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Unity and diversity - Cell structure

Grade 11IBBiology

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

🔑Concepts

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Cell Theory: All living organisms are composed of one or more cells. The cell is the basic unit of life, and all cells arise from pre-existing cells. Exceptions to this include striated muscle (multinucleated), giant algae (large size with single nucleus), and aseptate fungal hyphae (continuous cytoplasm).

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Surface Area to Volume Ratio (SA:VSA:V): As a cell increases in size, its volume (VV) increases at a faster rate than its surface area (SASA). Since the rate of metabolism is a function of volume and the rate of material exchange is a function of surface area, cells must remain small to maintain a high SA:VSA:V ratio.

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Prokaryotic Cell Structure: Prokaryotes (e.g., bacteria) lack a nucleus and membrane-bound organelles. They contain a cell wall (peptidoglycan), plasma membrane, cytoplasm, 70S70S ribosomes, and a nucleoid region containing naked circular DNA. They divide by binary fission.

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Eukaryotic Cell Structure: Eukaryotes have a compartmentalized structure with membrane-bound organelles. Key organelles include the nucleus (stores DNA), mitochondria (site of aerobic respiration), 80S80S ribosomes (protein synthesis), Rough Endoplasmic Reticulum (protein transport), Golgi apparatus (processing/packaging), and Lysosomes (digestion).

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Microscopy: Magnification is the ratio of an object's image size to its actual size. Resolution is the ability to distinguish two points as separate. Electron microscopes have a much higher resolution (0.10.1 nm) than light microscopes (200200 nm) because the wavelength of electrons is much shorter than that of light.

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Universal features: All cells share certain components, including a plasma membrane, cytoplasm, DNA as genetic material, and ribosomes for protein synthesis.

📐Formulae

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

Actual size=Image sizeMagnificationActual\ size = \frac{\text{Image\ size}}{\text{Magnification}}

SA:V=Surface AreaVolumeSA:V = \frac{\text{Surface Area}}{\text{Volume}}

1 mm=1000 μm1\ \text{mm} = 1000\ \mu\text{m}

1 μm=1000 nm1\ \mu\text{m} = 1000\ \text{nm}

💡Examples

Problem 1:

An electron micrograph shows a plant cell with a length of 40 mm40\ \text{mm}. The scale bar on the image represents 10 μm10\ \mu\text{m} and measures 20 mm20\ \text{mm} in length. Calculate the actual length of the plant cell in μm\mu\text{m}.

Solution:

  1. Find the magnification using the scale bar: M=Image size of barActual size of bar=20 mm10 μm=20000 μm10 μm=2000×M = \frac{\text{Image size of bar}}{\text{Actual size of bar}} = \frac{20\ \text{mm}}{10\ \mu\text{m}} = \frac{20000\ \mu\text{m}}{10\ \mu\text{m}} = 2000\times
  2. Calculate the actual size of the cell: A=Image size of cellM=40 mm2000=40000 μm2000=20 μmA = \frac{\text{Image size of cell}}{M} = \frac{40\ \text{mm}}{2000} = \frac{40000\ \mu\text{m}}{2000} = 20\ \mu\text{m}

Explanation:

First, convert all units to the same scale (usually μm\mu\text{m}). Use the scale bar to find the magnification of the image, then apply that magnification to the measured image size of the specimen.

Problem 2:

Compare the surface area to volume ratio of a cube-shaped cell with a side length of 2 μm2\ \mu\text{m} to one with a side length of 4 μm4\ \mu\text{m}.

Solution:

For 2 μm2\ \mu\text{m} cell: SA=6×(22)=24 μm2SA = 6 \times (2^2) = 24\ \mu\text{m}^2 V=23=8 μm3V = 2^3 = 8\ \mu\text{m}^3 SA:V=248=3.0SA:V = \frac{24}{8} = 3.0

For 4 μm4\ \mu\text{m} cell: SA=6×(42)=96 μm2SA = 6 \times (4^2) = 96\ \mu\text{m}^2 V=43=64 μm3V = 4^3 = 64\ \mu\text{m}^3 SA:V=9664=1.5SA:V = \frac{96}{64} = 1.5

Explanation:

The SA:VSA:V ratio decreases as the cell size increases. The 2 μm2\ \mu\text{m} cell has a ratio of 3.03.0, while the larger 4 μm4\ \mu\text{m} cell has a ratio of 1.51.5, making the smaller cell more efficient at diffusion.