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Form and function - Organelles and compartmentalization

Grade 11IBBiology

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

🔑Concepts

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Compartmentalization: The separation of the cell interior into distinct membrane-bound organelles, allowing different chemical reactions to occur simultaneously without interference.

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Advantages of compartmentalization: Enzymes and substrates can be concentrated in specific areas, pH levels can be optimized for specific reactions, and damaging substances (like hydrolytic enzymes in lysosomes) can be isolated.

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The Nucleus: Contains the cell's genetic material (DNADNA) and is surrounded by a double membrane called the nuclear envelope, which contains pores for the transport of mRNAmRNA and proteins.

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Mitochondria and Chloroplasts: Both are double-membrane bound organelles involved in energy transduction. Mitochondria perform aerobic respiration: C6H12O6+6O2→6CO2+6H2O+ATPC_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O + ATP. Chloroplasts perform photosynthesis: 6CO2+6H2O+light→C6H12O6+6O26CO_{2} + 6H_{2}O + light \rightarrow C_{6}H_{12}O_{6} + 6O_{2}.

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Endomembrane System: Includes the Rough Endoplasmic Reticulum (RER) for protein synthesis, Smooth Endoplasmic Reticulum (SER) for lipid synthesis, and the Golgi Apparatus for modifying, sorting, and packaging proteins into vesicles.

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Surface Area to Volume Ratio (SA:VSA:V): As a cell increases in size, its volume (r3r^{3}) increases faster than its surface area (r2r^{2}). This limits cell size because the rate of exchange of materials across the membrane cannot keep up with the metabolic demands of the cytoplasm.

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Endosymbiosis: The theory that mitochondria and chloroplasts originated as free-living prokaryotes that were engulfed by ancestral eukaryotic cells, evidenced by their own 70S70S ribosomes and circular DNADNA.

📐Formulae

Magnification(M)=Size of Image(I)Actual Size of Specimen(A)\text{Magnification} (M) = \frac{\text{Size of Image} (I)}{\text{Actual Size of Specimen} (A)}

Surface Area of a Sphere=4πr2\text{Surface Area of a Sphere} = 4\pi r^2

Volume of a Sphere=43πr3\text{Volume of a Sphere} = \frac{4}{3}\pi r^3

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

💡Examples

Problem 1:

Calculate the Surface Area to Volume ratio for a spherical cell with a radius of 3μm3\mu m.

Solution:

  1. Calculate Surface Area (SASA): SA=4×π×32=36π≈113.10μm2SA = 4 \times \pi \times 3^2 = 36\pi \approx 113.10\mu m^2
  2. Calculate Volume (VV): V=43×π×33=36π≈113.10μm3V = \frac{4}{3} \times \pi \times 3^3 = 36\pi \approx 113.10\mu m^3
  3. Calculate Ratio: SAV=36π36π=1\frac{SA}{V} = \frac{36\pi}{36\pi} = 1

Explanation:

In this specific case where r=3r=3, the numerical values for surface area and volume are equal, resulting in a ratio of 1:11:1. As rr increases further, the volume will grow much faster than the surface area, decreasing the ratio.

Problem 2:

A micrograph of a mitochondrion shows a length of 50mm50mm. If the magnification of the image is 25,000×25,000\times, what is the actual length of the mitochondrion in micrometers (μm\mu m)?

Solution:

  1. Use the formula A=IMA = \frac{I}{M}.
  2. Convert image size to micrometers: 50mm=50,000μm50mm = 50,000\mu m.
  3. Calculate: A=50,000μm25,000=2μmA = \frac{50,000\mu m}{25,000} = 2\mu m

Explanation:

To find the actual size, divide the measured image size by the magnification. Always ensure units are consistent (convert mmmm to μm\mu m by multiplying by 10001000).