Quiz: Biochemistry Master Exam Review — 50 questions

1. What distinguishes an enzyme from a substrate during a chemical reaction?

A. The enzyme increases reaction rate without being consumed, while the substrate is converted into product

B. The enzyme binds only to allosteric sites, while the substrate binds randomly

C. The enzyme supplies cellular energy, while the substrate functions as a cofactor

D. The enzyme is converted into product, while the substrate remains unchanged

Answer: (A)

Explanation

An enzyme acts as a biological catalyst and is not consumed, whereas the substrate is acted upon and converted into product.

2. Where does the substrate bind on an enzyme?

A. At a nonspecific surface region

B. At the enzyme’s cofactor-binding region

C. At the enzyme’s active site

D. At the enzyme’s allosteric regulatory site

Answer: (C)

Explanation

The substrate is the substance acted on by an enzyme and binds specifically at its active site. An allosteric regulator binds at a different site.

3. Which change can help overcome competitive inhibition?

A. Increasing the concentration of noncompetitive inhibitor

B. Raising the temperature far above the enzyme’s optimum

C. Increasing the concentration of substrate

D. Removing the enzyme’s cofactor

Answer: (C)

Explanation

A competitive inhibitor occupies the active site, so increasing substrate concentration can help the substrate outcompete the inhibitor. Noncompetitive inhibition occurs at another site and reduces catalytic activity.

4. What is a holoenzyme?

A. An active enzyme consisting of an apoenzyme and its required cofactor or coenzyme

B. An enzyme that has been denatured by excessive temperature

C. An inactive protein lacking its required helper molecule

D. A substrate bound temporarily to an allosteric site

Answer: (A)

Explanation

A holoenzyme is the active form made of an apoenzyme together with its necessary cofactor or coenzyme. The apoenzyme alone is inactive.

5. Which description correctly defines glycolysis?

A. A mitochondrial pathway that converts pyruvate into acetyl-CoA

B. A mitochondrial pathway that converts glucose directly into carbon dioxide

C. A cytosolic pathway of 10 reactions that converts glucose into two pyruvate molecules

D. A cytosolic pathway that converts two pyruvate molecules into one glucose molecule

Answer: (C)

Explanation

Glycolysis occurs in the cytosol and consists of 10 reactions that convert one six-carbon glucose molecule into two three-carbon pyruvate molecules.

6. What is the net yield of glycolysis from one molecule of glucose?

A. 2 ATP, 4 NADH, and 2 acetyl-CoA

B. 4 ATP, 2 NADH, and 1 pyruvate

C. 1 ATP, 1 NADH, and 2 pyruvate

D. 2 ATP, 2 NADH, and 2 pyruvate

Answer: (D)

Explanation

The net products of glycolysis per glucose are 2 ATP, 2 NADH, and 2 pyruvate.

7. A cell has high ATP levels and needs to reduce glycolytic flux. Which enzyme is directly inhibited according to the stated regulation of glycolysis?

A. Lactate dehydrogenase

B. Hexokinase

C. Phosphofructokinase-1

D. Pyruvate dehydrogenase

Answer: (C)

Explanation

PFK-1 is the major regulatory and rate-limiting enzyme of glycolysis, and ATP inhibits it. Hexokinase catalyzes the first step but is not the principal regulatory enzyme described here.

8. Why do red blood cells rely on anaerobic glycolysis and produce lactate from pyruvate?

A. They cannot convert glucose into pyruvate

B. They lack mitochondria needed for oxidative metabolism

C. They contain excessive mitochondria that rapidly consume oxygen

D. They lack cytosolic enzymes needed for glucose breakdown

Answer: (B)

Explanation

Red blood cells lack mitochondria, so they depend on anaerobic glycolysis and convert pyruvate to lactate.

9. What does gluconeogenesis primarily accomplish?

A. It stores glucose as highly branched glycogen

B. It converts acetyl-CoA directly into glucose

C. It breaks glucose into pyruvate for energy

D. It forms glucose from non-carbohydrate precursors

Answer: (D)

Explanation

Gluconeogenesis is the formation of glucose from non-carbohydrate precursors, occurring mainly in the liver and to a lesser extent in the kidney. Glycolysis, in contrast, breaks glucose down.

10. Which set correctly identifies the three gluconeogenic bypasses?

A. Pyruvate to acetyl-CoA, fructose-1,6-bisphosphate to pyruvate, and glucose-6-phosphate to glycogen

B. Pyruvate to PEP, fructose-1,6-bisphosphate to fructose-6-phosphate, and glucose-6-phosphate to glucose

C. PEP to pyruvate, fructose-6-phosphate to fructose-1,6-bisphosphate, and glucose to glucose-6-phosphate

D. Pyruvate to lactate, fructose-6-phosphate to fructose-1,6-bisphosphate, and glucose to glucose-6-phosphate

Answer: (B)

Explanation

Gluconeogenesis bypasses three irreversible glycolytic steps: pyruvate to PEP, fructose-1,6-bisphosphate to fructose-6-phosphate, and glucose-6-phosphate to glucose.

11. Which statement correctly compares pyruvate carboxylase with PEP carboxykinase?

A. Pyruvate carboxylase uses GTP in the cytosol, whereas PEP carboxykinase uses ATP in mitochondria

B. Pyruvate carboxylase uses ATP and biotin in mitochondria, whereas PEP carboxykinase uses GTP

C. Pyruvate carboxylase uses NADH in mitochondria, whereas PEP carboxykinase uses FADH₂ in the cytosol

D. Pyruvate carboxylase uses acetyl-CoA as an energy source, whereas PEP carboxykinase requires biotin

Answer: (B)

Explanation

Pyruvate carboxylase is a mitochondrial, biotin-dependent enzyme that uses ATP and is activated by acetyl-CoA. PEP carboxykinase uses GTP.

12. A rise in fructose-2,6-bisphosphate would be expected to have which effect?

A. It inhibits both glycolysis and gluconeogenesis

B. It stimulates both glycolysis and gluconeogenesis

C. It stimulates gluconeogenesis and opposes glycolysis

D. It stimulates glycolysis and opposes gluconeogenesis

Answer: (D)

Explanation

Increased fructose-2,6-bisphosphate stimulates glycolysis, whereas decreased levels stimulate gluconeogenesis.

13. Which bonding pattern characterizes glycogen structure?

A. β(1→4) bonds in linear chains and α(1→6) bonds at branch points

B. α(1→4) bonds in linear chains and α(1→6) bonds at branch points

C. α(1→2) bonds in linear chains and β(1→4) bonds at branch points

D. α(1→6) bonds in linear chains and α(1→4) bonds at branch points

Answer: (B)

Explanation

Glycogen is a highly branched glucose-storage polymer with α(1→4) linkages in its linear chains and α(1→6) linkages at branch points.

14. Which sequence correctly describes glycogenesis?

A. G6P to UDP-glucose, UDP-glucose to G1P, and G1P to glycogen

B. Glycogen to G1P, G1P to G6P, and G6P to UDP-glucose

C. G6P to G1P, G1P to UDP-glucose, and UDP-glucose to glycogen

D. G6P to pyruvate, pyruvate to acetyl-CoA, and acetyl-CoA to glycogen

Answer: (C)

Explanation

Glycogenesis proceeds from G6P to G1P, then to UDP-glucose, which is incorporated into glycogen. The enzymes involved are phosphoglucomutase, UDP-glucose pyrophosphorylase, and glycogen synthase.

15. Why can liver glycogen help maintain blood glucose, whereas muscle glycogen cannot directly do so?

A. Muscle lacks glucose-6-phosphatase, so its glycogen-derived glucose is used within muscle

B. The liver lacks glucose-6-phosphatase, so it retains glycogen-derived glucose

C. Liver glycogen is made from fatty acids, whereas muscle glycogen is made from amino acids

D. Muscle lacks glycogen synthase, so it cannot store glucose for later use

Answer: (A)

Explanation

The liver can release glucose into the blood, but muscle lacks glucose-6-phosphatase. Therefore, muscle glycogen primarily supplies energy for muscle itself.

16. Where does the TCA cycle occur, and what major metabolic role does it serve?

A. In the cytosol, where it converts glucose directly into glycogen

B. In the nucleus, where it produces glucose from non-carbohydrate precursors

C. In the mitochondrial matrix, where it supports oxidation of carbohydrates, fats, and amino acids

D. In the mitochondrial intermembrane space, where it synthesizes proteins

Answer: (C)

Explanation

The TCA cycle occurs in the mitochondrial matrix and serves as a central pathway for the oxidation of carbohydrate, fat, and amino-acid carbon skeletons.

17. Which statement correctly describes TCA-cycle intermediates and their sequence?

A. The cycle includes citrate, isocitrate, α-ketoglutarate, succinyl-CoA, succinate, fumarate, malate, and regenerated oxaloacetate

B. The cycle includes acetyl-CoA, citrate, glucose-6-phosphate, α-ketoglutarate, succinate, lactate, malate, and regenerated pyruvate

C. The cycle includes citrate, oxaloacetate, glycogen, succinate, fumarate, pyruvate, malate, and regenerated acetyl-CoA

D. The cycle includes glucose, pyruvate, lactate, acetyl-CoA, citrate, malate, fumarate, and regenerated glucose

Answer: (A)

Explanation

The eight TCA intermediates are citrate, isocitrate, α-ketoglutarate, succinyl-CoA, succinate, fumarate, malate, and oxaloacetate, with oxaloacetate regenerated at the end.

18. Which TCA-cycle enzyme is uniquely associated with production of FADH₂?

A. Isocitrate dehydrogenase

B. Malate dehydrogenase

C. α-Ketoglutarate dehydrogenase

D. Succinate dehydrogenase

Answer: (D)

Explanation

Succinate dehydrogenase is the only TCA-cycle enzyme that produces FADH₂. Malate dehydrogenase and several other dehydrogenases produce NADH instead.

19. Which reaction is catalyzed irreversibly by the pyruvate dehydrogenase complex in the mitochondrial matrix?

A. Pyruvate to acetyl-CoA, CO₂, and NADH

B. Acetyl-CoA to pyruvate, CO₂, and NADH

C. Oxaloacetate to pyruvate, CO₂, and NADH

D. Pyruvate to lactate, ATP, and NAD⁺

Answer: (A)

Explanation

The pyruvate dehydrogenase complex irreversibly converts pyruvate into acetyl-CoA, CO₂, and NADH in the mitochondrial matrix. Its irreversibility distinguishes it from many reversible downstream metabolic reactions.

20. Which set contains all five cofactors required by the pyruvate dehydrogenase complex?

A. TPP, biotin, carnitine, FAD, and NAD⁺

B. TPP, lipoic acid, CoA, FAD, and NAD⁺

C. Biotin, tetrahydrofolate, CoA, FAD, and NADP⁺

D. Lipoic acid, pyridoxal phosphate, CoA, FMN, and NADP⁺

Answer: (B)

Explanation

The five PDH cofactors are thiamine pyrophosphate, lipoic acid, CoA, FAD, and NAD⁺. The other choices substitute cofactors used by different enzymes or pathways.

21. A liver cell has abundant ATP, NADH, and acetyl-CoA. How does this condition affect pyruvate dehydrogenase activity?

A. PDH kinase is inhibited, phosphorylating and activating PDH

B. PDH phosphatase is activated, phosphorylating and inhibiting PDH

C. PDH kinase is activated, phosphorylating and inhibiting PDH

D. PDH is dephosphorylated by PDH kinase and becomes inhibited

Answer: (C)

Explanation

High ATP, NADH, and acetyl-CoA activate PDH kinase, which phosphorylates and inhibits the pyruvate dehydrogenase complex when energy is abundant. Dephosphorylation, not phosphorylation, activates PDH.

22. What are the principal products of the pentose phosphate pathway?

A. NADPH and a net yield of ATP

B. NADH and ribose-5-phosphate

C. NADPH and ribose-5-phosphate

D. FADH₂ and acetyl-CoA

Answer: (C)

Explanation

The cytosolic pentose phosphate pathway produces NADPH and ribose-5-phosphate. Unlike glycolysis, its defining products are not a net ATP yield and NADH.

23. How does the oxidative phase of the pentose phosphate pathway differ from its non-oxidative phase?

A. The oxidative phase is irreversible and produces two NADH per glucose-6-phosphate

B. The oxidative phase is irreversible and produces two NADPH per glucose-6-phosphate

C. The oxidative phase is reversible and consumes two NADPH per glucose-6-phosphate

D. The oxidative phase is reversible and produces a net ATP yield per glucose-6-phosphate

Answer: (B)

Explanation

The oxidative phase is irreversible and generates two NADPH molecules per glucose-6-phosphate. The non-oxidative phase is the reversible portion of the pathway.

24. Why can a person with G6PD deficiency develop hemolysis after taking an oxidant drug?

A. Excess NADPH blocks ribose production, causing red blood cells to rupture

B. Excess FADH₂ increases ketone production, causing oxidative damage to red blood cells

C. Reduced NADH prevents ATP production, directly damaging red blood cell membranes

D. Reduced NADPH limits glutathione regeneration, increasing oxidative damage to red blood cells

Answer: (D)

Explanation

G6PD deficiency lowers NADPH production, which reduces regeneration of reduced glutathione and weakens antioxidant protection. Oxidant drugs can therefore trigger oxidative damage and hemolysis, especially in red blood cells.

25. Which sequence correctly describes one cycle of β-oxidation?

A. Thiolysis, hydration, dehydrogenation producing FADH₂, and phosphorylation producing NADH

B. Dehydrogenation producing NADH, dehydration, hydration, and carboxylation producing acetyl-CoA

C. Dehydrogenation producing FADH₂, hydration, oxidation producing NADH, and thiolysis producing acetyl-CoA

D. Hydration, dehydrogenation producing NADH, thiolysis, and oxidation producing FADH₂

Answer: (C)

Explanation

Each β-oxidation cycle proceeds through dehydrogenation with FADH₂ production, hydration, a second oxidation with NADH production, and thiolysis yielding acetyl-CoA.

26. How do long-chain fatty acids enter the mitochondrial matrix, and what regulates this entry?

A. They diffuse through the inner membrane, and acetyl-CoA inhibits CPT-II

B. They use the citrate shuttle, and malonyl-CoA inhibits CPT-II

C. They use the carnitine shuttle, and malonyl-CoA inhibits CPT-I

D. They use the glycerol phosphate shuttle, and malonyl-CoA activates CPT-I

Answer: (C)

Explanation

Long-chain fatty acids require the carnitine shuttle to cross into the mitochondrial matrix. Malonyl-CoA inhibits CPT-I, limiting fatty-acid entry during fatty-acid synthesis.

27. What is the direct product set generated by each cycle of β-oxidation?

A. One acetyl-CoA, one NADH, and two FADH₂

B. Two acetyl-CoA, one NADH, and one FADH₂

C. One acetyl-CoA, two NADH, and one FADH₂

D. One acetyl-CoA, one NADH, and one FADH₂

Answer: (D)

Explanation

Every β-oxidation cycle produces one acetyl-CoA, one NADH, and one FADH₂. These products subsequently contribute to the citric acid cycle and oxidative phosphorylation.

28. Which set correctly lists the three ketone bodies?

A. Acetyl-CoA, citrate, and oxaloacetate

B. Acetoacetate, β-hydroxybutyrate, and acetone

C. Pyruvate, lactate, and acetone

D. Acetoacetate, citrate, and fumarate

Answer: (B)

Explanation

The three ketone bodies are acetoacetate, β-hydroxybutyrate, and acetone.

29. During prolonged fasting, why does the liver increase ketogenesis?

A. Fatty-acid oxidation raises acetyl-CoA while oxaloacetate is diverted to gluconeogenesis

B. Protein digestion increases glucose uptake while acetyl-CoA is converted to pyruvate

C. Glycogen synthesis raises oxaloacetate while fatty-acid oxidation declines

D. Ketone consumption by the liver increases acetyl-CoA production in mitochondria

Answer: (A)

Explanation

During fasting, fatty-acid oxidation produces abundant acetyl-CoA, while oxaloacetate is used for gluconeogenesis, favoring ketone-body formation in the liver.

30. Which enzyme is the rate-limiting step of ketogenesis?

A. Argininosuccinate lyase

B. Mitochondrial HMG-CoA synthase

C. Mitochondrial thiophorase

D. Hormone-sensitive lipase

Answer: (B)

Explanation

Mitochondrial HMG-CoA synthase is the rate-limiting enzyme of ketogenesis. Thiophorase is instead required for ketolysis in extrahepatic tissues.

31. Why can neither the liver nor red blood cells use ketone bodies as an energy source?

A. The liver lacks thiophorase, and red blood cells lack mitochondria

B. The liver lacks acetyl-CoA, and red blood cells cannot transport ketones

C. The liver lacks HMG-CoA synthase, and red blood cells lack hemoglobin

D. The liver cannot oxidize fatty acids, and red blood cells lack glycolytic enzymes

Answer: (A)

Explanation

The liver cannot perform ketolysis because it lacks thiophorase, while red blood cells cannot use ketone bodies because they have no mitochondria.

32. What is the primary function of the urea cycle?

A. Converting fatty acids into ketone bodies for brain metabolism

B. Converting urea into ammonia for reuse in protein synthesis

C. Converting toxic ammonia into urea for renal excretion

D. Converting glucose into glycogen for storage in the liver

Answer: (C)

Explanation

The urea cycle detoxifies ammonia by converting it into urea, mainly in the liver; urea then travels in blood to the kidneys and urine.

33. Which combination correctly identifies the sources of urea’s two nitrogen atoms and its carbon atom?

A. Ammonia, glutamine, and pyruvate

B. Ammonia, aspartate, and CO₂ or HCO₃⁻

C. Glutamate, ammonia, and oxaloacetate

D. Aspartate, glutamate, and acetyl-CoA

Answer: (B)

Explanation

One nitrogen comes from ammonia, the second from aspartate, and the carbon enters from CO₂ or HCO₃⁻.

34. Which enzyme sequence represents the five enzymes of the urea cycle in order?

A. CPS-I, arginase, OTC, argininosuccinate synthetase, argininosuccinate lyase

B. Argininosuccinate synthetase, CPS-I, OTC, arginase, argininosuccinate lyase

C. CPS-I, OTC, argininosuccinate synthetase, argininosuccinate lyase, arginase

D. OTC, CPS-I, arginase, argininosuccinate lyase, argininosuccinate synthetase

Answer: (C)

Explanation

The urea-cycle enzymes are CPS-I, ornithine transcarbamoylase, argininosuccinate synthetase, argininosuccinate lyase, and arginase, in that order.

35. Which statement correctly describes CPS-I and ornithine transcarbamoylase deficiency?

A. CPS-I uses two GTP and is activated by aspartate; OTC deficiency is mitochondrial and acquired

B. CPS-I uses no nucleotide and is activated by arginine; OTC deficiency is autosomal recessive and rarest

C. CPS-I uses two ATP and is activated by N-acetylglutamate; OTC deficiency is X-linked and most common

D. CPS-I uses one ATP and is inhibited by N-acetylglutamate; OTC deficiency is autosomal dominant

Answer: (C)

Explanation

CPS-I uses 2 ATP and requires activation by N-acetylglutamate. OTC deficiency is the most common inherited urea-cycle disorder and is X-linked.

36. Which statement best distinguishes fat-soluble from water-soluble vitamins?

A. Water-soluble vitamins include A, D, E, and K and are stored mainly in adipose tissue

B. Fat-soluble vitamins include A, D, E, and K and are stored more readily, increasing toxicity risk

C. Fat-soluble vitamins include the B-complex and C and are rapidly excreted in urine

D. Water-soluble vitamins are stored extensively in the liver and therefore have greater toxicity risk

Answer: (B)

Explanation

Vitamins A, D, E, and K are fat-soluble, stored in liver or adipose tissue, and carry greater toxicity risk. The B-complex vitamins and vitamin C are water-soluble.

37. A patient develops difficulty seeing in dim light before other major symptoms. Which deficiency is most consistent with this finding?

A. Vitamin A deficiency

B. Vitamin D deficiency

C. Vitamin B12 deficiency

D. Vitamin K deficiency

Answer: (A)

Explanation

Night blindness is the early deficiency sign associated with vitamin A. Vitamin D deficiency is associated with rickets or osteomalacia.

38. Which sequence correctly describes vitamin D activation?

A. Liver D3 to skin 25-hydroxy-D to kidney calcitonin

B. Skin D3 to kidney 25-hydroxy-D to liver 1,25-dihydroxy-D

C. Kidney D3 to liver 1,25-dihydroxy-D to skin 25-hydroxy-D

D. Skin D3 to liver 25-hydroxy-D to kidney 1,25-dihydroxy-D

Answer: (D)

Explanation

Vitamin D is activated sequentially in the skin, liver, and kidney, producing calcitriol, or 1,25-dihydroxy-D, in the kidney.

39. Which pairing correctly matches a vitamin with its function or active form?

A. Vitamin K—carboxylation of clotting factors; vitamin B1—TPP

B. Vitamin K—one-carbon metabolism; vitamin B1—PLP

C. Vitamin A—carboxylation of clotting factors; vitamin B1—NAD⁺

D. Vitamin A—homocysteine conversion; vitamin B1—CoA

Answer: (A)

Explanation

Vitamin K is required for carboxylation of clotting factors, while the active form of vitamin B1 is thiamine pyrophosphate (TPP).

40. In which direction is a new DNA strand synthesized during replication?

A. 5′→3′

B. Both directions on the same strand

C. 3′→5′

D. From the primer toward the template end only

Answer: (A)

Explanation

DNA polymerases synthesize DNA in the 5′→3′ direction while reading the antiparallel template strand.

41. Which enzyme-function pairing is correct during DNA replication?

A. Topoisomerase synthesizes the RNA primer

B. Helicase unwinds the DNA double helix

C. SSB proteins join Okazaki fragments

D. Primase relieves DNA supercoiling

Answer: (B)

Explanation

Helicase separates the DNA strands. Primase makes the RNA primer, topoisomerase relieves supercoiling, and SSB proteins prevent strand reannealing.

42. Why is the lagging strand synthesized as Okazaki fragments?

A. RNA primers can be used only on the lagging strand

B. DNA polymerase can extend DNA only in the 5′→3′ direction on an antiparallel template

C. The lagging-strand template cannot bind single-strand binding proteins

D. Helicase unwinds only the lagging-strand template

Answer: (B)

Explanation

Because the template strands are antiparallel and synthesis proceeds 5′→3′, the lagging strand must be made discontinuously as Okazaki fragments.

43. Which bacterial DNA polymerase pairing is correct?

A. DNA polymerase III performs main synthesis and proofreading, while DNA polymerase I removes RNA primers

B. DNA polymerase I joins DNA strands, while DNA polymerase III relieves supercoiling

C. DNA polymerase I performs main synthesis and proofreading, while DNA polymerase III removes RNA primers

D. DNA polymerase III makes RNA primers, while DNA polymerase I unwinds DNA

Answer: (A)

Explanation

In bacteria, DNA polymerase III carries out most DNA synthesis and proofreading, whereas DNA polymerase I removes RNA primers and fills the resulting gaps.

44. Which DNA repair pathway specifically removes bulky lesions such as UV-induced thymine dimers?

A. Nucleotide excision repair

B. Mismatch repair

C. Base excision repair

D. Nonhomologous end joining

Answer: (A)

Explanation

Nucleotide excision repair removes bulky DNA lesions, including UV-induced thymine dimers. Base excision repair instead removes individual abnormal bases.

45. Which sequence correctly describes base excision repair?

A. Mismatch proteins remove the entire damaged strand, followed by transcription and translation

B. AP endonuclease removes the abnormal base, DNA glycosylase seals the strand, and helicase completes repair

C. DNA polymerase removes the abnormal base, ligase cuts the AP site, and glycosylase fills the gap

D. DNA glycosylase removes the abnormal base, AP endonuclease cuts the AP site, and DNA polymerase and ligase complete repair

Answer: (D)

Explanation

Base excision repair begins with DNA glycosylase removing the abnormal base, followed by AP endonuclease cleavage and gap filling and sealing by DNA polymerase and ligase.

46. A person with extreme sensitivity to sunlight and a high risk of skin cancer most likely has a defect in which repair pathway?

A. Base excision repair

B. Homologous recombination

C. Mismatch repair

D. Nucleotide excision repair

Answer: (D)

Explanation

Xeroderma pigmentosum results from defective nucleotide-excision repair, impairing removal of UV-induced lesions and increasing UV sensitivity and skin-cancer risk.

47. Which statement correctly describes transcription?

A. It synthesizes RNA from DNA in the 5′→3′ direction without requiring a primer

B. It synthesizes RNA from proteins in the 5′→3′ direction using a primer

C. It synthesizes DNA from RNA in the 3′→5′ direction using a primer

D. It synthesizes DNA from DNA in both directions without a primer

Answer: (A)

Explanation

Transcription produces an RNA molecule from a DNA template in the 5′→3′ direction, and RNA polymerase initiates synthesis without a primer.

48. Which RNA polymerase produces the precursor of eukaryotic mRNA?

A. RNA polymerase II

B. The bacterial sigma holoenzyme

C. RNA polymerase I

D. RNA polymerase III

Answer: (A)

Explanation

In eukaryotes, RNA polymerase II produces the precursor to mRNA. Polymerase I produces major rRNAs, while polymerase III produces tRNA and 5S rRNA.

49. Which combination correctly describes eukaryotic pre-mRNA processing?

A. A 7-methylguanosine cap is added to the 3′ end, and translation removes the introns

B. Both ends receive poly-A tails, and introns are retained during splicing

C. A poly-A tail is added to the 5′ end, a 7-methylguanosine cap to the 3′ end, and exons are removed

D. A 7-methylguanosine cap is added to the 5′ end, a poly-A tail to the 3′ end, and introns are removed by splicing

Answer: (D)

Explanation

The 5′ cap contains 7-methylguanosine, the poly-A tail is added at the 3′ end after the AAUAAA signal, and splicing removes introns while retaining exons.

50. A single-nucleotide insertion occurs in a coding sequence and is not a multiple of three; what type of mutation is this?

A. Silent mutation

B. Missense mutation

C. Nonsense mutation

D. Frameshift mutation

Answer: (D)

Explanation

An insertion or deletion not in a multiple of three shifts the reading frame, making it a frameshift mutation. Silent, missense, and nonsense mutations describe different effects on codons or amino acids.