Practice Biological & Biochemical Foundations questions for the Medical College Admission Test. Every question includes a full explanation of why the correct answer is right and why the tempting distractors are wrong.
10 example questions with full explanations. Use the interactive practice above to work through the complete set.
Question 1medium
A researcher studying eukaryotic gene expression finds that a particular gene is being actively transcribed. Select all that apply: Which of the following events are required for a mature, translatable mRNA to be produced from this gene?
- A.Addition of a 7-methylguanosine cap to the 5' end of the pre-mRNA✓
- B.Removal of introns by the spliceosome✓
- C.Addition of a poly-A tail to the 3' end of the pre-mRNA✓
- D.Conversion of thymine bases to uracil bases in the coding sequence
- E.Binding of ribosomes to the pre-mRNA in the nucleus prior to export
Correct answer: A, B, C
Eukaryotic pre-mRNA processing requires three main steps before export to the cytoplasm: (A) addition of a 5' 7-methylguanosine cap that protects the mRNA and aids ribosome recognition, (B) splicing out of introns by the spliceosome so only exons remain in the mature mRNA, and (C) addition of a poly-A tail (~200 adenine nucleotides) to the 3' end that protects against degradation and aids export. Option D is incorrect because DNA already contains thymine; the RNA polymerase naturally incorporates uracil instead of thymine during transcription—no separate conversion step is needed. Option E is incorrect because ribosomes are cytoplasmic structures and do not bind pre-mRNA in the nucleus; translation occurs after export.
Question 2medium
An enzyme assay is performed with increasing concentrations of substrate at a fixed enzyme concentration. A competitive inhibitor is then added to a separate set of reactions at the same enzyme concentration. Select all that apply: Which of the following observations would be expected when comparing reactions with the inhibitor to those without it?
- A.The apparent Km increases in the presence of the competitive inhibitor✓
- B.The Vmax is reduced in the presence of the competitive inhibitor
- C.Adding excess substrate can overcome the effect of the competitive inhibitor✓
- D.The competitive inhibitor binds the active site of the enzyme✓
- E.The inhibitor permanently inactivates the enzyme by forming a covalent bond
Correct answer: A, C, D
Competitive inhibitors bind reversibly to the enzyme's active site, competing with the substrate. This raises the apparent Km (more substrate is needed to reach half-maximal velocity), making A correct. Because competitive inhibition is reversible and substrate can outcompete the inhibitor at high concentrations, Vmax is unchanged and adding excess substrate does overcome the inhibition—making C and D correct as well. Option B is incorrect because Vmax remains the same with competitive inhibitors (it is non-competitive inhibitors that reduce Vmax). Option E describes irreversible (covalent) inhibition, which is not a feature of competitive inhibitors by definition.
Question 3medium
During a sprint, skeletal muscle cells rapidly deplete their oxygen supply and shift to anaerobic metabolism. Select all that apply: Which of the following statements accurately describe what occurs in human skeletal muscle cells under these anaerobic conditions?
- A.Pyruvate is converted to lactate by lactate dehydrogenase✓
- B.NAD+ is regenerated, allowing glycolysis to continue✓
- C.Ethanol and CO2 are produced as fermentation byproducts
- D.A net yield of 2 ATP per glucose molecule is produced via glycolysis✓
- E.The TCA cycle operates at full capacity to maximize ATP production
Correct answer: A, B, D
In anaerobic conditions in human skeletal muscle, pyruvate accepts electrons from NADH via lactate dehydrogenase, producing lactate (A is correct). This reaction is critical because it regenerates NAD+, which is required for glycolysis to continue producing ATP (B is correct). Glycolysis yields a net of 2 ATP per glucose molecule (D is correct). Option C describes alcoholic fermentation, which occurs in yeast, not in human skeletal muscle. Option E is incorrect because the TCA cycle requires oxygen (as the final electron acceptor in the ETC); it cannot operate at full capacity under anaerobic conditions.
Question 4medium
A woman is a carrier of an X-linked recessive disorder. She mates with an unaffected man. Select all that apply: Which of the following statements regarding the inheritance of this disorder in their offspring are correct?
- A.50% of their sons are expected to be affected by the disorder✓
- B.50% of their daughters are expected to be carriers of the disorder✓
- C.Affected daughters could result from this cross
- D.The father's X chromosome is passed to all of his daughters✓
- E.Sons inherit their X chromosome from their father
Correct answer: A, B, D
The carrier mother has the genotype X^A X^a and the unaffected father has X^A Y. Sons receive the Y from their father and either X^A or X^a from their mother, so 50% of sons will be affected (X^a Y)—making A correct. Daughters receive the father's X^A along with either X^A or X^a from the mother, so 50% of daughters will be carriers (X^A X^a)—making B correct. The father passes his X chromosome to all daughters (never to sons), making D correct. Option C is incorrect because daughters would need two copies of X^a to be affected; since the father is unaffected and contributes X^A, no daughters can be affected in this cross. Option E is incorrect because sons receive their X chromosome from their mother, not their father.
Question 5medium
During DNA replication in eukaryotes, a researcher treats cells with a drug that specifically inhibits primase. Which of the following best describes the expected outcome?
- A.Both the leading and lagging strands will fail to be synthesized, because DNA polymerase cannot initiate new strands without an RNA primer.✓
- B.Only the leading strand will fail to be synthesized, because it requires a primer to initiate replication at the origin.
- C.Only the lagging strand will fail to be synthesized, because Okazaki fragments each require a new primer, while the leading strand needs only one.
- D.Neither strand will be affected, because helicase can substitute for primase in unwinding DNA to allow polymerase access.
Correct answer: A
Primase synthesizes the short RNA primers required by DNA polymerase III (in prokaryotes) or DNA polymerase delta/epsilon (in eukaryotes) to begin strand synthesis, since DNA polymerases can only extend existing strands—they cannot initiate de novo synthesis. Without primase, no primer is available for either the leading strand (which needs one primer at the origin) or the lagging strand (which needs a new primer for each Okazaki fragment), so both strands fail to be synthesized. Option C is tempting because the lagging strand requires many more primers, but the leading strand still requires at least one primer to initiate replication and thus is also blocked. Option D is incorrect because helicase unwinds the double helix but has no priming activity.
Question 6medium
A patient with a urea cycle deficiency accumulates excess ammonia in the blood (hyperammonemia). The urea cycle is most directly linked to which amino acid metabolic process that generates the ammonia requiring disposal?
- A.Transamination, which transfers amino groups to alpha-keto acids, directly producing urea as a byproduct
- B.Oxidative deamination of glutamate by glutamate dehydrogenase, which releases free ammonium ions (NH4+) that feed into the urea cycle✓
- C.Beta-oxidation of amino acid carbon skeletons, which produces NADH and acetyl-CoA along with ammonia
- D.Peptide bond hydrolysis by proteases, which directly generates ammonia as a product of peptide cleavage
Correct answer: B
Amino acid catabolism begins with transamination, which transfers amino groups to alpha-ketoglutarate to form glutamate; however, transamination itself does not release free ammonia. It is the subsequent oxidative deamination of glutamate by glutamate dehydrogenase that releases free NH4+, which then enters the urea cycle for disposal. Option A is incorrect because transamination transfers—not releases—amino groups and does not directly produce urea. Option C confuses beta-oxidation (a fatty acid pathway) with amino acid catabolism; amino acid carbon skeletons do enter various metabolic pathways, but beta-oxidation specifically refers to fatty acid degradation. Option D is incorrect because proteases hydrolyze peptide bonds to release amino acids, not ammonia directly.
Question 7medium
A researcher studies a population of beetles in which shell color is controlled by a single locus with two alleles (A and a), with allele frequencies p = 0.6 and q = 0.4. After confirming Hardy-Weinberg equilibrium, a new pesticide is introduced that kills only homozygous recessive (aa) beetles before they reproduce. What will happen to allele frequencies in the next generation?
- A.The frequency of the a allele will drop to zero in one generation because all aa individuals are eliminated.
- B.The allele frequencies will remain unchanged because the population remains in Hardy-Weinberg equilibrium after pesticide application.
- C.The frequency of the a allele will decrease, but not to zero, because heterozygous (Aa) carriers still survive and pass on the a allele.✓
- D.The frequency of the A allele will decrease because AA individuals are now at a selective disadvantage relative to Aa individuals.
Correct answer: C
When selection acts against the homozygous recessive genotype (aa), the a allele is not entirely eliminated in one generation because it is 'hidden' in heterozygous carriers (Aa), who survive and continue to pass on a alleles. This is a classic demonstration of why recessive alleles are difficult to eliminate through selection alone. Option A is incorrect because heterozygotes carry the a allele without expressing the selected-against phenotype. Option B is incorrect because the removal of aa individuals before reproduction violates the Hardy-Weinberg assumption of no natural selection, causing allele frequencies to shift. Option D is incorrect because AA individuals are not at a disadvantage—the selective pressure is specifically against aa individuals, so the A allele actually increases in frequency.
Question 8medium
During a sprint, skeletal muscle cells rapidly deplete oxygen and shift to anaerobic glycolysis, producing pyruvate faster than it can be processed by the mitochondria. Which of the following correctly explains why regenerating NAD+ is essential under these conditions, and how it is accomplished in human muscle?
- A.NAD+ is regenerated by the electron transport chain; without oxygen, the ETC is inhibited, so muscle cells import NAD+ from the liver via the bloodstream.
- B.NAD+ is regenerated when pyruvate is converted to ethanol and CO2 by pyruvate decarboxylase and alcohol dehydrogenase, as occurs in all anaerobic organisms.
- C.NAD+ is not required for anaerobic glycolysis; only NADH is consumed during substrate-level phosphorylation steps that produce ATP.
- D.NAD+ is regenerated when pyruvate is converted to lactate by lactate dehydrogenase, allowing glycolysis to continue producing ATP in the absence of oxygen.✓
Correct answer: D
Glycolysis requires NAD+ as an electron acceptor (in the glyceraldehyde-3-phosphate dehydrogenase step); if NAD+ is not regenerated, glycolysis halts and ATP production ceases. Under anaerobic conditions in human muscle, lactate dehydrogenase converts pyruvate to lactate while simultaneously oxidizing NADH back to NAD+, allowing glycolysis to continue. Option A is incorrect because NAD+ cannot be imported from the liver via the blood in meaningful quantities on this timescale, and the ETC does require oxygen but that is separate from the mechanism of NAD+ regeneration in muscle. Option B describes alcoholic fermentation, which occurs in yeast and some other microorganisms but not in humans. Option C is incorrect because NAD+ is specifically required as an oxidizing agent in step 6 of glycolysis, and NADH is produced (not consumed) during substrate-level phosphorylation steps.
Question 9medium
A cell biologist observes that a particular protein is synthesized on ribosomes in the cytosol but is ultimately found in the lumen of the endoplasmic reticulum. A mutation is then introduced that removes the protein's signal sequence. Where would the mutant protein most likely be found?
- A.In the cytosol, because the signal sequence is required for the ribosome to dock with the rough ER and translocate the protein into the lumen✓
- B.In the nucleus, because without a signal sequence the protein defaults to nuclear import
- C.In the mitochondrial matrix, because cytosolic proteins are preferentially imported into mitochondria
- D.On the outer leaflet of the plasma membrane, because secretory proteins are redirected to the cell surface when ER import is blocked
Correct answer: A
The N-terminal signal sequence is recognized by the Signal Recognition Particle (SRP), which directs the ribosome to the rough ER membrane and initiates co-translational translocation of the protein into the ER lumen. Without this signal sequence, SRP cannot recognize and dock the ribosome to the ER, so translation continues in the cytosol and the completed protein remains there. Option C is incorrect because nuclear import requires a nuclear localization signal (NLS), which is distinct from an ER signal sequence. Option A is incorrect because mitochondrial import requires a specific mitochondrial targeting sequence, not cytosolic default routing. Option D is incorrect because there is no mechanism to redirect proteins lacking an ER signal sequence to the plasma membrane; the secretory pathway requires ER entry as the first step.
Question 10medium
During meiosis II, homologous chromosomes separate from each other, reducing the chromosome number from diploid (2n) to haploid (n) in the resulting daughter cells.
Correct answer: B
This statement is False. It is meiosis I — not meiosis II — in which homologous chromosomes separate, reducing the cell from diploid (2n) to haploid (n). Meiosis II resembles mitosis: sister chromatids separate, producing four haploid daughter cells. A student integrating knowledge of both meiotic divisions must distinguish between homolog separation (meiosis I) and sister chromatid separation (meiosis II) to answer correctly.