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When fruits such as apples and pears are cut, the exposed areas begin to turn brown. This is the result of an enzymecatalyzed reaction. Often the browning can be prevented or slowed by adding a few drops of lemon juice. What is the chemical basis of this treatment?

Short Answer

Expert verified
Lemon juice lowers pH, inhibiting the browning enzyme.

Step by step solution

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01

Understanding the Browning Process

When fruits are cut and exposed to air, the enzyme polyphenol oxidase (PPO) catalyzes the oxidation of phenolic compounds in the fruit, leading to the formation of brown pigments called melanin. This process is commonly known as enzymatic browning.
02

Role of pH in Enzymatic Activity

Enzymatic activity, including the action of polyphenol oxidase, is sensitive to pH levels. Most enzymes have an optimal pH range where they function best. Outside of this range, the enzyme's activity can be inhibited, slowing down or stopping the reaction.
03

Effect of Lemon Juice

Lemon juice is acidic and contains citric acid. When added to the cut fruits, it decreases the pH of the exposed areas, creating an environment where the polyphenol oxidase enzyme is less active. This acidity helps to prevent or at least slow down the enzymatic browning process.
04

Conclusion

The addition of lemon juice alters the pH on the surface of the fruit, reducing the efficiency of the polyphenol oxidase enzyme and thereby slowing the browning process.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polyphenol Oxidase
When you cut fruits like apples or pears, have you noticed how they start turning brown? This happens because of an enzyme called Polyphenol Oxidase (PPO). Enzymes are proteins that speed up chemical reactions, and in this case, PPO interacts with certain compounds in the fruit called phenolics. When exposed to air, the enzyme helps turn these phenolics into brown pigments known as melanin. The overall process resulting in the browning is called enzymatic browning.
It's pretty interesting that while these brown pigments might not look too appetizing, they do not indicate that the fruit is spoiled or harmful to eat. They are just a natural product of oxidation catalyzed by PPO.
pH Levels
An important factor that affects how enzymes like Polyphenol Oxidase work is the pH level. You might know pH as a measure of acidity or alkalinity. Each enzyme has an optimal pH range, which is like its comfort zone where it works best. Outside this zone, the enzyme's ability to catalyze reactions decreases.
For enzymes such as PPO, if the pH level is not ideal, their activity drops. In essence, changing the pH level is like changing the engine speed in your car—it can slow down or speed up certain reactions within the fruit.
That's why sometimes you might add something acidic to the fruit to tweak the pH levels and control the enzyme's activity.
Citric Acid
When it comes to slowing down the browning of cut fruits, citric acid is a key player. Found in lemon juice and many other citrus fruits, citric acid works by lowering the pH level on the fruit's surface.
This drop in pH creates an environment that is less favorable for the activity of polyphenol oxidase. With the enzyme's activity reduced, it can't speed up the browning process as efficiently.
Besides affecting enzyme activity, citric acid has antioxidant properties. This means it can further help prevent the oxidation process that leads to the browning of cut fruits.
Enzyme Inhibition
In simple terms, enzyme inhibition is like putting brakes on an enzyme's activity. When we talk about enzyme inhibition in the context of enzymatic browning, we're referring to methods that slow or stop the activity of Polyphenol Oxidase.
This can be achieved through a variety of means, such as adjusting pH levels, temperature, or introducing compounds like citric acid. Each inhibitor works differently. For example, citric acid reduces the enzyme's activity by changing the pH of its environment.
  • By altering pH, enzyme action is stalled.
  • Chemicals like citric acid make the environment less suitable for enzymatic reactions.

Understanding enzyme inhibition is crucial for anyone looking to keep their fruits fresh and appealing for longer periods.

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Most popular questions from this chapter

Chlorine oxide \((\mathrm{ClO}),\) which plays an important role in the depletion of ozone, decays rapidly at room temperature according to the equation: $$ 2 \mathrm{ClO}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{O}_{2}(g) $$ From the following data, determine the reaction order and calculate the rate constant of the reaction. $$ \begin{array}{ll} \text { Time (s) } & {[\mathrm{ClO}](M)} \\ \hline 0.12 \times 10^{-3} & 8.49 \times 10^{-6} \\ 0.96 \times 10^{-3} & 7.10 \times 10^{-6} \\ 2.24 \times 10^{-3} & 5.79 \times 10^{-6} \\ 3.20 \times 10^{-3} & 5.20 \times 10^{-6} \\ 4.00 \times 10^{-3} & 4.77 \times 10^{-6} \end{array} $$

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ \(\begin{array}{ll}\text { fructose } & \text { glucose }\end{array}\) This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \text { Time (min) } & {\left[\mathbf{C}_{12} \mathbf{H}_{22} \mathbf{O}_{11}\right](\boldsymbol{M})} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \end{array} $$

Briefly comment on the effect of a catalyst on each of the following: (a) activation energy, (b) reaction mechanism, (c) enthalpy of reaction, (d) rate of forward reaction, (e) rate of reverse reaction.

When a mixture of methane and bromine is exposed to light, the following reaction occurs slowly: $$ \mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g) $$ Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

Consider the reaction: $$ \mathrm{A} \longrightarrow \mathrm{B} $$ The rate of the reaction is \(1.6 \times 10^{-2} \mathrm{M} / \mathrm{s}\) when the concentration of A is \(0.15 M\). Calculate the rate constant if the reaction is (a) first order in \(\mathrm{A}\) and \((\mathrm{b})\) second order in \(\mathrm{A}\).

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