Home Subjects Biochemistry

Biochemistry

Metabolic pathways, enzymes, proteins

278 Q 3 Topics Take Test
Advertisement
Difficulty: All Easy Medium Hard 161–170 of 278
Topics in Biochemistry
All Proteins & Enzymes 100 Carbohydrates 100 Lipids 78
Q.161 Easy Carbohydrates
Which carbohydrate classification includes both monosaccharides and oligosaccharides?
A Polysaccharides
B Simple carbohydrates
C Complex carbohydrates
D Reducing sugars
Correct Answer:  B. Simple carbohydrates
EXPLANATION

Simple carbohydrates include monosaccharides (glucose, fructose, galactose) and oligosaccharides (2-10 units like sucrose, maltose, lactose). Complex carbohydrates are polysaccharides with >10 units (starch, glycogen, cellulose).

Take Test
Q.162 Hard Carbohydrates
The Warburg effect describes increased glycolysis in cancer cells even in the presence of oxygen. Which enzyme is typically upregulated in cancer cells to support this?
A Phosphofructokinase-2 (PFK-2)
B Pyruvate kinase M2 (PKM2)
C Both A and B
D Pyruvate dehydrogenase
Correct Answer:  C. Both A and B
EXPLANATION

In the Warburg effect, PFK-2 and PKM2 are upregulated. PFK-2 produces fructose-2,6-bisphosphate (a potent PFK-1 activator), while PKM2 (pyruvate kinase isoform) is upregulated in cancer cells. PKM2 also has non-glycolytic functions in cancer metabolism.

Take Test
Q.163 Hard Carbohydrates
In glycolysis, which step is irreversible under physiological conditions and requires a different enzyme during gluconeogenesis?
A Glucose to Glucose-6-phosphate
B Fructose-6-phosphate to Fructose-1,6-bisphosphate
C Phosphoenolpyruvate to Pyruvate
D Glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate
Correct Answer:  B. Fructose-6-phosphate to Fructose-1,6-bisphosphate
EXPLANATION

The PFK reaction is highly exergonic (ΔG°' = -14.2 kJ/mol) and is essentially irreversible. During gluconeogenesis, fructose-1,6-bisphosphatase catalyzes the reverse reaction. This is a major control point in carbohydrate metabolism.

Take Test
Q.164 Hard Carbohydrates
Pasteur effect refers to the inhibition of glycolysis by oxidative phosphorylation. In terms of ATP and citrate, which statement is correct?
A High ATP and acetyl-CoA inhibit phosphofructokinase
B High ATP and citrate inhibit phosphofructokinase
C Low ATP activates phosphofructokinase
D Both B and C are correct
Correct Answer:  D. Both B and C are correct
EXPLANATION

PFK is inhibited by high ATP, citrate (signal of sufficient acetyl-CoA), and low pH. In aerobic conditions, efficient oxidative phosphorylation produces sufficient ATP, reducing need for glycolysis (Pasteur effect). Citrate is an allosteric inhibitor of PFK.

Take Test
Q.165 Hard Carbohydrates
A patient with galactosemia has a deficiency in which enzyme?
A Galactokinase
B Galactose-1-phosphate uridyltransferase
C UDP-galactose-4-epimerase
D Lactase
Correct Answer:  B. Galactose-1-phosphate uridyltransferase
EXPLANATION

Classic galactosemia results from deficiency of galactose-1-phosphate uridyltransferase (GALT), preventing conversion of galactose-1-phosphate to UDP-galactose. Leads to accumulation of galactose-1-phosphate and galactitol (in polyol pathway), causing cataracts, intellectual disability, and liver damage.

Take Test
Q.166 Medium Carbohydrates
In the pentose phosphate pathway, the oxidative phase primarily generates:
A ATP and GTP
B NADPH for reductive biosynthesis
C Pyruvate for TCA cycle
D Lactate for gluconeogenesis
Correct Answer:  B. NADPH for reductive biosynthesis
EXPLANATION

The oxidative phase of PPP generates 2 NADPH per glucose-6-phosphate, essential for fatty acid synthesis, cholesterol synthesis, and antioxidant defense (glutathione reduction). This is particularly important in adipose tissue, liver, and RBCs.

Take Test
Q.167 Medium Carbohydrates
Cellulose differs from starch in that it contains:
A α(1→4) glycosidic bonds instead of β(1→4)
B β(1→4) glycosidic bonds instead of α(1→4)
C More branching through α(1→6) bonds
D Fructose instead of glucose units
Correct Answer:  B. β(1→4) glycosidic bonds instead of α(1→4)
EXPLANATION

Cellulose contains β(1→4) glycosidic bonds linking glucose units, making it indigestible to humans (no cellulase enzyme). Starch contains α(1→4) bonds, which are digestible.

Take Test
Q.168 Medium Carbohydrates
HbA1c levels are used to monitor long-term glycemic control because glucose binds to hemoglobin through:
A Enzymatic glycosylation
B Non-enzymatic glycation (Maillard reaction)
C Disulfide bonding
D Hydrogen bonding
Correct Answer:  B. Non-enzymatic glycation (Maillard reaction)
EXPLANATION

HbA1c is formed through non-enzymatic glycation of hemoglobin by glucose, reflecting average blood glucose over 2-3 months (RBC lifespan). Normal HbA1c <5.7%, diagnostic for diabetes ≥6.5%.

Take Test
Q.169 Medium Carbohydrates
The anomeric carbon in glucose is:
A C1
B C2
C C3
D C6
Correct Answer:  A. C1
EXPLANATION

The anomeric carbon (C1 in glucose) is the carbon that becomes chiral when the open-chain form cyclizes to form the hemiacetal. This is the site of α/β isomerism.

Take Test
Q.170 Medium Carbohydrates
Cori cycle involves recycling of lactate from muscles to liver. Which enzyme is key in converting lactate back to glucose in the liver?
A Lactate dehydrogenase
B Pyruvate carboxylase
C Phosphofructokinase
D Aldolase
Correct Answer:  B. Pyruvate carboxylase
EXPLANATION

Pyruvate carboxylase catalyzes the first committed step of gluconeogenesis, converting pyruvate to oxaloacetate. Lactate is converted to pyruvate by LDH, then to glucose via gluconeogenesis.

Take Test
IGET
iget AI
Online · Ask anything about exams
Hi! 👋 I'm your iget AI assistant.

Ask me anything about exam prep, MCQ solutions, study tips, or strategies! 🎯
UPSC strategy SSC CGL syllabus Improve aptitude NEET Biology tips