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Dat Biology Section 2026 Complete Practice Test Bank With Detailed Explanations And Rationales Q&A| Instant Download Pdf

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This DAT Biology practice test bank gives you 150 questions with detailed answer explanations covering cell biology, genetics, metabolism, and molecular biology. Each rationale walks you through the reasoning so you can spot weak areas and build confidence for the real exam. Use it to practice under test-like conditions and review every topic thoroughly.

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PASS THE DAT BIOLOGY SECTION
2026 COMPLETE PRACTICE TEST
BANK WITH DETAILED

1. A mutation in the gene encoding the enzyme pyruvate dehydrogenase kinase (PDK)
results in a constitutively active kinase. Which metabolic consequence is most likely in
a well-oxygenated hepatocyte?

A. Increased conversion of pyruvate to acetyl-CoA
B. Enhanced flux through the citric acid cycle
C. Impaired conversion of pyruvate to acetyl-CoA
D. Stimulation of gluconeogenesis from lactate

Answer: C. Impaired conversion of pyruvate to acetyl-CoA
Rationale: PDK phosphorylates and inhibits pyruvate dehydrogenase (PDH), the enzyme
that converts pyruvate to acetyl-CoA. Constitutive PDK activity would keep PDH inhibited
even under aerobic conditions, reducing acetyl-CoA production. This would also limit
citric acid cycle flux and promote alternative fates of pyruvate such as gluconeogenesis, but
the direct and immediate consequence is impaired acetyl-CoA formation.


2. In a population of Drosophila, a recessive X-linked lethal allele causes death in
hemizygous males before eclosion. If the allele frequency is 0.1 in females, what
proportion of adult females will be carriers?
A. 0.09
B. 0.18
C. 0.42
D. 0.90

Answer: B. 0.18
Rationale: For an X-linked recessive lethal, adult males do not carry the allele. Assuming
Hardy-Weinberg equilibrium among adult females, let q be the frequency of the lethal allele
among X chromosomes in females. Given q = 0.1, the frequency of heterozygous carriers is
2pq = 2(0.9)(0.1) = 0.18. Homozygous recessive females (q² = 0.01) may also die or be
affected, but the question asks for carriers.




Page 1 of 76

,3. Which observation best supports the endosymbiotic theory for the origin of
mitochondria?

A. Mitochondria contain circular DNA and divide by binary fission.
B. Mitochondria are enclosed by a double membrane.
C. Mitochondria synthesize some proteins using their own ribosomes.
D. Mitochondria are inherited maternally in most eukaryotes.

Answer: A. Mitochondria contain circular DNA and divide by binary fission.
Rationale: The presence of circular DNA and binary fission is a strong indicator of
bacterial ancestry, as these features are characteristic of prokaryotes. While double
membranes and independent protein synthesis also support endosymbiosis, circular DNA
and division by fission are more direct evidence of a prokaryotic origin.


4. A patient with a mutation in the gene encoding the -subunit of the epithelial sodium
channel (ENaC) exhibits salt wasting and hypotension. Which renal tubule segment is
most directly affected?

A. Proximal convoluted tubule
B. Thick ascending limb of the loop of Henle
C. Distal convoluted tubule
D. Collecting duct

Answer: D. Collecting duct
Rationale: ENaC is primarily located in the apical membrane of principal cells in the
collecting duct, where it mediates sodium reabsorption under the regulation of aldosterone.
Mutations causing loss of function lead to salt wasting and hypotension. The other
segments have different sodium transport mechanisms.




Page 2 of 76

,5. During an experiment, a researcher adds a drug that inhibits the enzyme
dihydrofolate reductase (DHFR) to a culture of rapidly dividing cells. Which process
will be most immediately impaired?

A. DNA replication
B. Transcription
C. Translation
D. Glycolysis

Answer: A. DNA replication
Rationale: DHFR is essential for the synthesis of tetrahydrofolate, which is required for the
de novo synthesis of thymidylate and purines, precursors for DNA replication. Inhibition
rapidly halts DNA synthesis. Transcription and translation are less directly affected, and
glycolysis does not require folate.


6. Which of the following best explains why the theoretical maximum ATP yield from
complete oxidation of one glucose molecule is lower than the often-cited 36-38 ATP?

A. Some ATP is consumed during the transport of NADH into mitochondria.
B. The proton motive force is partially dissipated by uncoupling proteins.
C. Glycolysis produces fewer NADH than previously thought.
D. The citric acid cycle is not fully active under aerobic conditions.

Answer: A. Some ATP is consumed during the transport of NADH into mitochondria.
Rationale: The transport of cytosolic NADH into the mitochondrial matrix via the
glycerol-3-phosphate shuttle or malate-aspartate shuttle consumes energy or yields fewer
ATP, reducing the net yield. Uncoupling proteins can dissipate the proton gradient, but
under normal conditions they are not the primary reason for the lower yield. Glycolysis
and the citric acid cycle yields are well-established.




Page 3 of 76

, 7. A researcher performs a pulse-chase experiment using radioactive leucine to track
protein secretion in pancreatic acinar cells. After a short pulse, where would the
radioactivity be found immediately after the chase?

A. In the cytosol
B. In the rough endoplasmic reticulum
C. In the Golgi apparatus
D. In secretory vesicles

Answer: B. In the rough endoplasmic reticulum
Rationale: Immediately after a short pulse, newly synthesized secretory proteins are
translocated into the rough endoplasmic reticulum (RER) where they undergo initial
processing. The chase period allows tracking to later compartments, but the question asks
immediately after the chase, which is still the RER.


8. Which of the following is the most likely consequence of a mutation that prevents
the acrosome reaction in mammalian sperm?

A. Inability to penetrate the zona pellucida
B. Failure of the sperm to undergo capacitation
C. Premature fusion of sperm and egg plasma membranes
D. Loss of flagellar motility

Answer: A. Inability to penetrate the zona pellucida
Rationale: The acrosome reaction releases hydrolytic enzymes that digest a path through
the zona pellucida, allowing the sperm to reach the egg plasma membrane. Without it,
sperm cannot penetrate the zona pellucida. Capacitation precedes the acrosome reaction,
and flagellar motility is independent.




Page 4 of 76

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16 de octubre de 2026
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