Factors Affecting the Rate of Enzyme Reactions

Calculate the temperature coefficient (Q₁₀) for a biological reaction. At 25°C, the rate of reaction is 1.5 units/s. At 35°C, the rate of reaction is 3.0 units/s.

Step 1: State the formula for the temperature coefficient.

• $Q_{10} = \frac{\text{Rate at } (T + 10)^\circ\text{C}}{\text{Rate at } T^\circ\text{C}}$

Step 2: Substitute the known values into the equation.

• $Q_{10} = \frac{3.0}{1.5}$

Step 3: Calculate the final answer.

• $Q_{10} = 2$ (This confirms the general rule that enzyme reaction rates roughly double for every 10°C increase up to the optimum).

The Effect of pH on Enzyme Activity

Your stomach is a highly acidic environment, yet your small intestine is slightly alkaline. Different enzymes have adapted to work perfectly in these specific environments.

Every enzyme has an optimum pH, which is the exact pH level where its active site is the most effective shape. Most human enzymes peak at roughly pH 7, but there are notable site-specific exceptions. For example, pepsin in the stomach has an optimum of pH 2, while trypsin in the small intestine has an optimum of pH 8.

If the environment becomes too acidic or too alkaline, the excess of $H^+$ ions or $OH^-$ ions interferes with the amino acid chains. These ions disrupt and break the hydrogen and ionic bonds maintaining the active site's structure. Severe shifts away from the optimum pH cause permanent denaturation, meaning the active site is no longer complementary to the substrate. Unlike temperature, changes in pH do not alter kinetic energy or collision frequency; they only affect how well the substrate can bind to the active site.

Analysing pH Graphs

When analysing a pH graph, you will observe a characteristic bell-shaped curve. You must identify the following patterns:

1. The Rise: As pH moves toward the optimum, the active site shape becomes more effective, increasing the rate.
2. The Peak: The maximum rate of reaction occurs at the optimum pH.
3. The Sharp Decline: The rate drops rapidly on both sides of the peak. Because enzymes are so sensitive to $H^+$ and $OH^-$ ions, even a small move away from the optimum can cause denaturation.

Investigating pH (PAG 4: Amylase Practical)

• To test how pH affects amylase, you must mix the enzyme with a specific buffer solution to maintain a constant pH.
• To ensure a valid experiment, you must identify and maintain control variables. This includes using a water bath to keep the temperature constant and ensuring the volume and concentration of both the amylase and starch remain the same for every pH tested.
• Add starch, start a stopwatch, and extract a drop of the mixture every 10 seconds to test with an indicator on a spotting tile.
• In the presence of starch, iodine solution turns from orange-brown to blue-black.
• The reaction is complete when the iodine stops turning blue-black (remaining orange-brown), indicating all the starch has been digested.

In a PAG 4 experiment, it takes 80 seconds for amylase to completely break down starch at pH 6. Calculate the rate of reaction.

Step 1: State the reciprocal rate formula.

• $\text{Rate} = \frac{1}{\text{Time}}$

Step 2: Substitute the values.

• $\text{Rate} = \frac{1}{80}$

Step 3: Calculate the final answer with units.

• $\text{Rate} = 0.0125\text{ s}^{-1}$

The Effect of Substrate and Enzyme Concentration

Imagine trying to pair up dancers at a party — if there are too many people and not enough dancefloors, people have to wait. This is exactly what happens with enzyme and substrate molecules.

Substrate Concentration

When substrate concentration is low, increasing it will cause a directly proportional, linear increase in the rate of reaction. This is because there are plenty of empty active sites waiting to be filled, so adding more substrate increases the frequency of successful collisions and the formation of enzyme-substrate complexes (ESCs). In this linear phase, the substrate concentration is the limiting factor.

Eventually, the graph will level off into a plateau. This is called the saturation point, where every single available active site is occupied by a substrate molecule. Adding more substrate will no longer speed up the reaction because the enzymes are working as fast as they can — the reaction has reached its maximum rate ($V_{max}$). At this plateau, enzyme concentration becomes the new limiting factor.

Enzyme Concentration

Increasing enzyme concentration also increases the rate of reaction linearly at first. More enzyme molecules mean more available active sites, which increases the frequency of successful collisions and the formation of ESCs. In this phase, enzyme concentration is the limiting factor.

However, if the amount of substrate is kept constant, the rate will eventually plateau. This is because there are eventually more active sites than there are substrate molecules to fill them. At this point, adding more enzyme has no further effect because the substrate concentration has become the limiting factor.

Analysing Comparative Graphs (Higher Tier)

In exams, you may be shown a graph with two different lines (e.g., Line A and Line B) representing different concentrations of a variable.

• If Line A reaches a higher plateau ($V_{max}$) than Line B, it means Line A has a higher concentration of the limiting factor.
• For example, if you are investigating substrate concentration and Line A plateaus higher than Line B, then Line A must have a higher enzyme concentration, allowing more reactions to happen simultaneously once all sites are saturated.

Mathematical Analysis of Enzyme Rates

The highest rate of reaction almost always occurs right at the start ($t = 0$), known as the initial rate of reaction, because this is when substrate concentration is at its absolute highest. As the substrate is used up, the rate decreases.

To find the rate of reaction at a specific moment in time from a curved graph, you must calculate the gradient of a tangent line.

A student plots a curve showing the volume of oxygen produced by catalase over time. Calculate the rate of reaction at 20 seconds.

Step 1: Draw a tangent.

• Place a ruler on the curve at exactly $t = 20$ seconds and draw a straight line that touches the curve only at that point.

Step 2: Pick two easy-to-read points on your straight tangent line.

• Point 1: $(10\text{ s}, 15\text{ cm}^3)$
• Point 2: $(50\text{ s}, 75\text{ cm}^3)$

Step 3: State the formula for the gradient and substitute your values.

• $\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1}$

• $\text{Gradient} = \frac{75 - 15}{50 - 10}$

Step 4: Calculate the final rate with units.

• $\text{Gradient} = \frac{60}{40} = 1.5\text{ cm}^3/s$