BCH 210 Flashcards
Why are there two discrete vibration bands observed in the Frauenfelder B state occupied by CO upon photolysis from myoglobin at low temperatures?
The Frauenfelder B state is a specific location at the back of the heme pocket in van der Waals
contact with the heme. The CO is highly constrained rotationally and lies parallel to the heme
plane (90° to the heme normal), pointed toward the heme iron. There are two orientations, one
with the oxygen closer and the other with the carbon closer. These have slightly different
vibrational frequencies, hence the two B state bands
How was this FTIR of a photolyzed sample of Mb*CO interpreted?
As indicating that the angle of C-O bond relative to the heme normal changed from 0 degree to 90 degree upon photolysis
Describe what happens to CO upon being photolyzed from the L29F variant of myoglobin, in
which the bulky phenylalanine side chain fills the B state pocket
CO intitially pushes the the Phe 29 out of the way to occupy the B state, the Phe 29 pushes the distal histidine out of the way to accommodate the CO, but rapidly returns to its original position and swatts the CO out of the B state pocket. This means the CO departs the B state much faster than the wild type. Over time the CO is found further and further from the heme iron, in pockets later identified by Xe atoms in crystal structures.
Why does Intermediate II seen in the course of the oxidative (hydroxylating) half of the catalytic
cycle for p-hydroxybenzoate hydroxylase absorb so much more strongly than either Intermediate I or
III?
Intermediate II consists of the flavin-4a-hydroxide with the bound, unaromatized dieneone form
of product formed immediately upon hydroxylation. Unexpectedly, the dienone itself is rather
strongly absorbing in the same region of the near UV that the flavin-4a-peroxide of
Intermediate I and 4a-hydroxide of Intermediate III do, accounting for the greater absorbance
of Intermediate II.
What is involved in the breakdown of Intermediate II to III?
The conversion of Intermediate II to Intermediate III involves rearomatization of the product, which abolishes its absorbance, the flavin remaining as the 4a-hydroxide.
Why does p-hydroxybenzoate bind to and dissociate from reduced p-hydroxybenzoate
hydroxylase so much more slowly than to the oxidized enzyme?
In the reduced state, the flavin is found predominantly in the “in” position, where it
physically blocks substrate access to the substrate binding site. (By contrast, the flavin
exists predominantly in the “out” configuration in the oxidized enzyme, allowing much
easier access to the substrate binding site.
Discuss what happens when Tyr 201 in the active site of p-hydroxybenzoate hydroxylase is
mutated to a phenylalanine
The Y201F mutant is unable to ionize substrate, which is required for hydroxylation. As a
result, the flavin-4a-hydroxide breaks down directly to oxidized flavin and peroxide, with
little formation of hydroxylated product. While the rate of formation of Intermediate I in the
oxidative half-reaction is largely unaffected by the mutation, there is less substrate-induced
acceleration of the rate of enzyme reduction by NADPH in the reductive half-reaction with
the mutant as compared with the wild-type enzyme.
In the single-molecule study of cholesterol oxidase done by Lu et al., what was the evidence that
a wait time of a given duration was more likely than expected statistically to be followed by a
subsequent wait time of similar duration (implying that the two were somehow correlated)? How did
they determine the time scale on which the loss of this correlation occurred?
In the colored plots of likelihood that a wait time of a given duration will be succeeded by
wait times of various durations, it was found that the diagonal lit up. This means that it is
more likely than expected on statistical grounds that a wait time of a given duration is
followed by one of similar duration. (This is especially true of long wait times, which
otherwise very rare in the data trains).
Describe two experimental observations that indicate that at high substrate concentrations the
observed kcat for an individual immobilized molecule of b-galactosidase is not single-valued but
exhibits a distribution of values. Why is this behavior not seen at low substrate concentrations?
- the ES complex to E + P was rate-limiting, histograms of the frequency of the wait time
between successive catalytic events as a function of wait time duration became distinctly
nonexponential, reflecting a distribution of rate constants for decay of the ES complex. - When sets of 50 consecutive wait times were averaged over the course of the data train,
it was found that the averaged varied widely at high substrate concentrations, exhibiting a
Boltzmann-like distribution. At low substrate concentrations, when substrate binding - which
is not configuration dependent - is principally rate-limiting, the distribution of average wait
times was much narrower.
Describe the experimental setup used to probe the function of E. coli topoisomerase IV. In the eventuality, what is the principal reason that the enzyme acts more effectively on positively rather than negatively supercoiled DNA?
Magnetic beads were tethered to a coverslip by pairs of double-stranded DNA, and distended by a magnet placed above the coverslip. One or more crossovers (plectonemes) in the DNA were introduced by twisting the magnet, which shortened the distance of the magnetic bead off the cover slip. A solution containing topoisomerase was perfused over the cover slip, and upon binding to and uncrossing the plectoneme, the bead returned to its original position. Topoisomerase IV is more effective on positively supercoiled DNA because it can act processively, uncrossing successive plectonemes without having to dissociate from the gate strand of DNA. With negatively supercoiled DNA, it must dissociate and rebind after each uncrossing.
What is the effect of correct dNTP binding on the open-to-closed conformational change seen with T7 DNA polymerase when bound to DNA? Of binding one of the incorrect dNTPs?
Binding of a correct dNTP to the enzyme:DNA complex induces closure of the polymerase. Any of the incorrect dNTPs fail to do so.
Explain how the clamp loader component of the replisome ensures that the DNA polymerases bound to it are positioned at the replication fork? What is the effect of the DnaG RNA primase on binding to the processivity of the replisome? How does this make sense physiologically?
Upon binding ATP and the clamp, the clamp loader assumes a configuration that induces a lock-washer conformation in the clamp. The gap in clamp loader (missing one subunit of an otherwise hexameric ATPase-like structure) and clamp is only wide enough to allow single-stranded DNA through to the central cavity of the complex, after which the DNA slides through until double-stranded DNA is positioned in the central cavity. The complex is thus positioned at the point where double-stranded DNA changes to single-stranded DNA, i.e., the replication fork. DnaG binding to the helicase of the replisome counterintuitively reduces replisome processivity. This slows replication of the leading strand of DNA so that replication of the lagging strand can catch up, minimizing the amount of single-stranded DNA that accumulates in the cell during DNA replication.
Describe the different ways that secondary structural elements (principally double-helical stem loops) of rRNA can interact.
Any three of the following: Minor groove/minor groove; minor groove/phosphate ridge; end-on (intercalation of the loop from a stem loop); tetraloop/tetraloop receptor
What is the principal function of Domain V of the rRNA of the large ribosomal subunit? Of Domain IV?
Domain V harbors the peptidyl transferase center (with the opening of the exit tunnel for the nascent polypeptide), along with the P and A loops that bind to the 3’ CCA ends of the peptidyl- and aminoacyl-tRNAs, orienting them properly for the transferase reaction. Domain IV includes the “shelf” over which all three tRNAs hang, their 3’ ends extending down into the peptidyl transferase center and their anticodon stem-loops oriented toward the small ribosomal subunit and (in the A site) its decoding center. Domain IV also constitutes the major portion of the interface of the large ribosomal subunit with the small.
What is the structural evidence that A2451 does not function as an active site base in catalyzing peptidyl transfer in the ribosome?
It turns out that an induced fit conformational change occurs when a correct aminoacyl-tRNA (or suitable analog with components mimicking the CCA rather than just CA 3’ end of the tRNA) binds, and both the amino group of the aminoacyl-tRNA and the carbonyl bond of the 3’ ester of the peptidyl-tRNA reorient for optimal reaction leading to peptidyl transfer. This reorientation moves the amino group too far away from A2451 to make it an effective active site base. In the eventuality, it was shown that A2451 sits on the opposite side of the carbon of the tetrahedral intermediate than would be required for it to have a role in either the formation nor breakdown of the tetrahedral intermediate.
What are the universal common cores of the two ribosomal subunits? Include in your answer a discussion of the structural conservation seen in the decoding and peptidyl transferase centers.
The universal common cores are the structurally conserved parts of the small and large rRNAs that are universally conserved among bacteria, archaea and eukarya. In particular, the structures of the decoding center of the small rRNA and peptidyl transferase center of the large rRNA are extremely highly conserved structurally
