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Cell: Structure and Function - Structure of Prokaryotic and Eukaryotic Cells

Grade 11ICSEBiology

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

🔑Concepts

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Cell Theory: Proposed by Schleiden and Schwann, later modified by Rudolf Virchow stating Omnis cellula-e cellula\text{Omnis cellula-e cellula} (all cells arise from pre-existing cells).

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Prokaryotic Cells: Characterized by the absence of a membrane-bound nucleus. Genetic material is naked and lies in a region called the nucleoid. They possess 70S70S ribosomes (subunits 50S50S and 30S30S).

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Eukaryotic Cells: Possess a well-defined nucleus with a nuclear envelope and membrane-bound organelles like mitochondria, ER, and Golgi complex. They possess 80S80S ribosomes (subunits 60S60S and 40S40S).

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Cell Envelope in Prokaryotes: Consists of a glycocalyx, cell wall (made of peptidoglycan), and plasma membrane. Some bacteria possess small circular DNA called plasmids, which provide unique phenotypes like resistance to antibiotics.

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Fluid Mosaic Model: Proposed by Singer and Nicolson (19721972), describing the plasma membrane as a 'protein icebergs in a sea of lipids'. The quasi-fluid nature of lipids enables lateral movement of proteins.

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Ribosomes: The sites of protein synthesis. In eukaryotes, they are found in the cytoplasm (80S80S) and within organelles like mitochondria and chloroplasts (70S70S).

📐Formulae

Magnification (M)=Size of Image (I)Actual Size of Object (O)\text{Magnification (M)} = \frac{\text{Size of Image (I)}}{\text{Actual Size of Object (O)}}

Surface Area to Volume Ratio=6a2a3=6a (for a cuboidal cell)\text{Surface Area to Volume Ratio} = \frac{6a^2}{a^3} = \frac{6}{a} \text{ (for a cuboidal cell)}

Svedberg Unit (S)=10−13 s\text{Svedberg Unit (S)} = 10^{-13} \text{ s}

Resolution (d)=0.61λNA\text{Resolution (d)} = \frac{0.61 \lambda}{NA}

💡Examples

Problem 1:

Calculate the actual size of a bacterial cell if its image under a microscope is 2 cm2 \text{ cm} long at a magnification of 40,000×40,000\times.

Solution:

Actual size O=IM=2 cm40,000=2×10−2 m4×104=0.5×10−6 m=0.5μmO = \frac{I}{M} = \frac{2 \text{ cm}}{40,000} = \frac{2 \times 10^{-2} \text{ m}}{4 \times 10^4} = 0.5 \times 10^{-6} \text{ m} = 0.5 \mu\text{m}.

Explanation:

Using the magnification formula M=IOM = \frac{I}{O}, we rearrange to find the actual size of the specimen.

Problem 2:

Why is the 70S70S ribosome denoted by '70' even though its subunits are 50S50S and 30S30S?

Solution:

The 'S' stands for Svedberg unit, a measure of sedimentation rate during ultracentrifugation. 50S+30S→70S50S + 30S \rightarrow 70S.

Explanation:

The Svedberg unit is not additive; it depends on the mass, density, and shape of the particle. When subunits combine, the total surface area changes, resulting in a sedimentation coefficient of 70S70S for the whole prokaryotic ribosome.

Problem 3:

A spherical cell has a radius rr. If the radius increases by a factor of 33, how does the Surface Area to Volume (SA/VSA/V) ratio change?

Solution:

Initial ratio: 4πr243πr3=3r\frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}. New ratio: 33r=1r\frac{3}{3r} = \frac{1}{r}.

Explanation:

As the cell size increases, the volume increases much faster than the surface area, causing the SA/VSA/V ratio to decrease. This limits the efficiency of diffusion in larger eukaryotic cells.

Structure of Prokaryotic and Eukaryotic Cells Revision - Class 11 Biology ICSE