BCH5413 3 Practice Exam A+ Questions
and Answers Verified Solutions Latest
Update 2026/2027
Answer:
No. The steady state concentration is dependent on the rate of synthesis and the rate of degradation.
For example, ribosomal proteins are always needed and therefore have much more stable mRNAs
so the proteins can be consistently made. However, cytokines are condition dependent and are only
needed under certain conditions. Therefore, their mRNA is less stable because the proteins are only
needed conditionally.
■Deadenylation of an mRNA results in the loss of the polyA binding protein. What role does this
play in facilitating mRNA degradation?.
Answer:
De-adenylation of mRNA by the deadenylase complex begins at the 3' end of the poly-A tail and
removes the A residues until the poly-A tail is removed. This results in a loss of the PABP
(normally bound to poly-A and eIF4G to prevent degradation), which is a key protein to the
circularization of the mRNA. When the 3' end is exposed, the exosome can come in and degrade
the mRNA in the 3' to 5' direction. Alternatively, after the poly-A tail is removed, the mRNA can
be de-capped (via DCP1/DCP2) then degraded from the 5' to 3' end using XRN1 (predominant
pathway in yeast but not mammals).
■DNA synthesis- key conceptWhat catalyzed the extension of DNA?In what direction?.
Answer:
DNA Pol catalyzes the extension of DNA strands in 5' -> 3' direction
■Where is the deoxi on DNA?.
Answer:
2' position
■What is the significance of the ARE (AU-rich elements) found in the 3' non-translated region
(NTR) of some mRNAs?.
Answer:
,These regions contain a core AUUUA and are used to destabilize mRNA and lead to rapid
degradation. Specific proteins can bind to these regions and recruit de-adenylating enzymes and
exosome (rapid 3' to 5' degradation). Because of this recruitment function, AREs are usually found
in short-lived mRNAs.
■Compare and contrast the two deadenylation dependent mRNA decay pathways..
Answer:
The first pathway discussed involves the degradation of the mRNA in the 3' to 5' direction using
exosome. The second pathway discussed involves extra enzymes that have to remove the 5' cap
(DCP1 and 2) in order for a different enzyme to degrade the mRNA in the 5' to 3' direction
(XRN1).
■What happens to an mRNA that utilizes the endonuclease-mediated mRNA decay pathway?.
Answer:
This pathway does not rely on the de-adenylation of the mRNA prior to degradation. The mRNA
can be cut in the center by endonuclease, which provides an open 3' and 5' end. The exosome can
come in a degrade the 5' end of the mRNA in a 3' to 5' direction while XRN1 can degrade the 3' end
in the 5' to 3' direction.
■DNA synthesis- what are the two substrates.
Answer:
- template strand DNA with a primer (RNA or DNA) annealed by
base pairing- dNTP's
■What is XRN1?.
Answer:
A 5' to 3' exonuclease
■T.F Primer can be either RNA or DNA for DNA synthesis.
Answer:
True
■What is released when a dNTP is added to the chain.
Answer:
,PPi (pyrophosphate)
■What is the growing strands of DNA called?.
Answer:
Parental and daughter strands
■Reaction mechanism of DNA Pol- step- leads to....
Answer:
Oxygen at 3' end of elongating strand attacks alpha-phosphate of the triphosphate held in the active
site via Mg ions- leads to formation of phosphodiester linkage and elongation by one dNTP
■What is exosome?.
Answer:
The main 3' to 5' exonuclease in the cell
■What is deadenylation?.
Answer:
The shortening (and ultimate removal) of poly(A) tail of mRNA
■How many Mg+ does DNA pol have?.
Answer:
2
■Replication fork: what relieves torsional stress on helix.
Answer:
topoisomerase
■Replication fork- leading strand synthesizes...- lagging strand synthesizes...- what are okazaki
fragments?.
Answer:
- leading strand DNA synthesis occurs continuously 5' -> 3'-
lagging strand synthesis occurs in fragments as a series of 5' to 3' reactions (results in okazaki
fragments)
■DNA replication is (semi/full/dispersive) conservative.
, Answer:
SEMI- two parental strands generate two daughter DNA molecules that are then divided and
combined
■Conservative v. dispersive replication.
Answer:
Conservative- parental and daughter DNA strands are conserved to themselvesDispersive- two
parental and daughter strands combine/fuse
■Meselson and Stahl experiment- what it determined- what it used.
Answer:
- determined DNA is semiconservative- ultracentrifuged N14 and
N15 in CsCl (can separate two DNA's)
■Meselson and Stahl experiment- what would be seen if conservative, semi, or dispersive.
Answer:
Conservative- see two bands (heavy heavy, and light light)Semi- see two bands (heavy light, and
light heavy)Dispersive- two bands (mixed and mixed)
■What mRNA components are key to mRNA circularization?.
Answer:
PolyA tail, PABP, eIF4G, eIF4E, and the 5' cap
■What is endonuclease?.
Answer:
An enzyme which cleaves a polynucleotide chain by separating nucleotides other than the two end
ones
■Meselson and Stahl experiment- what the results showed.
Answer:
- single band after one generation- at generation two there are two
bands (heavy light, and light light); this is consistent only with semi conservative mode of
replication
and Answers Verified Solutions Latest
Update 2026/2027
Answer:
No. The steady state concentration is dependent on the rate of synthesis and the rate of degradation.
For example, ribosomal proteins are always needed and therefore have much more stable mRNAs
so the proteins can be consistently made. However, cytokines are condition dependent and are only
needed under certain conditions. Therefore, their mRNA is less stable because the proteins are only
needed conditionally.
■Deadenylation of an mRNA results in the loss of the polyA binding protein. What role does this
play in facilitating mRNA degradation?.
Answer:
De-adenylation of mRNA by the deadenylase complex begins at the 3' end of the poly-A tail and
removes the A residues until the poly-A tail is removed. This results in a loss of the PABP
(normally bound to poly-A and eIF4G to prevent degradation), which is a key protein to the
circularization of the mRNA. When the 3' end is exposed, the exosome can come in and degrade
the mRNA in the 3' to 5' direction. Alternatively, after the poly-A tail is removed, the mRNA can
be de-capped (via DCP1/DCP2) then degraded from the 5' to 3' end using XRN1 (predominant
pathway in yeast but not mammals).
■DNA synthesis- key conceptWhat catalyzed the extension of DNA?In what direction?.
Answer:
DNA Pol catalyzes the extension of DNA strands in 5' -> 3' direction
■Where is the deoxi on DNA?.
Answer:
2' position
■What is the significance of the ARE (AU-rich elements) found in the 3' non-translated region
(NTR) of some mRNAs?.
Answer:
,These regions contain a core AUUUA and are used to destabilize mRNA and lead to rapid
degradation. Specific proteins can bind to these regions and recruit de-adenylating enzymes and
exosome (rapid 3' to 5' degradation). Because of this recruitment function, AREs are usually found
in short-lived mRNAs.
■Compare and contrast the two deadenylation dependent mRNA decay pathways..
Answer:
The first pathway discussed involves the degradation of the mRNA in the 3' to 5' direction using
exosome. The second pathway discussed involves extra enzymes that have to remove the 5' cap
(DCP1 and 2) in order for a different enzyme to degrade the mRNA in the 5' to 3' direction
(XRN1).
■What happens to an mRNA that utilizes the endonuclease-mediated mRNA decay pathway?.
Answer:
This pathway does not rely on the de-adenylation of the mRNA prior to degradation. The mRNA
can be cut in the center by endonuclease, which provides an open 3' and 5' end. The exosome can
come in a degrade the 5' end of the mRNA in a 3' to 5' direction while XRN1 can degrade the 3' end
in the 5' to 3' direction.
■DNA synthesis- what are the two substrates.
Answer:
- template strand DNA with a primer (RNA or DNA) annealed by
base pairing- dNTP's
■What is XRN1?.
Answer:
A 5' to 3' exonuclease
■T.F Primer can be either RNA or DNA for DNA synthesis.
Answer:
True
■What is released when a dNTP is added to the chain.
Answer:
,PPi (pyrophosphate)
■What is the growing strands of DNA called?.
Answer:
Parental and daughter strands
■Reaction mechanism of DNA Pol- step- leads to....
Answer:
Oxygen at 3' end of elongating strand attacks alpha-phosphate of the triphosphate held in the active
site via Mg ions- leads to formation of phosphodiester linkage and elongation by one dNTP
■What is exosome?.
Answer:
The main 3' to 5' exonuclease in the cell
■What is deadenylation?.
Answer:
The shortening (and ultimate removal) of poly(A) tail of mRNA
■How many Mg+ does DNA pol have?.
Answer:
2
■Replication fork: what relieves torsional stress on helix.
Answer:
topoisomerase
■Replication fork- leading strand synthesizes...- lagging strand synthesizes...- what are okazaki
fragments?.
Answer:
- leading strand DNA synthesis occurs continuously 5' -> 3'-
lagging strand synthesis occurs in fragments as a series of 5' to 3' reactions (results in okazaki
fragments)
■DNA replication is (semi/full/dispersive) conservative.
, Answer:
SEMI- two parental strands generate two daughter DNA molecules that are then divided and
combined
■Conservative v. dispersive replication.
Answer:
Conservative- parental and daughter DNA strands are conserved to themselvesDispersive- two
parental and daughter strands combine/fuse
■Meselson and Stahl experiment- what it determined- what it used.
Answer:
- determined DNA is semiconservative- ultracentrifuged N14 and
N15 in CsCl (can separate two DNA's)
■Meselson and Stahl experiment- what would be seen if conservative, semi, or dispersive.
Answer:
Conservative- see two bands (heavy heavy, and light light)Semi- see two bands (heavy light, and
light heavy)Dispersive- two bands (mixed and mixed)
■What mRNA components are key to mRNA circularization?.
Answer:
PolyA tail, PABP, eIF4G, eIF4E, and the 5' cap
■What is endonuclease?.
Answer:
An enzyme which cleaves a polynucleotide chain by separating nucleotides other than the two end
ones
■Meselson and Stahl experiment- what the results showed.
Answer:
- single band after one generation- at generation two there are two
bands (heavy light, and light light); this is consistent only with semi conservative mode of
replication
