UWORLD B/B PT2 EXAM QUESTIONS
AND ANSWERS GRADED A+ 2025/2026
Malonyl-CoA is an intermediate in fatty acid synthesis and an inhibitor of long-chain fatty acid
transport to the mitochondria. Given this information, which of the following experimental
results would provide evidence that a certain drug causes cells to carry out increased beta
oxidation?
A. Adding malonyl-CoA to drug-treated cells causes an increase in the level of cytosolic acetyl-
CoA compared to cells without malonyl-CoA treatment.
B. Adding malonyl-CoA to drug-treated cells causes a decrease in the level of mitochondrial
acetyl-CoA compared to cells without malonyl-CoA treatment.
C. Treating cells with the drug results in a decreased concentration of cytosolic malonyl-CoA
compared to untreated cells.
D. Treating cells with the drug results in an increased concentration of cytosolic malonyl-CoA
compared to untreated cells. - ANS C. Treating cells with the drug results in a decreased
concentration of cytosolic malonyl-CoA compared to untreated cells.
Malonyl-CoA - ANS helps regulate fatty acid metabolism by inhibiting long-chain fatty acid
transport into the mitochondria. Processes that decrease malonyl-CoA levels are likely to
increase fatty acid transport to the mitochondria and, as a result, increase beta oxidation.
Two separate reactions are conducted in which a compound containing a ketone, an ester, and a
carboxylic acid is reacted with borane (BH3) in THF in one reaction and with NaBH4 in methanol
in the other. Which of the following explains why different products are observed?
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,A. BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic acids.
B. BH3 does not reduce ketones, and NaBH4 will selectively reduce esters.
C. BH3 reduces carboxylic acids and esters, and NaBH4 will only reduce esters.
D. BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones. -
ANS D. BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce
ketones.
Carboxylic acids can be selectively reduced to primary alcohols by the reducing agent BH3. They
cannot be reduced by NaBH4, which is selective for the reactive carbonyls ketones and
aldehydes.
The sequence below is a portion of exon 7 of the SMN1 mRNA transcript:
5′ - UCAAGUGAUUCUCCU - 3′
Which of the following represents the corresponding DNA coding strand sequence for this
transcript?
A. 5′ - AGGAGAATCACTTGA - 3′
B. 5′ - TCCTCTTAGTGAACT - 3′
C. 5′ - AGTTCACTAAGAGGA - 3′
D. 5′ - TCAAGTGATTCTCCT - 3′ - ANS D. 5′ - TCAAGTGATTCTCCT - 3′
Which step(s) are required to produce a mature FL-SMN transcript?
I. Addition of multiple adenine nucleotides to the 3′ end of SMN1 pre-mRNA
II. Removal of the 7-methylguanosine cap from mature mRNA
III. DNA polymerase binding to the SMN1 gene
A. I only
B. I and III only
C. II and III only
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,D. I, II, and III - ANS B. I and III only
Transcription - ANS is the process of synthesizing RNA from template DNA and begins with
RNA polymerase II binding to the gene promoter region. RNA polymerase II reads the DNA
strand in a 3′ to 5′ direction to generate a 5′ to 3′ pre-mRNA molecule. The pre-mRNA transcript
undergoes 5′ capping, the addition of a 3′ poly-A tail, and excision of noncoding regions
(introns) to be converted into mature mRNA.
Processing (which includes splicing) of the pre-mRNA - ANS transcript into mature mRNA
occurs in the nucleus.
Valproic acid (VA) is a histone deacetylase inhibitor that has been investigated for SMA therapy.
VA would most likely improve motor neuron survival by:
A. increasing SMN2 expression by making DNA more accessible.
B. increasing SMN2 expression by removing acetyl groups from histones.
C. decreasing SMN2 expression by modifying CpG sites in DNA.
D. decreasing SMN2 expression by inducing heterochromatin formation. - ANS A. increasing
SMN2 expression by making DNA more accessible.
Assume HNO2 dissociates in water with a Ka of 4 × 10−5 at 298 K.
HNO2(aq) ⇄ NO2−(aq) + H3O+(aq)
An aqueous solution containing a mixture of HNO2 and NO2− has a pH of 6.2 at 298 K.
Compared to the concentration of NO2−, the concentration of HNO2 must be:
A. greater, because the pH of the solution is greater than its pKa.
B. less, because the pH of the solution is greater than its pKa.
C. greater, because the pH of the solution is less than its pKa.
D. less, because the pH of the solution is less than its pKa. - ANS B. less, because the pH of
the solution is greater than its pKa.
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, If a naturally occurring sample of an unidentified element is found to contain three isotopes (A,
B, and C) and consists of 90.5% isotope A (mass number 20), 0.3% isotope B (mass number 21),
and 9.2% isotope C (mass number 22), the atomic weight of the element measured from the
sample will be:
A. greater than 21 amu.
B. less than 21 amu.
C. equal to 21 amu.
D. unrelated to the amount of isotope B. - ANS B. less than 21 amu.
Which of the following does NOT describe why a nucleophile will react more quickly with acetic
anhydride than with N,N-diisopropylisobutyramide?
A. N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride.
B. Acetic anhydride is a stronger electrophile than N,N-diisopropylisobutyramide.
C. Acetic anhydride is less sterically hindered than N,N-diisopropylisobutyramide.
D. N,N-diisopropylisobutyramide has a less stable leaving group than acetic anhydride. -
ANS A. N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride.
Reactivity of carboxylic acid derivatives - ANS depends on various factors, including inductive
effects, steric effects, and leaving group stability. The inductive effect is an electronic property in
which the electrons are donated through sigma bonds. Anhydrides contain stronger electron
withdrawing groups than amides; therefore, anhydrides experience a greater inductive effect
and react more readily than amides.
Of the four original solutions studied, which of the following pairs of solutions contain epimers?
Maltose and lactose
Galactose and glucose
Maltose and glucose
Lactose and galactose
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AND ANSWERS GRADED A+ 2025/2026
Malonyl-CoA is an intermediate in fatty acid synthesis and an inhibitor of long-chain fatty acid
transport to the mitochondria. Given this information, which of the following experimental
results would provide evidence that a certain drug causes cells to carry out increased beta
oxidation?
A. Adding malonyl-CoA to drug-treated cells causes an increase in the level of cytosolic acetyl-
CoA compared to cells without malonyl-CoA treatment.
B. Adding malonyl-CoA to drug-treated cells causes a decrease in the level of mitochondrial
acetyl-CoA compared to cells without malonyl-CoA treatment.
C. Treating cells with the drug results in a decreased concentration of cytosolic malonyl-CoA
compared to untreated cells.
D. Treating cells with the drug results in an increased concentration of cytosolic malonyl-CoA
compared to untreated cells. - ANS C. Treating cells with the drug results in a decreased
concentration of cytosolic malonyl-CoA compared to untreated cells.
Malonyl-CoA - ANS helps regulate fatty acid metabolism by inhibiting long-chain fatty acid
transport into the mitochondria. Processes that decrease malonyl-CoA levels are likely to
increase fatty acid transport to the mitochondria and, as a result, increase beta oxidation.
Two separate reactions are conducted in which a compound containing a ketone, an ester, and a
carboxylic acid is reacted with borane (BH3) in THF in one reaction and with NaBH4 in methanol
in the other. Which of the following explains why different products are observed?
1 @COPYRIGHT 2026 ALLRIGHTS RESERVED.
,A. BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic acids.
B. BH3 does not reduce ketones, and NaBH4 will selectively reduce esters.
C. BH3 reduces carboxylic acids and esters, and NaBH4 will only reduce esters.
D. BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones. -
ANS D. BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce
ketones.
Carboxylic acids can be selectively reduced to primary alcohols by the reducing agent BH3. They
cannot be reduced by NaBH4, which is selective for the reactive carbonyls ketones and
aldehydes.
The sequence below is a portion of exon 7 of the SMN1 mRNA transcript:
5′ - UCAAGUGAUUCUCCU - 3′
Which of the following represents the corresponding DNA coding strand sequence for this
transcript?
A. 5′ - AGGAGAATCACTTGA - 3′
B. 5′ - TCCTCTTAGTGAACT - 3′
C. 5′ - AGTTCACTAAGAGGA - 3′
D. 5′ - TCAAGTGATTCTCCT - 3′ - ANS D. 5′ - TCAAGTGATTCTCCT - 3′
Which step(s) are required to produce a mature FL-SMN transcript?
I. Addition of multiple adenine nucleotides to the 3′ end of SMN1 pre-mRNA
II. Removal of the 7-methylguanosine cap from mature mRNA
III. DNA polymerase binding to the SMN1 gene
A. I only
B. I and III only
C. II and III only
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,D. I, II, and III - ANS B. I and III only
Transcription - ANS is the process of synthesizing RNA from template DNA and begins with
RNA polymerase II binding to the gene promoter region. RNA polymerase II reads the DNA
strand in a 3′ to 5′ direction to generate a 5′ to 3′ pre-mRNA molecule. The pre-mRNA transcript
undergoes 5′ capping, the addition of a 3′ poly-A tail, and excision of noncoding regions
(introns) to be converted into mature mRNA.
Processing (which includes splicing) of the pre-mRNA - ANS transcript into mature mRNA
occurs in the nucleus.
Valproic acid (VA) is a histone deacetylase inhibitor that has been investigated for SMA therapy.
VA would most likely improve motor neuron survival by:
A. increasing SMN2 expression by making DNA more accessible.
B. increasing SMN2 expression by removing acetyl groups from histones.
C. decreasing SMN2 expression by modifying CpG sites in DNA.
D. decreasing SMN2 expression by inducing heterochromatin formation. - ANS A. increasing
SMN2 expression by making DNA more accessible.
Assume HNO2 dissociates in water with a Ka of 4 × 10−5 at 298 K.
HNO2(aq) ⇄ NO2−(aq) + H3O+(aq)
An aqueous solution containing a mixture of HNO2 and NO2− has a pH of 6.2 at 298 K.
Compared to the concentration of NO2−, the concentration of HNO2 must be:
A. greater, because the pH of the solution is greater than its pKa.
B. less, because the pH of the solution is greater than its pKa.
C. greater, because the pH of the solution is less than its pKa.
D. less, because the pH of the solution is less than its pKa. - ANS B. less, because the pH of
the solution is greater than its pKa.
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, If a naturally occurring sample of an unidentified element is found to contain three isotopes (A,
B, and C) and consists of 90.5% isotope A (mass number 20), 0.3% isotope B (mass number 21),
and 9.2% isotope C (mass number 22), the atomic weight of the element measured from the
sample will be:
A. greater than 21 amu.
B. less than 21 amu.
C. equal to 21 amu.
D. unrelated to the amount of isotope B. - ANS B. less than 21 amu.
Which of the following does NOT describe why a nucleophile will react more quickly with acetic
anhydride than with N,N-diisopropylisobutyramide?
A. N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride.
B. Acetic anhydride is a stronger electrophile than N,N-diisopropylisobutyramide.
C. Acetic anhydride is less sterically hindered than N,N-diisopropylisobutyramide.
D. N,N-diisopropylisobutyramide has a less stable leaving group than acetic anhydride. -
ANS A. N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride.
Reactivity of carboxylic acid derivatives - ANS depends on various factors, including inductive
effects, steric effects, and leaving group stability. The inductive effect is an electronic property in
which the electrons are donated through sigma bonds. Anhydrides contain stronger electron
withdrawing groups than amides; therefore, anhydrides experience a greater inductive effect
and react more readily than amides.
Of the four original solutions studied, which of the following pairs of solutions contain epimers?
Maltose and lactose
Galactose and glucose
Maltose and glucose
Lactose and galactose
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