MCAT Biochemistry Practice Questions

Question 1: Which amino acid is commonly found in the active site of enzymes that perform general acid-base catalysis around physiological pH?
A. Alanine
B. Valine
C. Histidine
D. Cysteine

Correct Answer: C
Explanation: Histidine has a side chain with a pKa (~6.0) near physiological pH, allowing it to both accept and donate protons during catalysis​ncbi.nlm.nih.gov. This makes histidine ideal for acid-base chemistry in enzyme active sites (choice C). Alanine and valine have nonpolar side chains that do not participate in proton transfer (choices A and B). Cysteine can engage in acid-base catalysis under some conditions, but its pKa (~8.3) is higher, making it less often used for general catalysis at neutral pH than histidine (choice D). Histidine’s imidazole ring can be protonated or deprotonated within the physiological pH range, facilitating enzymatic reactions.

Question 2: In the citric acid cycle, at which step is a molecule of GTP (or ATP) directly produced?
A. Conversion of fumarate to malate
B. Conversion of succinyl-CoA to succinate
C. Conversion of isocitrate to α-ketoglutarate
D. Conversion of malate to oxaloacetate

Correct Answer: B
Explanation: The only step of the TCA cycle that directly generates GTP (which can be converted to ATP) is the conversion of succinyl-CoA to succinate, catalyzed by succinyl-CoA synthetase (choice B). This substrate-level phosphorylation produces GTP in liver (or ATP in some tissues). The other steps listed do not directly produce a high-energy phosphate: fumarate to malate is a hydration step (choice A), isocitrate to α-ketoglutarate produces NADH and CO₂ (choice C), and malate to oxaloacetate produces NADH (choice D), but none of these directly generate GTP/ATP.

Question 3: Telomerase is an enzyme that adds DNA repeat sequences to the ends of chromosomes. What type of enzymatic activity does telomerase possess?
A. DNA-dependent RNA polymerase
B. RNA-dependent RNA polymerase
C. DNA ligase
D. RNA-dependent DNA polymerase

Correct Answer: D
Explanation: Telomerase carries its own RNA template and synthesizes DNA repeats onto chromosome ends, thus it is an RNA-dependent DNA polymerase, which is a type of reverse transcriptase (choice D). It uses an RNA template within the enzyme to add DNA (TTAGGG repeats in humans) to telomeres. DNA-dependent RNA polymerase (choice A) is just RNA polymerase (uses DNA to make RNA, as in transcription). RNA-dependent RNA polymerase (choice B) copies RNA from an RNA template (not the case for telomerase). DNA ligase (choice C) joins DNA strands together and is not involved in telomere elongation. Telomerase’s reverse transcriptase activity counteracts telomere shortening during DNA replication​en.wikipedia.org.

Question 4: Which cellular structure or process is primarily mediated by microfilaments (actin filaments) rather than microtubules?
A. Mitotic spindle formation
B. Cilia movement
C. Axonal transport of vesicles
D. Muscle fiber contraction

Correct Answer: D
Explanation: Muscle contraction is driven by the interaction of actin (microfilaments) and myosin, making it an actin-mediated process (choice D). Microfilaments (actin) also play key roles in cytokinesis and cell motility via lamellipodia. In contrast, the mitotic spindle is composed of microtubules (choice A). Cilia and flagella movement is powered by microtubules and dynein motors in a 9+2 arrangement (choice B). Axonal transport of vesicles occurs along microtubule tracks via motor proteins kinesin and dynein (choice C). Thus, among the options, only muscle contraction primarily involves microfilaments.

Question 5: A patient presents with dermatitis, diarrhea, and dementia. This triad of symptoms is most indicative of a deficiency in:
A. Niacin (Vitamin B₃)
B. Thiamine (Vitamin B₁)
C. Riboflavin (Vitamin B₂)
D. Vitamin C

Correct Answer: A
Explanation: The combination of dermatitis, diarrhea, and dementia is characteristic of pellagra, which is caused by niacin (vitamin B₃) deficiency​ncbi.nlm.nih.gov. Niacin is a precursor for NAD⁺/NADP⁺, and its deficiency leads to the “three D’s” of pellagra (choice A is correct). Thiamine (B₁) deficiency causes beriberi or Wernicke-Korsakoff syndrome, presenting with neurological and cardiac symptoms, not the pellagra triad (choice B). Riboflavin (B₂) deficiency can cause cheilosis and corneal vascularization, not classically the three D’s (choice C). Vitamin C deficiency (scurvy) leads to impaired collagen synthesis, causing bleeding gums and poor wound healing, not the symptoms described (choice D).

Question 6: Which type of DNA mutation is typically most likely to result in a completely nonfunctional protein product?
A. Silent mutation
B. Frameshift mutation
C. Missense mutation (conservative)
D. Nonsense mutation (early stop codon)

Correct Answer: B
Explanation: A frameshift mutation (choice B) results from insertion or deletion of a number of nucleotides not divisible by 3, altering the reading frame of the gene. This usually changes every amino acid downstream and often introduces a premature stop codon, yielding a severely dysfunctional protein. A silent mutation (choice A) does not change the amino acid sequence (no effect on protein function). A conservative missense mutation (choice C) substitutes one amino acid with a similar one, often preserving partial or full function of the protein. A nonsense mutation (choice D) introduces an early stop codon, which can truncate the protein. Nonsense mutations can be very damaging, but a frameshift early in the coding sequence is generally even more disruptive, as it alters the entire downstream sequence.

Questions 7–9 refer to the following passage:

Passage 1: A six-month-old male infant presents with hypotonia, developmental delay, and episodes of lactic acidosis. Laboratory tests show elevated blood lactate and alanine levels, especially after carbohydrate-rich meals. Genetic analysis reveals a mutation in the E1-alpha subunit of the pyruvate dehydrogenase complex (PDC). This complex links glycolysis to the citric acid cycle by converting pyruvate to acetyl-CoA. Some cases of PDC deficiency respond to high doses of thiamine. The care team begins the infant on a specialized ketogenic diet (high-fat, low-carbohydrate) to help manage his condition. The infant’s mother is also advised that this mutation is inherited in an X-linked manner.

Question 7: Which cofactor is most likely deficient or unable to bind effectively in this infant’s pyruvate dehydrogenase complex, contributing to the enzyme’s dysfunction?
A. Lipoic acid
B. FAD (Vitamin B₂)
C. Thiamine pyrophosphate (Vitamin B₁)
D. NAD⁺ (Vitamin B₃)

Correct Answer: C
Explanation: Pyruvate dehydrogenase (PDH) requires five cofactors: thiamine pyrophosphate (B₁), lipoic acid, CoA (derived from pantothenic acid, B₅), FAD (B₂), and NAD⁺ (B₃). Thiamine (vitamin B₁) is often a crucial limiting cofactor in PDH deficiency and some cases are responsive to high-dose thiamine​emedicine.medscape.com​step2.medbullets.com. A mutation in the E1-alpha subunit (the thiamine-dependent part) would impair PDH activity if thiamine cannot bind effectively. Thus, thiamine pyrophosphate is the cofactor of concern (choice C). Lipoic acid is another PDH cofactor but is not the target of thiamine therapy (choice A). FAD (riboflavin, B₂) and NAD⁺ (niacin, B₃) are also required by PDH, but the passage specifically notes thiamine responsiveness, pointing to B₁ (choices B and D are less likely).

Question 8: Why was the infant placed on a high-fat, low-carbohydrate ketogenic diet?
A. To provide an alternate fuel (ketone bodies) that bypasses the blocked PDH enzyme
B. To increase insulin release and drive glucose into cells
C. To build up muscle glycogen stores for energy
D. To stimulate gluconeogenesis from dietary protein

Correct Answer: A
Explanation: In pyruvate dehydrogenase (PDH) deficiency, pyruvate from carbohydrates cannot be efficiently converted into acetyl-CoA for the TCA cycle, leading to lactic acidosis. A ketogenic diet (high-fat, low-carb) provides fatty acids that the liver converts to ketone bodies, which can be used by the brain and other tissues as an alternative fuel that bypasses the PDH step (choice A). This reduces reliance on glucose and pyruvate metabolism​emedicine.medscape.com. Increasing insulin release (choice B) is not desired in PDH deficiency, as insulin promotes carbohydrate utilization, which the infant cannot handle well. Muscle glycogen (choice C) is not directly relevant here and storing more glycogen doesn’t solve the metabolic block. Stimulating gluconeogenesis (choice D) would create more glucose and pyruvate, which is counterproductive. Thus, the ketogenic diet helps by providing ketone bodies for energy and minimizing lactic acid production.

Question 9: The pyruvate dehydrogenase E1-alpha subunit gene is located on the X-chromosome. Given the information in the passage, what is the most likely inheritance pattern of this PDH deficiency?
A. Autosomal recessive
B. Autosomal dominant
C. X-linked recessive
D. Mitochondrial (maternal) inheritance

Correct Answer: C
Explanation: The passage indicates the mutation is inherited in an “X-linked manner” and the patient is a male infant, which is consistent with an X-linked recessive disorder (choice C). X-linked recessive conditions often affect males (who have only one X) and are transmitted by carrier mothers. Autosomal recessive (choice A) would typically affect males and females equally and often requires both parents to be carriers. Autosomal dominant (choice B) would usually appear in successive generations regardless of sex. Mitochondrial inheritance (choice D) is passed from mothers to all children and would not specifically involve an X-chromosomal gene. Since the PDH E1-alpha subunit gene (PDHA1) is on the X-chromosome, the disease manifests in this male and follows X-linked recessive inheritance (females would be carriers and less severely affected).

Question 10: A researcher adds an uncoupling agent (such as 2,4-dinitrophenol) to isolated mitochondria. Which of the following changes is most likely to occur as a result?
A. Decreased electron transport and increased ATP synthesis
B. Decreased electron transport and decreased oxygen consumption
C. Increased heat production and increased oxygen consumption
D. Increased ATP synthesis and increased proton gradient across the inner membrane

Correct Answer: C
Explanation: Uncoupling agents collapse the proton gradient across the inner mitochondrial membrane by allowing protons to flow back into the matrix without producing ATP. As a result, the electron transport chain speeds up (to try to re-establish the proton gradient) and oxygen consumption increases, but ATP synthesis drops. The energy of the proton gradient is released as heat​ncbi.nlm.nih.gov​ncbi.nlm.nih.gov. Thus, an uncoupler will increase heat production and increase O₂ consumption (choice C). Electron transport does not decrease – it either stays the same or increases because the block to flow (the proton gradient) is removed (eliminating choices A and B). ATP synthesis falls dramatically, and the proton gradient is dissipated, so choice D is incorrect (ATP synthesis would decrease, not increase).

Question 11: A Lineweaver–Burk plot of an enzyme is obtained with and without an inhibitor. In the presence of the inhibitor, the plot shows the same y-intercept (1/Vmax) as the control but a steeper slope (increased 1/V). What type of inhibition is most consistent with these data?
A. Noncompetitive inhibition
B. Uncompetitive inhibition
C. Competitive inhibition
D. Irreversible inhibition

Correct Answer: C
Explanation: If the y-intercept (1/Vmax) is unchanged by the inhibitor, it means Vmax is the same. A steeper slope with unchanged 1/Vmax indicates the x-intercept (−1/Km) is shifted, i.e. Km is increased. An inhibitor that increases apparent Km but does not change Vmax is a competitive inhibitor (choice C). Competitive inhibitors bind the active site and can be overcome by high substrate concentrations, so Vmax remains attainable​chem.libretexts.org. Noncompetitive inhibitors lower Vmax (different y-intercept) without changing Km (choice A is incorrect). Uncompetitive inhibitors lower both Vmax and Km (parallel Lineweaver–Burk lines, choice B is incorrect). Irreversible inhibition often looks like noncompetitive (Vmax reduced), but here the pattern fits classic competitive inhibition.

Question 12: Which feature is present in eukaryotic mRNA that is not found in prokaryotic mRNA?
A. 5′ 7-methylguanosine cap
B. Polycistronic transcripts
C. Shine–Dalgarno sequence
D. Simultaneous transcription and translation

Correct Answer: A
Explanation: Eukaryotic mRNA is processed with a 5′ methylguanosine cap and a 3′ poly-A tail, and usually contains introns that are spliced out. A 5′ cap (choice A) is unique to eukaryotic mRNA and is important for ribosome binding and mRNA stability. Prokaryotic mRNA lacks a 5′ cap. Polycistronic transcripts (one mRNA encoding multiple proteins) are typical of prokaryotes, not eukaryotes (choice B is a prokaryotic feature). The Shine–Dalgarno sequence is a ribosome-binding site on prokaryotic mRNA (choice C is prokaryotic). In prokaryotes, transcription and translation can occur simultaneously in the cytoplasm, whereas in eukaryotes transcription (nucleus) and translation (cytoplasm) are separated and cannot be simultaneous (choice D is a prokaryotic capability). Thus, the 5′ cap is the correct answer.

Question 13: Which of the following changes would most increase the fluidity of a cell’s plasma membrane at low temperature?
A. A higher proportion of unsaturated fatty acids in membrane phospholipids
B. Longer fatty acid tail chains in phospholipids
C. A higher proportion of saturated fatty acids in membrane phospholipids
D. Decreasing cholesterol content at low temperature

Correct Answer: A
Explanation: Unsaturated fatty acids (with one or more double bonds) have kinks in their tails that prevent tight packing of phospholipids, thereby increasing membrane fluidity at lower temperatures (choice A). In contrast, longer fatty acid chains and saturated fatty acids pack together more tightly and increase van der Waals interactions, reducing fluidity (choices B and C are opposite of what is asked). Cholesterol acts as a bidirectional buffer: at low temperatures, cholesterol actually increases fluidity by preventing fatty acid packing, and at high temperatures it stabilizes the membrane to reduce fluidity. However, removing cholesterol at low temperature (choice D) would make the membrane more rigid, not fluid. Thus, incorporating more unsaturated fatty acids is the best way to maintain membrane fluidity when it’s cold.

Question 14: Folate (vitamin B₉) and cobalamin (vitamin B₁₂) deficiencies can both cause megaloblastic anemia. Which of the following symptoms or signs would suggest a B₁₂ deficiency rather than folate deficiency?
A. High homocysteine levels
B. Neurological dysfunction (e.g. peripheral neuropathy)
C. Macrocytosis (enlarged red blood cells)
D. Glossitis (smooth, reddened tongue)

Correct Answer: B
Explanation: Vitamin B₁₂ (cobalamin) deficiency presents with megaloblastic anemia and neurological symptoms such as peripheral neuropathy, dorsal column spinal cord degeneration, and cognitive changes. Folate deficiency causes megaloblastic anemia without nerve involvement. Therefore, neurological dysfunction points toward B₁₂ deficiency (choice B). Both folate and B₁₂ deficiencies can elevate homocysteine levels (choice A is seen in both). Macrocytosis (large RBCs) is a feature of megaloblastic anemia in both deficiencies (choice C). Glossitis can occur in both B₁₂ and folate deficiencies as well (choice D). The distinguishing feature is the nerve damage, which is specific to B₁₂ deficiency and not seen in isolated folate deficiency​accesspediatrics.mhmedical.com​ncbi.nlm.nih.gov.

Questions 15–17 refer to the following passage:

Passage 2: Researchers are characterizing a new enzyme that breaks down a sugar. Kinetic data are collected in the presence and absence of an inhibitor (Inhibitor X). With Inhibitor X, the enzyme’s Michaelis–Menten curve still reaches the same maximal velocity V_max as the uninhibited enzyme, but a higher substrate concentration is required to reach half-maximal velocity. The researchers also find that adding a very high concentration of substrate can eventually overcome the effect of Inhibitor X. Inhibitor X is found to bind directly in the enzyme’s active site.

Question 15: What type of inhibition is Inhibitor X exhibiting?
A. Uncompetitive inhibition
B. Noncompetitive inhibition
C. Competitive inhibition
D. Allosteric (mixed) inhibition

Correct Answer: C
Explanation: Inhibitor X does not change the enzyme’s V_max (since the same maximal velocity can be reached) but increases the substrate concentration needed to reach ½ V_max (increases apparent K_m). This is characteristic of competitive inhibition​chem.libretexts.org. Competitive inhibitors bind the enzyme’s active site and can be overcome by sufficiently high substrate concentrations, consistent with the passage (choice C). Uncompetitive inhibitors bind only to the enzyme-substrate complex and lower both apparent V_max and K_m (choice A is not consistent with unchanged V_max). Noncompetitive inhibitors bind an allosteric site and decrease V_max (cannot be overcome by substrate), which is not the case here (choice B is incorrect). “Allosteric (mixed)” inhibition (choice D) generally also affects V_max. The data clearly point to competitive inhibition.

Question 16: If a very high concentration of substrate is added to the enzyme-inhibitor mixture described in the passage, what will happen to the reaction velocity?
A. The reaction will remain inhibited regardless of substrate concentration.
B. The reaction velocity will approach the same V_max as the uninhibited reaction.
C. The reaction will produce more product than the uninhibited enzyme.
D. The reaction velocity will only reach 50% of the uninhibited V_max.

Correct Answer: B
Explanation: In competitive inhibition, high concentrations of substrate can outcompete the inhibitor for the active site. Thus, at very high [S], the enzyme can still achieve the same V_max as it would without inhibitor (choice B). This is supported by the passage stating that a very high substrate concentration can overcome the effect of Inhibitor X. The reaction is not permanently inhibited – adding enough substrate will restore activity (so choice A is wrong). The enzyme won’t produce more product than normal; it will just reach its normal maximum rate (not exceed it, eliminating choice C). It will reach 100% of V_max, not just 50%, when substrate is saturating (choice D is incorrect). In summary, saturating substrate effectively “beats” a competitive inhibitor​chem.libretexts.org.

Question 17: Based on the information that Inhibitor X binds directly in the enzyme’s active site, which of the following is most likely true?
A. Inhibitor X binds the enzyme-substrate complex at an allosteric site.
B. Inhibitor X is structurally similar to the enzyme’s substrate.
C. Inhibitor X lowers the enzyme’s V_max by inducing a conformational change.
D. Inhibitor X can bind the enzyme simultaneously with the substrate.

Correct Answer: B
Explanation: Competitive inhibitors often resemble the substrate’s structure because they compete for binding in the active site​chem.libretexts.org. If Inhibitor X binds in the active site, it likely mimics the substrate (choice B). It would not be binding an allosteric site on the enzyme-substrate complex – that describes uncompetitive or noncompetitive inhibition (choice A is false for a competitive inhibitor). Competitive inhibitors do not change V_max and do not induce major conformational changes that reduce catalytic turnover; they simply block the active site (choice C is inaccurate given unchanged V_max). Inhibitor X cannot bind at the same time as the substrate if it occupies the active site – by definition, a competitive inhibitor precludes substrate binding (choice D is incorrect). Therefore, the most likely scenario is that Inhibitor X is structurally similar to the substrate and competes for the active site.

Question 18: Phosphofructokinase-1 (PFK-1) is a key regulatory enzyme in glycolysis. High levels of which of the following would directly inhibit PFK-1 activity?
A. AMP
B. ATP
C. ADP
D. Fructose-2,6-bisphosphate

Correct Answer: B
Explanation: PFK-1 is allosterically inhibited by high levels of ATP, the end product of glycolysis indicating abundant energy (choice B). When ATP is plentiful, PFK-1 slows glycolysis. In contrast, AMP is an allosteric activator of PFK-1, signaling low energy (choice A is opposite). ADP can also indicate low energy and tends to activate glycolysis (choice C is not inhibitory). Fructose-2,6-bisphosphate is a potent activator of PFK-1 in the liver, not an inhibitor (choice D is incorrect). Thus, high [ATP] signals that the cell’s energy charge is high and feeds back to inhibit PFK-1, preventing excess glycolytic flux.

Question 19: The E. coli lac operon is most strongly expressed under which conditions?
A. Lactose absent, glucose present
B. Lactose absent, glucose absent
C. Lactose present, glucose present
D. Lactose present, glucose absent

Correct Answer: D
Explanation: The lac operon is an inducible operon that is highly expressed when lactose is available (to serve as inducer) and glucose is scarce. Lactose present and glucose absent results in the strongest expression (choice D). Lactose binds the lac repressor, causing it to release from the operator, while low glucose leads to high cAMP, which binds CAP to activate transcription. If lactose is absent, the repressor stays bound and the operon is off regardless of glucose (choices A and B yield minimal expression). If glucose is present (especially with lactose present, choice C), catabolite repression occurs: cAMP levels are low, CAP is inactive, and the operon is only weakly expressed even if the repressor is lifted. Thus, lactose (+) and glucose (−) is the optimal condition for lac operon transcription.

Question 20: Red blood cells (erythrocytes) do not contain mitochondria. As a result, they:
A. cannot produce ATP at all.
B. produce ATP only via glycolysis.
C. utilize the citric acid cycle but not oxidative phosphorylation.
D. rely exclusively on fatty acid β-oxidation for energy.

Correct Answer: B
Explanation: Mature red blood cells have no mitochondria, so they cannot perform the citric acid cycle or oxidative phosphorylation. They produce ATP solely through anaerobic glycolysis (substrate-level phosphorylation) and lactic acid fermentation. Thus, they rely on glycolysis for ATP (choice B). RBCs do produce ATP, just not via mitochondrial pathways (choice A is wrong — they do make ATP, about 2 ATP per glucose in anaerobic glycolysis). They cannot utilize the TCA cycle at all without mitochondria (eliminating choice C). They also cannot perform fatty acid β-oxidation, as that occurs in mitochondria and RBCs lack them (choice D is incorrect). Therefore, glycolysis is the only ATP-generating process available to erythrocytes​sciencedirect.com.

Question 21: Vitamin C (ascorbic acid) is required as a cofactor for which post-translational modification in humans?
A. Hydroxylation of proline residues in collagen
B. γ-Carboxylation of glutamate residues in clotting factors
C. Addition of biotin to carboxylases
D. Phosphorylation of tyrosine residues in growth factor receptors

Correct Answer: A
Explanation: Vitamin C is essential for the hydroxylation of proline and lysine residues in collagen. Ascorbate is a cofactor for prolyl and lysyl hydroxylase, enzymes that stabilize the collagen triple helix by adding hydroxyl groups to proline and lysine (choice A). A deficiency (scurvy) impairs this process, leading to weak collagen and symptoms like bleeding gums and poor wound healing. γ-Carboxylation of glutamate in clotting factors is dependent on vitamin K, not vitamin C (choice B). Biotin addition (biotinylation) is a cofactor attachment for carboxylase enzymes and involves biotin (vitamin B₇), not vitamin C (choice C). Phosphorylation of tyrosines in receptors is performed by tyrosine kinase enzymes and does not require vitamin C (choice D). Thus, vitamin C’s classic role is in collagen proline hydroxylation​rarediseases.org.

Question 22: Statin drugs are used to lower cholesterol levels. They act by inhibiting the rate-limiting enzyme of cholesterol synthesis. Which enzyme is the target of statins?
A. HMG-CoA synthase
B. HMG-CoA reductase
C. Carnitine acyltransferase I
D. Cholesterol esterase

Correct Answer: B
Explanation: HMG-CoA reductase is the rate-limiting enzyme in cholesterol biosynthesis, converting HMG-CoA to mevalonate. Statins are competitive inhibitors of HMG-CoA reductase and thereby reduce cholesterol production​ncbi.nlm.nih.gov​my.clevelandclinic.org (choice B is correct). HMG-CoA synthase is involved in ketone body synthesis, not the target of statins (choice A is incorrect). Carnitine acyltransferase I is the rate-limiting enzyme of fatty acid β-oxidation (fat breakdown), not cholesterol synthesis (choice C is wrong). “Cholesterol esterase” is not the regulated step in cholesterol synthesis (it might refer to an enzyme hydrolyzing cholesterol esters, which is unrelated, choice D incorrect). Thus, statins lower cholesterol by inhibiting HMG-CoA reductase in the liver.

Question 23: Lactate dehydrogenase (LDH) catalyzes the conversion of pyruvate to lactate, coupled with conversion of NADH to NAD⁺. To which class of enzyme does LDH belong?
A. Lyase
B. Oxidoreductase
C. Transferase
D. Ligase

Correct Answer: B
Explanation: Lactate dehydrogenase facilitates a redox reaction: pyruvate is reduced to lactate while NADH is oxidized to NAD⁺. Enzymes that catalyze oxidation-reduction reactions are oxidoreductases (choice B). LDH specifically is an oxidoreductase, as it transfers electrons (hydride ions) between molecules (pyruvate and NADH). Lyases (choice A) break or form double bonds or ring structures without using water or redox (e.g., aldolase). Transferases (choice C) transfer functional groups from one molecule to another (e.g., kinases transfer phosphate). Ligases (choice D) join two molecules together typically with ATP hydrolysis (e.g., DNA ligase). Since LDH is involved in an oxidation-reduction process, oxidoreductase is the correct class.

Question 24: A protein has a native molecular weight of 100 kDa. On SDS-PAGE under non-reducing conditions, it runs as a single ~100 kDa band. Under reducing SDS-PAGE, it shows two bands at ~60 kDa and ~40 kDa. What does this indicate about the protein’s structure?
A. It is composed of two subunits (60 kDa and 40 kDa) linked by disulfide bonds
B. It is a homodimer of two 50 kDa subunits held by noncovalent forces
C. It is a single-chain protein that is glycosylated with 40 kDa of carbohydrate
D. The reducing agent broke the protein into fragments by cleaving peptide bonds

Correct Answer: A
Explanation: The appearance of two bands (60 kDa and 40 kDa) upon adding a reducing agent (like β-mercaptoethanol or DTT) implies that the protein consists of two different polypeptide chains (60k and 40k) that were connected by disulfide bonds. Under non-reducing SDS-PAGE, those subunits stay linked and migrate as one 100 kDa unit. Once reduced, the disulfide bonds are broken and the subunits separate, giving two bands. This matches a heterodimer held by disulfide bonds (choice A). A homodimer of 50 kDa subunits (choice B) would show a 50 kDa band upon reduction, not 60 and 40. A 40 kDa carbohydrate addition (choice C) is unlikely; SDS-PAGE under non-reducing vs reducing conditions wouldn’t split off carbohydrate (also glycosylation smears differently, not a sharp 40 kDa difference). Reducing agents break disulfide bonds, not peptide bonds, so they don’t cleave the protein’s backbone (choice D is wrong). Thus, the data indicate two subunits linked by disulfide bonds​en.wikipedia.org​en.wikipedia.org.

Question 25: A certain rare genetic disorder is passed from an affected mother to all of her children, sons and daughters alike. However, an affected father does not transmit the disorder to any of his children. Which type of inheritance does this pattern suggest?
A. Mitochondrial DNA inheritance
B. Autosomal dominant
C. Autosomal recessive
D. X-linked recessive

Correct Answer: A
Explanation: The described pattern – transmission through the mother to all offspring, with an affected father not passing it on – is characteristic of mitochondrial inheritance (choice A). Mitochondrial DNA is inherited only from the mother (through the ovum’s cytoplasm), so all children of an affected mother inherit the mitochondrial mutation, while children of an affected father do not inherit it​en.wikipedia.org. Autosomal dominant inheritance (choice B) would show transmission from either parent to ~50% of children on average, not exclusively maternal transmission. Autosomal recessive (choice C) requires both parents to carry the mutation and typically does not show an all-children pattern from one parent. X-linked recessive (choice D) can show mother-to-son transmission predominantly and wouldn’t affect all daughters and sons uniformly as described. Thus, the scenario strongly indicates mitochondrial inheritance.

Question 26: Steroid hormones differ from peptide hormones in their mechanisms of action. Which of the following is true of steroid hormones but not true of peptide hormones?
A. They bind to intracellular receptors that directly regulate gene transcription.
B. They bind to cell-surface receptors and activate second messenger cascades.
C. They are stored in secretory vesicles until needed.
D. They travel freely in the bloodstream without carrier proteins.

Correct Answer: A
Explanation: Steroid hormones (e.g. cortisol, estrogen) are lipophilic and diffuse through cell membranes to bind intracellular receptors (often in the cytosol or nucleus). The hormone-receptor complex then acts as a transcription factor, directly regulating gene expression (choice A). Peptide hormones cannot do this; they bind extracellular receptors. Peptide hormones (e.g. insulin, growth hormone) use cell-surface receptors and second messengers (choice B describes peptides, not steroids). Steroid hormones are not stored in vesicles — they are synthesized on demand and diffuse out of the cell (peptide hormones are pre-made and stored in vesicles, so choice C is a peptide trait). Steroids are hydrophobic and typically require carrier proteins in blood (like albumin); peptide hormones are water-soluble and travel freely (choice D is opposite — it’s true for peptides, not steroids). Thus, only choice A correctly describes steroid hormones uniquely.

Questions 27–29 refer to the following passage:

Passage 3: Investigators are studying gene expression differences between normal and cancerous cells for a gene called Gene Y. They isolate DNA, mRNA, and protein from both cell types. To assess gene copy number, they perform a Southern blot on genomic DNA using a probe for Gene Y. The cancer cells show a much darker band than normal cells, suggesting multiple copies of Gene Y in the cancer genome. To measure Gene Y mRNA, the researchers perform a Northern blot; the cancer cells have a markedly higher Gene Y mRNA level than normals. Correspondingly, a Western blot for the Gene Y protein shows overexpression in the cancer cells. These results indicate that Gene Y is amplified and overexpressed in the cancer. The scientists note that Gene Y likely functions as an oncogene. Finally, they use PCR to amplify the Gene Y coding sequence from both cell types for sequencing analysis.

Question 27: Which technique did the researchers use to determine that the cancer cells have extra copies of Gene Y in their genome?
A. Western blot
B. Northern blot
C. Southern blot
D. PCR

Correct Answer: C
Explanation: A Southern blot is the technique used to detect specific DNA sequences in a DNA sample​cytivalifesciences.com. In the passage, the researchers used a Southern blot on genomic DNA with a Gene Y probe and observed a darker band in cancer cells, indicating gene amplification. Western blot detects proteins, not DNA (choice A is wrong). Northern blot detects RNA (mRNA levels), not genomic copy number (choice B is wrong). PCR can amplify DNA and sometimes detect gene copy differences, but the passage specifically mentions a Southern blot for genomic analysis (choice D is not what was used to visualize the multiple gene copies). Therefore, the correct answer is Southern blot (choice C) for detecting amplified gene copies.

Question 28: Given the findings in the passage, Gene Y is most likely:
A. a tumor suppressor gene that is silenced in cancer cells.
B. a DNA repair gene with a loss-of-function mutation.
C. an oncogene that drives cancer when overexpressed.
D. inherited through mitochondrial DNA.

Correct Answer: C
Explanation: The data show that Gene Y is amplified (extra copies) and overexpressed in cancer cells, which is a hallmark of an oncogene. Oncogenes lead to cancer when they are overactive or overexpressed​ncbi.nlm.nih.gov​my.clevelandclinic.org (choice C is correct). In contrast, tumor suppressor genes typically are inactivated or deleted in cancer; you would not expect multiple copies and overexpression (choice A is opposite of the findings). A DNA repair gene with loss of function (choice B) would be more consistent with mutations or deletions, not increased copies and expression. Mitochondrial inheritance (choice D) is irrelevant to this somatic cell scenario. Gene amplification in tumors (e.g. HER2 in breast cancer) usually signifies an oncogene. Therefore, Gene Y is likely an oncogene activated by amplification and overexpression.

Question 29: When amplifying the Gene Y coding sequence by PCR for sequencing, which of the following components is essential for the PCR reaction but is not used in vivo during DNA replication?
A. DNA primers
B. DNA polymerase enzyme
C. Deoxynucleotide triphosphates (dNTPs)
D. Template DNA

Correct Answer: A
Explanation: PCR (polymerase chain reaction) requires DNA primers designed to anneal to the target sequence. In vivo DNA replication, however, uses RNA primers made by primase, not DNA primers (choice A is correct). Both PCR and cellular DNA replication need a DNA polymerase (Taq polymerase in PCR, DNA polymerase δ/ε in eukaryotes; choice B is used in both). Both require dNTPs as building blocks (choice C is used in both). Both require a template strand to copy (choice D is needed in both). The key difference is that PCR uses pre-made DNA primers to initiate replication of the target, whereas cells use RNA primers. Additionally, PCR uses a heat-stable DNA polymerase (from thermophilic bacteria) due to the high temperatures, but the question specifically highlights DNA vs RNA primers. Thus, DNA primers are essential in PCR and not found in natural DNA replication initiation.

Question 30: After about 48 hours of starvation, which of the following becomes the primary fuel source for the human brain?
A. Amino acids from muscle protein breakdown
B. Ketone bodies from fatty acid oxidation
C. Fatty acids mobilized from adipose tissue
D. Blood glucose exclusively from dietary sources

Correct Answer: B
Explanation: In prolonged fasting (beyond ~2 days), the liver converts fatty acids into ketone bodies (acetoacetate and β-hydroxybutyrate), which become a major fuel for the brain​emedicine.medscape.com. Ketone bodies can cross the blood-brain barrier and partly substitute for glucose, sparing muscle protein. So ketone bodies are the primary brain fuel at ~48 hours starvation (choice B). Amino acids from muscle breakdown are used for gluconeogenesis to supply some glucose, but the brain’s primary fuel shifts to ketones to reduce muscle proteolysis (choice A is not the main fuel, though some glucose from amino acids is still used). Fatty acids cannot significantly fuel the brain because they do not cross the blood-brain barrier efficiently (choice C is incorrect). Dietary glucose would be gone long before 48 hours (choice D is irrelevant after two days of no intake). Thus, ketone bodies are the key fuel for the brain in prolonged fasting.

Question 31: Cells of the adrenal cortex synthesize steroid hormones like cortisol and aldosterone. Which organelle would be abundant in these cells to accommodate this function?
A. Rough endoplasmic reticulum
B. Smooth endoplasmic reticulum
C. Lysosomes
D. Peroxisomes

Correct Answer: B
Explanation: The smooth endoplasmic reticulum (SER) is involved in lipid and steroid hormone synthesis. Adrenal cortex cells (which produce steroids) have an extensive SER (choice B is correct). The rough ER (choice A) is studded with ribosomes and is more involved in protein synthesis (especially secretory and membrane proteins). While adrenal cells do secrete hormones, steroid synthesis specifically takes place in SER and mitochondria. Lysosomes (choice C) handle catabolism of biomolecules and are not directly related to steroid production. Peroxisomes (choice D) are involved in very-long-chain fatty acid oxidation and other oxidative processes, not specifically steroid hormone synthesis. Therefore, an abundance of smooth ER is a hallmark of steroid-synthesizing cells such as those in the adrenal cortex.

Question 32: A 2-year-old child has bowing of the legs, widened growth plates, and a rachitic rosary on costochondral joints. Which vitamin deficiency is the most likely cause?
A. Vitamin A
B. Vitamin C
C. Vitamin D
D. Vitamin K

Correct Answer: C
Explanation: Bowed legs, widened growth plates, and rachitic rosary (prominent costochondral junctions) are classic signs of rickets, which is caused by vitamin D deficiency in children. Vitamin D is needed for calcium absorption and bone mineralization; deficiency leads to soft, poorly mineralized bones (choice C is correct). Vitamin A deficiency can cause night blindness and xerophthalmia, not rickets (choice A is incorrect). Vitamin C deficiency (scurvy) leads to bone problems due to poor collagen, but typically presents with fragile blood vessels and gum bleeding rather than the described bone deformities (choice B is wrong for rickets). Vitamin K deficiency causes bleeding problems, not bone deformities (choice D is unrelated). Thus, vitamin D deficiency is the cause of rickets in this child.

Question 33: Phenylketonuria (PKU) is an inherited metabolic disorder in which phenylalanine cannot be converted to tyrosine. As a result, tyrosine in individuals with PKU becomes:
A. an essential amino acid that must be obtained through the diet.
B. a nonpolar amino acid instead of a polar one.
C. toxic to neurons if it accumulates.
D. able to substitute for phenylalanine in protein synthesis.

Correct Answer: A
Explanation: In PKU, the enzyme phenylalanine hydroxylase is deficient, so phenylalanine cannot be converted into tyrosine. Normally, tyrosine is a nonessential amino acid (the body makes it from phenylalanine), but in PKU patients, tyrosine must be supplied by the diet, effectively making it an essential amino acid​step2.medbullets.com​step2.medbullets.com (choice A). The polarity of tyrosine (choice B) is not the issue; the problem is its synthesis. Tyrosine itself is not toxic in PKU – it’s the accumulation of phenylalanine (and its metabolites) that causes issues; tyrosine deficiency contributes to problems as well, but tyrosine doesn’t accumulate to toxic levels (choice C is incorrect). Tyrosine cannot substitute for phenylalanine in proteins; each amino acid is unique in the genetic code (choice D is incorrect). The key point is that tyrosine becomes conditionally essential in PKU because the body can’t make it.

Question 34: During DNA replication in E. coli, which enzyme is responsible for removing RNA primers and replacing them with DNA?
A. DNA polymerase I
B. DNA polymerase III
C. DNA ligase
D. Primase

Correct Answer: A
Explanation: In prokaryotes, DNA polymerase I has a 5′→3′ exonuclease activity that removes RNA primers and simultaneously fills in the gap with DNA​homework.study.com. Therefore, DNA Pol I carries out primer removal and replacement (choice A). DNA polymerase III is the main replicative polymerase synthesizing new DNA strands but cannot remove primers (choice B is incorrect for primer removal). DNA ligase is used after primer replacement to seal the nicks between Okazaki fragments, but it does not remove primers itself (choice C is wrong for removal). Primase is the enzyme that lays down the RNA primer in the first place, not remove it (choice D is incorrect). Thus, DNA Pol I is the enzyme that removes RNA primers in E. coli replication and fills in with DNA.

Question 35: Which of the following amino acids is essential in the human diet (i.e., cannot be synthesized by humans)?
A. Tyrosine
B. Alanine
C. Lysine
D. Glutamate

Correct Answer: C
Explanation: Lysine is an essential amino acid for humans, meaning it must be obtained from dietary sources because the body cannot synthesize it. Other essential amino acids include histidine, isoleucine, leucine, methionine, phenylalanine, threonine, tryptophan, and valine. Tyrosine (choice A) is nonessential because it can be made from phenylalanine (though in PKU patients phenylalanine can’t be converted, tyrosine becomes conditionally essential). Alanine (choice B) and glutamate (choice D) are nonessential as well; they can be synthesized by transamination of pyruvate and α-ketoglutarate, respectively. Lysine (choice C) cannot be made by human metabolic pathways and must come from protein in the diet.

Question 36: Proteins destined for secretion (export out of the cell) are synthesized by ribosomes located:
A. free in the cytosol.
B. on the rough endoplasmic reticulum.
C. in the mitochondrial matrix.
D. in the nucleus.

Correct Answer: B
Explanation: Secretory proteins are synthesized by ribosomes bound to the rough endoplasmic reticulum (RER) and enter the RER lumen during translation. From there, they are trafficked through the Golgi apparatus and eventually secreted via vesicles (choice B is correct). Cytosolic (free) ribosomes (choice A) synthesize proteins that remain in the cytosol or go to organelles like the nucleus or mitochondria, but not those destined for secretion. Mitochondrial matrix ribosomes (which resemble prokaryotic ribosomes) only synthesize a small number of mitochondrial proteins encoded by mitochondrial DNA (choice C is unrelated to secreted proteins). There are no ribosomes in the nucleus synthesizing secretory proteins (nuclear proteins are made in cytosol and imported); ribosomes function in the cytoplasm or on the ER (choice D is incorrect). Thus, the RER is the site of synthesis for secreted proteins (hormones, extracellular enzymes, etc.).

Question 37: Isoleucine acts as a negative feedback inhibitor of the enzyme threonine deaminase, the first step in isoleucine biosynthesis from threonine. This is an example of:
A. feed-forward activation.
B. feedback inhibition.
C. competitive inhibition.
D. positive cooperativity.

Correct Answer: B
Explanation: The scenario described is classic feedback inhibition (negative feedback). The end product of a pathway (isoleucine) accumulates and allosterically inhibits an earlier enzyme (threonine deaminase) in its own biosynthetic pathway (choice B). This regulation prevents overproduction of isoleucine once sufficient levels are present. Feed-forward activation (choice A) is the opposite concept, where an early metabolite activates a downstream enzyme (not the case here). Competitive inhibition (choice C) involves direct competition at an active site; feedback inhibition often involves allosteric regulation by the end product, not the product competing with the substrate at the active site. Positive cooperativity (choice D) refers to substrate binding increasing affinity at other sites (like hemoglobin O₂ binding), which is unrelated to end-product inhibition. Thus, isoleucine’s effect on threonine deaminase is a textbook example of feedback inhibition.

Questions 38–40 refer to the following passage:

Passage 4: Inclusion-cell (I-cell) disease is a rare lysosomal storage disorder in which a defect in the Golgi enzyme N-acetylglucosamine-1-phosphotransferase prevents addition of mannose-6-phosphate (M6P) tags to lysosomal enzymes. Without the M6P tag, lysosomal hydrolase enzymes are not trafficked to lysosomes and are instead secreted out of the cell. Patients with I-cell disease (mucolipidosis II) exhibit failure to thrive, developmental delays, coarse facial features, and accumulation of macromolecules within lysosomes (forming “inclusion bodies”). Laboratory tests show high levels of lysosomal enzymes in the blood. These enzymes are inactive in the extracellular fluid but are missing from where they should be active (inside lysosomes).

Question 38: In I-cell disease, a defect in a Golgi apparatus enzyme leads to failure of adding the mannose-6-phosphate tag. Which organelle is primarily affected by this tagging defect?
A. Nucleus
B. Golgi apparatus
C. Lysosome
D. Endoplasmic reticulum

Correct Answer: C
Explanation: Mannose-6-phosphate tags are required for proper targeting of enzymes to the lysosome. In I-cell disease, lysosomal enzymes do not reach the lysosome and instead are secreted. The organelle that ultimately lacks its functional enzymes is the lysosome, leading to accumulation of undegraded substrates​en.wikipedia.org​en.wikipedia.org (choice C). The Golgi apparatus is where the tagging process fails (the site of the defective enzyme), but the question asks which organelle is primarily affected by the defect – the consequence is dysfunctional lysosomes (not functioning properly due to missing enzymes). The nucleus (choice A) is not involved in this tagging process. The ER (choice D) is upstream in protein processing but not the site of M6P tagging (which occurs in the Golgi). Thus, lysosomes suffer the most from the failure to tag hydrolases, as they don’t receive the enzymes they need.

Question 39: In I-cell disease, lysosomal enzymes are absent from the lysosome and instead found in high levels in the bloodstream. Why are these enzymes inactive when present in blood and extracellular fluid?
A. They require proteolytic activation by trypsin, which is absent in blood.
B. They are bound by inhibitors in the bloodstream.
C. They need a low pH environment (around 5) to be active.
D. They are immediately ubiquitinated and deactivated outside the cell.

Correct Answer: C
Explanation: Lysosomal enzymes (acid hydrolases) are optimally active at the acidic pH (~5) found inside lysosomes. In the bloodstream and extracellular fluid, the pH is ~7.4, at which these enzymes have little to no activity​en.wikipedia.org. Thus, even though the enzymes are secreted into blood in I-cell disease, they remain largely inactive because they are not in the acidic environment they require (choice C). They do not need trypsin activation (that’s relevant for some digestive enzymes, choice A is incorrect). There aren’t specific inhibitors in blood universally binding all lysosomal enzymes (choice B is not the reason). They are not ubiquitinated and degraded immediately outside (ubiquitination targets proteins for proteasomal degradation inside cells, choice D is irrelevant). The key issue is the pH mismatch: outside the lysosome, the enzymes cannot function effectively.

Question 40: Which of the following best explains the molecular cause of I-cell disease described in the passage?
A. A failure of the rough ER to synthesize lysosomal enzymes
B. A mutation that prevents proper folding of lysosomal enzymes
C. Inability to add a phosphate to a mannose on lysosomal enzyme proteins
D. A defect in the proteasome that degrades lysosomal enzymes prematurely

Correct Answer: C
Explanation: I-cell disease is caused by a defect in the phosphotransferase enzyme in the Golgi, which fails to add mannose-6-phosphate tags to lysosomal enzyme proteins​en.wikipedia.org. Without the M6P tag, enzymes aren’t sorted to lysosomes. Thus, the molecular cause is the inability to phosphorylate the mannose residues on those enzymes (choice C). It’s not a synthesis problem in the rough ER – the enzymes are made, but not properly tagged (choice A is wrong). The enzymes fold and function (they’re just sent to the wrong place), so it’s not primarily a folding mutation in the enzymes themselves (choice B is incorrect). The proteasome isn’t the issue here; the enzymes aren’t being degraded, they’re being secreted (choice D is unrelated). So the lack of M6P tagging (phosphate addition to mannose) in the Golgi is the cause of I-cell disease.

Question 41: The Na⁺/K⁺ ATPase pump in cell membranes transports:
A. 3 Na⁺ out of the cell and 2 K⁺ into the cell per ATP hydrolyzed.
B. 2 Na⁺ out of the cell and 3 K⁺ into the cell per ATP hydrolyzed.
C. 3 Na⁺ into the cell and 2 K⁺ out of the cell per ATP hydrolyzed.
D. 1 Na⁺ out and 1 K⁺ in, using one phosphate group.

Correct Answer: A
Explanation: The Na⁺/K⁺ ATPase (sodium-potassium pump) is a P-type ATPase that, for each ATP hydrolyzed, exports 3 Na⁺ ions from the cell and imports 2 K⁺ ions into the cell. This creates a sodium gradient (high outside) and potassium gradient (high inside) across the plasma membrane (choice A is correct). Choice B reverses the ions (it pumps 3 Na out, not 2; and 2 K in, not 3). Choice C has the directions wrong (Na is pumped out, not in). Choice D is incorrect because the stoichiometry is 3:2, not 1:1, and it specifically uses the energy from ATP hydrolysis. The 3 Na⁺ out / 2 K⁺ in ratio also results in a net charge export (electrogenic pump), contributing to the resting membrane potential.

Question 42: Which hormone primarily acts to increase blood glucose levels by stimulating glycogenolysis and gluconeogenesis in the liver during fasting?
A. Insulin
B. Glucagon
C. Thyroid hormone
D. Aldosterone

Correct Answer: B
Explanation: Glucagon, a peptide hormone from pancreatic α-cells, is released during fasting and low blood glucose. It raises blood glucose by stimulating glycogen breakdown (glycogenolysis) and gluconeogenesis in the liver (choice B). Insulin (choice A) does the opposite: it lowers blood sugar by promoting glucose uptake and storage (glycogen synthesis, fat synthesis). Thyroid hormone (choice C) can increase basal metabolic rate and potentiate effects of other hormones, but it’s not the acute regulator of blood glucose during fasting. Aldosterone (choice D) regulates salt and water balance (sodium retention, potassium excretion) and has no direct role in glucose metabolism. Thus, glucagon is the key hormone that signals the liver to release glucose into the blood in the fasting state.

Question 43: Ultraviolet (UV) light exposure causes the formation of thymine dimers in DNA. These lesions are typically repaired in human cells by:
A. base excision repair.
B. nucleotide excision repair.
C. mismatch repair.
D. homologous recombination.

Correct Answer: B
Explanation: Thymine dimers (covalent links between adjacent thymidines) are bulky DNA lesions usually repaired by nucleotide excision repair (NER). In NER, a segment of the DNA strand containing the dimer is cut out and replaced with new DNA​sciencedirect.com​ncbi.nlm.nih.gov (choice B is correct). Base excision repair (choice A) typically fixes small non-bulky lesions, such as deaminated bases or single-base damage, by removing a single base and patching it. Mismatch repair (choice C) fixes replication errors (mismatched bases) shortly after DNA replication, not UV-induced damage. Homologous recombination (choice D) is a pathway for repairing double-strand breaks using a sister chromatid as a template. Therefore, the dedicated pathway for thymine dimer removal is NER (defects in NER cause conditions like xeroderma pigmentosum, characterized by UV sensitivity).

Question 44: Insulin promotes which of the following processes in the liver?
A. Glycogen synthesis and fatty acid synthesis
B. Glycogen breakdown and ketone body formation
C. Gluconeogenesis from amino acids
D. Ketone utilization as an energy source

Correct Answer: A
Explanation: Insulin, released after a carbohydrate-rich meal, signals the liver (and other tissues) to uptake glucose and store energy. In the liver, insulin activates glycogen synthase, increasing glycogen synthesis, and it also promotes fatty acid synthesis from excess glucose (which can later be stored as triglycerides)​en.wikipedia.org. Thus, choice A is correct. Insulin inhibits glycogen breakdown and ketogenesis, so choice B is the opposite of insulin’s effect. Insulin also inhibits gluconeogenesis – instead it favors glycolysis and storage, so choice C is wrong. Ketone utilization (choice D) is not a process the liver does (the liver produces but does not use ketones for itself) and insulin in any case reduces ketone production by promoting glucose usage. So the best answer is that insulin stimulates glycogen synthesis and lipogenesis in the well-fed state.

Question 45: Night blindness (nyctalopia) and xerophthalmia (dry, keratinized conjunctiva) can result from a deficiency of which vitamin?
A. Vitamin K
B. Vitamin A
C. Vitamin E
D. Vitamin B₁ (Thiamine)

Correct Answer: B
Explanation: Vitamin A (retinol/retinal) is required for normal vision (it’s a component of rhodopsin in rod cells) and for maintaining epithelial tissues. Deficiency of vitamin A can cause night blindness and xerophthalmia (dry eyes and Bitot’s spots), and in severe cases, corneal degeneration. Thus, choice B is correct. Vitamin K deficiency causes bleeding tendencies (choice A wrong). Vitamin E deficiency can cause neurological problems but not typically night blindness (choice C wrong). Thiamine (B₁) deficiency causes beriberi or Wernicke’s encephalopathy, not vision or eye epithelial issues (choice D wrong). Therefore, the fat-soluble vitamin A is the one whose lack leads to the eye symptoms described.

Question 46: In an X-linked dominant condition, an affected male will pass the trait:
A. to all of his daughters and none of his sons.
B. to all of his sons and none of his daughters.
C. to half of his sons and half of his daughters.
D. only if the mother is a carrier of the trait.

Correct Answer: A
Explanation: In X-linked dominant inheritance, an affected father (who has the mutant allele on his single X chromosome) will transmit that X chromosome to all of his daughters (who each receive his X), so all daughters will inherit the condition. He will transmit his Y (not X) to all of his sons, so none of his sons inherit the X-linked allele from him (choice A is correct). Therefore, no sons are affected by an affected father alone. Choice B is exactly the opposite and is incorrect. Choice C describes an autosomal dominant pattern from an affected father (50% to each sex child on average), not X-linked. Choice D confuses it with X-linked recessive – in X-linked dominant, an affected father doesn’t require the mother to have anything for all daughters to get it. Thus, all daughters and no sons is the hallmark of an X-linked dominant male-to-offspring transmission.

Question 47: A patient experiences hemolytic anemia after eating fava beans. This reaction is most likely due to a deficiency in which enzyme?
A. Pyruvate kinase
B. Glucose-6-phosphate dehydrogenase
C. Hexokinase
D. Lactate dehydrogenase

Correct Answer: B
Explanation: Hemolysis after fava bean ingestion is classic for glucose-6-phosphate dehydrogenase (G6PD) deficiency, an X-linked recessive disorder. G6PD is the first enzyme of the pentose phosphate pathway and generates NADPH in red blood cells, which is needed to reduce glutathione and protect RBCs from oxidative damage. Fava beans contain oxidants that, in G6PD-deficient individuals, cause oxidative stress leading to hemolysis (choice B is correct). Pyruvate kinase deficiency can also cause hemolytic anemia (impaired glycolysis in RBCs) but is not specifically triggered by fava beans (choice A is less likely here). Hexokinase deficiency is not a common cause of hemolysis (choice C is wrong). Lactate dehydrogenase deficiency is rare and would affect exercise tolerance, not cause acute hemolysis with fava beans (choice D wrong). G6PD deficiency is known for causing episodic hemolysis with triggers like fava beans, sulfa drugs, or infections​homework.study.com.

Question 48: Chloramphenicol is an antibiotic that binds the 50S ribosomal subunit and inhibits bacterial protein synthesis. Why does chloramphenicol have minimal effect on eukaryotic protein synthesis?
A. Eukaryotic ribosomes have 60S and 40S subunits, which differ from bacterial 50S/30S subunits.
B. Eukaryotic cells inactivate chloramphenicol with enzymes.
C. Eukaryotic ribosomes are located in the nucleus where chloramphenicol cannot enter.
D. Eukaryotes use different tRNAs that chloramphenicol cannot affect.

Correct Answer: A
Explanation: Eukaryotic ribosomes are 80S, composed of 60S and 40S subunits, whereas prokaryotic ribosomes are 70S (50S and 30S subunits). Chloramphenicol specifically targets the bacterial 50S subunit and does not bind effectively to the structurally different eukaryotic 60S subunit​sciencedirect.com. Thus, eukaryotic cytosolic protein synthesis is largely unaffected (choice A). Although eukaryotic mitochondria have 55–60S ribosomes somewhat similar to bacteria, chloramphenicol’s concentration and specificity spare most eukaryotic function. Eukaryotic cells do not universally inactivate chloramphenicol enzymatically (choice B is not the main reason; bacteria actually can inactivate it via acetyltransferases). Eukaryotic ribosomes are in the cytoplasm (or RER), not the nucleus (choice C is false). The tRNAs used are not the reason for selectivity (choice D is irrelevant). The key is the structural difference in ribosomes between bacteria and eukaryotes, making choice A correct.

Question 49: All of the following metabolic processes take place (at least partially) in the mitochondria EXCEPT:
A. the citric acid cycle.
B. β-oxidation of fatty acids.
C. cholesterol synthesis.
D. fatty acid synthesis.

Correct Answer: D
Explanation: Fatty acid synthesis is the one listed process that occurs in the cytosol (choice D). The citric acid cycle (Krebs cycle) takes place entirely in the mitochondrial matrix (choice A is in mitochondria). β-Oxidation of fatty acids occurs in the mitochondrial matrix (with very long chains first shortened in peroxisomes, but primarily mitochondria, choice B is mitochondrial). Cholesterol synthesis begins in the cytosol but has key steps in the smooth ER; however, if considering subcellular localization, cholesterol synthesis is not mitochondrial either. The question likely expects “fatty acid synthesis” as the cytosolic process (choice D). (Note: Cholesterol synthesis is a bit tricky as it’s cytosol/ER, but clearly fatty acid synthesis is cytosolic). Therefore, fatty acid synthesis is the process that does not occur in mitochondria.

Question 50: Vitamin K is necessary for:
A. proper vision in low-light conditions.
B. calcium absorption from the gut.
C. post-translational modification of clotting factors.
D. preventing oxidative damage in cell membranes.

Correct Answer: C
Explanation: Vitamin K is required for the γ-carboxylation of glutamate residues on clotting factors II, VII, IX, and X (and proteins C, S). This post-translational modification allows the clotting proteins to bind calcium and function in coagulation​accesspediatrics.mhmedical.com. So, choice C is correct. Vision in low light is related to vitamin A (choice A is vitamin A’s role). Calcium absorption in the gut is facilitated by vitamin D (calcitriol) not K (choice B is vitamin D’s function). Preventing oxidative damage in membranes is a role of vitamin E (antioxidant), not K (choice D is vitamin E). Thus, vitamin K’s primary role is to act as a cofactor for the enzyme γ-glutamyl carboxylase in the liver, modifying clotting factors for proper blood clotting.

Question 51: DNA with a high percentage of guanine-cytosine (GC) base pairs has a higher melting temperature than DNA with a high adenine-thymine (AT) content. This is because:
A. G≡C base pairs form three hydrogen bonds, while A=T pairs form two.
B. GC-rich DNA forms a triple helix structure.
C. AT-rich DNA is more prone to mutations.
D. GC pairs cause DNA strands to wrap tighter around histones.

Correct Answer: A
Explanation: GC base pairs are connected by three hydrogen bonds, compared to the two hydrogen bonds of AT base pairs. The extra hydrogen bond (and stronger base stacking interactions of GC) means more energy (higher temperature) is required to separate GC-rich DNA strands​frontiersin.org. Thus, choice A is correct. GC-rich DNA does not form a triple helix (choice B is incorrect; triple helices are unusual structures not simply caused by GC content). AT-rich DNA is not inherently more prone to mutations; the question is about melting temperature, not mutation rate (choice C is irrelevant). GC content does not directly relate to DNA wrapping around histones (which involves histone-DNA interactions, choice D is unrelated). In summary, the greater number of hydrogen bonds in G≡C pairs raises the melting temp of GC-rich DNA.

Questions 52–54 refer to the following passage:

Passage 5: Hemoglobin (Hb) is a tetrameric protein in red blood cells that exhibits cooperative binding of oxygen. Myoglobin, in contrast, is a monomeric oxygen-binding protein in muscle. Researchers compare oxygen-binding curves for hemoglobin and myoglobin. Hemoglobin’s O₂ saturation curve is sigmoidal, reflecting cooperative binding: as one O₂ binds, hemoglobin’s affinity for additional O₂ increases. Myoglobin’s O₂ binding curve is hyperbolic (non-cooperative). The scientists also examine factors that affect hemoglobin’s oxygen affinity. They observe that in tissues with high CO₂ and low pH (conditions of vigorous respiration), hemoglobin’s O₂ affinity decreases, causing O₂ to be released (the Bohr effect). Additionally, they note that fetal hemoglobin (HbF) has a higher O₂ affinity than adult hemoglobin (HbA) because HbF binds 2,3-bisphosphoglycerate (2,3-BPG) less strongly.

Question 52: What structural property of hemoglobin is responsible for its sigmoidal oxygen-binding curve?
A. Its heme prosthetic group
B. Cooperative interactions between its subunits
C. The planar structure of the porphyrin ring
D. Its ability to undergo the Bohr effect

Correct Answer: B
Explanation: Hemoglobin’s sigmoidal O₂ binding curve is due to cooperative binding among its four subunits. When one subunit binds O₂, it induces a conformational change (to the R state) that increases the affinity of the remaining subunits for O₂​ncbi.nlm.nih.gov. This inter-subunit cooperativity produces the characteristic sigmoidal (S-shaped) curve (choice B). The heme prosthetic group (choice A) is present in both hemoglobin and myoglobin; it allows O₂ binding, but by itself doesn’t explain cooperativity (myoglobin has heme but is not cooperative). The planar porphyrin ring (choice C) is part of heme structure; again, both proteins have heme rings, that’s not the cause of sigmoidality. The Bohr effect (choice D) is related to pH/CO₂ influence on O₂ affinity, not the cause of the sigmoidal shape (though cooperative structure is also required for Bohr effect to manifest fully). Thus, hemoglobin’s quaternary structure with multiple subunits is what allows cooperative binding and a sigmoidal curve.

Question 53: In actively respiring tissues, CO₂ production and low pH promote O₂ release from hemoglobin. This phenomenon is known as:
A. the Bohr effect.
B. cooperative binding.
C. allosteric inhibition by O₂.
D. the Haldane effect.

Correct Answer: A
Explanation: The description given (high CO₂ and low pH leading hemoglobin to release O₂) is the classic Bohr effect​ncbi.nlm.nih.gov​ncbi.nlm.nih.gov. Protons (H⁺) and CO₂ stabilize the deoxy (T) form of hemoglobin, reducing its affinity for oxygen and causing oxygen unloading in tissues (choice A is correct). Cooperative binding (choice B) is about how O₂ binding affects affinity for more O₂ – that underlies the Bohr effect’s impact but the term “Bohr effect” specifically refers to pH/CO₂ influence. “Allosteric inhibition by O₂” (choice C) is not a standard term (O₂ is actually an allosteric activator of other subunits in Hb, not an inhibitor). The Haldane effect (choice D) is related but opposite: it describes how O₂ binding to hemoglobin in the lungs promotes release of CO₂ and H⁺ from hemoglobin. Here we’re dealing with CO₂/H⁺ promoting O₂ release (Bohr effect). So choice A is the correct term.

Question 54: Fetal hemoglobin (HbF) has a higher affinity for oxygen than adult hemoglobin (HbA). This is largely because:
A. HbF has a higher number of heme groups per molecule.
B. HbF does not exhibit the Bohr effect.
C. HbF binds 2,3-BPG less strongly than HbA.
D. HbF subunits are incapable of cooperative binding.

Correct Answer: C
Explanation: Fetal hemoglobin (α₂γ₂) has a lower affinity for 2,3-bisphosphoglycerate (2,3-BPG) compared to adult hemoglobin (α₂β₂). 2,3-BPG binds to deoxy-hemoglobin and stabilizes the T (tense, low affinity) state, thus promoting O₂ release. Because HbF binds 2,3-BPG less well, it remains in a higher affinity state, resulting in a left-shifted O₂ dissociation curve (choice C is correct)​researchgate.net​frontiersin.org. HbF and HbA both have four heme groups per tetramer (choice A is wrong). HbF does exhibit a Bohr effect, although slightly different; the Bohr effect is present in fetal hemoglobin as well (choice B is not the primary reason for higher affinity). HbF is cooperative (sigmoidal binding curve, though slightly different shape than adult); it’s still a tetramer capable of cooperativity (choice D is incorrect). The key difference is the reduced binding of 2,3-BPG due to differences in the γ subunit, which raises HbF’s O₂ affinity so that fetal blood can extract O₂ from maternal blood across the placenta.

Question 55: An acid has a pKₐ of 6.0. Approximately what is the ratio of deprotonated to protonated forms of this acid at pH 7.0?
A. 1:1
B. 1:10
C. 10:1
D. 2:1

Correct Answer: C
Explanation: The Henderson–Hasselbalch equation is pH = pKₐ + log([A⁻]/[HA]), where [A⁻] is the deprotonated (base) form and [HA] is the protonated form. If pH = 7.0 and pKₐ = 6.0, then pH – pKₐ = 1.0. So, log([A⁻]/[HA]) = 1.0, which means [A⁻]/[HA] = 10^1 = 10. Thus, the ratio of deprotonated to protonated is 10:1 (choice C). In other words, at pH one unit above its pKₐ, an acid will be ~90% deprotonated (ten times more in the base form than acid form). A 1:1 ratio occurs when pH = pKₐ (choice A would be at pH 6.0, not 7.0). 1:10 (choice B) would be at pH one unit below pKₐ. 2:1 (choice D) would correspond to pH only slightly above pKₐ (about 0.3 units above, since log(2)=0.3). Here, with pH exactly 1 unit above pKₐ, 10:1 is the correct ratio.

Question 56: In uncontrolled diabetes mellitus, oxaloacetate in the liver is depleted due to gluconeogenesis. As a result, acetyl-CoA from fat breakdown builds up and is diverted to form:
A. lactate.
B. ketone bodies.
C. glycogen.
D. pyruvic acid.

Correct Answer: B
Explanation: In diabetic ketoacidosis or prolonged fasting, oxaloacetate is consumed for gluconeogenesis and is not available in sufficient amount to condense with acetyl-CoA in the TCA cycle. The excess acetyl-CoA (from fatty acid β-oxidation) in the liver is therefore used to produce ketone bodies (acetoacetate, β-hydroxybutyrate)​emedicine.medscape.com (choice B). These ketones are released into the blood and can be used by other tissues (like the brain) as fuel. Lactate (choice A) is produced from pyruvate in anaerobic glycolysis (or from RBCs and exercising muscle), not directly from acetyl-CoA. Glycogen (choice C) formation would not result from high acetyl-CoA in this context; in fact, in uncontrolled diabetes, glycogen is broken down rather than synthesized. Pyruvic acid (pyruvate, choice D) is upstream of acetyl-CoA; acetyl-CoA won’t be converted back to pyruvate (that step is irreversible in human metabolism). Thus, ketone body formation is the pathway used for excess acetyl-CoA when OAA is low.

Question 57: If the pH of a solution is changed from 7.0 to 5.0, how has the hydrogen ion concentration [H⁺] changed?
A. It has increased by a factor of 100.
B. It has increased by a factor of 2.
C. It has decreased by a factor of 100.
D. It has decreased by a factor of 2.

Correct Answer: A
Explanation: The pH scale is logarithmic. A decrease in pH by 2 units (from 7 to 5) means the hydrogen ion concentration has increased by 10^2 = 100 times. Lower pH indicates higher [H⁺]. Specifically, pH 7 corresponds to [H⁺] = 1×10^(-7) M, and pH 5 corresponds to [H⁺] = 1×10^(-5) M; the latter is 100 times greater [H⁺] than the former. Therefore, [H⁺] increased 100-fold (choice A). (For completeness: a pH drop of 1 unit = 10× [H⁺] increase; 2 units = 100×. Choice B, 2-fold, would be incorrect as it’s nowhere near the log scale change. Choices C and D describe decreases, which would correspond to raising pH.)

Question 58: Skeletal muscle glycogen cannot directly contribute to raising blood glucose because skeletal muscle lacks which enzyme?
A. Glycogen phosphorylase
B. Debranching enzyme
C. Glucose-6-phosphatase
D. Phosphoglucomutase

Correct Answer: C
Explanation: Muscle glycogen cannot be used to maintain blood glucose because muscle cells lack glucose-6-phosphatase. This enzyme, present in liver (and kidney), converts glucose-6-phosphate to free glucose that can exit the cell. In muscle, glycogen is broken down to glucose-6-phosphate which enters glycolysis for local energy, but without glucose-6-phosphatase, it cannot be dephosphorylated to release glucose into the bloodstream​ncbi.nlm.nih.gov​accesspediatrics.mhmedical.com. Glycogen phosphorylase (choice A) and debranching enzyme (choice B) are present in muscle and are active in breaking down glycogen, but the breakdown stops at glucose-6-P. Phosphoglucomutase (choice D) is also present in muscle; it converts glucose-1-P to glucose-6-P during glycogenolysis. The key missing enzyme is glucose-6-phosphatase (choice C), which is why muscle glycogen is used for muscle energy but not for blood glucose maintenance.

Question 59: Which amino acid has a side chain that is negatively charged at physiological pH (~7.4)?
A. Lysine
B. Arginine
C. Glutamate
D. Histidine

Correct Answer: C
Explanation: Glutamate (glutamic acid) has a carboxyl group in its side chain with a pKₐ around 4.1. At pH 7.4, which is well above this pKₐ, the side chain is deprotonated and carries a negative charge (–COO⁻) (choice C). Aspartate (aspartic acid) similarly would be negatively charged at pH 7.4. Lysine and arginine (choices A and B) have basic side chains (pKₐ ~10.5 and ~12, respectively) that are protonated (positively charged) at pH 7.4, not negative. Histidine (choice D) has an imidazole side chain with pKₐ ~6.0; at pH 7.4, histidine will be mostly deprotonated (uncharged neutral form). It is not negatively charged; rather, about 90-95% of histidine is neutral and a small fraction is positively charged at physiological pH. Therefore, glutamate is the clear example of a negatively charged side chain at pH 7.4.

Question 60: Low-density lipoprotein (LDL) is often referred to as “bad cholesterol.” High levels of LDL are associated with an increased risk of atherosclerosis because LDL particles:
A. transport cholesterol from the liver to peripheral tissues.
B. transport cholesterol from tissues back to the liver.
C. carry mostly triglycerides and very little cholesterol.
D. lack apolipoproteins and cannot be cleared from circulation.

Correct Answer: A
Explanation: LDL particles are rich in cholesterol and cholesterol esters. They function to deliver cholesterol from the liver to peripheral tissues. High levels of LDL lead to deposition of cholesterol in artery walls, contributing to plaque formation (choice A). In contrast, high-density lipoprotein (HDL) carries cholesterol from peripheral tissues back to the liver for excretion (reverse cholesterol transport), which is protective (choice B describes HDL). LDL carries much more cholesterol relative to triglycerides (choice C is false; chylomicrons and VLDL carry mostly triglycerides). LDL particles do have apolipoproteins (notably ApoB-100) which allow them to bind LDL receptors and be cleared by the liver; the problem in atherosclerosis is often an overload of LDL or dysfunctional uptake, not a complete lack of apolipoproteins (choice D is incorrect). Thus, LDL is “bad” because it deposits cholesterol in arteries (choice A).