Enzymes Questions in English

(D) Enzymes are classified into six major classes based on the type of reaction they catalyze.
Ligases are the class of enzymes that catalyze the joining or covalent bonding of two substrate molecules.
For example,$DNA$ ligase joins two fragments of $DNA$ by forming a phosphodiester bond.
Invertase,Amylase,and Glutamate pyruvate transaminase belong to other classes (Hydrolases and Transferases,respectively).

(D) Statement $(a)$ is correct: The substrate binds to the active site of the enzyme to form an enzyme-substrate complex.
Statement $(b)$ is incorrect: Enzymes from thermophilic organisms are stable and remain functional at high temperatures (often up to $80^{\circ}C-90^{\circ}C$), not denatured at $50^{\circ}C$.
Statement $(c)$ is incorrect: The active site of the enzyme breaks the chemical bonds of the substrate to convert it into the product, not the bonds of the product.
Statement $(d)$ is correct: Prosthetic groups are organic compounds that are tightly bound to the apoenzyme (the protein part) to form a functional holoenzyme.
Therefore, statements $(a)$ and $(d)$ are correct.

(D) Most enzymes are proteinaceous in nature. However,there are certain $RNA$ molecules that possess catalytic activity and are known as ribozymes. Ribonuclease-$P$ is a well-known example of a ribozyme,which is an $RNA$ enzyme that cleaves $RNA$ molecules. Therefore,ribozyme and ribonuclease-$P$ are exceptions to the rule that all enzymes are proteins.

(B) Enzymes that catalyze oxidation-reduction reactions,which involve the transfer of electrons from one molecule (the reductant) to another (the oxidant),are classified as $Oxidoreductases$.
Examples include dehydrogenases,reductases,and cytochrome oxidases,which are essential components of the electron transport chain.

(C) Assertion $(A)$ is correct: Hormones are chemical messengers that regulate physiological processes and can stimulate the production or release of enzymes.
Reason $(R)$ is incorrect: Hormones are not necessarily 'used up' in metabolism in the same way as substrates; they act as signaling molecules. Furthermore,the statement that 'enzymes can act over and over again' is true,but the claim that 'hormones are used up in metabolism' is a generalization that does not accurately describe their catalytic or signaling nature compared to enzymes. Therefore,the reason provided is scientifically inaccurate.

(D) Non-competitive inhibitors bind to an allosteric site (a site other than the active site) on the enzyme,which changes the conformation of the enzyme such that the active site becomes less effective. This binding does not prevent substrate binding but reduces the catalytic rate,thereby decreasing the maximum velocity $(V_{max})$ of the reaction. Therefore,the Assertion is incorrect. The Reason is also incorrect because,in non-competitive inhibition,the inhibitor and substrate bind at different sites,not the same site. Thus,both Assertion and Reason are incorrect.

(C) Assertion $(A)$ is correct: Allosteric enzymes exhibit sigmoidal kinetics rather than the hyperbolic kinetics described by the Michaelis-Menten equation. They do not follow standard Michaelis-Menten behavior because their activity is regulated by the binding of effectors at sites other than the active site.
Reason $(R)$ is incorrect: Enzymes are highly specific and have different optimal $pH$ values depending on their structure and the environment in which they function (e.g.,pepsin works at acidic $pH$,while trypsin works at alkaline $pH$).

(C) Assertion: $\alpha$-amylase in wheat endosperm is known to exist in multiple isoenzymic forms,specifically $16$ isoenzymes,which help in the mobilization of starch during germination.
Reason: In competitive inhibition,the inhibitor resembles the substrate and competes for the active site. This increases the $K_m$ value,but the $V_{max}$ remains unchanged because the inhibition can be overcome by increasing the substrate concentration. Therefore,the statement that $V_{max}$ decreases is incorrect.

(B) In non-competitive inhibition,the inhibitor does not resemble the substrate in structure and does not compete for the active site.
Instead,the inhibitor binds to an allosteric site (a site other than the active site) on the enzyme.
This binding induces a conformational change in the enzyme's structure,rendering the active site non-functional.
Cyanide acts as a non-competitive inhibitor of cytochrome oxidase in the electron transport chain,preventing the final step of cellular respiration.

(A) Malonate,an analogue of succinate,is a strong competitive inhibitor of succinate dehydrogenase.
It binds to the active site of the enzyme,thereby blocking the activity of the citric acid cycle.

(B) Sucrose is a disaccharide composed of glucose and fructose. The enzyme sucrase (also known as invertase) catalyzes the hydrolysis of sucrose into one molecule of glucose and one molecule of fructose.

(B) Lactose is a disaccharide composed of glucose and galactose. The enzyme lactase catalyzes the hydrolysis of lactose into its constituent monosaccharides,glucose and galactose. The reaction is represented as:
Lactose $\xrightarrow{\text{Lactase}}$ Glucose + Galactose.

(B) Lipases,proteases,carbohydrases,and nucleases are enzymes that catalyze the hydrolysis of their respective substrates (lipids,proteins,carbohydrates,and nucleic acids) by adding water molecules.
These enzymes belong to the class of $Hydrolases$ in the enzyme classification system.
$Hydrolases$ are enzymes that catalyze the cleavage of bonds such as $C-O$,$C-N$,$C-C$,etc.,by the addition of water.

(A) Most enzymes are proteins in nature. However,$Ribozymes$ are catalytic $RNA$ molecules that function as enzymes without being composed of proteins. They are involved in processes like $RNA$ splicing and protein synthesis within the ribosome.

(A) The active site of an enzyme is a specific pocket or cleft formed by the tertiary structure of the protein.
It is the region on the surface of the enzyme molecule where the substrate binds to form the enzyme-substrate complex.
Therefore,the active site is present on the enzyme,not on the substrate or the product.

(D) Inorganic catalysts typically function efficiently under conditions of high temperature and high pressure. These conditions are often damaging to biological systems,which is why inorganic catalysts are generally not used in biochemical reactions. Furthermore,compared to enzymes (biological catalysts),inorganic catalysts are less efficient and lack specificity. Therefore,all the given options are correct.

(A) The rate of a chemical reaction is defined as the change in the concentration of a product $(p)$ or a substrate $(s)$ per unit time $(t)$.
Mathematically,the rate of reaction can be expressed as the change in product concentration over time,which is $\frac{\delta p}{\delta t}$.
Alternatively,it can be expressed as the decrease in substrate concentration over time,which is $-\frac{\delta s}{\delta t}$.
Among the given options,$\frac{\delta p}{\delta t}$ represents the rate of formation of the product,which is a standard way to express the reaction rate.

(D) In the absence of any enzyme, the reaction between $CO_2$ and $H_2O$ to form $H_2CO_3$ is very slow, with about $200$ molecules being formed in an hour.
However, in the presence of the enzyme carbonic anhydrase, the reaction is significantly accelerated.
With the enzyme, the rate of reaction increases dramatically, resulting in the formation of about $600,000$ molecules of $H_2CO_3$ per second.
Therefore, the correct rate of product formation in the presence of the enzyme is $600,000$ molecules per second.

(D) In the absence of any enzyme,the reaction between carbon dioxide and water to form carbonic acid is very slow,with about $200$ molecules being formed in an hour.
However,in the presence of the enzyme carbonic anhydrase,the reaction rate increases dramatically.
With the enzyme,the rate of formation of carbonic acid is about $600,000$ molecules per second.
Therefore,the correct rate of formation of products in the presence of the enzyme is $600,000/\text{second}$.

(B) In a biochemical reaction catalyzed by an enzyme:
$1$. The substance that undergoes chemical change is called the $Reactant$ $(P)$.
$2$. The substance formed after the reaction is called the $Product$ $(Q)$.
$3$. The enzyme itself facilitates the reaction but remains chemically unchanged at the end of the process,acting as a catalyst $(R)$.
Therefore,$P$ is $Reactant$,$Q$ is $Product$,and $R$ is $Enzyme$.

(D) In the first graph,the energy level of the substrate (reactant) is higher than the energy level of the product. This indicates that energy is released during the reaction,which characterizes an Exergonic process.
In the second graph,the energy level of the substrate (reactant) is lower than the energy level of the product. This indicates that energy is absorbed during the reaction,which characterizes an Endergonic process.
Therefore,the correct sequence is Exergonic process,Endergonic process.

(D) The catalytic cycle of an enzyme action can be described in the following steps:
$1$. First,the substrate binds to the active site of the enzyme,fitting into the active site $(I)$.
$2$. The binding of the substrate induces the enzyme to alter its shape,fitting more tightly around the substrate,leading to the formation of the transition state structure $(II)$.
$3$. The active site of the enzyme,now in close proximity to the substrate,breaks the chemical bonds of the substrate and the new enzyme-product complex is formed $(IV)$.
$4$. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate $(III)$.
Therefore,the correct sequence is $I \rightarrow II \rightarrow IV \rightarrow III$.

(A) Activation energy $(E_a)$ is defined as the minimum amount of extra energy required by a reactant molecule to get converted into the product.
In a chemical reaction, the reactants must reach a high-energy state known as the transition state (or activated complex) before they can form products.
The activation energy is calculated as the difference between the energy of the transition state and the average energy of the reactants.
Mathematically, $E_a = \text{Energy of transition state} - \text{Energy of reactants}$.

(D) Enzymes are biocatalysts that accelerate the rate of biochemical reactions by lowering the activation energy required for the reaction to proceed.
Option $A$ is correct as enzymes increase the reaction rate.
Option $B$ is correct as the active site is a specific pocket or cleft in the enzyme structure where the substrate binds.
Option $C$ is correct as most enzymes are proteins,which are polymers of amino acids.
Option $D$ is incorrect because enzymes function by decreasing,not increasing,the activation energy level of the reaction.

(C) The mechanism of enzyme action follows these steps:
$1$. The enzyme $(E)$ binds with the substrate $(S)$ to form the enzyme-substrate complex $(ES)$: $E + S \rightarrow ES$ $(IV \rightarrow II)$.
$2$. The enzyme converts the substrate into the product,forming the enzyme-product complex $(EP)$: $ES \rightarrow EP$ $(II \rightarrow III)$.
$3$. The enzyme releases the product and becomes free to bind with another substrate molecule: $EP \rightarrow E + P$ $(III \rightarrow I)$.
Therefore,the correct sequence is $IV \rightarrow II \rightarrow III \rightarrow I$.

(C) The graph represents the relationship between substrate concentration $[S]$ and reaction velocity $V$.
In Michaelis-Menten kinetics,$V_{max}$ is the maximum velocity of the reaction.
$K_m$ (Michaelis constant) is defined as the substrate concentration at which the reaction velocity is half of its maximum value $(V = \frac{1}{2} V_{max})$.
In the standard plot of $V$ versus $[S]$,the point $P$ on the $y$-axis corresponds to half of the maximum velocity,which is $\frac{1}{2} V_{max}$.

(A) An enzyme is often composed of a protein portion and a non-protein portion.
The protein portion of an enzyme is known as the $Apoenzyme$.
The non-protein portion is referred to as the $Cofactor$.
When the $Apoenzyme$ and $Cofactor$ combine,they form the active enzyme known as the $Holoenzyme$.

(C) Digestive enzymes catalyze the breakdown of complex food molecules into simpler forms by adding water molecules to the chemical bonds. This process is known as hydrolysis. Therefore,digestive enzymes belong to the class of enzymes called Hydrolases.

(D) The enzymes that catalyze the linking together of two compounds are known as ligases. These enzymes are also commonly referred to as synthetases because they facilitate the synthesis of new molecules by joining two substrates together,typically coupled with the hydrolysis of $ATP$ or a similar triphosphate.

(A) Enzymes are classified into six main classes based on the type of reaction they catalyze.
$Lactate \text{ } Dehydrogenase$ catalyzes the oxidation-reduction reaction where $Lactate$ is converted to $Pyruvate$ by removing hydrogen atoms.
According to the International Union of Biochemistry and Molecular Biology $(IUBMB)$, enzymes that catalyze oxidation-reduction reactions are classified as $Oxidoreductases$ (or $Dehydrogenases$).
Therefore, $Lactate \text{ } Dehydrogenase$ belongs to the class of $Dehydrogenases$ (a subclass of $Oxidoreductases$).

(C) Enzymes are classified into six main classes based on the type of reaction they catalyze.
$1$. Hydrolases are enzymes that catalyze the hydrolysis (breakdown) of bonds such as $C-O, C-N, C-C$,etc.,by the addition of water.
$2$. Dehydrogenases catalyze oxidation-reduction reactions.
$3$. Transferases catalyze the transfer of a group between two substrates.
$4$. Ligases catalyze the joining of two compounds.
Therefore,the correct class for enzymes that break down compounds in the presence of water is Hydrolases.

(C) Digestion involves the breakdown of complex macromolecules (like proteins,carbohydrates,and fats) into simpler absorbable forms by the addition of water.
This reaction is catalyzed by enzymes that break chemical bonds using water,which are classified as Hydrolases.
Therefore,digestive enzymes such as amylases,proteases,and lipases belong to the class of Hydrolases.

(D) Most enzymes are proteins in nature. However,certain $RNA$ molecules,known as ribozymes,possess catalytic activity and act as enzymes. For example,$23S$ $rRNA$ in bacteria acts as a ribozyme during peptide bond formation. Therefore,both proteins and $RNA$ can act as enzymes.

(B) The reaction $2H_2O_2 \rightarrow 2H_2O + O_2$ is catalyzed by the enzyme catalase.
Catalase is a conjugated protein that requires a prosthetic group to function.
The prosthetic group present in catalase is $Haem$ (an iron-containing porphyrin ring).
Therefore,the correct option is $B$.

(B) Option $B$ is the correct answer because malonate is a competitive inhibitor of the enzyme succinate dehydrogenase.
Inhibition of succinate dehydrogenase by malonate occurs due to the close structural resemblance between malonate and the substrate succinate.
Competitive inhibitors are often used in the control of bacterial pathogens.

(B) The correct answer is option $B$ because malonate shows close structural similarity with the substrate (succinate) and it competes with the substrate for the active site of the enzyme succinic dehydrogenase.
This type of inhibition,where the inhibitor resembles the substrate in molecular structure and binds to the active site,is known as competitive inhibition.
Options $A$,$C$,and $D$ are incorrect because they do not describe the mechanism where an inhibitor competes with the substrate due to structural similarity.

(B) The correct matching is as follows:

$A.$ Lipase	$II.$ Ester bond (Digests lipids)
$B.$ Nuclease	$IV.$ Phosphodiester bond (Digests nucleic acids)
$C.$ Protease	$I.$ Peptide bond (Digests proteins)
$D.$ Amylase	$III.$ Glycosidic bond (Digests carbohydrates/starch)

Therefore,the correct sequence is $A-II, B-IV, C-I, D-III$,which corresponds to option $B$.

(D) The correct sequence of the catalytic cycle of an enzyme is as follows:
$(1)$ First,the substrate binds to the active site of the enzyme $(E)$.
$(2)$ This leads to the formation of the enzyme-substrate complex $(A)$.
$(3)$ The active site of the enzyme,now in close proximity to the substrate,breaks the chemical bonds of the substrate,forming the enzyme-product complex $(D)$.
$(4)$ The enzyme then releases the products of the reaction $(C)$.
$(5)$ Finally,the free enzyme is ready to bind to another molecule of the substrate to repeat the catalytic cycle $(B)$.
Thus,the correct order is $E, A, D, C, B$.

(C) The reaction $S-G + S^* \rightarrow S + S^*-G$ involves the transfer of a functional group $G$ from one substrate $S$ to another substrate $S^*$.
Enzymes that catalyze the transfer of a group other than hydrogen between a pair of substrates are classified as Transferases.
Therefore,the correct class of enzyme is Transferase.

(C) An enzyme is often composed of a protein part and a non-protein part.
The protein portion of an enzyme is known as the $Apoenzyme$.
The non-protein part is called the $Cofactor$.
When the $Apoenzyme$ and $Cofactor$ combine,they form the active enzyme known as the $Holoenzyme$.
Therefore,the correct answer is $Apoenzyme$.

(D) Enzymes are classified into six major classes based on the type of reaction they catalyze.
Ligases are the enzymes that catalyze the linking together of two compounds,such as $C-O$,$C-S$,$C-N$,or $P-O$ bonds.
Therefore,the enzyme that catalyzes the joining of $C-S$ bonds is a Ligase.

(D) In the given graph representing the progress of an enzymatic reaction:
$1$. The $y$-axis represents the potential energy of the molecules. Thus,$a$ corresponds to $(i)$ Potential energy.
$2$. The peak of the curve represents the transition state of the reaction. Thus,$b$ corresponds to (ii) Transition state.
$3$. The activation energy is the energy required to reach the transition state from the substrate level. The curve with the higher peak represents the reaction without an enzyme,and the curve with the lower peak represents the reaction with an enzyme.
$4$. Therefore,$c$ represents the activation energy without enzyme (iv),and $d$ represents the activation energy with enzyme (iii).
$5$. Matching these: $a-(i), b-(ii), c-(iv), d-(iii)$. This corresponds to option $D$.

(B) Enzymes are classified into six major classes based on the type of reaction they catalyse.
$1$. Oxidoreductases: Catalyse oxidation-reduction reactions.
$2$. Transferases: Catalyse the transfer of a group between substrates.
$3$. Hydrolases: Catalyse the hydrolysis of ester,ether,peptide,or glycosidic bonds.
$4$. Lyases: Catalyse the removal of groups from substrates by mechanisms other than hydrolysis,leaving double bonds.
$5$. Isomerases: Catalyse the interconversion of optical,geometric,or positional isomers.
$6$. Ligases: Catalyse the linking together of two compounds,such as joining $C-O, C-S, C-N, P-O$ etc. bonds.
Therefore,the correct class is Ligases.

(A) The Michaelis-Menten equation describes the rate of enzymatic reactions by relating the reaction rate $V$ to the substrate concentration $[S]$.
The equation is given by: $V = \frac{V_{\text{max}}[S]}{K_m + [S]}$.
In this equation,$K_m$ (the Michaelis constant) is defined as the substrate concentration at which the reaction rate is exactly half of the maximum velocity $(V_{\text{max}})$.
Looking at the provided graph,the x-axis represents the substrate concentration $[S]$.
The point $X$ on the x-axis corresponds to the velocity $\frac{V_{\text{max}}}{2}$ on the y-axis.
Therefore,$X$ represents the Michaelis constant $(K_m)$,which is the substrate concentration at which the reaction rate is $\frac{1}{2} V_{\text{max}}$.

(C) The Enzyme Commission $(EC)$ classifies enzymes into six major classes based on the type of reaction they catalyze.
The first digit of the $EC$ number represents the class:
$1.$ Oxidoreductases
$2.$ Transferases
$3.$ Hydrolases
$4.$ Lyases
$5.$ Isomerases
$6.$ Ligases
In the given $EC$ number $1.2.1.4$,the first digit is $1$,which corresponds to the class Oxidoreductases.
Dehydrogenases are a sub-type of Oxidoreductases that catalyze the removal of hydrogen atoms from a substrate.
Therefore,the enzyme belongs to the class of Dehydrogenases.

(D) $1$. Transferases shift functional groups from one molecule to another.
$2$. Ligases join two large molecules using the energy obtained from the hydrolysis of an energy-rich compound like $ATP$ or $GTP$.
$3$. Hydrolases use water to break bonds.
$4$. Lyases are enzymes that catalyse the removal of groups of atoms from substrates by mechanisms other than hydrolysis,leaving double bonds.

(C) $i$. Most enzymes are proteinaceous in nature,except for ribozymes ($RNA$ enzymes).
$ii$. Enzymes are generally denatured at high temperatures like $60^{\circ}C$ to $70^{\circ}C$,leading to a loss of activity.
$iii$. Enzymes possess specific active sites where the substrate binds to form the enzyme-substrate complex.
$iv$. Enzymes function optimally at a specific $pH$. Any significant increase or decrease in $pH$ from this optimum value leads to a decrease in enzyme activity,not an enhancement.
$v$. Enzymes are not consumed in the reaction; they remain active and can catalyze subsequent reactions after the first one is completed.
Therefore,statements $i, iii,$ and $v$ are correct.

(C) The correct matching is as follows:
$1$. Purely proteinaceous enzymes are made entirely of amino acids. Protease is an example of such an enzyme $(i-c)$.
$2$. Conjugated enzymes (holoenzymes) consist of a protein part (apoenzyme) and a non-protein part (cofactor/coenzyme). $FMN$ (Flavin mononucleotide) acts as a coenzyme in such systems $(ii-d)$.
$3$. Transferase enzymes catalyze the transfer of a functional group from one molecule to another. Glucokinase is a transferase that transfers a phosphate group from $ATP$ to glucose $(iii-a)$.
$4$. Isomerase enzymes catalyze structural changes within a molecule. Epimerase is an isomerase that catalyzes the interconversion of epimers $(iv-b)$.
Therefore,the correct sequence is $i-c, ii-d, iii-a, iv-b$.

(C) Enzymes are classified based on their site of action into two types: endoenzymes and exoenzymes.
$1$. Endoenzymes: These are enzymes that function within the cell in which they are synthesized. They catalyze intracellular metabolic reactions.
$2$. Exoenzymes: These are enzymes that are secreted by the cell and function outside the cell in which they are synthesized (e.g.,digestive enzymes).
Therefore,the correct term for enzymes acting within the cell is endoenzymes.

(C) An increase in substrate concentration generally leads to an increase in the velocity of enzyme activity until a certain point,called the saturation point,where all enzyme molecules are occupied. This is known as the saturation kinetics of enzyme activity. Therefore,the statement that an increase in substrate concentration decreases the velocity of enzyme activity is incorrect.