CB 1 Overarching Concepts P 11 19 Enzymes

CB 1 – Overarching Concepts (P 11 -19) Enzymes - Label the diagram: Transport - complete the table Diagram Enzyme graphs Questions Q 1) Why is there an increase in activity from 0 o. C and 40 o. C? Osmosis Active Transport Core Practical – Q 2) What does the term ‘optimum temperature’ mean? 1. What equipment would you need to investigate osmosis in carrots and sugar solutions of different concentrations? Q 3) Why is there a decrease in enzyme activity between 40 o. C and 63 o. C? Convert the following to standard form: 158000000 9350000 0. 00000034 0. 006624 Convert the following to normal form: 2. 45 x 105 6. 10 x 10 -3 5. 55 x 107 8. 18 x 10 -12 Convert the following to metres: 10, 345 mm 12 km 145, 000 nm 754, 000 pm 12, 000µm 13, 333 nm Example in plants Diffusion Explain in detail what is meant by the lock and key theory of enzymes. Applying maths Description Cells - complete the table about specialised cells 2. Use the graph to estimate the concentration of sugar in the carrot. Microscopes – complete the table Calculate the actual size of an image if it appears to be 4 cm with a magnification of 100 x.

CB 1 – Overarching Concepts (P 11 -19) Enzymes - Label the diagram: Transport - complete the table Diagram Enzyme graphs Questions Q 1) Why is there an increase in activity from 0 o. C and 40 o. C? Explain in detail what is meant by the lock and key theory of enzymes. The particles have more kinetic energy and therefore there will be more successful collisions Q 2) What does the term ‘optimum temperature’ mean? The temperature that enzymes work best at Q 3) Why is there a decrease in enzyme activity between 40 o. C and 63 o. C? Because enzymes get to hot and their active sites change shape. This means the substrate doesn’t fit in. This is denaturing. Applying maths Convert the following to standard form: 158000000 = 1. 58 x 108 9350000 = 9. 35 x 106 0. 00000034 = 3. 40 x 10 -7 0. 006624 = 6. 62 x 10 -3 Convert the following to normal form: 2. 45 x 105 = 245, 000 6. 10 x 10 -3 = 0. 00610 5. 55 x 107 = 55, 500, 000 8. 18 x 10 -12 = 0. 000000818 Convert the following to metres: 10, 345 mm = 10. 345 m 12 km = 12, 000 m 145, 000 nm = 0. 000145 m 754, 000 pm = 0. 000754 m 12, 000µm = 0. 012 m 13, 333 nm = 0. 000013333 m The active site is similar to a lock and the substrate is similar to a key. They have complimentary shapes. This shows how different enzymes can only break down specific substrates. Description Diffusion Movement of particles from a high to low concentration Osmosis Movement of water particles from a high to low concentration across a partially permeable membrane Active Transport Core Practical – Movement of particles form a low to a high concentration. Requires energy. Root Absorbs hair cellwater & minerals Sperm Fertilises egg cell Large surface area Small Absorbs intestinal nutrients cell Microvilli for large surface area Removes dust from trachea Moving hairs (cilia) Ciliated cell 2. Use the graph to estimate the concentration of sugar in the carrot. 0. 3 mol/dm 3 Microscopes – complete the table - Tail - Acrosome (releases enzymes) Gases exchange in leaves Water in roots Minerals in roots 1. What equipment would you need to investigate osmosis in carrots and sugar solutions of different concentrations? Apple corer, electronic balance, paper towel, different concentrations of sucrose, test tubes. Cells - complete the table about specialised cells Example in plants x 1, 500 x 100, 000 light 2 D electrons 3 D Thin slice on slide Coated in gold and in a vacuum Calculate the actual size of an image if it appears to be 4 cm with a magnification of 100 x. 4 ÷ 100 = 0. 04 cm