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Engineering Life: Miracles in Biotechnology - Introduction to Biotechnology-advanced

Grade 9CBSE

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

🔑Concepts

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Biotechnology is the use of living organisms or their products to modify human health and the human environment. It involves techniques like genetic engineering and tissue culture.

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Tissue Culture: A technique of maintaining and growing plant cells, tissues, or organs especially on an artificial nutrient medium in sterile conditions. The initial tissue used is called the ExplantExplant, which develops into an unorganized mass of cells called a CallusCallus.

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Genetic Engineering: The deliberate modification of the characteristics of an organism by manipulating its genetic material (DNADNA). This leads to the production of Genetically Modified Organisms (GMOsGMOs).

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Stem Cells: Special types of cells that have the ability to divide and give rise to similar stem cells or differentiate into various specialized cell types. They are classified into EmbryonicEmbryonic stem cells and AdultAdult (Somatic) stem cells.

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Applications in Agriculture: Includes the production of BtBt CottonCotton (using genes from BacillusBacillus thuringiensisthuringiensis), bio-fertilizers like RhizobiumRhizobium and AzotobacterAzotobacter, and the development of high-yielding, disease-resistant crop varieties.

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Applications in Health: Production of vaccines, antibiotics, and human hormones like insulin through recombinant DNADNA technology.

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Bio-fertilizers: These are substances containing living microorganisms which, when applied to seeds or soil, colonize the rhizosphere and promote growth by increasing the supply of primary nutrients to the host plant.

📐Formulae

N=2nN = 2^n (where NN is the total number of cells produced after nn successive mitotic divisions in tissue culture)

Efficiency of Bio-fertilizer=Yield with Bio-fertilizer−Yield withoutYield without×100\text{Efficiency of Bio-fertilizer} = \frac{\text{Yield with Bio-fertilizer} - \text{Yield without}}{\text{Yield without}} \times 100

💡Examples

Problem 1:

In a laboratory experiment, a scientist starts with a single explantexplant cell in a nutrient-rich medium. If the cell undergoes 66 successive mitotic divisions to form a calluscallus, calculate the total number of cells present in the calluscallus.

Solution:

Given the number of divisions n=6n = 6. Using the formula for mitotic cell division: N=2nN = 2^n N=26N = 2^6 N=64N = 64

Explanation:

In mitosis, one cell divides to form two identical daughter cells. Therefore, the number of cells doubles with every division, following a geometric progression 2n2^n.

Problem 2:

A farmer used bio-fertilizers on a patch of land and obtained a crop yield of 500500 kg. On an identical patch without bio-fertilizers, the yield was 400400 kg. Calculate the percentage increase in yield.

Solution:

Increase in yield = 500−400=100500 - 400 = 100 kg. Percentage Increase=100400×100\text{Percentage Increase} = \frac{100}{400} \times 100 Percentage Increase=25%\text{Percentage Increase} = 25\%

Explanation:

The use of bio-fertilizers like AzotobacterAzotobacter improves nutrient availability, leading to a significant increase in productivity compared to traditional methods.

Introduction to Biotechnology-advanced Class 9 Notes & Examples