Ch 1 · Enzymes · O Level Biology 5090
Topic 1 of 1 in Enzymes — O Level Biology 5090 — 48 lessons.
Introduction
Almost every chemical reaction in a living cell, from digesting a meal to copying DNA, is controlled by an enzyme. Without enzymes these reactions would be far too slow at body temperature to keep an organism alive. Learn the definitions and the lock-and-key model word for word; temperature and pH investigations are also a favourite Paper 6 context.
Key ideas — Catalysts and enzymes
A catalyst is a substance that increases the rate of a chemical reaction and is not changed by the reaction.
🧑🏫 Teacher's explanation
Imagine you are trying to move a heavy pile of bricks from one side of a yard to another. If you carry them one by one, it will take all day. Now, imagine a person (the catalyst) who helps you by providing a wheelbarrow. The wheelbarrow makes the job much faster, but once the pile is moved, the wheelbarrow is still there, ready to be used again. It hasn't been 'used up' by the job.
📖 Definitions
🔍 Worked example — Easy
A student investigates the effect of temperature on the rate of an enzyme-catalyzed reaction. At 20^° C, the rate is 5 units/min. At 40^° C, the rate is 15 units/min. Calculate the percentage increase in the rate of reaction.
🔍 Worked example — Standard
Describe the effect of increasing temperature from 10^° C to 40^° C on the rate of an enzyme-catalyzed reaction.
🔍 Worked example — Exam-style
Explain why an enzyme stops functioning when the temperature is raised to 70^° C.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — How enzymes work: the lock-and-key model
In the lock-and-key model the substrate is the key and the active site is the lock. Just as a lock accepts only the correctly shaped key, an enzyme catalyses only the reaction of the substrate that fits its active site. The enzyme is unchanged and can be reused.
🧑🏫 Teacher's explanation
Imagine you are trying to open a door. You have a specific key, and the door has a specific lock. If you try to use a key that is slightly too large or a different shape, it simply won't turn. This is exactly how enzymes work. They aren't just random blobs; they are highly specialized tools with a very specific 'socket' called an active site.
📖 Definitions
🔍 Worked example — Easy
A student observes that the enzyme amylase breaks down starch into maltose. If the substrate is starch, identify the enzyme-substrate complex and explain why amylase cannot break down proteins.
🔍 Worked example — Standard
Calculate the percentage increase in reaction rate if the rate increases from 0.5 units/sec to 2.5 units/sec due to an increase in substrate concentration.
🔍 Worked example — Exam-style
An experiment shows that at 20circC, an enzyme produces 10 products per minute. At 40circC, it produces 50 products per minute. Explain the increase in rate using kinetic theory.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Effect of temperature
A typical investigation uses amylase and starch. Equal volumes of the two solutions are placed in separate tubes in water baths at a range of temperatures, then mixed. Every 30 seconds a drop is tested with iodine solution on a spotting tile. The time taken for the iodine to stay orange-brown (no starch left) is recorded; a shorter time means a higher rate. Volumes, concentrations and pH are kept constant and each temperature is repeated.
🧑🏫 Teacher's explanation
Think of an enzyme as a specialized key and the substrate as a specific lock. For the lock to turn, the key must fit perfectly into the keyhole. In a cold room, you might move very slowly, making it hard to find the keyhole even if you have it. As you speed up, you bump into the lock more often, making the process faster. However, if you get too hot, the lock itself melts and warps; even if you have the right key, it no longer fits the distorted keyhole.
📖 Definitions
🔍 Worked example — Easy
A student measures the time taken for starch to be completely broken down by amylase at 20 ^circC as 300 seconds and at 40 ^circC as 100 seconds. Calculate the factor by which the rate of reaction has increased.
🔍 Worked example — Standard
Explain why the rate of reaction decreases when the temperature is raised from 40 ^circC to 60 ^circC in an investigation involving salivary amylase.
🔍 Worked example — Exam-style
An experiment shows that the rate of reaction for an enzyme is 0.5 units/min at 35 ^circC and 0.0 units/min at 70 ^circC. Interpret this data regarding the enzyme's properties.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Effect of pH
To investigate pH, the same amylase and starch method is used at constant temperature, with a few drops of a different buffer solution (for example pH 3, 5, 7, 9 and 11) added to each tube. Rate is again judged from the time taken for starch to disappear.
🧑🏫 Teacher's explanation
Think of an enzyme like a specialized key and the substrate like a lock. For the key to work, it must fit perfectly into the lock's shape. Now, imagine if the lock was made of soft wax. If the environment becomes too acidic or too alkaline, it's like applying heat or chemicals that melt and warp that wax. The lock's shape changes, and the key no longer fits.
📖 Definitions
🔍 Worked example — Easy
A student measures the time taken for starch to be digested by amylase at different pH levels. At pH 7, starch disappears in 120 seconds. At pH 4, starch disappears in 300 seconds. Calculate the factor by which the rate of reaction decreased when moving from pH 7 to pH 4.
🔍 Worked example — Standard
Describe the effect of moving from pH 7 to pH 11 on the activity of an enzyme that has an optimum pH of 7.
🔍 Worked example — Exam-style
An experiment is conducted to find the optimum pH of pepsin. The results are: pH 2 (fastest), pH 4 (slow), pH 7 (no activity). Explain why pepsin shows no activity at pH 7.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Question. A student investigated the effect of temperature on catalase, which breaks down hydrogen peroxide, by measuring the volume of oxygen released in one minute. The volume rose from 4 cm³ at 10 °C to 22 cm³ at 40 °C, then fell to 1 cm³ at 60 °C. Explain these results. [5]
🔍 Worked example — Easy
Calculate the percentage increase in the volume of oxygen released when the temperature increases from 10 °C to 40 °C (from 4 cm^3 to 22 cm^3).
🔍 Worked example — Standard
A student observes that an enzyme's activity is highest at 37 °C and decreases significantly at 50 °C. Explain the decrease in activity using the concept of kinetic energy and denaturation.
🔍 Worked example — Exam-style
An experiment shows that the rate of reaction for an enzyme is 0.5 units/min at 20 °C and 1.5 units/min at 40 °C. Explain why the rate increases.
⚠️ Common errors
Common mistakes
Practice
Answers A substance that increases the rate of a chemical reaction [B1] and is not changed by the reaction [B1]. Enzymes speed up the chemical reactions of metabolism [B1]; without them, reactions would be too slow at body temperature to sustain life [B1]. The substrate fits into the complementary active site [B1]; an enzyme-substrate complex forms [B1]; the substrate is broken down and the products leave the active site, which is unchanged [B1]. Add amylase and starch solutions, each with a buffer of known pH, to tubes [M1]; use at least five pH values, keeping volumes, concentrations and temperature constant [M1]; test drops with iodine solution at regular intervals [M1]; record the time for the blue-black colour to stop appearing, and repeat each pH [M1]. The active site of lipase is complementary in shape to fat only [B1]; starch has a different shape so cannot bind and no enzyme-substrate complex forms [B1].