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Engineering Life: Miracles in Biotechnology - Food processing-advanced

Grade 9CBSE

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

🔑Concepts

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Biotechnology in Food: It involves the use of living organisms like bacteria, yeast, or enzymes to perform specific processes to improve food quality, taste, and shelf life.

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Fermentation: A metabolic process where microorganisms like yeast and bacteria convert carbohydrates (sugars) into alcohol or organic acids under anaerobic conditions. For example, conversion of glucose: C6H12O6→2C2H5OH+2CO2C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2.

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Genetically Modified Organisms (GMOs): Organisms whose genetic material has been altered using genetic engineering techniques. In food, this is used to enhance nutritional value, such as Golden Rice which is rich in Vitamin A (C40H56C_{40}H_{56}).

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Bio-fortification: The process of increasing the density of vitamins and minerals in a crop through plant breeding or genetic engineering.

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Food Preservation - Pasteurization: A heat-treatment process that destroys pathogenic microorganisms. High-Temperature Short-Time (HTST) involves heating milk to 72∘C72^{\circ}C for 1515 seconds.

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Lyophilization (Freeze-drying): A process where water is removed from food by sublimation (converting ice directly into water vapor) under low pressure, preserving the food's structure and nutrients.

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Single Cell Protein (SCP): Protein extracted from pure cultures of algae, fungi, or bacteria (e.g., Spirulina) used as a high-protein food supplement.

📐Formulae

C6H12O6→Yeast/Anaerobic2C2H5OH+2CO2+EnergyC_{6}H_{12}O_{6} \xrightarrow{\text{Yeast/Anaerobic}} 2C_{2}H_{5}OH + 2CO_{2} + \text{Energy}

C6H12O6→Lactic Acid Bacteria2CH3CH(OH)COOH+EnergyC_{6}H_{12}O_{6} \xrightarrow{\text{Lactic Acid Bacteria}} 2CH_{3}CH(OH)COOH + \text{Energy}

Shelf Life Extension=Standard Shelf Life+Treatment Gain\text{Shelf Life Extension} = \text{Standard Shelf Life} + \text{Treatment Gain}

💡Examples

Problem 1:

In a fermentation process, a technician starts with 180180 g of Glucose (C6H12O6C_6H_{12}O_6). If the theoretical yield of Ethanol (C2H5OHC_2H_5OH) is 51%51\%, calculate the mass of ethanol produced.

Solution:

Given mass of Glucose = 180180 g. Molar mass of Glucose (C6H12O6C_6H_{12}O_6) ≈180\approx 180 g/mol. According to the equation C6H12O6→2C2H5OH+2CO2C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2, 11 mole of Glucose produces 22 moles of Ethanol. Molar mass of Ethanol (C2H5OHC_2H_5OH) ≈46\approx 46 g/mol. Theoretical mass for 100%100\% yield: 2×46=922 \times 46 = 92 g. Since the yield is 51%51\%, the mass produced is: Mass=92×51100=46.92 g\text{Mass} = 92 \times \frac{51}{100} = 46.92 \text{ g}

Explanation:

The fermentation equation dictates the stoichiometric ratio, and the percentage yield accounts for the efficiency of the biological process.

Problem 2:

Determine the temperature in Fahrenheit for the pasteurization of milk if it is kept at 63∘C63^{\circ}C for 3030 minutes (Batch method). Use the formula F=95C+32F = \frac{9}{5}C + 32.

Solution:

Substitute C=63C = 63 into the formula: F=(95×63)+32F = \left(\frac{9}{5} \times 63\right) + 32 F=113.4+32F = 113.4 + 32 F=145.4∘FF = 145.4^{\circ}F

Explanation:

Pasteurization requires specific temperature-time combinations to ensure the safety of dairy products by killing bacteria like Mycobacterium tuberculosis.

Problem 3:

A food processing plant uses a probiotic culture that doubles every 2020 minutes. If the initial count is 1×1031 \times 10^3 cells, what will be the cell count after 11 hour?

Solution:

Total time = 6060 minutes. Number of doubling periods (nn) = 6020=3\frac{60}{20} = 3. Final Population (NN) is given by N=N0×2nN = N_0 \times 2^n. N=1×103×23N = 1 \times 10^3 \times 2^3 N=1000×8=8000N = 1000 \times 8 = 8000 N=8×103 cellsN = 8 \times 10^3 \text{ cells}

Explanation:

The growth of microorganisms in food processing (like yogurt making) follows an exponential growth pattern during the log phase.