Carbohydrate Metabolism — MCQs

Carbohydrate Metabolism Indian Medical PG Practice Questions and MCQs

Question 271: Cellulose is not broken down due to beta anomerism at which carbon atom?

A. C1 (Correct Answer)
B. C2
C. C5
D. C6

Explanation: **Explanation:** The correct answer is **A. C1**. **Why C1 is correct:** Cellulose is a linear homopolysaccharide composed of D-glucose units linked by **β(1→4) glycosidic bonds**. In glucose, the **C1 carbon** is the anomeric carbon. In cellulose, the hydroxyl group at C1 is in the **beta (β) configuration** (pointing upwards). Human digestive enzymes, specifically **α-amylase** (found in saliva and pancreatic juice), are stereospecific; they can only hydrolyze **α(1→4)** linkages (found in starch and glycogen). Humans lack the enzyme **cellulase** (β-glucosidase), which is required to break the β(1→4) bond at the C1 anomeric position. Consequently, cellulose remains undigested and serves as dietary fiber. **Why the other options are incorrect:** * **C2:** This carbon carries a hydroxyl group but is not involved in the glycosidic linkage or the determination of anomerism. * **C5:** This carbon is part of the pyranose ring structure and determines the D or L isomerism, but it does not form the glycosidic bond. * **C6:** This is the primary alcohol group ($CH_2OH$) located outside the ring. While it can be modified in other sugars, it is not the site of anomerism or the glycosidic bond in cellulose. **High-Yield Facts for NEET-PG:** * **Dietary Fiber:** Because cellulose cannot be digested, it adds bulk to the stool, promotes peristalsis, and prevents constipation. * **Ruminants:** Animals like cows can digest cellulose because they harbor symbiotic bacteria in their gut that secrete the enzyme cellulase. * **Starch vs. Cellulose:** Starch (amylose) has **α(1→4)** bonds and is digestible; Cellulose has **β(1→4)** bonds and is indigestible. * **Iodine Test:** Cellulose does not give a color with iodine, unlike starch (blue) or glycogen (reddish-brown).

Question 272: Glucose 6-phosphatase deficiency occurs in which condition?

A. Gaucher's disease
B. Von Gierke's disease (Correct Answer)
C. Pompe's disease
D. Hurler's disease

Explanation: **Explanation:** **Von Gierke’s Disease (Glycogen Storage Disease Type I)** is the correct answer. This condition is caused by a deficiency of the enzyme **Glucose 6-phosphatase**, which is responsible for the final step of both glycogenolysis and gluconeogenesis: converting Glucose 6-phosphate into free glucose in the liver and kidneys. Without this enzyme, glucose cannot be released into the bloodstream, leading to severe fasting hypoglycemia, hepatomegaly (due to glycogen accumulation), and lactic acidosis. **Analysis of Incorrect Options:** * **Gaucher's disease:** This is a Lysosomal Storage Disease (Sphingolipidosis) caused by a deficiency of **$\beta$-glucocerebrosidase**. It presents with hepatosplenomegaly and "crinkled paper" cytoplasm cells, but does not involve glucose metabolism. * **Pompe's disease (GSD Type II):** This is caused by a deficiency of **Lysosomal $\alpha$-1,4-glucosidase** (Acid Maltase). It primarily affects cardiac and skeletal muscle, leading to hypertrophic cardiomyopathy. * **Hurler's disease:** This is a Mucopolysaccharidosis (MPS I) caused by a deficiency of **$\alpha$-L-iduronidase**, leading to the accumulation of dermatan and heparan sulfate. It presents with coarse facial features and corneal clouding. **High-Yield Clinical Pearls for NEET-PG:** * **Biochemical Hallmarks of Von Gierke’s:** Hyperuricemia (leading to gout), Hyperlipidemia, and Hyperlactatemia. * **Diagnostic Clue:** Hypoglycemia that does **not** respond to glucagon administration (because the block is at the final step of glucose release). * **Type Ib:** A variant caused by a deficiency in **Glucose 6-phosphate translocase**, often associated with neutropenia and recurrent infections.

Question 273: NADPH+ and H+ is generated in the reaction catalyzed by which enzyme?

A. Lactate dehydrogenase (LDH)
B. Glucose-6-phosphate dehydrogenase (G-6-PD) (Correct Answer)
C. Glyceraldehyde-3-phosphate dehydrogenase (G-3-PD)
D. Alcohol dehydrogenase

Explanation: **Explanation:** The correct answer is **Glucose-6-phosphate dehydrogenase (G-6-PD)**. This enzyme catalyzes the first and rate-limiting step of the **Hexose Monophosphate (HMP) Shunt** (Pentose Phosphate Pathway). In this reaction, Glucose-6-phosphate is oxidized to 6-phosphogluconolactone, and **NADP+ is reduced to NADPH + H+**. This pathway is the primary source of NADPH in the body, which is essential for reductive biosynthesis (e.g., fatty acids, steroids) and maintaining reduced glutathione to protect cells against oxidative stress. **Analysis of Incorrect Options:** * **Lactate dehydrogenase (LDH):** Involved in anaerobic glycolysis; it interconverts pyruvate and lactate using **NADH/NAD+**, not NADPH. * **Glyceraldehyde-3-phosphate dehydrogenase (G-3-PD):** A key enzyme in glycolysis that converts Glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate, producing **NADH**. * **Alcohol dehydrogenase:** Catalyzes the oxidation of ethanol to acetaldehyde in the cytosol, utilizing **NAD+** as a coenzyme to produce **NADH**. **High-Yield Clinical Pearls for NEET-PG:** * **G6PD Deficiency:** The most common enzymopathy worldwide. Since RBCs lack mitochondria, they rely solely on the HMP shunt for NADPH. Deficiency leads to inadequate reduced glutathione, resulting in **hemolysis** triggered by oxidative stress (e.g., Fava beans, Primaquine, or infections). * **Heinz Bodies & Bite Cells:** Classic peripheral smear findings in G6PD deficiency. * **Tissue Distribution:** The HMP shunt is highly active in the liver, lactating mammary glands, adrenal cortex, and RBCs.

Question 274: All of the following processes are increased during fasting except:

A. Lipolysis
B. Ketogenesis
C. Gluconeogenesis
D. Glycogenesis (Correct Answer)

Explanation: **Explanation:** The metabolic state during fasting is governed by a **low Insulin-to-Glucagon ratio**. This hormonal shift aims to maintain blood glucose levels and provide alternative energy sources for the brain and muscles. **Why Glycogenesis is the correct answer:** **Glycogenesis** is the process of synthesizing glycogen from glucose for storage. This occurs during the **fed state** (high insulin) when glucose is abundant. During fasting, the body needs to mobilize glucose, not store it. Therefore, glycogenesis is inhibited, while its opposite—glycogenolysis—is activated. **Analysis of Incorrect Options:** * **Lipolysis (A):** During fasting, hormone-sensitive lipase is activated in adipose tissue to break down triglycerides into free fatty acids and glycerol, providing fuel and substrates for gluconeogenesis. * **Ketogenesis (B):** As fasting prolongs, the liver converts excess Acetyl-CoA (from fatty acid oxidation) into ketone bodies (acetoacetate, β-hydroxybutyrate) to serve as an alternative fuel for the brain. * **Gluconeogenesis (C):** Once hepatic glycogen stores are depleted (usually after 12–18 hours), the liver synthesizes glucose de novo from non-carbohydrate precursors like lactate, glycerol, and amino acids to maintain glycemia. **NEET-PG High-Yield Pearls:** * **Key Regulatory Enzyme:** Glycogen synthase (inhibited during fasting via phosphorylation by Protein Kinase A). * **Timeline:** Glycogenolysis is the primary source of glucose for the first 12–18 hours of fasting; thereafter, gluconeogenesis becomes the dominant pathway. * **Organ Specificity:** The liver performs gluconeogenesis and glycogenolysis to maintain blood glucose, whereas muscle glycogen is used only for local muscular contraction (due to lack of Glucose-6-Phosphatase).

Question 275: Which of the following substrates cannot contribute to net gluconeogenesis in mammalian liver?

A. Alanine
B. Glutamate
C. Palmitate (Correct Answer)
D. Pyruvate

Explanation: **Explanation:** The correct answer is **Palmitate**. In mammals, even-chain fatty acids like palmitate cannot contribute to net gluconeogenesis. **1. Why Palmitate is correct:** Palmitate is a 16-carbon saturated fatty acid. Through beta-oxidation, it is broken down into **Acetyl-CoA**. In humans, the **Pyruvate Dehydrogenase (PDH) complex** reaction (Pyruvate → Acetyl-CoA) is irreversible. There is no metabolic pathway to convert Acetyl-CoA back into Pyruvate or Oxaloacetate for net glucose synthesis. While Acetyl-CoA enters the TCA cycle, its two carbons are lost as $CO_2$ before reaching Oxaloacetate, resulting in zero net gain of glucose. **2. Why the other options are incorrect:** * **Alanine:** The primary glucogenic amino acid. It undergoes transamination to form **Pyruvate**, a direct precursor for gluconeogenesis via the Glucose-Alanine cycle. * **Glutamate:** A glucogenic amino acid that is converted to **$\alpha$-ketoglutarate** (a TCA cycle intermediate), which eventually forms Oxaloacetate to enter the gluconeogenic pathway. * **Pyruvate:** The central substrate for gluconeogenesis. It is carboxylated by **Pyruvate Carboxylase** (the first rate-limiting step) to form Oxaloacetate. **NEET-PG High-Yield Pearls:** * **Exception to the rule:** While even-chain fatty acids are not glucogenic, **Odd-chain fatty acids** are. Their final breakdown product is **Propionyl-CoA**, which enters the TCA cycle as Succinyl-CoA and can contribute to net glucose synthesis. * **Glycerol:** The glycerol backbone of triglycerides *is* glucogenic, entering the pathway at the level of Dihydroxyacetone phosphate (DHAP). * **Leucine and Lysine:** These are the only two amino acids that are **purely ketogenic** and cannot contribute to gluconeogenesis.

Question 276: The pentose phosphate pathway produces which of the following?

A. ATP
B. ADP
C. NADH
D. NADPH (Correct Answer)

Explanation: **Explanation:** The **Pentose Phosphate Pathway (PPP)**, also known as the Hexose Monophosphate (HMP) Shunt, is an alternative pathway for glucose oxidation. Unlike glycolysis, its primary purpose is not energy production but the generation of biosynthetic precursors. **Why NADPH is the correct answer:** The oxidative phase of the PPP is the body's primary source of **NADPH** (Reduced Nicotinamide Adenine Dinucleotide Phosphate). This occurs via the rate-limiting enzyme **Glucose-6-Phosphate Dehydrogenase (G6PD)**. NADPH is essential for: 1. **Reductive Biosynthesis:** Synthesis of fatty acids, cholesterol, and steroid hormones. 2. **Antioxidant Defense:** Maintaining reduced glutathione to protect cells (especially RBCs) against reactive oxygen species (ROS). **Why other options are incorrect:** * **ATP & ADP:** The PPP is unique because it **neither consumes nor produces ATP**. It is an energy-neutral pathway focused on carbon shuffling and redox potential. * **NADH:** While NADH is structurally similar to NADPH, it is primarily used in the **Electron Transport Chain (ETC)** for ATP generation. The PPP specifically produces NADPH, which has a phosphate group that directs it toward anabolic (building) reactions rather than catabolic (breaking down) energy production. **NEET-PG Clinical Pearls:** * **Rate-limiting enzyme:** Glucose-6-Phosphate Dehydrogenase (G6PD). * **Non-oxidative phase:** Produces **Ribose-5-Phosphate**, essential for nucleotide (DNA/RNA) synthesis. * **Clinical Correlation:** G6PD deficiency leads to **hemolytic anemia** because RBCs cannot generate enough NADPH to combat oxidative stress (e.g., after eating fava beans or taking Primaquine), leading to the formation of **Heinz bodies** and **Bite cells**. * **Tissue Distribution:** Highly active in the liver, adrenal cortex, lactating mammary glands, and RBCs.

Question 277: How many net ATPs are generated per glucose molecule upon undergoing the TCA cycle?

A. 10 ATPs
B. 20 ATPs (Correct Answer)
C. 7 ATPs
D. 2 ATPs

Explanation: **Explanation:** The generation of ATP from glucose occurs in stages. To understand why **20 ATPs** are generated specifically during the **TCA cycle**, we must look at the products yielded per molecule of glucose. One molecule of glucose produces **two molecules of Acetyl-CoA** (via the Link Reaction). Each Acetyl-CoA entering the TCA cycle generates: * **3 NADH** (3 × 2.5 = 7.5 ATP) * **1 FADH₂** (1 × 1.5 = 1.5 ATP) * **1 GTP/ATP** (Substrate-level phosphorylation) * **Total per Acetyl-CoA:** 10 ATPs. Since one glucose molecule provides two Acetyl-CoA molecules, the total yield is **10 × 2 = 20 ATPs**. **Analysis of Incorrect Options:** * **Option A (10 ATPs):** This is the yield for a single turn of the TCA cycle (one Acetyl-CoA). The question asks per glucose molecule. * **Option C (7 ATPs):** This represents the net ATP yield from Aerobic Glycolysis alone (using the Malate-Aspartate shuttle). * **Option D (2 ATPs):** This is the net ATP yield from Anaerobic Glycolysis. **High-Yield Clinical Pearls for NEET-PG:** * **Total ATP Yield:** Under aerobic conditions, one glucose molecule yields **30 or 32 ATPs** (depending on the shuttle used). This includes Glycolysis (7 or 9), the Link Reaction (5), and the TCA cycle (20). * **Rate-Limiting Enzyme:** Isocitrate Dehydrogenase is the key regulatory enzyme of the TCA cycle. * **Inhibitor:** Fluoroacetate inhibits Aconitase, while Arsenite inhibits the α-Ketoglutarate Dehydrogenase complex. * **Amphibolic Nature:** The TCA cycle is both catabolic (energy production) and anabolic (providing intermediates for gluconeogenesis and amino acid synthesis).

Question 278: A male patient presents with pain in his calf muscles upon exercise. Muscle biopsy reveals the presence of glycogen. Which enzyme deficiency is most likely responsible for this condition?

A. Branching enzyme
B. Phosphofructokinase I (Correct Answer)
C. Debranching enzyme
D. Glucose 6 phosphatase

Explanation: ### **Explanation** The clinical presentation of calf muscle pain during exercise (exercise intolerance) combined with the accumulation of glycogen on biopsy points toward a **Glycogen Storage Disease (GSD)** affecting the muscles. **1. Why Phosphofructokinase-1 (PFK-1) is correct:** Deficiency of the muscle isoform of PFK-1 leads to **Tarui Disease (GSD Type VII)**. PFK-1 is the rate-limiting enzyme of glycolysis. When it is deficient, glucose-6-phosphate and fructose-6-phosphate accumulate. High levels of glucose-6-phosphate allosterically activate **glycogen synthase**, leading to increased glycogen synthesis and storage in muscle tissues. Because glycolysis is blocked, muscles cannot generate ATP rapidly during exercise, causing pain and cramping. **2. Why the other options are incorrect:** * **Branching enzyme (GSD Type IV/Andersen disease):** Deficiency leads to the accumulation of abnormal glycogen with long outer chains (polyglucosan bodies). It typically presents with infantile liver failure and cirrhosis, not isolated exercise-induced muscle pain. * **Debranching enzyme (GSD Type III/Cori disease):** While it can cause muscle weakness, it primarily presents with hepatomegaly and fasting hypoglycemia. Biopsy would show "limit dextrin" (abnormally short outer branches). * **Glucose-6-phosphatase (GSD Type I/Von Gierke disease):** This enzyme is absent in muscles. Its deficiency affects the liver and kidneys, causing severe fasting hypoglycemia, lactic acidosis, and hyperuricemia, but does not cause direct exercise-induced muscle pain. **3. Clinical Pearls for NEET-PG:** * **Tarui Disease vs. McArdle Disease:** Both present with exercise-induced cramps and myoglobinuria. However, Tarui disease (PFK-1 deficiency) often shows **hemolytic anemia** (due to partial deficiency of PFK in RBCs) and hyperuricemia. * **Ischemic Forearm Exercise Test:** In both McArdle and Tarui diseases, there is a **failure of blood lactate to rise** after exercise. * **High-Yield Fact:** PFK-1 is inhibited by ATP and Citrate; it is activated by Fructose-2,6-bisphosphate and AMP.

Question 279: What is the number of optical isomers possible for Glucose?

A. 2
B. 4
C. 8
D. 16 (Correct Answer)

Explanation: ### Explanation The number of optical isomers (stereoisomers) for a carbohydrate is determined by the number of **asymmetric carbon atoms (chiral centers)** in its structure. **1. Why 16 is the Correct Answer:** The formula to calculate the number of optical isomers is **$2^n$**, where **$n$** is the number of chiral centers. * **Glucose** is an aldohexose ($C_6H_{12}O_6$). * In its open-chain form, carbons 2, 3, 4, and 5 are asymmetric (each is attached to four different groups). * Applying the formula: $2^4 = 2 \times 2 \times 2 \times 2 = \mathbf{16}$. These 16 isomers include 8 L-forms and 8 D-forms (including mannose and galactose, which are epimers of glucose). **2. Why Other Options are Incorrect:** * **A (2):** This represents the number of **anomers** ($\alpha$ and $\beta$) formed when glucose cyclizes, or the pair of **enantiomers** (D and L) for a single specific sugar. * **B (4):** This would be the number of isomers for a sugar with only 2 chiral centers (e.g., Erythrose). * **C (8):** This is the number of isomers for an aldopentose (like Ribose) which has 3 chiral centers ($2^3 = 8$). **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Van’t Hoff’s Rule:** The $2^n$ formula is known as Van’t Hoff’s rule. * **Keto-sugars:** Fructose (a ketohexose) has only **3 chiral centers**, so it has $2^3 = \mathbf{8}$ optical isomers. * **Epimers:** Glucose and Galactose are **C-4 epimers**; Glucose and Mannose are **C-2 epimers**. This is a frequent "match the following" topic. * **Biological Significance:** Only **D-isomers** of sugars are naturally metabolized by the human body (the "D" stands for the position of the -OH group on the penultimate carbon).

Question 280: In glycogen metabolism, which enzyme acts on the 1-6 glycosidic bond of glycogen to produce glucose?

A. Glucose-1-phosphate
B. Glucose-6-phosphate
C. Maltose
D. Glucose (Correct Answer)

Explanation: **Explanation:** In glycogenolysis, the breakdown of glycogen involves two primary enzymes: **Glycogen Phosphorylase** and the **Debranching Enzyme**. The **Debranching Enzyme** is a bifunctional protein. Its second activity, **$\alpha$-1,6-glucosidase**, specifically hydrolyzes the $\alpha$-1,6-glycosidic bond at the branch point. Unlike the action of phosphorylase (which uses inorganic phosphate to release Glucose-1-Phosphate), the $\alpha$-1,6-glucosidase activity uses water (hydrolysis) to release a **free Glucose** molecule directly. For every branch point, one molecule of free glucose is produced, accounting for approximately 8-10% of the total glucose released from glycogen. **Analysis of Options:** * **Option A (Glucose-1-phosphate):** This is the product of Glycogen Phosphorylase acting on $\alpha$-1,4-linkages. It is the major product of glycogenolysis but is not released from the 1-6 bond. * **Option B (Glucose-6-phosphate):** This is formed in the liver by the isomerization of Glucose-1-phosphate via *Phosphoglucomutase*. It is not a direct product of the debranching enzyme. * **Option C (Maltose):** This is a disaccharide produced during the digestion of starch by amylase, not during intracellular glycogen metabolism. **NEET-PG High-Yield Pearls:** * **Debranching Enzyme Deficiency:** Leads to **Cori’s Disease (GSD Type III)**, characterized by the accumulation of "limit dextrins" (abnormal glycogen with short outer branches). * **Ratio:** Glycogenolysis yields Glucose-1-Phosphate and free Glucose in a ratio of approximately **10:1**. * **Key Difference:** Phosphorylase performs *phosphorolysis* (saves ATP), while debranching enzyme performs *hydrolysis*.

Practice by Chapter

Carbohydrate Chemistry and Classification

Practice Questions

Glycolysis: Reactions and Regulation

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Gluconeogenesis: Reactions and Regulation

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Glycogen Metabolism: Synthesis and Breakdown

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Glycogen Storage Diseases

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Pentose Phosphate Pathway

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Metabolism of Fructose and Galactose

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Disorders of Fructose and Galactose Metabolism

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Blood Glucose Regulation

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Diabetes Mellitus: Biochemical Aspects

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Glycosylation and Glycoproteins

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Lactose Intolerance and Galactosemia

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