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Cell Biology - Specialized Cells

Grade 9IB

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

🔑Concepts

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Cell Differentiation: The process by which a stem cell changes to become specialized for a specific function. This involves the expression of specific genes within the cell's DNADNA.

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Surface Area to Volume Ratio (SA:VSA:V): Many specialized cells, such as root hair cells and red blood cells, increase their surface area to maximize the rate of diffusion or absorption relative to their volume.

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Red Blood Cells (Erythrocytes): Specialized for transporting oxygen. They have a biconcave shape to increase SA:VSA:V, contain hemoglobin to bind oxygen, and lack a nucleus to provide more space for hemoglobin.

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Sperm Cells: Specialized for reproduction. They possess a flagellum (tail) for motility, a middle piece packed with mitochondria for ATPATP production via aerobic respiration, and an acrosome containing enzymes to digest the egg cell membrane.

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Nerve Cells (Neurons): Specialized for rapid transmission of electrical impulses. They have long axons to carry signals over distances and branched dendrites to form connections (synapses) with other cells.

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Root Hair Cells: Plant cells specialized for the absorption of water and mineral ions. They have long cytoplasmic projections to increase surface area and a large permanent vacuole to maintain a water potential gradient.

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Xylem and Phloem: Xylem cells are specialized for water transport; they are lignified and dead, forming hollow tubes. Phloem cells are specialized for the translocation of sugars and have sieve plates and companion cells.

📐Formulae

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

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

1 mm=1000 μm1\ mm = 1000\ \mu m

SA:V=Total Surface AreaTotal VolumeSA:V = \frac{Total\ Surface\ Area}{Total\ Volume}

💡Examples

Problem 1:

A specialized ciliated epithelial cell has an actual length of 40 μm40\ \mu m. If a student draws the cell with a length of 20 mm20\ mm, calculate the magnification used for the drawing.

Solution:

Magnification=Image sizeActual sizeMagnification = \frac{Image\ size}{Actual\ size} First, convert mmmm to μm\mu m: 20 mm×1000=20000 μm20\ mm \times 1000 = 20000\ \mu m Magnification=20000 μm40 μm=500×Magnification = \frac{20000\ \mu m}{40\ \mu m} = 500\times

Explanation:

To calculate magnification, units must be consistent. We converted the drawing size from mmmm to μm\mu m before dividing by the actual size.

Problem 2:

Explain how the structure of a muscle cell is adapted to its function.

Solution:

Muscle cells contain a high density of mitochondria to provide ATPATP (energy) for contraction. They also contain specialized protein filaments (actin and myosin) that slide over each other to facilitate movement.

Explanation:

Specialization involves structural adaptations (more organelles like mitochondria) to meet the high energy demands of the cell's specific task (contraction).