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Diseases and Immunity - Pathogens and transmission

Grade 12A LevelBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A pathogen is a disease-causing organism, typically including bacteria, viruses, fungi, and protoctists (e.g., PlasmodiumPlasmodium which causes malaria).

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Transmissible diseases are those where the pathogen can be passed from one host to another via direct contact (blood, body fluids) or indirect contact (contaminated surfaces, H2OH_2O, food, or air).

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The body maintains mechanical barriers (skin, hairs in the nose) and chemical barriers (mucus, stomach acid containing HClHCl at a pH≈2.0pH \approx 2.0) to prevent pathogen entry.

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Active immunity is the defense against a pathogen by antibody production in the body, often stimulated by VaccinationVaccination which introduces weakened or dead pathogens containing specific AntigensAntigens.

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Passive immunity is a short-term defense against a pathogen by antibodies acquired from another individual, such as IgGIgG antibodies passing across the placenta or through breast milk.

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Phagocytosis is the process where phagocytes (white blood cells) engulf and digest pathogens using enzymes, while lymphocytes produce AntibodiesAntibodies that bind to specific antigens.

📐Formulae

N=N0×2nN = N_0 \times 2^n

Magnification=Image sizeActual sizeMagnification = \frac{\text{Image size}}{\text{Actual size}}

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

💡Examples

Problem 1:

A population of bacteria in a contaminated water sample starts with N0=500N_0 = 500 cells. If the bacteria divide every 3030 minutes, calculate the total number of bacteria (NN) after 33 hours of incubation.

Solution:

n=3 hours0.5 hours=6n = \frac{3 \text{ hours}}{0.5 \text{ hours}} = 6 generations. Using N=500×26N = 500 \times 2^6, we get N=500×64=32,000N = 500 \times 64 = 32,000 cells.

Explanation:

The growth of bacteria follows an exponential pattern where nn is the number of divisions that occur in the given time frame.

Problem 2:

Calculate the actual size of a virus particle if the image size under an electron microscope is 20 mm20 \text{ mm} and the magnification is ×100,000\times 100,000.

Solution:

Actual size=Image sizeMagnification=20 mm100,000=0.0002 mm=200 nmActual \ size = \frac{Image \ size}{Magnification} = \frac{20 \text{ mm}}{100,000} = 0.0002 \text{ mm} = 200 \text{ nm}.

Explanation:

To find the actual size, divide the measured image length by the magnification factor, ensuring units are converted to scientific notation or micrometers/nanometers for clarity.

Problem 3:

Explain the role of the H2OH_2O cycle in the transmission of Vibrio choleraeVibrio \ cholerae.

Solution:

Vibrio choleraeVibrio \ cholerae is transmitted via the faecal-oral route through contaminated H2OH_2O or food.

Explanation:

In areas with poor sanitation, infected faeces enter the water supply. When individuals ingest this water, the bacteria release toxins in the small intestine, leading to severe dehydration.