&\text{11}&\text{27}&\text{60}&\text{98}&\text{70}&\text{36}&\text{17}\\ \hline\end{array} (a) State the day with the most new cases. [1] (b) Calculate the percentage decrease in new cases from day 5 to day 8. [2] (c) Suggest two reasons for the fall in cases after day 5. [2] (d) Evaluate the claim that the mask rules caused the fall. [3]
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(a) Day 5 (1). (b) (98-17)/98\times100=82.7\approx83\% decrease (1 for the method, 1 for the value). (c) Any two: masks reduced transmission (1); infected people isolated (1); many people became immune, so fewer susceptible people remained (1). (d) The fall begins the day after the rule, which fits the claim (1); but the rise had already slowed and the data show an association in time only, with no control town without the rule (1); herd immunity and isolation could produce the same pattern, so the data cannot show that masks caused the fall (1).

Final answer

(a) Day 5. (b) 83 %. (c) Fewer susceptible people; masks; isolation. (d) Timing fits but is not proof; other causes; no comparison town.

Question C2

[8 marks]
Patients with a bacterial infection were given antibiotic X or a placebo. Treatment lasted 7 days.\begin{array}{lcc}\hline \text{Group}&\text{Patients}&\text{Cured after 7 days}\\ \hline \text{Antibiotic X}&\text{180}&\text{153}\\ \hline \text{Placebo}&\text{175}&\text{70}\\ \hline\end{array} (a) Calculate the percentage of each group that was cured. [2] (b) Calculate the difference between the two percentages, in percentage points. [1] (c) Explain the purpose of the placebo group. [2] (d) Evaluate whether these data justify giving antibiotic X to all patients with this infection. [3]
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(a) 153/180\times100=85\% (1); 70/175\times100=40\% (1). (b) 85-40=45 percentage points (1). (c) The placebo group shows how many recover without the drug, so the effect of X can be separated (1); it controls for the effect of patients believing they have been treated (1). (d) The cure rate is much higher with X and the groups are fairly large (1); but the data do not say how patients were allocated, how severe the infections were or whether X had side effects (1); only one infection and one 7-day period were tested, so use for all patients is not yet justified (1).

Final answer

(a) 85 % and 40 %. (b) 45 points. (c) Shows how many recover without the drug; controls expectation effects. (d) Large effect but unknown allocation, one infection, 7 days, side effects.

Question C3

[8 marks]
In a trial, 20\,000 adults received a vaccine and 20\,000 received a placebo. Over three months, 40 vaccinated and 160 placebo participants developed the disease. (a) Calculate the percentage of each group that developed the disease. [2] (b) Calculate the percentage by which the vaccine reduced the risk of disease. [2] (c) State two features of the design that make the result more reliable. [2] (d) Suggest two limits on what the trial shows. [2]
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(a) 40/20\,000\times100=0.2\% (1); 160/20\,000\times100=0.8\% (1). (b) (0.8-0.2)/0.8\times100=75\% (1 for the method, 1 for the value). (c) Any two: a placebo control group (1); large groups of the same size (1); the same follow-up period for both groups (1). (d) Any two: follow-up was only three months, so long-term protection is unknown (1); only adults took part (1); the data do not show whether the disease was less severe, or whether side effects occurred (1).

Final answer

(a) 0.2 % and 0.8 %. (b) 75 %. (c) Placebo control; large equal groups. (d) Only three months; adults only; severity not given.

Question C4

[6 marks]
A bacterium enters the body through a cut in the skin of the hand. (a) Explain how the body defends itself against this bacterium, from the cut to the formation of immunity. [6]
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(a) Any six: platelets release clotting factors and thrombin converts fibrinogen to fibrin, so a clot seals the cut (1); phagocytes engulf bacteria by endocytosis (1) and lysosome enzymes digest them (1); a helper T-cell activates a B-cell that has bound the bacterial antigen (1); the B-cell divides by mitosis to form a clone (1); plasma cells secrete the specific antibody, which marks the bacteria for phagocytes (1); memory cells persist and give faster protection if the bacterium returns (1).

Final answer

(a) Clot seals the cut; phagocytes engulf and digest; helper T-cell activates the B-cell with the antigen; mitosis forms clone; plasma cells secrete antibody; memory cells give immunity.

Question C5

[6 marks]
Antibiotic prescriptions (per 1000 people per year) and the percentage of resistant bacterial samples in three countries.\begin{array}{lcc}\hline \text{Country}&\text{Prescribed per 1000}&\text{Resistant / \%}\\ \hline \text{P}&\text{300}&\text{8}\\ \hline \text{Q}&\text{550}&\text{15}\\ \hline \text{R}&\text{900}&\text{36}\\ \hline\end{array} (a) Describe the relationship shown by the data. [1] (b) Explain the relationship in terms of natural selection. [2] (c) Evaluate whether the data prove that prescribing causes resistance. [3]
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(a) As prescriptions rise from 300 to 900, the percentage of resistant samples rises from 8 to 36 (1). (b) Heavier antibiotic use kills susceptible bacteria and the resistant ones survive (1); they divide and pass on the gene, so the resistant share of the population rises (1). (c) The data show a positive association (1); only three countries are compared and other factors such as hygiene, sampling methods or use in farming are not controlled (1); a correlation does not prove a cause, though the mechanism of selection makes the link plausible (1).

Final answer

(a) More prescriptions, more resistance. (b) More antibiotic use selects resistant bacteria, which survive and reproduce. (c) Only three countries and no control of other factors, so correlation not proof.

Question C6

[6 marks]
A patient with untreated HIV infection had the helper T-cell count measured. A count below 200 cells per mm^3 is taken to mean that AIDS has developed.\begin{array}{lccccc}\hline \text{Year}&\text{0}&\text{2}&\text{4}&\text{6}&\text{8}\\ \hline \text{Helper T-cells / mm}^3&\text{900}&\text{780}&\text{560}&\text{320}&\text{140}\\ \hline\end{array} (a) Calculate the percentage decrease in the helper T-cell count from year 0 to year 8. [2] (b) Explain why the patient would make less antibody against a new infection in year 8 than in year 0. [3] (c) Deduce the period in which AIDS developed. [1]
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(a) (900-140)/900\times100=84.4\approx84\% decrease (1 for the method, 1 for the value). (b) HIV destroys helper T-cells, and by year 8 few remain (1); fewer B-cells are activated by helper T-cells, so fewer plasma cells form (1); less antibody is secreted against the new infection (1). (c) The count falls from 320 to 140 between year 6 and year 8, crossing 200 (1).

Final answer

(a) 84 %. (b) Fewer helper T-cells, so fewer B-cells activated, fewer plasma cells and less antibody. (c) Between year 6 and year 8.

Connected questions

Question X1

Connected questions[8 marks]
C3.2 + A2.3 + D1.2. A virus passes from an animal to people and then between people. A vaccine for it contains mRNA that codes for a protein on the virus surface. (a) State two features of a virus that explain why antibiotics do not harm it. [2] (b) Outline how the mRNA vaccine leads to the formation of memory cells. [3] (c) Explain how a mutation that changes the shape of the surface protein could reduce the protection given by the vaccine. [3]
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(a) Any two: it has no metabolism of its own (1); it has no cell wall to block (1); it makes its proteins on host ribosomes, so there is no bacterial process to inhibit (1). (b) Ribosomes in the person's cells translate the mRNA into the viral surface protein (1); the protein acts as an antigen that activates helper T-cells and B-cells (1); the B-cells divide and some of the clone become memory cells (1). (c) A mutation changes the base sequence and so the amino acid sequence of the surface protein (1); the shape of the antigen changes, so antibodies and memory cells made against the old shape no longer fit (1); the virus can infect cells without being bound, so protection falls (1).

Final answer

(a) No metabolism; no cell wall or ribosomes of its own. (b) Cells translate the mRNA into the protein; it acts as antigen; helper T-cells and B-cells activate; memory cells form. (c) The antibodies fit the old shape; the new shape is not bound; the virus escapes.

Question X2

Connected questions[8 marks]
C3.2 + D4.1. In a hospital the frequency of a gene that gives resistance to an antibiotic in one species of bacterium rose from 0.02 to 0.14 over six years. (a) Calculate the percentage increase in the frequency of the gene. [2] (b) Explain the rise using the steps of natural selection. [3] (c) Suggest why resistance can spread through bacteria more quickly than a resistance allele through an insect population. [3]
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(a) (0.14-0.02)/0.02\times100=600\% (1 for the method, 1 for the value). (b) The population varied because a mutation produced the resistance allele (1); the antibiotic acted as the selection pressure, so resistant bacteria survived and reproduced more (1); they passed the allele to their offspring, so its frequency in the population rose (1). (c) Bacteria divide in minutes, so there are many generations in a short time (1); they reproduce by binary fission, so each survivor founds a resistant clone (1); resistance genes can also pass between bacteria, which insects do not do (1).

Final answer

(a) 600 %. (b) Variation from mutation; antibiotic is the selection pressure; resistant survive and breed more; frequency rises. (c) Short generation time; binary fission; genes can pass between bacteria.

Question X3

Connected questions[8 marks]
C3.2 + B1.2 + B2.2. An antibody is a protein with two identical binding sites, made by plasma cells and released from the cell. (a) Explain how the amino acid sequence of an antibody determines the shape of its binding site. [2] (b) Plasma cells have extensive rough endoplasmic reticulum and a large Golgi apparatus. Explain why. [3] (c) Explain why all the plasma cells in one clone make the same antibody. [3]
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(a) The order of amino acids (primary structure) decides how the polypeptide folds (1); the folded shape forms a binding site that fits one antigen shape (1). (b) Ribosomes on rough ER make the antibody protein in large amounts (1); the Golgi apparatus processes it and packs it in vesicles (1); the vesicles fuse with the plasma membrane and release the antibody by exocytosis (1). (c) The clone arises by mitosis from one activated B-cell (1), so every cell has identical genes, including the gene for the same antibody (1); each cell therefore makes the antibody that fits the same antigen (1).

Final answer

(a) The sequence sets how the chain folds; the folded shape forms the binding site that fits one antigen. (b) Rough ER makes the protein; Golgi modifies and packs it into vesicles for secretion. (c) The clone is made by mitosis, so every cell has the same genes and expresses the same antibody gene.

Question X4

Connected questions[8 marks]
C3.2 + B2.1. A phagocyte takes in a bacterium by endocytosis, and the vesicle then joins a lysosome. (a) Explain why endocytosis depends on the fluidity of the phagocyte membrane. [3] (b) Explain why the digestive enzymes are kept inside lysosomes and not free in the cytoplasm. [2] (c) Contrast the way a phagocyte takes in a whole bacterium with the way oxygen enters the cell. [3]
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(a) The membrane has to bend round the bacterium and pinch off a vesicle (1); phospholipids can move sideways in the fluid bilayer, so it can change shape (1); the vesicle membrane can later fuse with the lysosome membrane (1). (b) Digestive enzymes break down proteins and other molecules, so free in the cytoplasm they would digest the cell itself (1); a lysosome keeps them enclosed until a vesicle with a target joins it (1). (c) Oxygen is small and non-polar and crosses by simple diffusion, down its concentration gradient, directly through the bilayer (1); a whole bacterium is far too large, so it is enclosed in a vesicle by endocytosis (1), which uses energy and changes the shape of the membrane (1).

Final answer

(a) The membrane must bend, pinch off and fuse; phospholipids move sideways so the bilayer can reshape. (b) Free enzymes would digest the cell. (c) Oxygen: simple diffusion of a small non-polar molecule down a gradient; bacterium: vesicle by endocytosis, needs energy.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Defence against disease from the core ideas to full exam questions, with question types, worked examples and graded practice.

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