MCAT Biology Practice Questions
Master MCAT Biology using focused practice problems, essential topic breakdowns, and proven techniques to improve accuracy and retention.
(Note: This resource also appears in our MCAT Ultimate Guide.)
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Introduction
MCAT Biology Practice Passage #1
MCAT Biology Practice Passage #2
MCAT Biology Practice Passage #3
MCAT Biology Practice Questions (Standalone)
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Introduction
You’ve taken all of your premed classes and are ready to start prepping for the MCAT. You already made a great study schedule, have started content review, and are ready to start taking some more practice problems.
Unlike most tests you’ve taken before in undergrad, the MCAT is an entirely different challenge. The exam is difficult and long, especially if you don’t prepare the right way. As you progress further in your medical training, your method of studying will shift from content review to practice problems. The MCAT is no exception—practice problems will help to significantly improve your score.
Don’t worry, though. We have you covered. Biology is an important subject on the MCAT, and you’ll see biology in the chemistry/physics, biology/biochemistry, and psychology/sociology sections of the exam. Therefore, you’ll need to be ready to answer bio questions asked in a variety of contexts.
Here, we’ll test your biology knowledge using MCAT-style passages written by a 528 scorer. The MCAT relies heavily on repurposing scientific articles and then asking you questions about those articles. As a result, you need to be able to sift through scientific data and answer questions that combine information from the passage with your outside knowledge.
Use the following three biology passages and five standalone questions to test your ability to apply your biology knowledge to real, MCAT-style passages. Each explanation for the passage-based questions will have suggestions for what you should review if you miss a question. Good luck!
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MCAT Biology Practice Passage #1
Postoperative acute kidney injury (AKI) is a severe complication after liver transplantation (LT) as its deterioration and magnification can lead to an increase in mortality. Connexin43 (Cx43) mediates direct transmission of intracellular signals between neighboring cells, always considered to be the potent biological basis of organ damage deterioration and magnification.
Researchers obtained donor liver tissues of patients in order to explore the effects of Cx43 on organ damage. Reperfusion is a biological process used to refill blood in AKI. Donor livers displayed more severe lobular distortion with necrosis, apparent edema, hemorrhage, and neutrophil infiltration after reperfusion, compared with donor liver tissues obtained before reperfusion or normal liver tissue (from patients with hepatic hemangioma). After reperfusion, Cx43 mRNA and protein levels were determined as shown in Figure 1.
Figure 1 Cx43 mRNA and protein levels in normal or donor livers, before reperfusion, and after reperfusion.
Autologous orthotopic liver transplantation (AOLT) rats were built by researchers in order to further study AKI. Researchers hypothesized that the change in Cx43 expression might play an important role in AKI following AOLT. Researchers used heptanol, a well-known inhibitor of Cx43 without hepatotoxicity, to alter the function of junctions composed of Cx43. Heptanol had no effects on Cx43 expression in kidneys or livers.
Heptanol or a sham control were injected intravenously into the rats, and the pathological score of kidney cells was determined to analyze disease progression. The results are shown in Figure 2.
Figure 2 Pathological score of AOLT mice treated with Cx43 inhibitor.
Researchers also believe that the RIP1 may be involved in Cx43 function, and they confirmed the interaction of the proteins through co-immunoprecipitation. RIP1 is a typical marker of necroptosis.
1. What type of junction does Cx43 most likely form?
A) Gap junction
B) Tight junction
C) Desmosome
D) Hemidesmosome
2. A researcher discovers a new Cx43 inhibitor made of a glycerol backbone and three saturated fatty acid tails. Which of the following describes the path by which the inhibitor enters the cells?
A) The inhibitor is recognized by a specific receptor.
B) The inhibitor is taken up by a non-specific receptor.
C) The inhibitor diffuses through the cell membrane.
D) The inhibitor passes through the cell membrane via a channel.
3. Researchers decide to interrupt the Cx43-RIP1 interaction in the AOLT rat model without using a small molecule inhibitor. Which of the following will most likely disrupt the interaction?
A) SiRNA for RIP1
B) Radiation against heptanol
C) Acetyltransferase specific for β-actin
D) Plasmid producing Cx43
4. Which conclusion about Cx43 regulation is best supported by Figure 1?
A) All expression control occurs at the transcriptional level.
B) All expression control occurs at the translational level.
C) Donor livers have higher levels of Cx43 after reperfusion when compared to normal liver tissue.
D) Normal livers treated with heptanol have lower levels of Cx43 compared to donor livers treated with heptanol.
Answer key for practice passage #1
1. Answer choice A is correct. Cx43 is a connexin protein as stated by the passage, and connexins form gap junctions (choice A is correct).
Review junction types.
2. Answer choice C is correct. The inhibitor described in the question stem is very hydrophobic, and this means it will pass through the membrane via diffusion (choice C is correct).
Review plasma membrane composition and properties.
3. Answer choice A is correct. SiRNAs bind to mRNAs for certain genes and lead to degradation. If the RIP1 mRNA is degraded, Cx43 will no longer be able to interact with RIP1 (choice A is correct). Radiation is highly non-specific and would likely affect other processes (choice B is incorrect). β-actin is not involved in the Cx43-RIP1 interaction based on information from the passage (choice C is incorrect). A plasmid producing Cx43 would likely not be taken up by the rat cells, and an increase in Cx43 expression should not disrupt the interaction (choice D is incorrect).
Review siRNAs.
4. Answer choice C is correct. The graph in Figure 1 shows that donor livers have higher expression levels of Cx43 mRNA when compared to normal liver tissue (choice C is correct). The mRNA expression levels given by the graph and the protein expression levels given by the blot are different, meaning that regulation occurs at both the transcriptional and translational levels (choices A and B are incorrect). Heptanol is not addressed in Figure 1 (choice D is incorrect).
Review transcription versus translation. Practice interpreting experimental data, and make sure to understand the figure completely.
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MCAT Biology Practice Passage #2
Proopiomelanocortin (POMC) neurons in the arcuate nucleus of the hypothalamus (ARH) control energy homeostasis by sensing hormonal and nutrient cues and activating secondary melanocortin sensing neurons. Researchers identified the expression of a G protein-coupled receptor, Gpr17, in the ARH and hypothesized that it contributes to the regulatory function of POMC neurons on metabolism.
In order to test this hypothesis, researchers generated POMC neuron-specific Gpr17 knockout (PGKO) mice and determined their energy and glucose metabolic phenotypes on normal chow diet (NCD) and high-fat diet (HFD).
In Experiment 1, researchers measured body weight and weight gain of WT and PGKO mice fed a high-fat diet. Body weight was assessed at 5 months and 10 months after the diet change for adult mice. The results from this experiment are shown in Figure 1.
Figure 1 Change in body weight and weight gain for WT versus PGKO mice on an HFD.
Neuropeptides from POMC neurons are known to play a role in appetite regulation. α-MSH is anorexigenic, while the effect of β-endorphin (β-EP) on satiety is context-dependent. The bioavailability of α-MSH and β-EP is partially determined by the expression of POMC and subsequent proteolytic processing by PC1, PC2, and Cpe. Researchers measured α- MSH, β-EP, and POMC neuropeptides in the medio- basal hypothalamic samples from mice fed HFD for 2 weeks. The results are shown in Figure 2.
Figure 2 Changes in levels of POMC neuropeptides.
1. Researchers hypothesize that PGKO mice maintain a low weight on an HFD diet by increasing β-oxidation. If this hypothesis is correct, which molecule might they expect to find at higher levels than WT mice?
A) Pyruvate
B) Acetyl-CoA
C) Glucose-6 phosphate
D) Glucose-1 phosphate
2. Based on Figure 2, which of the following conclusions is true?
A) Females have higher levels of POMC neurons.
B) PKGO mice produce higher levels of neuropeptides.
C) WT females have higher levels of α-MSH than WT males.
D) WT males have higher levels of intracellular β-EP than WT females.
3. Researchers are interested in studying a new G-protein coupled receptor (GPCR) called GPCR84. Which of the following mutations may researchers introduce to the protein to decrease GPCR84 signaling?
A) Mutate the alpha subunit of GPCR84 so that it is locked in a GDP-bound form.
B) Mutate the alpha subunit of GPCR84 so that it is locked in a GTP-bound form.
C) Add an agonist ligand that binds to GPCR84.
D) Express GPCR84 on cellular surfaces that are spatially close to agonist ligands.
4. Researchers observe a surge in α-MSH immediately after eating along with which other hormone?
A) Insulin
B) Glucagon
C) Somatostatin
D) Norepinephrine
5. In the HFD, researchers feed mice unsaturated fats. Which of the following represents a fat with the most double bonds?
A) 18:1∆-9
B) 18:2∆-9,12
C) 28:1∆-17
D) 28:3∆-9,17,23
Answer key for practice passage #2
1. Answer choice B is correct. β-oxidation converts fatty acids into acetyl-CoA (choice B is correct). Fatty acids cannot be converted to carbohydrates to produce energy in humans (choices A, C, and D are incorrect).
Review β-oxidation.
2. Answer choice C is correct. Looking at the left graph in Figure 2, there is a significant difference (indicated by ***) between α-MSH levels in male and female WT mice (choice C is correct). The figure provides no information about the number of POMC neurons (choice A is incorrect). PKGO mice do not produce more neuropeptides in male mice (choice B is incorrect). WT males have lower levels of β-EP (choice D is incorrect).
Review significant differences. Practice interpreting experimental data, and make sure to understand the figure completely.
3. Answer choice A is correct. GPCR signaling exchanges the GDP bound to the alpha subunit for a GTP to carry out signaling. If the GDP form is locked, signaling would not be carried out (choice A is correct; choice B is incorrect). An agonist ligand would increase GPCR signaling (choices C and D are incorrect).
Review the GPCR signaling pathway.
4. Answer choice A is correct. α-MSH is anorexigenic, meaning it decreases appetite, often after eating a meal. Insulin is also released after eating a meal in order to promote the uptake of glucose from the blood (choice A is correct). Glucagon is released during periods of starvation to promote the release of glucose into the blood (choice B is incorrect). Somatostatin blocks both insulin and glucagon (choice C is incorrect). Norepinephrine is released during the flight or fight response during which digestion is halted (choice D is incorrect).
Review hormones, especially insulin and glucagon.
5. Answer choice D is correct. The first number indicates the number of carbons, and the second number indicates the number of double bonds. The numbers after the “∆-“ indicate the positions of the double bonds. So, in the correct answer, there are 28 carbons with 3 double bonds starting at carbons 9, 17, and 23 (choice D is correct). Answer choices A and C have one double bond while answer choice B has two double bonds (choices A, B, and C are incorrect).
Review lipid nomenclature.
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MCAT Biology Practice Passage #3
Oculocutaneous albinism (OCA) is a genetically heterogeneous group of disorders characterized by absent or reduced pigmentation of the skin, hair, and eyes from the time of birth. OCA type II (OCA2) is the most common form of the disease, but no effective treatments currently exist for the disease. Researchers interested in studying OCA2 decided to develop a mouse model of the disease as a model through which potential therapeutics could be evaluated.
To identify the causative gene in the Z0015 mouse cell line, researchers performed a family-based genome-wide linkage study (GWLS) to map the chromosome regions co-segregating with the mutant phenotypes. The results showed a single missense mutation c.2228 C > T that resulted in a serine transition to leucine (OCA2S743L/S743L hereafter referred to as S743L) in exon 21. Further data revealed that the S743L mutation was likely the causative mutation for OCA2 in mice. This mutation is orthologous to the human S736L mutation.
The S743L mutation is located in the predicted trans- membrane domain 10 of the porcine OCA2 protein and is evolutionarily conserved among distinct mammals indicating its vital role within this domain for the function of OCA2. The S743L mutation did not alter OCA2 expression at the transcript level. Researchers then hypothesized that protein expression might be affected, and they measured protein expression in the eyes and scalp of mutant pigs. The results are shown in Figure 1.
Figure 1 Measuring OCA2 expression in S743L mutant mice by Western blotting.
Next, researchers sought to investigate whether the S743L mutation induced growth or fertility aberrations in OCA2S743L/S743L pigs. Researchers characterized the growth and reproductive traits of the pigs in comparison to WT pigs by measuring litter size, the number of weaned pigs, and mean litter birth weight. The results from this set of experiments are shown in Figure 2.
Figure 2 Total litter size, number weaned, and mean litter birth weight for WT vs. S743L mutants.
1. The change in properties of amino acids in the causative mutation for OCA2 can be best described as changing a:
A) Charged amino acid to a hydrophobic amino acid
B) Polar, uncharged amino acid to a hydrophobic amino acid.
C) Hydrophobic amino acid to a polar, uncharged amino acid.
D) Aromatic amino acid to a charged amino acid.
2. Based on Figure 1, which of the following conclusions is most likely to be correct?
A) The S743L mutation decreases OCA2 transcription.
B) WT mice contain higher levels of OCA2 transcripts.
C) S743L mutants have lower levels of OCA2 protein.
D) WT mice translationally upregulate OCA2 protein.
3. Researchers find a second mutant that produces a UAG codon in the DNA. Which of the following best characterizes the nature of the mutation?
A) Nonsense
B) Point
C) Frameshift
D) Missense
4. A researcher concludes that S473 mice produce heavier offspring than WT mice. Is this conclusion reasonable?
A) Yes; S473 mice produce a litter with a higher birth weight.
B) Yes; S473 mice without functional OCA2 have lighter organs.
C) No; the difference in mean offspring weight between WT and S473 mice is not significant.
D) The data in Figure 2 contains too many outliers to make a conclusion.
Answer key for practice passage #3
1. Answer choice B is correct. The passage indicates that the mutation is an S to an L. Serine is a polar, uncharged amino acid whereas leucine is hydrophobic (choice B is correct).
Review amino acids.
2. Answer choice C is correct. The passage states that “the S743L mutation did not alter OCA2 expression at the transcript level.” Therefore, choices A and B are incorrect. Western blots measure protein expression, and Figure 1 shows that in both the eyes and scalp, S743L mice have less OCA2 protein (choice C is correct). Though WT mice do have more OCA2 protein, Figure 1 does not provide enough information to say that this is due to translational upregulation of OCA2. In reality, it is likely translational downregulation of OCA2 in the S743L mice (choice D is incorrect).
Review transcription versus translation. Practice interpreting experimental data, and make sure to understand the figure completely.
3. Answer choice A is correct. UAG is one of three stop codons, and this leads to a nonsense mutation (choice A is correct). A point mutation is a change in a single nucleotide, but the original codon is not known so we cannot state that the UAG is a point mutation (choice B is incorrect). A frameshift mutation is due to the insertion or deletion of nucleotides that are not a multiple of three, but once again, more information would be needed to determine if the UAG is a frameshift mutation in this case (choice C is incorrect). A missense mutation results in a different amino acid (choice D is incorrect).
Review types of mutations in proteins. Review the triplet code and know the three codons encoding for a stop codon.
4. Answer choice C is correct. There are no asterisks (* or ** or ***) indicating significance in Figure 2. Therefore, there is no significant difference in mean birth weight (choice C is correct; choice A is incorrect). No conclusion about organ weight can be made (choice B is incorrect). The conclusion from Figure 2 is that there is no significant difference (choice D is incorrect).
Review significant differences, p-values, and null hypotheses. Practice interpreting experimental data, and make sure to understand the figure completely.
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MCAT Biology Practice Questions (Standalone)
1. Researchers are studying a potential novel oncoprotein implicated in gastric cancer. RT-qPCR results show an upregulation of the mRNA transcripts for the novel protein. Which of the following is a potential mechanism by which transcription of the protein is upregulated?
A) Opening of chromatin structure
B) Histone deacetylation
C) Increased ribosome synthesis
D) Increased amino acid availability
2. A red flower is mated with a white flower, resulting in pink progeny. This is an example of:
A) Co-dominance
B) Heterodominance
C) Complete dominance
D) Incomplete dominance
3. A non-diabetic individual eats a large, carbohydrate-rich meal. Which of the following glycolysis enzymes is insulin most likely to upregulate?
A) Phosphoglucose isomerase
B) Phosphofructokinase-1
C) Phosphoglyceromutase
D) Enolase
4. Cells are treated with a drug that dissipates the proton gradient across the inner mitochondrial membrane. What is the net result?
A) Cancer progression is halted
B) ATP cannot be produced via ATP synthase
C) The cell will use exocytosis to dispose of the nonfunctional mitochondria
D) Cells will rapidly evolve a new ATP synthesis mechanism
5. Following a viral infection, a patient is not showing any respiratory symptoms or fever. What cycle is the virus most likely in?
A) Lytic
B) Lysogenic
C) Invasion
D) Protein production
Answer key for standalone practice questions
1. Answer choice A is correct. Opening of the chromatin structure allows transcription factors to more readily access the DNA, leading to an increase in transcription, which produces mRNA transcripts. Histone deacetylation leads to a closing of the chromatin structure (choice B is incorrect). Ribosome synthesis and amino acid availability might indicate increased protein translation, not transcription (choices C and D are incorrect).
Review Histone methylation and acetylation. Review euchromatin and heterochromatin.
2. Answer choice D is correct. Incomplete dominance describes the “mixing” of two phenotypes, which is what occurs when a red and white flower are mated to form a pink (intermediate phenotype). Co-dominance describes both phenotypes being shown, which would mean daughter progeny have both red and white splotches (choice A is incorrect). Complete dominance describes normal dominance, which would be red flowers in this case if the red flowers are dominant (choice C is incorrect). Heterodominance is a made-up term (choice B is incorrect).
Reivew methods of inheritance.
3. Answer choice B is correct. Phosphofructokinase-1 (PFK-1) is an irreversible enzyme, which means it is an important point for regulating flux, or flow, through glycolysis. After eating a large meal, insulin can upregulate PFK-1 activity through a series of steps, allowing glycolysis to proceed more quickly. Answer choices A, C, and D are reversible enzymes, so they are not good points for regulation and are not used as control points for flux through a pathway.
Review the steps of glycolgysis.
4. Answer choice B is correct. The proton gradient is generated so that protons flow from the intermembrane space into the matrix based on their chemical gradient. The passage of protons through the inner mitochondrial membrane then allows ATP synthase the necessary energy to produce ATP. The question stem does not address cancer progression (choice A is incorrect). The cells will not get rid of the mitochondria when they stop working, and they are also unlikely to rapidly evolve a new ATP synthesis mechanism as the first one likely took millions of years to develop in the first place (choices C and D are incorrect).
Review ATP synthesis and the electron transport chain.
5. Answer choice B is correct. The lysogenic cycle is where the virus integrates its own genome into the host genome, but it does not lyse the cell or produce lots of progeny yet. As a result, it may be undetected by the immune system. The lytic cycle occurs when the virus begins producing progeny and lysing host cells (choice A is incorrect). Invasion and protein production describe processes that the virus may use, but they are not actual cycles (choices C and D are incorrect).
Review lytic vs. lysogenic viral cycles
