Enzymology Revision (Multiple Choice Questions With Complete Explanations)

Q.1: Glycogen phosphorylase, which mobilizes glycogen for energy, requires which of the following as a cofactor?

1. Adenosyl Cobalamine
2. Coenzyme A
3. Pyridoxal phosphate
4. Tetrahydrofolate
5. Thiamine pyrophosphate

Correct Answer: C) Pyridoxal phosphate

Explanation: Glycogen phosphorylase is instrumental in the process of glycogenolysis, where it facilitates the breakdown of glycogen into glucose-1-phosphate, thereby playing a vital role in energy mobilization during physical activity or between meals. This enzyme’s activity hinges on the presence of pyridoxal phosphate, a vitamin B6 derivative, serving as its cofactor. Pyridoxal phosphate’s role is crucial in stabilizing the transition state of the enzyme-substrate complex, enhancing the enzyme’s efficiency.

The other options, such as Adenosyl Cobalamine, Coenzyme A, Tetrahydrofolate, and Thiamine pyrophosphate, are involved in various other biochemical pathways but do not act as cofactors for glycogen phosphorylase. Thiamine pyrophosphate, for instance, is primarily involved in decarboxylation reactions within the energy metabolism but does not participate in glycogenolysis.

Q.2: Which one of the following statements is not true about an active site?

1. The active site is a three-dimensional cleft
2. The active site takes up a large part of the total volume of an enzyme
3. Substrates are bound to enzymes by multiple weak attractions
4. The specificity of binding depends on the precisely defined arrangement of atoms in an active site
5. Active sites are adaptable that do change shape during the reaction process

Correct Answer: B) The active site takes up a large part of the total volume of an enzyme

Explanation: The active site of an enzyme is a highly specialized and small area designed for substrate binding and catalysis, contradicting statement B, which inaccurately suggests it occupies a significant portion of the enzyme’s total volume. Instead, the active site forms a minor part of the enzyme’s structure, where substrates bind through various weak interactions such as hydrogen bonds, hydrophobic interactions, and van der Waals forces. This specificity and efficiency in substrate binding are due to the precise arrangement of atoms within the active site, ensuring high specificity for the enzyme’s substrate(s). Furthermore, active sites are not rigid structures; they can undergo conformational changes to facilitate substrate binding—a concept known as induced fit. This flexibility allows enzymes to adapt their shape to bind substrates more effectively, enhancing catalytic efficiency. Therefore, statement B is incorrect as it misrepresents the size of the active site relative to the enzyme.

Q.3: Any of the following processes except one are involved at the active site of an enzyme to accelerate the rate of reaction?

1. Acid-base catalysis
2. Catalysis by Bond Strain
3. Catalysis by Proximity and Orientation
4. Covalent catalysis
5. Thermal denaturation

Correct Answer: E) Thermal denaturation

Explanation: The active site of an enzyme is designed to specifically bind substrates and catalyze reactions at a rate much faster than would occur without the enzyme. Several mechanisms are employed by enzymes to achieve this:

Acid-base catalysis (A): Involves the transfer of protons (H+) to or from the substrate to stabilize the transition state and lower the activation energy of the reaction.

Catalysis by Bond Strain (B): The enzyme binds to the substrate and induces strain or distortion in the substrate’s bonds, making it more reactive towards the catalytic process.

Catalysis by Proximity and Orientation (C): By binding substrates closely together in the correct orientation, enzymes greatly increase the likelihood of reaction between the substrates, enhancing the rate of the reaction.

Covalent catalysis (D): A transient covalent bond forms between the enzyme and the substrate, lowering the activation energy needed for the reaction and thus speeding up the reaction process.

Thermal denaturation (E) is not a mechanism that accelerates the rate of enzyme-catalyzed reactions. Instead, it refers to the loss of the enzyme’s three-dimensional structure caused by heat (or other factors such as pH changes), leading to a loss of enzyme activity. When enzymes are denatured, their active sites are distorted, preventing substrate binding, and thus stopping the catalytic activity. Therefore, thermal denaturation does not contribute to the acceleration of reaction rates at the active site but is a process that inhibits or destroys enzyme activity.

Q.4: Which of the following terms refers to a set of enzymes that can act upon the same substrate(s) and produce the same product(s) but are different in their amino acid sequence?

1. Allosteric enzymes
2. Group-specific enzymes
3. Isoenzymes
4. Substrate specific enzymes
5. Zymogens

Correct Answer: C) Isoenzymes

Explanation: Isoenzymes, or isozymes, are variants of enzymes that catalyze the same chemical reaction but differ in their amino acid composition and often in their regulatory properties. These variants allow for the fine-tuning of metabolism in different tissues or under different physiological conditions. For example, isoenzymes can have variations in their kinetic properties or in their response to regulators, enabling an organism to adapt its metabolic processes to meet diverse demands.

Allosteric enzymes (A) are characterized by their ability to change their catalytic activity in response to the binding of regulatory molecules at sites other than the active site. This mechanism allows them to play key roles in the regulation of metabolic pathways.

Group-specific enzymes (B) are those that act on molecules that contain specific functional groups, but they don’t necessarily imply the existence of multiple forms of the enzyme with the same function.

Substrate-specific enzymes (D) are highly specific for their substrates, but this term does not denote the presence of multiple enzyme forms with the same function across different tissues or conditions.

Zymogens (E) are inactive enzyme precursors that require a biochemical modification, such as cleavage of a peptide segment, to become active. This is a common regulatory mechanism for enzymes, particularly those involved in digestion and blood coagulation.

Isoenzymes facilitate the nuanced regulation of biochemical pathways, allowing the same reaction to be efficiently carried out in different parts of the body or under different physiological conditions, making option C the correct answer.

Q.5: A recently diagnosed hypertensive patient has been prescribed an ACE inhibitor (Angiotensin Converting Enzyme inhibitor), which is known to act by lowering Vmax. What is the possible mechanism of inhibition of this drug?

A. Allosteric inhibition
B. Competitive
C. Non-competitive
D. Uncompetitive
E. Reversible inhibition

Correct Answer: C) Non-competitive

Explanation: ACE inhibitors are used primarily in the treatment of hypertension and certain types of heart failure, acting by inhibiting the angiotensin-converting enzyme (ACE). This inhibition prevents the conversion of angiotensin I to angiotensin II, a potent vasoconstrictor, leading to lower blood pressure. The specific action of lowering the maximal velocity (Vmax) of the enzyme-catalyzed reaction without altering the enzyme’s affinity for its substrate (as indicated by an unchanged Km) is indicative of non-competitive inhibition.

In non-competitive inhibition:

• The inhibitor binds to an enzyme at a site other than the active site. This can occur whether or not a substrate is already bound to the enzyme.
• This mode of inhibition alters the enzyme’s activity by changing its shape or functional capacity, resulting in a decrease in the enzyme’s maximum rate (Vmax) of catalysis.

For the other options:

Competitive inhibition (B) involves the inhibitor competing with the substrate for binding to the enzyme’s active site, affecting the apparent Km but not the Vmax.

Uncompetitive inhibition (D) occurs when the inhibitor binds only to the enzyme-substrate complex, affecting both Vmax and Km.

Allosteric inhibition (A) and Reversible inhibition (E) describe broader categories of enzyme inhibition. Allosteric inhibitors bind to sites other than the active site and can influence enzyme activity in various ways, not necessarily by lowering Vmax alone. Reversible inhibition encompasses any inhibition that can be reversed, either by removing the inhibitor or by other means and includes competitive, non-competitive, and uncompetitive mechanisms.

Thus, non-competitive inhibition (C) best describes the action of ACE inhibitors in the context provided, as it directly relates to the reduction in enzyme activity (lowering Vmax) independent of substrate concentration, aligning with the clinical goal of reducing angiotensin II production to manage hypertension.

Q.6: Which statement out of the following is incorrect about the effect of altering temperature on enzyme activity?

A. A ten-degree Centigrade rise in temperature will increase the activity of most enzymes by 50 to 100%.
B. Most animal enzymes rapidly become denatured at temperatures above 40°C.
C. Raising the temperature increases the kinetic energy of molecules.
D. Storage of enzymes at five °C or below is generally not suitable.
E. Enzyme activity peaks and then declines as temperature increases beyond a certain point.

Correct Answer: D) Storage of enzymes at five °C or below is generally not suitable.

Explanation: Temperature changes can significantly impact enzyme activity, primarily through effects on molecular kinetic energy and enzyme stability.

Raising the temperature increases the kinetic energy of molecules (C), which typically enhances the rate of enzyme-catalyzed reactions up to a point. This increase in kinetic energy facilitates more frequent and effective collisions between enzyme and substrate.

A ten-degree Centigrade rise in temperature will increase the activity of most enzymes by 50 to 100% (A), is a general statement that holds true up to an enzyme’s optimal temperature. Beyond this optimal temperature, enzyme activity begins to decline due to denaturation.

Most animal enzymes rapidly become denatured at temperatures above 40°C (B), reflecting the thermal sensitivity of these proteins. Enzyme denaturation involves the unfolding and loss of the three-dimensional structure essential for catalytic activity, rendering the enzyme inactive.

Storage of enzymes at five °C or below is generally not suitable (D), is the incorrect statement given the context of enzyme preservation. In reality, storing enzymes at low temperatures, including five °C or below (often in a freezer), is a common practice to maintain their stability and prolong their shelf life by slowing down denaturation processes and the activity of any contaminating proteases.

Enzyme activity peaks and then declines as temperature increases beyond a certain point (E) correctly describes the general behavior of enzyme activity in relation to temperature, where each enzyme has an optimal temperature range for activity, and excessive heat can lead to denaturation.

Thus, the incorrect statement among the options given about the effects of temperature on enzyme activity is related to the unsuitability of storing enzymes at five °C or below. In practice, low-temperature storage is often preferred to preserve enzyme activity over time.

Q.7: A 54-year-old male was rushed to the emergency when he collapsed in the middle of a business meeting. Examination revealed excessive sweating and high blood pressure. ECG chest was conclusive of Acute Myocardial Infarction. Which biochemical investigation out of the following would be of no help in confirming the diagnosis?

1. Cardiac Troponins
2. Creatine Phosphokinase-MB (CPK-MB)
3. Lactate Dehydrogenase
4. Serum Myoglobin
5. Serum Electrolytes

Correct Answer: C) Lactate Dehydrogenase

Explanation: In the context of Acute Myocardial Infarction (AMI), certain biomarkers are crucial for early diagnosis and management:

Cardiac Troponins (A) are highly specific and sensitive to myocardial injury. They become elevated hours after the onset of AMI and remain elevated for days, making them crucial for both early and late diagnosis.

Creatine Phosphokinase-MB (CPK-MB) (B) is specific to cardiac tissue and rises within a few hours after myocardial damage, peaking around 24 hours, making it valuable for early diagnosis.

Serum Myoglobin (D) is one of the earliest markers to rise following myocardial injury within 1-4 hours and falls rapidly. It is useful for early detection but is less specific to cardiac tissue.

Lactate Dehydrogenase (C), while it does increase in cases of myocardial infarction, is a late marker, peaking 2-3 days after the onset of symptoms. Its lack of specificity and delayed elevation make it less useful for the immediate confirmation of an AMI diagnosis, particularly in the acute setting when early intervention is critical.

Serum Electrolytes (E), although essential in managing patients, do not directly confirm myocardial infarction and are more related to overall metabolic status and complications arising from AMI.

Given the emphasis on immediate confirmation of AMI for prompt management, Lactate Dehydrogenase is considered of less immediate help in the acute phase of diagnosis due to its late rise and lower specificity compared to other markers listed. It’s more useful for indicating the extent of myocardial injury rather than for early diagnosis.

Q.8: A 42-year-old obese female presented to the emergency center with complaints of worsening nausea, vomiting, and abdominal pain. Her pain was located in the mid-epigastric area and right upper quadrant. Blood biochemistry revealed a high serum amylase level. What is the probable diagnosis for this patient?

A. Acute Gastritis
B. Acute Pancreatitis
C. Renal Colic
D. Viral Hepatitis
E. Peptic Ulcer Disease

Correct Answer: B) Acute Pancreatitis

Explanation: The clinical presentation of worsening nausea, vomiting, and specifically localized abdominal pain in the mid-epigastric and right upper quadrant area, coupled with elevated serum amylase levels, strongly points toward Acute Pancreatitis. This condition is marked by the inflammation of the pancreas, leading to the leakage of pancreatic enzymes like amylase into the bloodstream, thereby elevating serum amylase levels. This enzymatic elevation is a key diagnostic marker for pancreatitis.

Acute Gastritis (A) and Peptic Ulcer Disease (E) can cause abdominal pain and nausea but typically do not lead to significantly elevated serum amylase levels, making them less likely in this scenario.

Renal Colic (C) is associated with kidney stones and presents with a distinct pattern of severe pain, usually radiating from the back down to the groin, rather than primarily elevating amylase levels.

Viral Hepatitis (D) tends to present with symptoms such as jaundice, fatigue, and liver enzyme elevation rather than elevated amylase levels.

Given the symptoms described and the specific laboratory finding of elevated serum amylase, Acute Pancreatitis emerges as the most probable diagnosis, highlighting the importance of considering the entirety of the clinical picture along with laboratory results for accurate diagnosis.

Q.9: A 60-year-old chronic alcoholic was brought to the hospital with complaints of a protuberant abdomen (ascites) and edema feet. He also had a history of hemorrhages. Blood biochemistry revealed – High serum transaminases, low Serum total proteins, Albumin, and a prolonged prothrombin time. Urine analysis was normal. What could be the possible diagnosis?

1. Cirrhosis of the liver
2. Heart Failure
3. Protein Malnutrition
4. Renal Failure
5. Chronic Pancreatitis

Correct Answer: A) Cirrhosis of the liver

Explanation: The clinical presentation of ascites, peripheral edema, history of hemorrhages, and specific laboratory findings such as elevated serum transaminases, hypoalbuminemia (low serum albumin), and prolonged prothrombin time are indicative of Cirrhosis of the Liver. Cirrhosis is a late stage of scarring (fibrosis) of the liver caused by many forms of liver diseases and conditions, such as chronic alcoholism. The high serum transaminases indicate liver injury, low albumin and total proteins reflect the liver’s diminished synthetic function, and prolonged prothrombin time signifies impaired production of coagulation factors due to liver dysfunction.

Heart Failure (B) can cause peripheral edema and ascites but typically does not lead to elevated liver enzymes, hypoalbuminemia, or prolonged prothrombin time as the primary findings.

Protein Malnutrition (C) might cause hypoalbuminemia but does not explain the high serum transaminases or prolonged prothrombin time, which are specific markers of liver dysfunction.

Renal Failure (D) could explain the edema but would likely show abnormalities in the urine analysis and not typically cause elevated liver enzymes or prolonged prothrombin time.

Chronic Pancreatitis (E), often related to chronic alcoholism, might share some clinical features with cirrhosis but would not directly cause hypoalbuminemia, elevated liver enzymes, or prolonged prothrombin time in the absence of concurrent liver disease.

Given these clinical signs and laboratory findings, Cirrhosis of the Liver is the most probable diagnosis, emphasizing the need for a comprehensive evaluation and management plan addressing the underlying liver disease and its complications.

Q.10: A 2-week-old child was brought to the emergency. The parents were fearful that the child had been given some poison as they noted black discoloration on the diaper. A diagnosis of Alkaptonuria was made, and the child was given Vitamin C as a supplement. Alkaptonuria occurs due to reduced activity of the Homogentisic acid oxidase enzyme. What is the role Vitamin C plays in this defect?

A. Antioxidant
B. Coenzyme
C. Inducer
D. Oxidant
E. Positive allosteric modifier

Correct Answer: B) Coenzyme

Explanation: In the treatment of Alkaptonuria, a condition characterized by a deficiency in homogentisic acid oxidase, leading to the accumulation of homogentisic acid, Vitamin C’s role transcends its well-known antioxidant capabilities. It is utilized as a coenzyme, a non-protein chemical that assists in enzyme activity. Specifically, Vitamin C is thought to aid in the functioning of the homogentisic acid oxidase enzyme. This role is crucial because, in Alkaptonuria, the deficient activity of this enzyme leads to the accumulation of homogentisic acid, which can cause damage to various body tissues, including the darkening of the urine when exposed to air. By functioning as a coenzyme, Vitamin C can potentially enhance the residual activity of homogentisic acid oxidase in Alkaptonuria patients, thus helping to mitigate the accumulation of homogentisic acid.

Antioxidant (A) describes Vitamin C’s general function in the body, protecting cells from the damage caused by free radicals. While beneficial, this is not the primary mechanism through which Vitamin C aids in Alkaptonuria.

Inducer (C) and Positive allosteric modifier (E) are terms that describe other mechanisms by which compounds can increase enzyme activity, but in the context of Alkaptonuria and Vitamin C’s role, “Coenzyme” is the accurate description.

Oxidant (D) is incorrect, as Vitamin C acts predominantly as an antioxidant, not promoting oxidation.

Therefore, the correct answer is B) Coenzyme, highlighting Vitamin C’s supportive role in enhancing the activity of homogentisic acid oxidase in the management of Alkaptonuria.

Q.11: A 67-year-old army officer in good health previously presented with sudden pain in the great toe. The serum uric acid level was high, and a diagnosis of gouty arthritis was made. He was advised bed rest, painkillers, and Allopurinol. What is the mechanism of action of Allopurinol in lowering serum uric acid levels?

A. Allosteric inhibition
B. Feedback inhibition
C. Non-competitive inhibition
D. Suicidal inhibition

Correct Answer: D) Suicidal inhibition

Explanation: Allopurinol is a medication commonly used in the treatment of gout, a condition characterized by elevated levels of uric acid in the blood, leading to the formation of urate crystals in joints and surrounding tissues, causing intense pain and inflammation. Allopurinol works by inhibiting xanthine oxidase, an enzyme involved in the purine metabolism pathway that converts hypoxanthine to xanthine and xanthine to uric acid.

The mechanism by which Allopurinol lowers serum uric acid levels is known as suicidal inhibition or mechanism-based inhibition. In this process, Allopurinol acts as a substrate analog that is recognized by xanthine oxidase. However, once bound, it is converted into Alloxanthine (oxipurinol), which remains tightly bound to the enzyme, thereby inactivating it. This inactivation prevents the formation of uric acid from its precursors, effectively lowering the levels of uric acid in the serum.

Allosteric inhibition (A), Feedback inhibition (B), and Non-competitive inhibition (C) describe different mechanisms by which compounds can inhibit enzyme activity. However, these do not accurately describe how Allopurinol works. Allopurinol’s action is more specifically categorized as suicidal because it permanently inactivates xanthine oxidase by being itself transformed into a metabolite that binds irreversibly to the enzyme.

Thus, D) Suicidal inhibition is the correct answer, reflecting Allopurinol’s unique mechanism of reducing uric acid synthesis by rendering the xanthine oxidase enzyme inactive.

Q.12: Which one out of the following enzymes has absolute specificity for its substrate?

A. Carboxypeptidase
B. Lipoprotein lipase
C. Pancreatic lipase
D. Ribonuclease
E. Urease

Correct Answer: E) Urease

Explanation: Absolute specificity is the characteristic of an enzyme to catalyze only one specific substrate to a particular product. Among the options:

Urease (E) is known for its absolute specificity as it uniquely catalyzes the hydrolysis of urea into carbon dioxide and ammonia, demonstrating a strict preference for urea over any other potential substrate.

Carboxypeptidase (A) works by removing amino acids from the carboxyl end of peptides and proteins, showing specificity towards the end residues of proteins but not absolute specificity for a single substrate.

Pancreatic lipase (C) and Lipoprotein lipase (B) are enzymes involved in lipid metabolism. Pancreatic lipase hydrolyzes dietary triglycerides into monoglycerides and free fatty acids in the small intestine, while lipoprotein lipase breaks down triglycerides present in circulating lipoproteins. Both exhibit specificity for lipid substrates but do not have absolute specificity for a single substrate.

Ribonuclease (D) catalyzes the degradation of RNA into smaller components. It shows specificity for RNA substrates but, like the others, does not demonstrate the absolute specificity seen with urease.

Given these descriptions, E) Urease stands out as the enzyme with absolute specificity for its substrate, urea, making it the correct choice among the given options.

Q.13: Which out of the following is a substrate-specific enzyme?

A. Decarboxylase
B. Hexokinase
C. Lactase
D. Thiokinase
E. Transferases

Correct Answer: C) Lactase

Explanation: Substrate specificity refers to the ability of an enzyme to preferentially catalyze the reaction of a particular substrate among many possibilities. Among the options given:

Lactase (C) exhibits high substrate specificity, as it specifically catalyzes the hydrolysis of lactose into glucose and galactose. This enzyme is tailored to act on lactose, demonstrating a classic example of substrate specificity.

Hexokinase (B) can phosphorylate several six-carbon sugars, including glucose, fructose, and others. Though it has a preference for glucose, it is not as substrate-specific as lactase because it can act on a variety of hexoses.

Thiokinase (D), also known as acyl-CoA synthetase, activates fatty acids by attaching them to Coenzyme A, preparing them for degradation or synthesis pathways. It acts on a range of fatty acids, thus showing specificity but not absolute substrate specificity.

Decarboxylase (A) refers to a group of enzymes that catalyze the removal of carboxyl groups from various substrates. Given the variety of decarboxylase enzymes, each acting on different substrates, this category does not exhibit the same level of substrate specificity as lactase.

Transferase(E) is a class of enzymes that catalyze the transfer of functional groups (such as methyl, acyl, glycosyl, or phosphate groups) from one molecule to another. These enzymes play essential roles in various metabolic pathways, including biosynthesis, degradation, and modification of biomolecules.

Therefore, C) Lactase is the correct answer as it specifically targets lactose, making it an exemplar of a substrate-specific enzyme among the options provided.

Q.14: Which out of the following is not a substrate-specific enzyme?

A. Fructokinase
B. Glucokinase
C. Hexokinase
D. Phosphofructokinase
E. Urease

Correct Answer: C) Hexokinase

Explanation: Substrate specificity refers to an enzyme’s ability to catalyze a reaction for a specific substrate or a very narrow range of substrates.

Hexokinase (C) has a broad substrate specificity within its category, as it is capable of phosphorylating several different hexoses (six-carbon sugars), including glucose, fructose, and others. This broad range makes it less substrate-specific than the other listed enzymes, which have more narrowly defined substrate preferences.

Glucokinase (B) is a variant of hexokinase found primarily in the liver and pancreatic β-cells. It has a high specificity for glucose and plays a significant role in regulating carbohydrate metabolism by facilitating glucose phosphorylation when glucose levels are high.

Fructokinase (A) specifically catalyzes the phosphorylation of fructose to fructose-1-phosphate in the liver, demonstrating a high degree of substrate specificity toward fructose.

Phosphofructokinase (D) is an enzyme that specifically catalyzes the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate in glycolysis. It is highly specific for its substrate, fructose-6-phosphate, and plays a crucial role in regulating the rate of glycolysis.

Urease (E) is known for its absolute specificity to urea. It is highly specific to its substrate, contrasting with hexokinase’s broader specificity.

Therefore, among the options, C) Hexokinase is the enzyme that is not substrate-specific, as it acts on a wider range of hexoses, making it less specific compared to the other enzymes listed, which have a narrower substrate range.

Q.15: In which of the following reactions do Group 1 coenzymes participate?

A. Carboxylation
B. Oxidation-reduction
C. Phosphorylation
D. Transamination
E. Hydrolysis

Correct Answer: B) Oxidation-reduction

Explanation: Group 1 coenzymes, which include notable members such as NAD⁺ (Nicotinamide Adenine Dinucleotide), NADP⁺ (Nicotinamide Adenine Dinucleotide Phosphate), FAD (Flavin Adenine Dinucleotide), and FMN (Flavin Mononucleotide), play a pivotal role in oxidation-reduction (redox) reactions across various metabolic processes. These coenzymes act as electron carriers, facilitating the transfer of electrons between molecules, which is fundamental to the cellular processes of energy production, metabolism, and the maintenance of redox balance.

Carboxylation (A), Phosphorylation (C), and Transamination (D) involve different sets of coenzymes. For instance, biotin is involved in carboxylation reactions, ATP in phosphorylation as the phosphate donor and pyridoxal phosphate (vitamin B6) in transamination processes.

Hydrolysis (E) is a reaction that involves the cleavage of bonds in molecules through the addition of water. While hydrolysis is fundamental to metabolism, Group 1 coenzymes are not directly involved in hydrolytic reactions; these processes typically do not require electron transfer facilitated by these coenzymes.

Given the specific roles of these coenzymes in facilitating electron transfer during metabolic reactions, B) Oxidation-reduction accurately reflects the primary function of Group 1 coenzymes in biological systems, underscoring their importance in energy production and metabolic regulation.

Q.16: Which of the following coenzymes takes part in hydrogen transfer reactions in the electron transport chain?

A. Biotin
B. Coenzyme Q
C. Methyl Cobalamin
D. Tetrahydrofolate
E. Vitamin K

Correct Answer: B) Coenzyme Q

Explanation: The electron transport chain (ETC) is a series of complexes that transfer electrons from electron donors to electron acceptors via redox reactions, coupled with the transfer of protons (H⁺ ions) across the mitochondrial membrane. This process is crucial for the production of ATP, the cell’s energy currency. Among the listed coenzymes, Coenzyme Q (B), also known as Ubiquinone, plays a pivotal role in the ETC by participating in hydrogen transfer reactions. It acts as a mobile electron and proton carrier within the mitochondrial inner membrane, shuttling electrons between complex I or II and complex III of the electron transport chain, facilitating the creation of a proton gradient used by ATP synthase to produce ATP.

Biotin (A) and Tetrahydrofolate (D) are involved in various carboxylation and one-carbon transfer reactions, respectively, and are not directly involved in the electron transport chain.

Methyl Cobalamin (C), a form of Vitamin B12, serves as a cofactor in enzymatic reactions involving methyl group transfers but is not a participant in the electron transport chain.

Vitamin K (E) is essential for the post-translational modification of certain proteins involved in blood clotting and bone metabolism but does not participate in the ETC’s hydrogen transfer reactions.

Therefore, B) Coenzyme Q is the correct answer, as it is specifically involved in hydrogen transfer reactions within the electron transport chain, contributing to the generation of ATP in cells.

Q.17: Which of the following participates as a coenzyme in the conversion of pyruvate to oxaloacetate?

A. Biotin
B. FAD
C. NAD+
D. NADPH
E. Thiamine pyrophosphate

Correct Answer: A) Biotin

Explanation: The conversion of pyruvate to oxaloacetate is a crucial anaplerotic reaction that replenishes the citric acid cycle (TCA cycle) intermediates. This reaction is catalyzed by the enzyme pyruvate carboxylase, which requires Biotin (A) as a coenzyme. Biotin acts as a carrier of activated CO₂ (in the form of carboxybiotin) and is essential for the carboxylation of pyruvate to form oxaloacetate. This step is vital for maintaining the balance of TCA cycle intermediates and for gluconeogenesis, the metabolic pathway that results in the generation of glucose from certain non-carbohydrate carbon substrates.

FAD (B) and NAD⁺ (C) are involved in various oxidation-reduction reactions in metabolic pathways. NAD⁺ is particularly important as a coenzyme in dehydrogenase reactions, where it acts as an electron acceptor, but it is not directly involved in the carboxylation of pyruvate to oxaloacetate.

NADPH (D) plays a key role in biosynthetic reactions, providing the reducing power for anabolic processes, and is not involved in the conversion of pyruvate to oxaloacetate.

Thiamine pyrophosphate (E) is a coenzyme in the decarboxylation of alpha-keto acids, including the decarboxylation of pyruvate to acetyl-CoA in the pyruvate dehydrogenase complex, but not in the carboxylation reaction that forms oxaloacetate from pyruvate.

Thus, A) Biotin is the correct answer, highlighting its specific role as a coenzyme in the enzymatic conversion of pyruvate to oxaloacetate.

Q.18: The drug Fluorouracil is recommended for the treatment of cancers. It undergoes a series of changes and then binds to the thymidylate synthase enzyme, resulting in its inhibition and blockage of cell division. What is the most probable mode of inhibition by this drug?

A. Allosteric inhibition
B. Competitive inhibition
C. Non-competitive Inhibition
D. Suicidal inhibition
E. Reversible inhibition

Correct Answer: D) Suicidal inhibition

Explanation: Fluorouracil (5-FU) is utilized in cancer treatment due to its ability to interfere with DNA synthesis, crucial for cell division. After being metabolized, it becomes an active compound that binds to and inhibits thymidylate synthase, an enzyme essential for DNA replication. This inhibition occurs through the formation of a stable, covalently bound complex between the metabolized drug and the enzyme, effectively inactivating thymidylate synthase and preventing the synthesis of thymidine monophosphate (dTMP) from deoxyuridylate (dUMP). This blockage leads to a shortage of thymidine triphosphate (dTTP), necessary for DNA synthesis and repair, thereby hindering cell division in cancer cells.

This action is classified as suicidal inhibition (D) because the inhibitor, once bound, permanently inactivates the enzyme through a covalent bond, making this process irreversible and distinct from other forms of enzyme inhibition such as allosteric, competitive, non-competitive, and reversible inhibition. Suicidal inhibition is specific because the inhibitor is chemically transformed into a reactive species within the active site, leading to the enzyme’s permanent inactivation.

Thus, D) Suicidal inhibition accurately describes Fluorouracil’s mechanism in inhibiting thymidylate synthase, showcasing its effectiveness in cancer treatment by halting the proliferation of cancer cells.

Q.19: Many enzymes, membrane transporters, and other proteins can have their activities rapidly altered through the addition or removal of a phosphate group to specific amino acid residues. This essential regulatory mechanism is known as which of the following?

1. Allosteric Modification
2. Covalent Modification
3. Feedback Inhibition
4. Induction
5. Repression

Correct answer: B) Covalent Modification. This option is correct because the addition or removal of a phosphate group (phosphorylation or dephosphorylation) involves the formation or breaking of covalent bonds. This process can rapidly activate or inactivate enzymes and other proteins, making it a crucial regulatory mechanism in cells. Phosphorylation is one of the most common types of covalent modifications that affects protein function by altering its structure and, thus, its activity.

Explanation for Incorrect Options:

A) Allosteric Modification: This involves the regulation of an enzyme’s activity through the binding of a molecule at a site other than the enzyme’s active site (allosteric site). While this can change the enzyme’s activity, it does not involve the covalent addition or removal of groups from the protein and, thus, is not the process described in the question.

C) Feedback Inhibition: This regulatory mechanism involves the end product of a metabolic pathway inhibiting an enzyme that acts earlier in the pathway to prevent the overproduction of the end product. While it is a form of regulation, it is not related to the covalent modification of proteins through phosphorylation or dephosphorylation.

D) Induction: This term is commonly used in gene expression, where the presence of a substrate or other molecule induces the expression of genes related to the metabolism of that molecule. It’s a regulatory mechanism at the genetic level rather than a post-translational modification of proteins.

E) Repression: Opposite to induction, repression is the process by which the presence of a certain molecule inhibits the expression of certain genes. This mechanism controls the amount of enzyme or protein produced by the cell but does not directly involve the modification of proteins after they are produced.

In summary, the question specifically refers to the regulation of protein activity through the covalent addition or removal of phosphate groups, which is accurately described by “Covalent Modification.” Other options represent different regulatory mechanisms that do not involve this specific type of post-translational modification.

Q.20: A 55-year-old male with symptoms of difficult breathing and swollen ankles is diagnosed with heart failure, leading to pulmonary congestion. He is given a drug that inhibits Angiotensin-Converting Enzyme (ACE). By inhibiting this enzyme, which of the following aspects of the reaction it catalyzes is primarily affected?

1. Energy of Activation
2. Equilibrium Concentration of Substrate
3. Equilibrium Concentration of Product
4. Net Free Energy Change
5. Rate of Reaction

Correct answer: E. Rate of Reaction. ACE inhibitors primarily affect the rate at which the enzyme converts angiotensin I to angiotensin II by blocking the enzyme’s active site. This slows down the reaction without altering the energy of activation, the net free energy change, or the equilibrium concentrations of substrate and product.

Explanation:

A. Energy of Activation: While the enzyme’s activity can influence how quickly a reaction reaches its activation energy, inhibitors like ACE inhibitors mainly function by reducing the enzyme’s activity, thereby affecting the rate of reaction.

B. Equilibrium Concentration of Substrate: Inhibitors do not alter the equilibrium concentrations of substrates or products; they affect how quickly the system reaches equilibrium.

C. Equilibrium Concentration of Product: The equilibrium concentration of products, like that of substrates, is not changed by enzyme inhibitors. These agents impact the kinetics, not the thermodynamics, of the reaction.

B. Net Free Energy Change: This is a thermodynamic property that remains unchanged by the presence of an enzyme or inhibitor. It depends solely on the properties of the reactants and products.

In the context of ACE inhibition, the primary effect is a decrease in the rate at which angiotensin I is converted to angiotensin II, leading to lowered blood pressure and alleviating symptoms associated with heart failure.

Q. 21: During prolonged exercise, muscle tissue actively breaks down glucose via the glycolytic pathway to supply energy, while the liver synthesizes glucose through the gluconeogenic pathway to maintain blood glucose levels. Considering these conditions, which enzyme from the pathways mentioned would most likely follow Michaelis–Menten kinetics, displaying a hyperbolic curve when plotting substrate concentration against reaction velocity?

1. Fructose-1, 6-Bisphosphatase
2. Glucokinase
3. Lactate Dehydrogenase
4. Phosphofructokinase 1
5. Pyruvate Kinase

Correct Answer: C. Lactate Dehydrogenase

Lactate dehydrogenase (LDH) operates under Michaelis–Menten kinetics, displaying a hyperbolic curve when plotting substrate concentration against reaction velocity. This is because LDH, unlike some of the other enzymes listed, is not allosterically regulated and, therefore, adheres more closely to the classical Michaelis–Menten model.

Justifications for Incorrect Answers Given the Allosteric Nature:

A. Fructose-1, 6-Bisphosphatase: Although important in gluconeogenesis, the enzyme is regulated allosterically by a number of small molecules, including AMP and fructose-2,6-phosphate, which are negative regulators, and ATP is a positive regulator. It is less likely to exhibit the simple Michaelis–Menten kinetics.

B. Glucokinase: While glucokinase does exhibit Michaelis–Menten kinetics under certain conditions, it has a high Km for glucose, making it act more efficiently at higher glucose concentrations. It’s not the best example of an enzyme strictly following Michaelis–Menten kinetics under all physiological conditions, especially compared to lactate dehydrogenase.

D. Phosphofructokinase 1 (PFK-1): This enzyme is a classic example of allosteric regulation, displaying a sigmoidal activity curve as it is sensitive to various effector molecules. Its activity is crucially regulated by cellular energy levels, making it less likely to exhibit the simple Michaelis–Menten kinetics.

E. Pyruvate Kinase: Also subject to complex regulation, including allosteric effects, and thus its kinetics may deviate from the simple Michaelis–Menten model. It plays a critical role in the last steps of glycolysis and is regulated in a manner that integrates it into the cell’s overall metabolic state.

Lactate Dehydrogenase is the correct answer if we’re distinguishing based on the absence of allosteric regulation and the presence of a hyperbolic versus sigmoidal response curve to substrate concentration. LDH catalyzes the conversion of pyruvate to lactate (and vice versa), a reaction crucial in many tissues, especially under conditions where oxygen levels are insufficient for aerobic metabolism. Its activity can be directly related to substrate concentration in a manner consistent with Michaelis–Menten kinetics, making it the most suitable choice.

Q. 22:Which enzyme catalyzes the relocation of a functional group from one position to another within the same molecule?

1. Hydrolase
2. Isomerase
3. Ligase
4. Mutase
5. Transferase

Correct Answer: D. Mutase. Mutases specifically catalyze the shifting of a functional group from one position to another within the same molecule. This type of reaction is a subset of isomerization, where the structure of a molecule is changed, but its molecular formula remains the same. Mutases are essential for facilitating intramolecular rearrangements, making them crucial in various metabolic pathways.

Explanations for Incorrect Answers:

A. Hydrolase: Hydrolases catalyze the cleavage of bonds within molecules through the addition of water. These enzymes are involved in hydrolysis reactions, not in the relocation of functional groups within the same molecule.

B. Isomerase: Isomerases facilitate the rearrangement of atoms within a molecule, which can include a broad range of reactions, including the relocation of functional groups. While mutases are a specific type of isomerase, the question specifically asks for the enzyme that catalyzes the relocation of functional groups, making “Mutase” the more precise answer.

C. Ligase: Ligases are enzymes that catalyze the joining of two molecules with the concurrent hydrolysis of a diphosphate bond in ATP or a similar triphosphate. They are not involved in the relocation of functional groups within the same molecule.

E. Transferase: Transferases catalyze the transfer of functional groups from one molecule to another. Unlike mutases, they do not specialize in moving functional groups within the same molecule.

In summary, while isomerases, in general, can catalyze the rearrangement of atoms within molecules, mutases are specifically known for their role in relocating functional groups within the same molecule, making them the correct answer to this question.

Q.23: What kind of Km value characterizes an enzyme with low affinity for its substrate?

1. High Km
2. Low Km
3. Moderate Km
4. Variable Km
5. Zero Km

Correct Answer: A) High Km. A high Km value indicates that a higher concentration of substrate is necessary to achieve half of the enzyme’s maximum reaction velocity (Vmax). This reflects a low affinity of the enzyme for its substrate because it requires a greater amount of substrate to effectively catalyze the reaction.

Explanations for Incorrect Answers:

B. Low Km: A low Km value suggests a high affinity between the enzyme and its substrate, as only a small amount of substrate is needed to significantly engage the enzyme in the catalytic process. This is opposite to what is seen in enzymes with low substrate affinity.

C. Moderate Km: This term is relative and does not specifically address the affinity of the enzyme for its substrate. Km values are absolute measures that indicate specific enzyme-substrate interaction characteristics, not moderate or intermediate affinity levels.

D. Variable Km: The Km value is a constant for a given enzyme-substrate pair under specific conditions. It does not vary for that pair unless the conditions change. This option does not accurately describe the relationship between enzyme affinity and Km value.

E. Zero Km: A Km value of zero would imply that the enzyme has an infinite affinity for its substrate, which is not practical in biological systems. Every enzyme-substrate interaction has a definable and finite affinity.

In essence, an enzyme that exhibits low affinity for its substrate is characterized by a high Km value, indicating that larger substrate concentrations are required to approach the enzyme’s maximal catalytic activity.

Q. 24: What effect do competitive inhibitors have on the Vmax and Km of an enzymatic reaction?

1. Decrease Vmax
2. Decrease Km
3. Increase Km
4. Increase Vmax
5. Keep Vmax constant

Correct Answers: C) Increase Km E) Keep Vmax constant

Explanations  for the Correct Answers:

C. Increase Km: Competitive inhibitors increase the Km value of an enzyme-catalyzed reaction. This is because competitive inhibition involves the inhibitor binding to the active site of the enzyme, thereby competing with the substrate for binding. As a result, a higher concentration of substrate is needed to achieve the same reaction velocity that would be observed in the absence of the inhibitor, indicating a lower apparent affinity of the enzyme for the substrate (higher Km).

E. Keep Vmax constant: Competitive inhibitors do not change the maximum velocity (Vmax) of the reaction because the inhibition can be overcome by increasing the concentration of the substrate. With enough substrate, it becomes more likely that the substrate rather than the inhibitor will bind to the enzyme, allowing the enzyme to reach its maximum catalytic rate despite the presence of the inhibitor.

Explanations for Incorrect Answers:

A. Decrease Vmax: This effect is more characteristic of non-competitive inhibitors, which bind to an allosteric site on the enzyme (not the active site) and decrease the enzyme’s activity regardless of the substrate concentration.

B. Decrease Km: Decreasing the Km would imply an increase in the apparent affinity of the enzyme for its substrate, which is not the case with competitive inhibition. Competitive inhibitors effectively reduce the enzyme’s affinity for the substrate, as reflected by an increased Km.

D. Increase Vmax: Competitive inhibitors do not increase the maximum velocity of an enzymatic reaction. The Vmax can remain constant in the presence of a competitive inhibitor but cannot be increased by it.

In summary, competitive inhibitors increase the Km of an enzymatic reaction (indicating a decrease in enzyme affinity for its substrate due to competition) and leave the Vmax constant (since the inhibition effect can be overcome by increasing substrate concentration).

Q. 25: In what way do non-competitive inhibitors affect the Vmax and Km of an enzymatic reaction?

1. Decrease Km
2. Decrease Vmax
3. Increase Km
4. Increase Vmax
5. Keep Km constant

Correct Answers: B) Decrease Vmax E) Keep Km constant

Explanations for the Correct Answers:

A. Decrease Vmax: Non-competitive inhibitors bind to the enzyme at a location other than the active site, which can alter the enzyme’s shape or function in such a way that it reduces the enzyme’s catalytic efficiency. This results in a decreased maximum velocity (Vmax) because the enzyme’s ability to convert substrate to product at high substrate concentrations is compromised, regardless of how much substrate is available.

E. Keep Km constant: The Km value, which indicates the substrate concentration at which the reaction rate is half of Vmax, remains unchanged in the presence of non-competitive inhibitors. This type of inhibition does not affect the enzyme’s affinity for its substrate, as it does not interfere with the substrate’s ability to bind to the active site. Therefore, the Km remains constant because the enzyme’s binding affinity for the substrate is not altered by the inhibitor.

Explanations for Incorrect Answers:

A. Decrease Km: Decreasing the Km would suggest an increase in the enzyme’s affinity for the substrate, which does not occur with non-competitive inhibition. Non-competitive inhibitors affect the enzyme’s catalytic function without changing its affinity for the substrate.

C. Increase Km: Increasing the Km would imply a decrease in the enzyme’s affinity for the substrate. Non-competitive inhibitors do not affect the enzyme’s affinity for the substrate; thus they do not increase the Km value.

D. Increase Vmax: Non-competitive inhibitors cannot increase the Vmax of an enzymatic reaction. By affecting the enzyme’s catalytic ability, they only have the potential to decrease the Vmax or leave it unchanged in cases not applicable here.

In summary, non-competitive inhibitors lead to a decrease in Vmax, reflecting a reduced rate of catalysis, and they do not alter the Km value, indicating that the enzyme’s substrate binding affinity remains unaffected.

Q. 26: What type of control is represented by the process of enzyme induction in the regulation of enzyme activity?

1. Coarse control
2. Direct control
3. Fine control
4. Indirect control
5. No control

Correct Answer: A. Coarse control. Enzyme induction, which involves the increase in enzyme concentration through the synthesis of new enzyme molecules in response to specific stimuli, is considered a form of coarse control. This type of regulation affects the overall capacity of a metabolic pathway by altering the amount of enzymes available to catalyze reactions rather than adjusting the activity of enzymes already present. Coarse control mechanisms operate over longer timescales and are typically less sensitive and less rapid than fine control mechanisms, which modulate the activity of existing enzyme molecules through allosteric regulation or covalent modification.

Explanations for Incorrect Answers:

B. Direct control: While enzyme induction directly influences the level of enzyme expression, the term “direct control” is not specifically used to describe this regulatory mechanism within the context of metabolic regulation. Direct control more commonly refers to mechanisms that directly affect enzyme activity, not enzyme quantity.

C. Fine control: This refers to the regulation of enzyme activity through rapid, reversible modifications like allosteric regulation or covalent modification (e.g., phosphorylation). Fine control allows for quick adjustments to enzyme activity in response to immediate cellular needs, contrasting with the longer-term adjustments seen in enzyme induction.

D. Indirect control: Indirect control can refer to regulatory mechanisms that influence enzyme activity through secondary messengers or other intermediates. While enzyme induction could be considered an indirect means of regulating the overall metabolic flux, it is more accurately described as coarse control because it involves changes in enzyme concentration rather than modulation of activity.

E. No control: This option is incorrect because enzyme induction is a significant form of metabolic regulation, allowing cells to adapt to changes in their environment by synthesizing more or less of specific enzymes as needed.

In summary, the induction of an enzyme is a form of coarse control in metabolic regulation, focusing on the synthesis of enzyme molecules in response to environmental or cellular signals, thereby adjusting the metabolic pathway’s capacity over longer periods.

Q. 27: Arsenate, which binds to the SH group of an enzyme to inhibit its activity, exhibits what mode of inhibition?

1. Competitive
2. Irreversible
3. Non-competitive
4. Reversible
5. Uncompetitive

Correct Answer: C. Non-competitive: Arsenate binding to the SH group of an enzyme exemplifies non-competitive inhibition because it binds to a site other than the enzyme’s active site. This action can alter the enzyme’s shape or its functional state in a way that decreases its activity, regardless of the presence of the substrate. Non-competitive inhibition does not prevent the substrate from binding but instead reduces the overall number of enzyme molecules available for catalysis or the efficiency of those enzymes, thereby decreasing the reaction rate.

Explanations for incorrect options:

A. Competitive: Incorrect because competitive inhibition involves direct competition with the substrate for the active site, which is not the case with arsenate binding to the SH group.

B. Irreversible: Although arsenate binding could lead to long-lasting effects, calling it irreversible might require specific evidence of a permanent covalent bond that cannot be undone under physiological conditions. Non-competitive inhibition, as a category, does not inherently specify whether the binding is reversible or irreversible, focusing instead on the effect on enzyme activity.

D. Reversible: This option is not necessarily incorrect but incomplete without specifying how the inhibitor affects enzyme activity (competitively, non-competitively, etc.). Reversibility pertains to the inhibitor’s ability to dissociate from the enzyme, which can apply to various types of inhibition.

E. Uncompetitive: Incorrect for this context, as uncompetitive inhibitors specifically bind to the enzyme-substrate complex, reducing the activity without affecting the substrate’s ability to bind.

Given this understanding, non-competitive inhibition more accurately describes the mode of action for arsenate inhibiting an enzyme by binding to its SH group, affecting the enzyme’s activity without directly competing for the active site.

Q. 28: Which enzyme is typically found in higher concentrations in plasma under normal health conditions?

1. Alkaline Phosphatase
2. LDH-1
3. LDH-2
4. SGOT (AST)
5. SGPT (ALT)

Correct Answer: C. LDH-2. Lactate dehydrogenase (LDH) exists in several isoenzyme forms, with LDH-1 and LDH-2 being the most common. Under normal health conditions, LDH-2 is typically found in greater concentration in plasma than LDH-1. LDH-2 is predominant in the serum because it is more prevalent in the heart, red blood cells, and kidneys. The relative distribution and concentration of LDH isoenzymes can indicate specific organ damage when their usual balance is disrupted, but in a healthy individual, LDH-2 levels are generally higher in plasma compared to LDH-1.

Explanations for Incorrect Answers:

A. Alkaline Phosphatase: While alkaline phosphatase levels can provide important information about liver and bone health, it is not typically present in higher concentrations than LDH-2 under normal conditions.

B. LDH-1: LDH-1 is found in high concentrations in heart muscle and red blood cells, but under normal conditions, LDH-2 is more abundant in plasma.

D. SGOT (AST) and E. SGPT (ALT): Both SGOT (AST) and SGPT (ALT) are enzymes associated with liver function. While important for diagnosing liver damage, under normal health conditions, their concentrations in plasma do not exceed that of LDH-2.

In summary, under normal health conditions, LDH-2 is the enzyme typically found in greater concentration in plasma compared to the other enzymes listed. This distribution reflects the enzyme’s widespread presence in various tissues and its role in metabolic processes.

Q.29:What do transferases catalyze the transfer of?

1. Electrons
2. Groups other than reducing equivalents
3. Protons
4. Reducing equivalents
5. Water molecules

Correct Answer: B) Groups other than reducing equivalents. Transferases are a class of enzymes that catalyze the transfer of functional groups from one molecule to another. These groups can include methyl, glycosyl, amino, and phosphate groups, among others. The key characteristic of transferases is their ability to facilitate the movement of these functional groups without the transfer of electrons as reducing equivalents.

Explanations for Incorrect Answers:

A. Electrons: The transfer of electrons is typically the function of oxidoreductases, not transferases. Oxidoreductases are involved in oxidation-reduction reactions where electrons are transferred between molecules.

C. Protons: The transfer of protons is also a characteristic of oxidoreductases, specifically in reactions where hydrogen atoms (which include protons) are moved between molecules.

D. Reducing equivalents: Reducing equivalents (such as NADH, NADPH, and FADH₂) are involved in redox reactions, which are catalyzed by oxidoreductases rather than transferases.

E. Water molecules: The movement of water molecules is typically associated with hydrolases, which catalyze the hydrolytic cleavage of chemical bonds, including the addition of water.

In summary, transferases specialize in the transfer of functional groups between molecules, excluding the transfer of reducing equivalents, which distinguishes them from other classes of enzymes, such as oxidoreductases and hydrolases.

Q.30: How much does enzyme activity increase with every ten-degree rise in temperature?

1. 1 fold
2. 2 fold
3. 3 fold
4. 4 fold
5. 5 fold

Enzyme activity typically increases with temperature due to the increased kinetic energy that speeds up molecular interactions. However, this increase continues only up to a certain point, beyond which the enzyme begins to denature and lose activity. The general rule of thumb is that enzyme activity doubles or increases by a factor of 2 for every ten °C increase in temperature up to the enzyme’s optimal temperature.

Correct Answer: B) 2 fold

This rule applies within a certain temperature range that does not exceed the enzyme’s optimal temperature. Beyond this optimal temperature, the enzyme’s structure begins to destabilize, leading to a rapid decrease in activity.

Q.31: When enzyme activity is directly proportional to substrate concentration, how is the reaction rate classified?

1. First order
2. Non-linear
3. Second order
4. Third order
5. Zero-order

Correct Answer: A. First order. In a first-order reaction, the rate of the reaction is directly proportional to the concentration of one reactant. In the context of enzyme kinetics, when the enzyme activity (reaction rate) is directly proportional to substrate concentration, it implies that the rate of the enzyme-catalyzed reaction increases linearly with an increase in substrate concentration, characteristic of first-order kinetics. This behavior is typically observed at low substrate concentrations, where the enzyme is not saturated, and each increase in substrate concentration leads to a proportional increase in the reaction rate.

Explanations for Incorrect Answers:

B. Non-linear: While enzyme reactions can display non-linear behavior, especially at higher substrate concentrations where the enzyme becomes saturated (following Michaelis-Menten kinetics), the direct proportionality of enzyme activity to substrate concentration describes a linear relationship, which is not “non-linear.”

C. Second order: Second-order reactions depend on the concentrations of two reactants or two molecules of the same reactant. In the context given, where enzyme activity is directly proportional to the concentration of the substrate alone, the scenario does not fit the definition of a second-order reaction.

D. Third order: Third-order reactions involve the simultaneous concentration dependence on three reactant molecules. This does not apply to the scenario where enzyme activity is directly proportional to substrate concentration, indicating a first-order reaction.

E. Zero order: In zero-order kinetics, the rate of the reaction is constant and does not depend on the concentration of the substrate. This is observed at high substrate concentrations where the enzyme is saturated and operates at its maximum rate, Vmax, which contrasts with the scenario described.

In summary, when enzyme activity is directly proportional to substrate concentration, indicating that each increase in substrate concentration leads to a proportional increase in reaction rate, the reaction is classified as first-order kinetics.