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Form and function - Cell specialization

Grade 11IBBiology

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

🔑Concepts

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Cell differentiation is the process by which unspecialized cells develop into cells with distinct structures and functions. This occurs through the selective expression of specific genes within the cell's genome while others remain dormant.

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The genome of an organism consists of the entire set of genetic instructions. All diploid cells in a multicellular organism share the same genome, but the proteome (the set of expressed proteins) differs between specialized cells.

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Stem cells are characterized by two key properties: Self-renewal (the ability to divide repeatedly) and Potency (the capacity to differentiate into various cell types).

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Potency levels are classified as: Totipotent (can form any cell type, e.g., zygote), Pluripotent (can form any body cell type, e.g., embryonic stem cells), Multipotent (can form a limited range of related cell types, e.g., umbilical cord blood stem cells), and Unipotent (can only form one cell type).

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The Surface Area to Volume ratio (SA:VSA:V) is a critical limiting factor for cell size. As a cell grows, its volume (V∝r3V \propto r^3) increases much faster than its surface area (SA∝r2SA \propto r^2), making it harder to exchange materials and heat with the environment.

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Emergent properties arise from the interaction of component parts in a multicellular organism; the whole is greater than the sum of its parts (1+1>21 + 1 > 2).

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Specialized cell examples include: Red Blood Cells (lack a nucleus to maximize hemoglobin space), Sperm Cells (contain a flagellum for motility and many mitochondria for energy), and Root Hair Cells (large SASA for water absorption).

📐Formulae

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

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

Surface Area of a Cube=6s2\text{Surface Area of a Cube} = 6s^2

Volume of a Cube=s3\text{Volume of a Cube} = s^3

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

💡Examples

Problem 1:

Calculate the magnification of a micrograph where a cell measuring 20 μm20\text{ }\mu m in actual size is drawn with a length of 4 cm4\text{ cm}.

Solution:

I=4 cm=40,000 μmI = 4\text{ cm} = 40,000\text{ }\mu m A=20 μmA = 20\text{ }\mu m M=40,00020=2000×M = \frac{40,000}{20} = 2000\times

Explanation:

To calculate magnification, first ensure the image size and actual size are in the same units. Converting 4 cm4\text{ cm} to micrometers involves multiplying by 10,00010,000. Then apply the formula M=IAM = \frac{I}{A}.

Problem 2:

Compare the SA:VSA:V ratio of two cubic cells. Cell A has a side length of 1 units1\text{ units} and Cell B has a side length of 3 units3\text{ units}. Which cell is more efficient at diffusion?

Solution:

For Cell A (s=1s=1): SA=6(1)2=6SA = 6(1)^2 = 6 V=13=1V = 1^3 = 1 SA:V=6:1=6SA:V = 6:1 = 6

For Cell B (s=3s=3): SA=6(3)2=54SA = 6(3)^2 = 54 V=33=27V = 3^3 = 27 SA:V=54:27=2:1=2SA:V = 54:27 = 2:1 = 2

Explanation:

Cell A has a higher SA:VSA:V ratio (66) compared to Cell B (22). A higher ratio indicates that for every unit of volume that requires nutrients, there is more surface area available for exchange, making Cell A more efficient at diffusion.