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Cell Biology - Ultrastructure of Cells

Grade 11IBBiology

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

🔑Concepts

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Prokaryotic cells have a simple structure without compartmentalization. They lack a nucleus and membrane-bound organelles, containing instead a nucleoid with naked DNADNA and 70S70S ribosomes.

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Eukaryotic cells have a compartmentalized structure, which increases efficiency by isolating chemical reactions. Key organelles include the nucleus, 80S80S ribosomes, mitochondria, and the Golgi apparatus.

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The resolution of a microscope is the ability to distinguish between two close points. Electron microscopes have a much higher resolution (approximately 0.1 nm0.1\text{ nm}) compared to light microscopes (approximately 200 nm200\text{ nm}) because the wavelength of electrons is much shorter than that of light.

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Binary fission is the method of asexual reproduction in prokaryotes, involving DNADNA replication followed by the division of the cytoplasm.

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The mitochondria are the sites of aerobic respiration, producing ATPATP. They possess a double membrane with the inner membrane folded into cristae to increase surface area.

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Exocrine gland cells of the pancreas are specialized for protein secretion, featuring an extensive Rough Endoplasmic Reticulum (rERrER) and numerous Golgi vesicles.

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Palisade mesophyll cells in plant leaves are specialized for photosynthesis, containing a high density of chloroplasts and a large central vacuole for turgor pressure.

📐Formulae

Magnification=Measured Size of ImageActual Size of Specimen\text{Magnification} = \frac{\text{Measured Size of Image}}{\text{Actual Size of Specimen}}

Actual Size=Image SizeMagnification\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}

1 mm=103 μm=106 nm1\text{ mm} = 10^3\text{ }\mu\text{m} = 10^6\text{ nm}

💡Examples

Problem 1:

A micrograph of a mitochondrion shows its length to be 50 mm50\text{ mm}. The magnification of the image is ×20,000\times 20,000. Calculate the actual length of the mitochondrion in micrometers (μm\mu\text{m}).

Solution:

Actual Size=50 mm20,000=0.0025 mm=2.5 μm\text{Actual Size} = \frac{50\text{ mm}}{20,000} = 0.0025\text{ mm} = 2.5\text{ }\mu\text{m}

Explanation:

To find the actual size, divide the measured image length by the magnification. Since the question asks for the answer in micrometers, multiply the result in millimeters by 1,0001,000 (1 mm=103 μm1\text{ mm} = 10^3\text{ }\mu\text{m}).

Problem 2:

Calculate the magnification of a drawing if a scale bar labeled 5 μm5\text{ }\mu\text{m} measures 20 mm20\text{ mm} on the paper.

Solution:

Magnification=20 mm5 μm=20,000 μm5 μm=×4,000\text{Magnification} = \frac{20\text{ mm}}{5\text{ }\mu\text{m}} = \frac{20,000\text{ }\mu\text{m}}{5\text{ }\mu\text{m}} = \times 4,000

Explanation:

First, convert both values to the same unit. 20 mm20\text{ mm} is equal to 20,000 μm20,000\text{ }\mu\text{m}. Then, divide the measured length of the scale bar by its indicated value to find the magnification.

Ultrastructure of Cells - Revision Notes & Key Diagrams | IB Grade 11 Biology