Welcome back. So, we were discussing about
those electron transfers following photoexcitation.
So, if we had P-700 stared; that
means we have the excited species at the photosystem
one. So, this particular one is residing at
PS 1. So, at PS 1 we get P-700
stared. So, that P-700 stared we have some
electron from there and at the next point
it is reacting with the ferredoxin what we
have shown in the Z-scheme is P-430. It is
very interesting to level all this species;
P stands for your pigment and this is the
corresponding wave length. Because whatever
species you are handling; that means where
you have this Fe 4 S 4 species as the
ferredoxin, and these are very good species
we know as electron transfer material. So,
we have the corresponding two forms
depending upon it is E 0 for the oxidation
and the reduction reactions.
And all these iron sulfur proteins or we know
that they can expand a very wide range of
E 0 values; that is why what we studied
earlier also we know that all these ferredoxin
type of molecules can be present in photosynthesis,
can also be present in
respiration. So, depending upon your protein
environment around these cluster molecules,
you can have the oxidized form where
iron centers are present in the ferric form
and the reduced form where the iron centers
are present in the reduced form. So, this
particular one; that means this Fe 4 S 4 which
is our well known ferredoxin molecule and
that ferredoxin molecule is coupled
with that of your P-700 stared one, and this
ferredoxin protein molecule is attached to
this one and this particular level which is
its corresponding electronic absorption. So,
electronic absorption behavior at 430 nanometer
for the oxidized form; that means
the ferric form of the ferredoxin molecule.
Then it is coupled with some flavoprotein,
FP is our flavoprotein. These are all our
well known redox mediators because we
cannot connect all these directly to the species
what we are trying to attach; that means NADP
plus. So, it has one particular
redox potential value, it has also one redox
potential value and between these two we have
the intermediate redox mediators
which can couple with P-700 star two NADP
plus. And at this point when you have PS I
and how we studied this also; that means
you have one such species; we just studied
in our previous two three classes, we read
about plastocyanin which are copper
proteins. So, blue copper protein is responsible
for electron transfer. So, this plastocyanin
is taking that electron. So, electron is
basically entering in PS I from this plastocyanin.
So, plastocyanins in the reduced form can
put the electron to PS I and in that
particular one when you have this plastocyanin.
So, sometimes we find that whether your corresponding
redox couple. So, thermodynamics will tell
us whether the reaction of
plastocyanin which is a copper based one and
if it is copper in reduced form it can react
to it the corresponding ferredoxin in
oxidized form. So, it can react with ferredoxin
in oxidized form. So, this redox couple what
we studied earlier also in case of
cytochrome chain and cytochrome c oxidases
that whenever you have the pair of redox partners
you can find out the
corresponding changing in e 0 values, the
delta e you can find out and the corresponding
gives free energy change will tell you
whether you reaction is facile or not. So,
in this particular case what we find that
you have this plastocyanin potential is in
the
range of 0.37 volt and ferredoxin one is 0.45
volt.
So this gives us that oxidized plastocyanin,
PC is your plastocyanin in the oxidized form
and ferredoxin in the reduced form
which is nothing but Fe 2 plus. So, when they
are basically connected we can calculate the
delta E difference of these two which
is minus 0.82 volt and you can calculate the
corresponding delta G 0 prime as plus 79 kilo
Joule per mole. So, this particular case;
that means whether you have at a particular
wave length, so this is basically this level
is your 700 nanometer wave length and
from that wave length and from the corresponding
values for this, we find that only this much
amount of energy change is
available for this electron transfer.
So, some processes we can have is a favored
processes and some are not favored depending
upon these values, but if we just put
the radiation the eliminated portion of this
can change the particular electron transfer
pathways in a different way and that we
can also see in case of P-680 as well where
you have certain other species connected to
that and one of them is cytochrome b,
then one such species which is not based on
copper but it is a quinone based species which
is plastoquinone and the
corresponding redox potential is similar to
that of your catechol quinone potential and
this then clubbed with another cytochrome
b type of different version. F is feuille
which is a French term for leaf f e u i l
l e feuille. So, which is nothing but the
French type
related to leaf.
So, this are then is also connected to plastocyanin
and this is then responsible for generation
of chlorophyll a 1 cation radical. So,
all these are involved in the different types
of electron transfer chain and if we can find
out the corresponding values and how
this particular, say, plastoquinone because
we know the corresponding potential for water
and oxygen couple and if this PQ and
PQ H 2; that means the quinone form and the
catechol form, PQ H 2 is the catechol and
if this redox couple is this much and we
know the corresponding values for O 2 and
H 2 O which is 0.82 volt. So, in this region
whether you can have. So, you can
differentiate out these two values and you
find out the corresponding E 0 and that E
0 we can correlate with that of our h nu,
nothing else we are doing.
So, h nu for one photosystem is 680 nanometer
and another is above 700. So, these are the
two values only. So, at this value what
we get is 176 kilo Joule and we calculate
the corresponding E 0 prime; from there we
can calculate the corresponding delta G 0
prime and what is the corresponding value
of that kilo Joule per mole, you can find
out that in that particular region. So, this
typical excitation can give you and you have
a corresponding delta G calculated value which
will be positive in that particular
case. So, these all thermodynamic parameters
and all these values will talk us something
related to your excitation of magnesium,
how we get the electron, how we derive those
electrons from those centers and how we can
put those electrons to several redox
mediators, say, plastocyanin, plastoquinone
and all these things.
But what we get that the excitated electron
and its chemical energy we have to transfer
to our CO2 molecule and which we
cannot lose any kind of energy due to molecular
vibration as well as florescence. So, the
oxidation of this water molecule; that
means when we have water in photosystem II
we get something what we call also as oxygen
evolving center because from water
we are getting O 2 molecule. So, it is known
as also oxygen evolving center where we can
go for typical oxygen evolution and is
a 4 one electron state. So, this 4 one electron
state is required and your water we are consuming
and analysis of metal centers,
because it is very easy to know that how people
discovered that magnesium is present in chlorophyll
and manganese is present in
photosystem II. So, it is manganese analysis.
So, any manganese analysis technique it can
be a spectroscopic technique also that identification
of manganese from its
corresponding atomic form will tell us that
this is basically tetranuclear manganese.
And idea behind knowing the tetranuclear
manganese is very simple because we need four
electron transfer and in all these cases though
we all know that manganese can
have the stable oxidation state from plus
two to plus seven. So, you can argue that,
okay, you can one mononuclear center, you
can take out one after another electron for
this corresponding oxidation as well as reduction.
But in all these cases what you have
if you have a mononuclear manganese center
and if you can take out one after another
electron to go for a change in oxidation
state from plus 2 to, say, plus 6 or plus
3 to plus 7, you will encounter with different
E 0 values, E 1, E 2, E 3 and E 4.
So, most of the cases where we find that between
these two; that mean the first manganese center
and its second immediate
oxidized form. They are interrelated and we
can write the corresponding electron transfer
equilibrium in a reversible sign and
these electron transfer property should always
be reversible; that is why we always rely
on something in biological system also
the corresponding measurements using cyclic
voltammetry; that means whatever you do; that
means this is the anodic or the
cathodic big potential then you get something
which is your E 0; that means whatever you
do. So, you can subtle between these
two forms, one form and the immediate next
one. We are not going from here to here, here
to here; otherwise your reversibility
will be lost.
So, when we go for these oxidation for one
manganese center, say, if it is a manganese
three you immediately can convert it to
manganese four and whatever E values we get
for the corresponding transfer for the manganese
three and manganese four. So,
what happens there that when we go and when
we just oxidize it, then we reverse the codes
of its scanning; that means on the
electrode surface what you have, you have,
say, a platinum electrode. So, platinum is
there and is a connected. So, on this
electrode surface what you have this manganese
and manganese is formed in the plus four.
Then you are switching back the
corresponding cycle the corresponding direction
of the scan. So, this is the current access,
this is the voltage access.
So voltage, say, you are scanning from 0.0
volt to, say, 1.5 volt. Now you are scanning
back; that means 1.5 volt to you are
coming back to 0.0 volt. So, now what is happening?
Your manganese four which has been produced
from the oxidation of
manganese three is absorbed on this electrode
and that is now reduced back. So, this manganese
four will be reduced back to
manganese three because you are switching
back. So, only some little bit difference
is there between these two values, but all
the
manganese four what is forming over there
will be reduced back to manganese three. So,
the stability or the life span or the life
period for this form; that means the oxidized
form is important.
So, whatever species you have even if you
just talk about the corresponding ferredoxin
one in the reduced form and another in
the oxidized form because it has a very good
co-relation with that of the species what
we know and well known species is the Fe
4 ferredoxin molecule the iron sulfur proteins
Fe 4 S 4 species. So, it has certain similarity
with that manganese four species and
there also we get in this particular case
we get only one electron transfer. So, it
is the oxidation and it is the reduction;
not that
you are going from a Fe 4 species from one,
two, three, all these states because to stay
within the reversible domain that whatever
you are doing; that means you are just simply
oxidizing one particular manganese iron center
on the ferredoxin cluster. If you go
for oxidizing the other centers as well, your
structure will not be stable.
So, you will just break the structure. You
can force it because this is electrochemical
oxidation; with the help of electrodes you
are forcing the oxidation. But biological
reducing agents and oxidizing agents are available;
it will not provide you the sufficient E
0 value for this second step or for the third
step. This is one reason and another reason
is that that you have this absorbed species
and which is coming back like this manganese
four is coming back to manganese three but
if this is not stable; that means your
manganese four should have certain stability,
otherwise it will break; that means its structure
can change. It can have a changed
structure if it is a cluster, a cube type
of structure. Suppose it is a four, means,
very easy to put four of this metal centers
like this,
one here, one here, one here and one here.
So, you have so much connectivity’s. Some
of these ligands because these other corners
are occupied by the ligands either sulfur
or in case of Mn 4 we will see by oxygen.
So, if they are connected for that then the
structure will change and it will not be stable
enough to go back to the original position.
After this shuttling what you are getting
back, you can compare with that of your
original structure. So, if it is ferredoxin
one after one electron oxidation, you cycle
it back, you go for the reduction, you should
get back the original ferredoxin reduced form.
Because you are doing that particular electron
transfer in solution, you have water;
you have some other groups also.
If this has less stability the oxidized form
has less stability, it can react with some
other water molecule or it can break from
their
structure and the iron centers, the ferrous
ion or ferric ion can leach; that means it
can leave your Fe 3 plus from the cluster.
So
the entity, it should have certain stability
that when you switch back to the original
position; that means when you reduce it back,
the oxidized ferredoxin when it is reduced
back, it should produce the same ferredoxin
reduced from what you have originally
what you have used for the oxidation processes.
So, if this is not sufficiently stable what
happens, how you know this cyclic
voltagram can show you that whatever you are
producing on the electrode surface which is
absorbed there which is not stable
which is not the same species having the same
values for this, what will happen after here;
that means you are oxidizing it and it
is transforming to some other species, you
are getting back not with the current value
of this position.
So, you will be losing the current value.
So, your thing will go away and basically
you are getting for this; that means this
is for
oxidizing this one, but this you are not getting
what your suppose to get for the reduction
of this manganese four species. So, that
is the typical signature whether your structure
is unstable whether you are not getting back
the same structure. So, if you just go
for one step to this, this step to that; that
means most of the cases the safest way of
thinking this particular arrangement this
multicenter arrangement is that, you can have
one particular center and four times of that
should give you some arrangement.
So, this four manganese center which is your
water evolving center and this then handling
the water and immediately this has
attachment for one group which is the plastoquinone;
that means the immediate species which is
responsible for electron transfer
and we are utilizing photon for this. So photo-oxidoreductases,
this nomenclature we know already. So, these
are very important
things and very simple one also because this
is nothing but your quinine, this is a big
Q, it is a quinone form; plastoquinone is
nothing but quinone. So, if you have a catechol
quinone conversion. So, the catechol quinone
type of species is responsible to
interact with water molecule and which is
controlled by eradiation and we are looking
for some oxidation reduction reaction. So,
this is very big name, but what are these
though, you have water.
So, from the very beginning of our classes
we are just talking about two water molecules
and two of this quinone form which is
oxidized form the plastoquinone in the oxidized
form eradiated giving O 2 plus twice of the
catechol form of this. So, this
particular one we get but for that particular
transfer we need to transfer four electrons.
So, like that of your particular metal
center where your manganese is there in plus
three oxidation state and stepwise you are
oxidizing by one electron, then another
electron, then third electron and the fourth
electron. So, manganese plus three can go
up to manganese seven but similarly if you
have this particular, so we need the four
electron species and we can have arrangement
of a tetranuclear manganese complex. So,
this tetranuclear manganese complex, so one
particular oxidation state we can consider
it has S 0 state.
So, s is related to the spin sate. So one
oxidation state, then what people have done
that you change it to another state. So, you
are just leveling; the cluster is your Mn
4 cluster. So, this cluster complex when you
get this. So, S 0 to S 1 we get through
ejection of electron; that means through oxidation.
So through oxidation, at the same time it
is also leveraging one proton. So, that
we all know that whenever there is electron
transfer when some system is getting oxidized
it can be your amide backbone also.
So, CONH backbone when it is bound to the
metal center, if you go for the oxidation
of the metal center and the nitrogen is
bound to the metal center. But at the same
time when you take out the electron from the
metal center, the metal center should be
electron greedy and that will go for the deprotonation
of the amide function. So, your CONH will
be CON minus.
So, that can be very readily stabilized in
the high oxidation state. So, whatever you
have in the system when you go for one
electron oxidation, simultaneously you will
be losing one proton. It can be from any other
bound ligand or it can be from
sometime from the bound water molecule or
the hydroxide group also. So, this particular
thing is activated by because all the
states are activated by h nu. So, these are
the states like the dark phases and light
phases of photosystem II; knowing this
particular photosystem II for water oxidation,
whether it has to be eliminated or not that
we should know. So, at this particular
step you put h nu and you get S 1. And in
the second step we have other step; that means
S 2 and S 2 is again by loss of second
electron and it is also triggered by h nu
but no proton loss at this step. So this can
be differentiated from your S 1, then we have
the S 3 step; again one more electron loss
and one more proton loss and is also h nu.
This S 3 by simple electron loss can go to
S 4. So, this is the step where you get this
particular unit. So, you have one, two, three,
four, 4 electrons have been lost from the
system. So, by doing so; that means by doing
or taking the electron one after another,
we
can move from S 0 to S 4. So, S 0 you can
immediately say that this is the super-reduced
form of this M 4 complex and this is the
super-oxidized form. So, either you put all
together the four electrons to the system
or you take out four electrons from the
system but interestingly among all these 5
species. So, you have 5 species and you can
immediately say that this we are talking
about four electron transfer. So, four electron
transfer plus one starting species; so one
starting species plus four electron transfer
giving you 5 states. So, 5 of this form and
out of this form S 1 is the most stable form.
So, this is the stable form 
in the dark and this particular one. So, since
it is stable it is very easy to identify also
this corresponding
oxidation state of the manganese and this
particular one has been identified as a manganese
three four complex, where you have
two of the manganese centers in the trivalent
state and other two in the tetravalent state;
sorry manganese four two. So, it is a
mixed balanced species where you have two
of the manganese in the plus 3 state and two
other in plus 4 oxidation state. So,
basically what we are talking about here that
the most readily accessible oxidation states
what we are talking about here is that it
can go from three to four or it can go down
from three to two. So by doing so this particular
step you get, 2 plus 2 4 electrons, So,
this is the simplest possible picture what
you can have and in this step when you have
this oxidized form.
So, oxidized form basically will liberate
two protons from here and this is the step
where you get your dioxygen. So, whatever
cycle we write over here in this particular
state because we are talking about only the
photo excitation. You have given photon,
you have taken out electron and the protons
are coming out. So, this particular S 4 basically
collapsing to S 0 when it is reacting
with two molecules of water giving you that
required dioxygen molecule, so collapsing
of that thing to S 0. So, now the challenge
is that how we can identify slowly that what
are the species. So, manganese is well known.
So, manganese is there and we can
have the knowledge about this manganese that
manganese has corresponding affinity for oxidizes
is not a corresponding iron
sulfur type of thing.
So, it is very easy to know also once we know
the tetranuclear ferredoxin molecule, the
four iron ferredoxin molecule; that means
you can have these two plus you can have sometime
loosely bound water molecules. So, these are
basically from oxygen
environment and these oxygen environments
are further supplemented by the corresponding
protein side chains; that mean the
carboxyl side chains or histidine side chains.
So, these are the other ligating groups and
those groups can give you all these values
and this particular conversion from water
to hydroxido group to hydroxido to oxide group
will also be responsible for the
elimination of your required number of protons
during electron transfer.
So, the very basic assembly for this tetranuclear
system what we can see know that once you
have this; that means if you just go
for a mononuclear to a binuclear system, we
all know that you can have some connectivity
at least one connectivity to get a
binuclear one and you can have certain connectivity
such that if you have a group like this, this
oxo or the hydroxido function
can do very easily in a mu 3 form hydroxido
bridge mu 3 that can give you a trinuclear
one and also this O 2 minus. When these
are there this O 2 minus if it is bridging
four metal centers, you get very quickly a
tetranuclear assembly. But for this sort of
proton transfer; that means if you have a
hydroxido bridging or oxido bridging, this
is not a stable structure. Once you can go
for
protonation, immediately we will lose the
entire structure because it is the group;
that means the oxido function you have and
this
is basically the nucleus of the cluster because
it is a tetrahedral arrangement.
This oxygen is in tetrahedral geometry and
this tetrahedral geometry is basically holding
all four metal centers. It can be your 4
manganese also but if you immediately go for
protonation. So, one protonation step for
this O 2 minus can change this to a
hydroxide. This X is nothing but your hydroxide
group and its affinity for coordinating four
metal centers will be lost. So, it
cannot be a mu four ligand; immediately when
you go for protonation, it would be a mu 3
ligand. If not a mu 2 one; this is a mu 2
one. So, what is the other alternative formulation
what you can have to get this is if you have
this dinuclear assembly and if you
put another dinuclear assembly close to it.
So, that should be our right choice for a
tetranuclear assembly and that we all know
comparing the corresponding Fe 4 S 4 ferredoxin
structure.
So, that structure we have and not that this
Fe 4 you need which can give you a corresponding
manganese 4 unit, but also there
are certain analysis and that analysis tells
us that this particular Mn 4 is there as well
as you can have a nearby calcium ion, like
the superoxide dismutase or like the magnesium
and all these. So, this presence of this calcium
it is not a redox active metal ion.
So, it will not take part in any kind of electron
transfer behavior, but this as always you
will find whenever any complicated
structure you find either it is synthetic
one or is a biological one. So, it has some
important role to play in catalysis, how?
Because
it can have large number of water bound molecules
or any other group bound molecules. So, it
can provide some activated water
molecules instead of free water molecule and
also it can contribute to a large extent to
the structure of the system; that means if
you have a cubic type of structure or cube-like
structure, it has something to play with that
cubic structure.
So, if we level all the four manganese centers
around this one particular central oxido function,
this is manganese one, this is
manganese two and the third one is manganese
three. So, this particular arrangement gives
us something that it is not a typical
cubic type of arrangement cube-like arrangement
but this particular one which is there as
a mu three one. So, if we have
something; that means the mu three group is
there and in a cube type of arrangement if
you have all the alternate coordinates are
occupied by the metal centers. If it is a
four metal unit, it can be your ferredoxin
F 4 or this water evolving center Mn 4. So,
these
are the four metal centers and this other
groups these are your bridging groups. It
can be your hydroxido group or the oxido group
and the nature of these groups are mu 3. All
these X groups are binding three metal centers
together.
So, when they are occupying the four coordinates
of the cube they are in mu 3mode; the binding
is in mu 3 mode. So, we have
satisfied this particular mode in the mu 3
mode. So, when you have mu 3 mode and in this
particular case that if you have a
system which is Mn 4 and O 4, forget about
the nature of this O, because we get different
spectroscopic analysis and x ray
structure analysis to some extent. We get
this arrangement and we can differentiate
only that whether these particular groups
are
oxygen or sulfur or any halogen group, but
sometime we are not able to precisely determine
whether this is oxido group or
hydroxido group whether this oxygen is attached
to a proton or not, that sometimes is very
difficult to identify.
So this mu 3 group, so it can be your hydroxido
function or it can be your oxido function
and this mu 3 group when you are
viewing from other direction; that means depending
upon these; that means if you have four manganese
center and four oxo
center, you get the regular cube structure;
that means all the eight corners of the cube
are occupied. But if you try to have
something; that means four manganese are there
we know. But if you make it Mn 4 O 3, what
will happen? If you make it Mn O
3; that means your one of the vertex is missing,
but still you have this cuboidal arrangement.
If you take out this vertex and if
your another bound distance and bound angle
remains same, you will have basically a cuboidal
structure; we call those structures
as defective cubes or partial cube; we call
them as defective cube or partial cube.
So in this arrangement, so we have these groups.
So, one of these; that means if you take out
this and if you view from here, if
you try to view from this direction, you will
find that this particular oxygen is like this
attaching to manganese one, manganese
two and manganese three being this one. So,
this is your this one. Then you have another
oxygen. So, this side oxygen. So, this I
made. So, how you would nicely you can draw
you should be able to draw it also very nicely
that within hexagonal arrangement.
So, this is basically this one because you
have all; you have one, two, three, four,
4 bridging atoms, 4 of these oxygen one, two,
three, four, 4 these and three metals. So,
you have this; four manganese you have and
sorry the reverse one, Mn 3 O 4 this one.
So, whether you have taken out this; that
means in this particular case you can take
out one metal center or you can take out one
bridging group. So, you can make it a manganese
three system or you can make it a manganese
four system with three bridging
group. So, this oxygen is there. So, if you
find that this one is immediately is only
mu two and if it can extent its binding to
the
other direction, because it is a highly distorted
geometry. So, once you start from this particular
code structure. So, code structure
is a partial Cuban, structure then you have
this manganese four and in between, between
these things this is not directly
connected. You have the calcium close to manganese
three. So, what we have in our hand is a Mn
3 from this unit a calcium four
oxygen and a distant manganese. So, this is
your distinct manganese and this O 2 little
bit we should know about the environment.
So it is bound to nitrogen of histidine 332,
then it is bound to glutamate.
So, only oxygen and nitrogen environment glutamate
189 and this one is bound to another carboxyl
function of amino acid is
aspartate 342 and since calcium is not a transaction
metal ion, sometimes it so happen which is
not so common for transaction
metal ion that the same carboxyl function
like acetate function, what we all know that
it is very difficult to go for a corresponding
binding of acetate group to copper; acetate
group most of the time it can function as
a bridging group only. So, these two oxygen
when they are utilized for coordination, it
should go for two different metal centers,
for copper, for copper iron and all but
calcium it is loosely bound because both the
two bounds are not same, it is loosely bound.
So this is alanine, this ala 344 and this
again to a carboxyl and nuclear bound and
another.
So, this basic arrangement what we can have
in our hand that during this transformation
when we go for the S 4 state; so electron
transfer is taking place from one after another
from this manganese, two of them are manganese
in the trivalent state and two of
them are in tetravalent state, but how this
calcium is playing some important role not
only for this structural chain but also for
a
different type of arrangement whether this
calcium can go out for another form. So, another
form we can get the another
structure will see; we will not show all the
detailed structure of these; how this particular
arrangement can immediately change if
we can have this manganese oxygen is here,
if we bring down this one over here that basically
is taking place. So, change in the
structure and within this hexagon we are trying
to draw what that positions are changed; that
means if you have a cube like
structure and all the four positions what
are occupied by metal center is now occupied
by oxygen; that means their positions are
interchanged.
The position of manganese and position of
oxygen can change. So, that is a typical structural
change for the entire complex. So,
this one is connected with this central oxygen
and two is now over here and three is here.
This is this oxygen. You see that it is
not that we are making all these structures
in the laboratory but only for one simple
electron transfer, the thing has changed from
this group to that group where you have now
close to that partial cube structure, your
calcium is close to number three and it is
within the cube structure. Now this same calcium
due to that electron transfer has come out
to this position. Here is now your
calcium which was earlier close to manganese
number three.
This was your manganese number three and this
two is basically this group which is attached
to the other manganese. So, that
means this particular group ala 344. So, basically
what we are now leveling that your manganese
center which was here has
moved from here to here. This is the manganese
position now. You see this original manganese
position is now occupied by the
oxygen and oxygen position is now occupied
by the manganese. So, that much changes, so
basically it is moving. So, if you
consider that as a clockwise movement. So
if you consider this as a regular hexagon,
all the regular hexagon you know has a
seismic symmetry. So, you just simply move
in a angle of 60 degree; just you move for
a angle of sixty degree. But for that the
driving force for that is that now two is
there and people can identify not only the
metal but also these groups; the ala 344 can
be
identified very nicely.
So, this ala 344 is not moving away much because
that is why I pointed out that this is a typical
binding typical coordination of
the entire carboxyl function to the metal
centre which we do not see in case of any
metal salt like copper acetate, iron acetate,
any metal acetate; the acetate groups are
mostly bridging. So, copper acetate is basically
a dimer because your acetate groups are
bridging two copper centers. So, this is rather
a different type of coordination where the
entire acetate group is bound only to the
calcium and this is not showing any interaction
with that available group because you do not
have any manganese over here. But
if you move one manganese from here to here,
you have available manganese over here and
this can establish the bridging group.
So, you see that is the role of the manganese
and as well as the calcium, why calcium is
there and calcium is playing something,
the role of the calcium what is being played
for its structural change.
So, calcium was there which is this acetate
coordination. Now calcium and this manganese
are now bridged. So, whatever entity
we can have; you have a heterodinuclear system,
we can consider it as like the superoxide
dismutase what we know that it has
copper and zinc center; in super oxide dimutases
you have copper and zinc center. Now it is
a system where you have a acetate
bridged calcium manganese compound. It is
very difficult to make also if you go for
making some synthetic molecule out of
these. If you have a binucleating ligand and
if you have a binucleating ligand like this
having two different pockets, you can have
a phenol unit over here and this phenol is
coordinating to calcium and manganese and
you are using some salt as the metal salt
manganese acetate as the salt, calcium acetate
can also be made.
But it is not possible to make this hetero
dimetallic system because all the time you
will be ending with either a dicalcium
compound or a dimanganese compound because
your coordination environment for these two
if they are symmetrical in nature.
So, whatever binucleating system you can have,
one particular part should favor manganese
coordination and other particular
part should favor the calcium coordination.
So, that is why this sort of chemistry is
pretty difficult to do because if you have
a
huge database to know that what are the groups
which can bind very easily to manganese center
and what are the other groups
which can favor the calcium binding, then
only you can get a simple binuclear compound
having calcium in one pocket and
manganese in other pocket but here if you
have this particular group and is already
available, but only thing what is happening
over there is the translocation; that means
the movement of the metal center, so within
the cluster arrangement.
Since you are putting electron you are getting
out electron, your structure is getting changed.
Why it is getting changed? It is not
that you have the oxidation or reduction,
you are changing what? You are changing manganese
oxygen distance, you are moving
from manganese three to manganese four, you
are going from one particular case level of
protonation to other level of
protonation, that also can change the manganese
oxygen distance as well as you can change
the manganese oxygen manganese
angle depending upon your nature of this group
whether it is oxido function or hydroxido
function as well as the nature of the two
oxidation states.
So, these are the other contributing factors
which can contribute for the structural change,
but the calcium is responsible for
binding because it do not have any preference
for regular coordination geometry; that is
why calcium. If you think that
manganese is there or any other transaction
metal is there and if they are extra coordinated,
immediately we will say no, this can
only pay for an octahedral geometry and for
octahedral geometry you know this bound distance,
bound angle are all fixed. But if
you go for a main group element or a non-transaction
element like zinc also, it has no crystal
field stabilization; it has no special
preference for coordination geometry. If you
have metal center like this calcium or cadmium,
suppose you have calcium, what
happens there?
Since it has no such preference for this binding,
only the available donor groups from this
protein chain or some biological chain;
only the available donor groups can come and
is such that it can have all sorts of arbitrary
coordination. So, this one such
arbitrary coordination can make this particular
group as a bridging model; that is why you
have very distorted one because one
bound is short and another bound can be long.
Because initially you can think of that this
acetate function is coordinating to
calcium center in a monodentate fashion because
these two bounds are not same. So, it is not
a symmetric binding of the acetate
group as a bidentate ligand because acetate
cannot be a bidentate ligand. It is a forming
a 4 membered ring. So, one bound is
short and another bound is long which is in
the other direction.
So, this calcium can change the structure
in a large way. You have this nitrogen coordination;
you have this acid coordination
from the glutamate group because these groups
are not moving. This is still the glutamate
189 but what change is taking place. So,
this is one part of change; that means your
system is moved from this direction to 60
degree and you have this and this oxygen
was mu 3. Now this is not there but what will
happen. This manganese number four is there;
you will have this manganese, you
have this. So, earlier you have this manganese
which is connected to a mu 3 oxido function.
So it will try to establish, it will try to
remain there with the help of some other bridging
group. So, this is the part which is very
different from the other structure.
So, this particular group which was there
on this number four can stabilize your arrangement
like this. So, you have this group.
So, it is moved but your number four manganese
cannot move that much; that is why it is the
system where you have Mn 3 plus
one manganese which is different from an arrangement
where you have a regular cube structure. So,
once it is there and you
move basically what we are moving. We are
moving one manganese over here. So, once you
move that manganese. So, this
particular new bridging function can take
place because you have immediately you can
take over water; after deprotonation you
can go for hydroxido bridging or further deprotonation
you can go for oxido bridging.
So, this particular unit which is a special
one which can go for bridging and this particular
distance we can measure which is 3.3
angstorm; that means it is tightly bound to
the m three unit. It is not that very loosely
bound. Earlier we can consider this a
dangling one which is the pendant one, this
fourth manganese is a dangler; that is why
call it as dangler but when it is forming
a
firm bridging, it is within this cluster and
that basically tells us that how this particular
one is responsible for accepting water
molecule and it goes for the removal of this
O 2. Once you get one group of this and another
group of this from the other unit to
make O-O bond. So, OO bond formation can take
place and these two manganese groups basically
can activate this particular
oxygen and this particular O 2 unit from there.
So, basic idea behind this thing is that how
you go for the change in structure
during electron transfer.
Thank you.
