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Cell Biology - Ultrastructure of cells

Grade 12IBBiology

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

🔑Concepts

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Prokaryotic cells have a simple cell structure without compartmentalization. They lack a nucleus and membrane-bound organelles. Their genetic material is located in a nucleoid region as circular, naked DNADNA.

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Eukaryotic cells have a compartmentalized cell structure. Organelles such as the nucleus, mitochondria, and chloroplasts are enclosed by double membranes, while others like the Golgi apparatus and endoplasmic reticulum have single membranes.

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Ribosomes are essential for protein synthesis; prokaryotes possess 70S70S ribosomes, while eukaryotes possess larger 80S80S ribosomes.

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The Endosymbiotic Theory explains the origin of eukaryotic cells, suggesting that mitochondria and chloroplasts were once free-living prokaryotes that were engulfed by a larger cell. Evidence includes their own 70S70S ribosomes and circular DNADNA.

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Electron microscopes have a much higher resolution (0.10.1 nmnm to 11 nmnm) than light microscopes (200200 nmnm), enabling the visualization of the ultrastructure (fine detail) of organelles.

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The Golgi apparatus consists of flattened membrane sacs called cisternae and functions in the modification, sorting, and packaging of proteins for secretion via vesicles.

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Mitochondria are the site of aerobic respiration, producing ATPATP through the oxidation of glucose and other organic compounds.

📐Formulae

Magnification=Measured length of imageActual length of specimenMagnification = \frac{\text{Measured length of image}}{\text{Actual length of specimen}}

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

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

💡Examples

Problem 1:

An electron micrograph of a chloroplast has a scale bar of 22 μm\mu\text{m} which measures 4040 mm\text{mm} in length. Calculate the magnification of the image.

Solution:

First, convert units to be consistent: 40 mm=40,000 μm40\ \text{mm} = 40,000\ \mu\text{m}. Use the formula Magnification=Image SizeActual SizeMagnification = \frac{\text{Image Size}}{\text{Actual Size}}. Magnification=40,000 μm2 μm=20,000×Magnification = \frac{40,000\ \mu\text{m}}{2\ \mu\text{m}} = 20,000 \times

Explanation:

To find magnification, the length of the scale bar (image size) is divided by the value indicated on the scale bar (actual size). Both values must be in the same units.

Problem 2:

A cell organelle is observed to be 0.50.5 μm\mu\text{m} in actual length. If the magnification is ×50,000\times 50,000, what is the length of the organelle in the drawing in millimeters (mm\text{mm})?

Solution:

Image Size=Actual Size×MagnificationImage\ Size = Actual\ Size \times Magnification Image Size=0.5 μm×50,000=25,000 μmImage\ Size = 0.5\ \mu\text{m} \times 50,000 = 25,000\ \mu\text{m} Convert to mm\text{mm}: 25,0001,000=25 mm\frac{25,000}{1,000} = 25\ \text{mm}

Explanation:

Rearrange the magnification formula to solve for Image Size. The result is initially in μm\mu\text{m}, which is then divided by 1,0001,000 to convert to mm\text{mm}.

Ultrastructure of cells - Revision Notes & Key Diagrams | IB Grade 12 Biology