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Cells and Life Processes - Cell Theory

Grade 7IB

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

🔑Concepts

The Cell Theory consists of three main principles: 1. All living organisms are composed of one or more cells. 2. The cell is the basic unit of structure and organization in organisms. 3. Cells arise from pre-existing cells.

All living things perform seven basic life processes, often remembered by the acronym MR H GRENMR\ H\ GREN: Metabolism, Reproduction, Homeostasis, Growth, Response (Sensitivity), Excretion, and Nutrition.

Metabolism refers to the sum of all chemical reactions within a cell, including the release of energy through ATPATP production.

Cells are measured in micrometers (μm\mu m) and nanometers (nmnm). Conversion: 1 mm=1,000 μm1\ mm = 1,000\ \mu m and 1 μm=1,000 nm1\ \mu m = 1,000\ nm.

The surface area to volume ratio (SA:VSA:V) is a limiting factor for cell size. As a cell grows, its volume (VV) increases faster than its surface area (SASA), making it harder to move materials like O2O_2 and CO2CO_2 efficiently across the membrane.

Unicellular organisms, such as ParameciumParamecium or ChlamydomonasChlamydomonas, must perform all functions of life within a single cell.

Multicellular organisms exhibit emergent properties, where the whole organism is more than the sum of its cellular parts due to cellular differentiation and specialization.

📐Formulae

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

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

Surface Area to Volume Ratio=Surface AreaVolume\text{Surface Area to Volume Ratio} = \frac{\text{Surface Area}}{\text{Volume}}

💡Examples

Problem 1:

A student views a plant cell under a microscope. The image of the cell measures 30 mm30\ mm in length. If the actual size of the cell is 60 μm60\ \mu m, calculate the magnification used.

Solution:

Magnification=30,000 μm60 μm=500×Magnification = \frac{30,000\ \mu m}{60\ \mu m} = 500 \times

Explanation:

First, convert the image size from millimeters to micrometers so the units match: 30 mm×1,000=30,000 μm30\ mm \times 1,000 = 30,000\ \mu m. Then, divide the image size by the actual size (30,000/6030,000 / 60) to get a magnification of 500×500 \times.

Problem 2:

Explain why a cube-shaped cell with a side length of 2 μm2\ \mu m is more efficient at waste removal than a cube-shaped cell with a side length of 4 μm4\ \mu m.

Solution:

For the 2 μm2\ \mu m cell, SA:V=24 μm28 μm3=3:1SA:V = \frac{24\ \mu m^2}{8\ \mu m^3} = 3:1. For the 4 μm4\ \mu m cell, SA:V=96 μm264 μm3=1.5:1SA:V = \frac{96\ \mu m^2}{64\ \mu m^3} = 1.5:1.

Explanation:

The smaller cell has a higher surface area to volume ratio (3:13:1 vs 1.5:11.5:1). This means it has more membrane surface available relative to its internal volume to transport waste products out of the cell efficiently.

Cell Theory - Revision Notes & Key Formulas | IB Grade 7 Science