Page 1 of 64
“UWORLD B/B PT2 QUESTIONS & ANSWERS
EXAM 2026 ”LATEST EXAM 2026 – 2027 SOLVED
QUESTIONS & ANSWERS VERIFIED 100%
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UWorld B/B pt2
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.
C. Treating cells with the drug results in a decreased concentration of cytosolic
malonyl-CoA compared to untreated cells.
Malonyl-CoA
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?
A. BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic
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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.
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′
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
D. I, II, and III
B. I and III only
Transcription
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
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.
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:
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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.
B. less, because the pH of the solution is greater than its pKa.
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.
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.
A. N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride.
Reactivity of carboxylic acid derivatives
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
A. I only
B. II only
C. I and III only
D. II and IV only
B. II only
What is the total number of anomeric carbons in disaccharides that are linked
by β-glycosidic bonds, such as lactose?
A. 0
B. 1
C. 2
D. 10
C. 2
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If the researchers used the form of galactose most common in nature (ie, d-
galactose), the sequence of stereochemical configurations from carbon 2 to
carbon 5 must be:
A. S, R, R, S
B. R, S, R, S
C. R, S, S, R
D. R, R, R, S
c. R, S, S, R
The infrared stretching frequencies for lactam carbonyl groups are shown
below.
Which of the following best explains the difference in observed IR stretching
frequencies?
A. Molecular weight
B. The inductive effect
C. Acidity
D. Ring strain
D. Ring strain
The spin quantum number ms describes the quantum state of an electron in an
orbital with respect to its angular momentum. According to the Pauli exclusion
principle, what are the values of ms for each of the electrons in the 4s orbital
of a calcium atom?
A. +½ and +½
B. +½ and −½
C. −½ and −½
D. +½, −½, and +½
B. +½ and −½
If lower phospholipid levels in ME patients are due to inhibition of fatty acid
synthesis, researchers will most likely observe which of the following?
A. Increased cytosolic NADPH
B. Decreased mitochondrial acetyl-CoA
C. Decreased mitochondrial fumarate
D. Increased cytosolic lysine
A. Increased cytosolic NADPH
(Choice B) Fatty acid synthesis takes place in the cytosol, not the mitochondria. In
addition, cytosolic acetyl-CoA levels would most likely increase when fatty acid
synthesis is inhibited because less acetyl-CoA is consumed.
(Choice C) Fumarate concentration is not directly related to fatty acid synthesis and
would most likely be unaffected.
(Choice D) Lysine is not directly involved in fatty acid synthesis. In addition, Figure 1
shows that Category II amino acids, including lysine (K), had decreased levels in the
plasma of ME patients and would likely be decreased in the cytosol as well.
Educational objective:Acetyl-CoA, NADPH, and ATP are the reactants needed to
generate fatty acid chains during lipid synthesis. The pentose phosphate pathway
generates the NADPH needed to reduce the carbonyl groups from each molecule of
acetyl-CoA that is added to a fatty acid chain.
To confirm that β-oxidation was NOT impaired in healthy controls, the
researchers most likely measured which metabolic step in PBMCs?
A. The export of citrate to the cytoplasm to produce acetyl-CoA
“UWORLD B/B PT2 QUESTIONS & ANSWERS
EXAM 2026 ”LATEST EXAM 2026 – 2027 SOLVED
QUESTIONS & ANSWERS VERIFIED 100%
GRADED A+ (LATEST VERSION) WELL REVISED
100% GUARANTEE PASS
UWorld B/B pt2
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.
C. Treating cells with the drug results in a decreased concentration of cytosolic
malonyl-CoA compared to untreated cells.
Malonyl-CoA
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?
A. BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic
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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.
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′
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
D. I, II, and III
B. I and III only
Transcription
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
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.
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:
, Page 3 of 64
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.
B. less, because the pH of the solution is greater than its pKa.
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.
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.
A. N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride.
Reactivity of carboxylic acid derivatives
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
A. I only
B. II only
C. I and III only
D. II and IV only
B. II only
What is the total number of anomeric carbons in disaccharides that are linked
by β-glycosidic bonds, such as lactose?
A. 0
B. 1
C. 2
D. 10
C. 2
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If the researchers used the form of galactose most common in nature (ie, d-
galactose), the sequence of stereochemical configurations from carbon 2 to
carbon 5 must be:
A. S, R, R, S
B. R, S, R, S
C. R, S, S, R
D. R, R, R, S
c. R, S, S, R
The infrared stretching frequencies for lactam carbonyl groups are shown
below.
Which of the following best explains the difference in observed IR stretching
frequencies?
A. Molecular weight
B. The inductive effect
C. Acidity
D. Ring strain
D. Ring strain
The spin quantum number ms describes the quantum state of an electron in an
orbital with respect to its angular momentum. According to the Pauli exclusion
principle, what are the values of ms for each of the electrons in the 4s orbital
of a calcium atom?
A. +½ and +½
B. +½ and −½
C. −½ and −½
D. +½, −½, and +½
B. +½ and −½
If lower phospholipid levels in ME patients are due to inhibition of fatty acid
synthesis, researchers will most likely observe which of the following?
A. Increased cytosolic NADPH
B. Decreased mitochondrial acetyl-CoA
C. Decreased mitochondrial fumarate
D. Increased cytosolic lysine
A. Increased cytosolic NADPH
(Choice B) Fatty acid synthesis takes place in the cytosol, not the mitochondria. In
addition, cytosolic acetyl-CoA levels would most likely increase when fatty acid
synthesis is inhibited because less acetyl-CoA is consumed.
(Choice C) Fumarate concentration is not directly related to fatty acid synthesis and
would most likely be unaffected.
(Choice D) Lysine is not directly involved in fatty acid synthesis. In addition, Figure 1
shows that Category II amino acids, including lysine (K), had decreased levels in the
plasma of ME patients and would likely be decreased in the cytosol as well.
Educational objective:Acetyl-CoA, NADPH, and ATP are the reactants needed to
generate fatty acid chains during lipid synthesis. The pentose phosphate pathway
generates the NADPH needed to reduce the carbonyl groups from each molecule of
acetyl-CoA that is added to a fatty acid chain.
To confirm that β-oxidation was NOT impaired in healthy controls, the
researchers most likely measured which metabolic step in PBMCs?
A. The export of citrate to the cytoplasm to produce acetyl-CoA
