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>>INSTRUCTOR: Okay. Let's get started then. A few quick comments, so you'll be reminded
by bioluminescent colony, to make sure next week when we do PCR when we select bacteria
from that colony that it was founded by one initial bacterium so we know it's a pure strain
of bacteria. Because we'll do PCR I'll go over it again and we'll be PCR two gene loci,
the ribosomal RNA and the B gene, we'll do photosynthesis, this will tag the light reactions
that you learned about earlier in this morning. To next week, it's, meiosis, I would encourage
you to read about mites and meiosis, I think you've heard about it before but please make
sure you focus on the meiosis aspect, we'll focus on a couple of things there.
And we'll talk about sequencing next week. Office hours, I have office hours this week
on Wednesday, 10 to 11 and then 2 to 4 I'm going to have an extra office hour, if wow
want your iClicker registered you come to office hours registered for that. We'll be
having lab exam 1 coming up there's reviews online I'll hold extra office hour, in particular
students have been asking me what's a good way to study for the material, do not reread
the lab material five times that's a not the way, read it once, outline it, make drawings,
diagrams, ask each other questions and go through the exam in exam reader you have 4
exams in exam reader set aside the time, 80 minutes, time yourself, literally time yourself,
80 minutes, after the timer goes off, if you didn't make it far that's not a good sign.
Okay. You want to get the practice of doing these, go back, figure out what you missed
what concepts you. And to know to answer each question and do
that repeatedly and you'll see there's this familiar pattern and that is to understand
the genetics questions you need to understand meiosis, you need to know how jam meets are
made, few concepts but keywords, standard curves some people have been asking me how
do I figure out the amylase activity in the saliva, if you remember, part 1 all students
did in part 1 what we did was we varied the source of the amylase, so we used a 50nanogram
or 200, 400nanogram, source of enzyme and we plotted the OD that we got. We should have
come up with knight nice straight [Indiscernible] like this because as you double the amount
of enzyme, the amount of product formed, product formed, should be double.
Now, this is not a rate per se per enzyme molecule. Okay. So make sure when you take
the lab exam that you understand the difference between the units on the Yaxis.
So, if this is the case, and I have this question always every semester, and people always ask,
wait I don't get this, so if I have a 50nanogram per mil enzyme and we're not going to say
it gives me [Indiscernible] in a 5 minute time course, if I have another solution that's
100 enzyme and a it's excess substrate, I would expect this to be double to be 0.6 OD
so if I'm look at the actual amount of product produced not rate not product per enzyme but
product over all, this indeed is double. But remember, we have excess substrate, which
means each enzyme molecule should be working as fast as possible.
So when we have a 50nanogram per mil source, each is working as fast as possible. When
I do 100nanogram each is working as fast as possible. Which is why this doubled but the
activity, that is the OD, per nanogram of enzyme, would be the same value.
Because this would be 0.3 divided by 50, and this would be 0.6 divided by 100.
Okay. So realize there's a difference here, this
is just total product form. A rate is different. It's the OD per nanogram
of enzyme. So think about units, units are critical,
what do you have on that Yaxis or the Xaxis so a lot of people are forgetting that point.
So what we did is part 1 we made the standard curve and then we took some saliva we did
various solutions of it and identical experiment but the source of the enzyme was not known
concentrations but the delusion, so if we have a delusion of 1 and 1,000 gives a is
of 0 nanograms per mil equivalent we have to multiply that by a thousand fold to get
the activity in the actual undiluted saliva, does that make sense? Anyone not get that?
So that's going to one of your questions ton worksheet. Standard curves are used a lot.
What we do is take known values, we generate a curve like and then we can use that to determine
unknown in the future, okay. With t respect to the vibrio, PCR stuff, hopefully
you had bioluminescent colonies, and or take another colony and streak it that's fine what
we're going to do is eventually, we're going to PCR to loci we're going to do another streak,
objective to that get isolated colony. And then with the PCR, remember, we're working
in this case with a large bacterial chromosome. And that is double stranded.
Forms that double helix like this, various proteins in there. So there's a couple of
locations where those primers should be specific. So people have asked, why do you use primers?
Primers target the location where you will PCR, one primer there, if that's the fore
ward primer, what would be the primer for the other strand?
This is the forward what's the other one? Reverse, they come in pairs.
Okay. That's one pair forward and reverse, and there's
another pair here, forward and reverse. They come in pairs, the purpose of the primers,
identify these start site for PCR, and they will base pair, following specific base pair
rules once this strand is separated. So, once we separate that, here's that one
primer there, with the other strand, there would be that primer.
Now, this is the 3 prime end of the primer, there's also a 5prime end.
What does this make this end of this strand realize it goes on and on and on, and I'm
only showing you a subset of it, but still talking about polarity, this verses this,
is this the 5 prime or 3 prime end? 5 prime because they're anti parallel. It turns out
all DNA polymerase have a requirement, that requirement is they can only attach to a 3
prime hydroxyl group, so it's to locate the place where we start replication, making the
pop copy, plus it provide that 3 prime hydroxyl group with we is add one by one when we do
that extension. So the steps were denaturation to separate the two strands, annealing of
the primers, extension, and then this will go for a period of time and we'll heat it
up and repeat the cycle. Okay. Now, you could actually figure out how often
a primer would occur, we actually know some sequencer formation so that we pretty know
this stretch of DNA that flanks the target, flanks the target and is universal means it
should work for most bacteria that we have. But if I had a piece of DNA and it's huge,
and I would say the percentage of A is 1/4 do you know what percent T is? 1/4.
Or I should say 25%. And 25% T.
And of course that makes it 25% C and 25% G.
So if my primer, you know, this huge background, if my primer is only an A, that's going to
occur a lot of times in that large background. If that primer was too long, A, let's a, it
would curl less frequently and you can calculate the probability of this, so if this primer
was 5 nucleotides long, that would occur 1/4 to the 5th in that sequence.
That's the probability. Because the probability of it being T in the
template here is 1/4, 1/4, 1/4, and it could be anything like that.
The numbers are going to make it easy. Reality is 1/4, 1/4, but we can change that and it
make it is more difficult. But you can calculate the probability of a primer binding so if
I a have a huge piece of DNA that's a million long, and I want that primer to appear once,
in that sequence. So the templates a million base pairs long
and I want the primer to appear once what should be the probability of that primer event?
One over a million. That's giving me once so you can actually
calculate how big a primer has to be to hit that target that would be in a general situation
so be familiar with that and we'll come back to this again. Trying to plant the seed right
now letting it germinate a little bit. Okay.
Any questions about that probability aspect? All right.
So, on the lab exam I often times will have here's the piece of DNA, here's the percentage
of this, how long should the primer be to appear once or five times or instead you have
a piece of DNA the that's so long here's the primer, or the sequence, how often does it
appear in the background? Does this make sense, raise your hand if it does not. Question?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: It does not mean it will definitely
show up at all. It's the probability.
That's what statistics are about. Probabilities it doesn't mean it's going to be guaranteed
one time it could show up 7 times or none. It's a probability okay.
Yeah? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: Okay. So, yes, typically. It's very difficult the questions easier
to do in some directions, if I'm going have you do something like this how often will
it appear? Usually I'm going do 25% 25%, but let's do a different problem.
So, what would you like the percentage of G to be? Make it like 10 or 20 or 30 or 40,
what was the question? What would you like the percentage of G to be?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: 5%. Can I make it 10? Okay.
Thank you. You'll see why. It's 10%.
What's the percent C then? 10%. So this is the same as 1/10, so this is same
as 1/10 so what's the percentage of C? We've used up 20, we have 80 left so this has to
be 40% or 4/10 and then the percent A has to be 40% or 4/10.
So, given that information, I'm just going to tell you the sequence is an A, a C, a T,
A, G, T. How often would that appear? We do the probability?
How often should the A appear? 4/10. Someone said 2 out of 5 we'll keep it 4/10.
Hold up often should the C appear? 1/10. How often should the T appear? 4/10.
And the A? 1/10. And the C? Whoops, 4/10. And the G? 1/10.
And I'm going to look now, the G 4/10. So this becomes a million on the bottom.
So 10 to the 6 and this is 256. So that would be the probability of the event.
Okay. If you know how big the DNA is you can tell
me, statistically how often that should appear, statistically, not defined as exact number.
Make sense? So if it can go both ways, I just have to give you the data, okay.
All right. So, PCR, be very familiar with how it works.
If you draw this out for 3 rounds yourself color coding things it really makes more sense
so get some pens, label it, color code it, just draw it for 3 round it makes a huge difference.
Again just illustrate this we take the double strand DNA, we denature it to make it double
stand, heat it up, we anneal our primers they bind specific locations that starts the location,
and generates the 3 prime end and then we extend.
Okay. And in the end, we're going to get a large
number of products. And let's go through the formula one more
time. Since we're doubling at the end, this is at
the end of a round, 2 to the Nth, is the total number of product.
Since we always have those original two templates so this is with a large background template
we have the original double strand template we always increase by 2 the number of nonunits
made so that means number of nonunits, is always N x 2, so the units is 2 to the N this
number. We'll just do an example if it was the 6th round, 2 to the 6 is 2, 4, 8, 16,
32 did I do this right? 64?
2 x 6 is 12. Means we have 52 units.
And nonunits. So the number of units are increasing exponentially, with each round, the number
of nonunits always increase by 2. Yes? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: So, remember the unit that I defined was our target, and our product always
has to include the primers that flanked it. So, whenever we make a product that is this...
Plus or target, that is unit. But if we had the primer and it extends past
that unit, the target, this is nonunit. Okay.
So go back and look at the figures from last time. But unit will always be our target +
the primers because the primers have to be a part of the product.
They have to be a part of the product because they extend from this.
So in all cases the product will have the primer + the extension product.
Okay. In all cases.
All right. Question?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: I'm sorry?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: How do you make it stop? You
heat it up, which would separate the two strands, so if you heat this up, so imagine this
is proceeding along and you heat this up, then you know you're going to have this strand
bound to this, you have another strand, you start another cycle. Or conversely, let's
see if I can show you this drawing here, we'll look at this image here. What we see here,
let's see if I I don't have this one. It turns out that what happens is after a few
rounds, you actually have some that are unit length.
So it's the primer + this, primer + this. Okay.
So that's the unit. When I separate this, I'm going to call this
strand, strand A and strand B, strand A, strand B.
When I (No sound). Primer to this, and my primer to that, I can
only go that far and that far. So what stops synthesis in that case in extension is, you
run out of length of template? So you can stop at one of two ways you run out of template,
which is why we get that large increase in number of units every time, or it's extending
and you separate the two strands before it can keep going much further. Does that answer
your question? And that's that formula so for photosynthesis. Every time I open this
file it change it is arrows, every time. So anyway, I don't know why, but it does. Magnification
factor anyway, in terms of photosynthesis, when we have the reaction 6 C O 2 + 12 waters
the water is being oxidized to oxygen. That is an extremely unusual reaction.
Takers duct loads of energy. Where do those bucket loads of energy come from? Sunlight.
So I had a student who was in the navy he was on a nuclear submarine and he was an engineer
and approximately 1/3% of all of the energy expenditure for the nuclear submarine is to
do that reaction. Oxidize water so they can breathe. And everyone then they have fairly
low oxygen concentrations, which is why people in submarines which is why they all get varicose
veins really bad. But the bottom line [Indiscernible] very unusual reaction.
Plants can go that. Some bacteria can do that.
But like the sulfur bacteria don't use water they use H 2 S and in this case the C O 2
is being reduced to the sugar, what's the source of reductant there?
What's the actual in vitro, what is in the CO2? NADH. So we'll talk two pathways we'll
talk about the light dependent reactions, which convert the capture light and convert
it into chemical energy. Okay. So we'll talk about the light reactions where the energy
is photons are captured and then we make oxidation reaction, where we donate electron, and lose
electron. And then we'll talk about the Calvin Benson
cycle, this use to be called the dark reaction, horrible name for them.
Of the enzymes in the Calvin Benson cycle are activated by light, and they use the product
from the light reactions. So don't have those product and they're not
activated they can't proceed. So we'll talk about this in plants. As you know it occurs
in leaves, here's a cross section through a leaf showing you the upper and lower surface,
on the lower surface they have more stomata for gas exchange the book shows,CO2 and oxygenase
it's focused on photosynthesis and CO2 being captured, due the oxidation of water but,
what they fail to show is the release of water that's a major concern is water loss through
the stomata we'll talk about alternative forms to capture CO2 and those alternative forms
are to min rise water laws. So we allow gas exchange in here we have the mesophyll cells
we have chloroplasts within the mesophyll cell, in the chloroplast here's this, I want
to go over this very fairfully with you. Here is one phospholipid by layer, here is a second
one. This is called the outer and inner. These are close together, usually when you have
two membranes close together it's called an envelope membrane. Like the nuclear envelope
membrane, in this case it's chloroplast, there's an additional third membrane that is a [Indiscernible],
they come in stacks like this and they're there are thylakoids that connect the stacks
as well, we're making various compartments, critical that you know the compartments and
what's occurring in the compartments. Here's the thylakoid lumen or space. That's by the
thylakoid membranes and the stroma is by the inner membrane and the outer membrane that's
the compartment there. There's indeed a intermembrane space but we'll ignore that for the most part.
I'm going to go over it. Stroma is where the Calvin Benson cycle will be occurring and
the thylakoid membranes where the pigments will be embedded they're hydrophobic they're
embedded in that the thylakoid membranes that's where the light is our can go that where we're
going to form NADH which will occur right on the surface of this.
So we'll have NADH in the stroma and we make a proton gradient and pump the protons into
the lumen and in the process of that we have a gradient, proton will move from the lumen
to the stroma following the gradient and make ATP. So ATP productions in the stroma, NADH
productions in the stroma, the Calvin Benson cycle is in the stroma.
So let's just run through this. We have the light dependent reactions, light
we'll talk about two photosystems we'll talk about the pigments that capture the light.
I'll mention it now before I forget and mention it again, repeated, chlorophyll A there's
about 200 chlorophyll A molecules per photosystem. Okay.
All right. About 200 chlorophyll A molecules and [Indiscernible]
a bunch of carotenoids. And these chlorophylls are not naked they're associated with proteins.
So when I have chlorophyll A, molecule 1, if it's with protein 1, associated with it,
it's going have a slightly different absorbed spectra than chlorophyll A that's protein
2 and A 3 with 3. So by having all of these chlorophylls with
different proteins and different neighbors with them they have slightly different absorption
spectra. And that's going to be a key concept.
Because what we'll see is, this chlorophyll A, if it's excited, will absorb light, and
then basically have florescence and it will emit the light at a longer shorter wavelength.
When it emits light, shorter or longer wavelength? Longer it has to be less image. It can't be
more it has to be less. And then this pigment will be able to absorb that and this pigment
to that and what we'll do is funnel all that light energy to the reaction center chlorophylls.
That's special dimer. It's because we have these various proteins
that they're differently absorption spectra so we can use more of the wavelength of light.
So more area to capture the light. So the light reactions occur in the thylakoids we'll
be taking and capturing light, we will be reducing NADH to NADPH. NADPH is a very strong
reductant so this is a pretty unusual reaction. Photosystem of 1 is capable of reducing ferredoxin
[Indiscernible] very unusual it takes a lot of energy to do that. We also make ATP because
we have a proton gradient form, so the NADPH out in the stroma are used in the Calvin Benson
cycle to reducing CO2 to sugar. Okay.
So we need light, we need water, we need CO2, by product, O 2 NADPH and ATP which are used
by the Calvin Benson cycle. There's quite a few questions you can ask yourself, I won't
do any iClicker question. Why are most plants green? Because they have more chlorophyll,
they don't absorb in the green range they transmit it, reflects it and that's why they're
green. What's the role of the soil? It's magnesium, elements like that it's not the CO2 that comes
from the air, okay. Of course water comes from the soil too.
So you know, food doesn't really come from the soil it's those essentially minerals that's
from the soil. Okay. Due chlorophyll... (Reading). If so how many? It is not electrons that are
being transferred here to the reaction center it is not electrons, it's common a misconception
it is the energy of the photon that was captured by this pigment, that gets transferred to
this pigment gets transferred to this and the funnel, the bottom of the funnel is the
reaction center. That's why this has to be a very long wavelength because you have to
be able to funnel energy to this, so in case of the photosystem 2, that's also called P
680 and the photosystem 1 that reaction is called P 700. Notice very long wavelengths
that allows it to be at the bottom of that energy funnel, funnel energy trap. Okay.
Again, it's not electrons that are getting transferred in this antenna it's the photons
it's the light energy. Okay. On the bright sunny day would you expect plants
to undergo cellular respiration yes. So they'll have that. When the stomata close can you
have any CO2 going in? No. So you're going to start limiting for CO2 if you do light
reactions what happens to the amount of light reaction in time when the stomata are closed
oxygen will start increasing when we talk about the Calvin Benson cycle you will see
that's a problem because it will promote photorespiration so that's the concept of the plants closing
stomata, but you Crete this other problem of CO2 and oxygen balance.
Dr. Pauly has more time to talk about those. Different wavelengths of light we know that
by having different pigments and there's associated with different proteins we can capture more
of the light, in terms of the various wavelengths so we can increase efficiency there.
And if you think about this, chloroplasts from plants that grow on the surface of a
pond verses deep, the light quality the wavelengths amount of light would be different at the
top of the water verses bottom and they will have adapted to change the pigment composition
within the photosystems to match that. So, let's just run through a little bit about
the light. We know that light consists of wavelengths, you can take the visible light
from short to long, higher energy lower energy so we know that happens.
You know that what happens is for a chemical to absorb a photon it has to be of a specific
wavelength to match the electron orbitals and things like this you know that from O
chem and all of that MNR, etc., various things happen. So only specific wavelengths can be
captured. And onto a certain extent. And when a pigment
is excited that photon takes and moves electron to an excited state and then something can
happen. All right.
So the 3 things that can happen are as follows: So we'll start on the left you can give off
light, and heat and by the way, heat should be in every one of these things right? Because
no energy transformation is 100% efficient. Heat should be on every one, but this the
case we can have light released and heat. We could have what's called decay by residence
transfer. What happens is this excited pigment here, is going to fluoresces, release light,
and I have slide, so this pigment 1 it will release light like that, pigment 2, will be
like this, it will capture the light, pigment 3, will be here, and we'll just transfer the
light down this sort of in this complex, eventually to the reaction center.
Okay. So that's what residence transfer is. I'll
show you a slide that shows the absorption spectrum doing that. And the third thing is
we can have electron transfer so we again we have capture the light, excitation but
this in case, we have a chemical reaction occurring, we take and transfer the electron
to transfer to something. Acceptor and if that chlorophyll has donated electron is it
now oxidized or reduced? Oxidized. What's its charge on it plus or
minus? Plus because its lost electron so for the process you better reduce it again back
to the ground state. Okay. Question? Okay, so here's what happens the
photon, chlorophyll gets excited say we're at the reaction center that electron can then
go to if it's in tax system be donated through the electron transport chain we'll go over
this in a moment. Or if you take an isolate that pigment out, so this is a bunch of basically
an acetone extract, those pigments are now separated from they're proteins.
So, they all have the same absorption spectrum, what happens is the they can't transfer electrons,
because there's no proteins that are needed anymore and they can't do residence transfer
because there's nothing to accept, so they can only florescence so, this is an [Indiscernible]
extract they absorb the light, and release heat as well.
This illustrates that residence energy transfer, this is the antenna complex.
Basically, what happens is in a given photosystem, it's sort of lookalikes this schematically.
In a given photosystem we have a couple hundred chlorophyll As so I'm just going to draw it
like this, scatter through here, couple hundred chlorophyll As each with the various proteins,
chlorophyll Bs with they're various proteins. We have that special reaction center that
dimer. Of A and A.
And we have some carotenoids in there. And since carotenoids have different absorption
spectra, we have a large number of pigment molecules here.
So we effectively increase the surface aerographer capturing so we're doing two things here by
having antenna like this, we increase the number, so I'm going to ask during office
hours is it possible for reaction center to absorb light directly? It is but this is such
a small percent in terms of the number actual surface area it wouldn't happen very often,
okay. So we increase the number. And we increase the wavelengths we can use.
All right. So the illustration to the right there shows
that. Here's pigment 1, pigment 2, pigment 3, emits
light gets absorbed by this pigment, this light gets absorbed by that pigment and we're
transferring the energy here, to the reaction center.
This occurs very rapidly. The order of these reactions light transfer
is about 1 x 12 to the minus seconds. It's all physics you can drop this down to 5 degrees
Calvin you can't measure changing the rate so it's just physics at this point. Where
it is chemical is when you get that electron donation. Okay.
So let's look at this a little bit carefully. We have the antenna, capture the photon, transfer
the energy to that reaction center, the specialized pairs, chlorophyll A and they have proteins
with them. Gets excited and now that electron gets donated to what's called a primary electron
acceptor. Primary is first one here. Once this has occurred, this dimer here is oxidized
it's a plus state. It can no longer do this reaction again until it gets electron and
goes back to its ground state, [Indiscernible] charge. That's true for both photosystem 2
and 1 it's similar like this, antenna, special pair.
Here's two absorption spectra of chlorophyll A and B, and these are isolated from the proteins.
So all of the chlorophyll A molecules would look the same you see they absorb in the blue
range, the red range, but notice, if this was the actuality reaction center chlorophyll
with they're proteins it would absorb here at 680, everything gets funneled as a captured
to that reaction center. Longer wavelength.
Here it shows what's called an action spectrum we measure some photosynthesis, we can measure
oxygen release, so we measure photosynthesis by amount of oxygen released and at the different
wavelength we see different amounts. In lab we will measure the rate of photosynthesis
using the reduction of DCPIP, but with can measure other things.
Okay. On the lab exam I specifically asked you to figure out which graph would correspond
and it's no longer measuring DCPIP but some other like ATP or oxygen release so you have
to understand what's happening so look at the structure of chlorophyll has this porphyrin
ring, these to absorb the light. What membrane? Inner, outer or thylakoid?
Thylakoid. Chlorophyll A has a methyl group, chlorophyll
B as an aldehyde. The way I remember that is bad.
B aldehyde. Now, I won't typically can you on the lab
exam does chlorophyll A have a methyl group or is it chlorophyll B, but I would ask you
which would be more hydrophobic? So A or B more hydrophobic? So again, general concept, but we have two reactions centers.
Two photosystem, we have photosystem 2 and 1. Which refer to talk about photosystem 1,
most people talk about photosystem 2 first. Let's work with photosystem 1 first. So we
have excitation in the antenna, the energy gets funneled through residence energy transfer
to the reaction center, electron gets donated to the primary electron acceptor we reduce
ferredoxin, okay. So ferredoxin gets reduced.
So we electrons and hydrogen going to NADP + to form NADPH + proton.
Absolutely critical to know that hydrogen a proton + an electron.
I know you know this, I'm going to write on the board it’s that important.
The reason why it's is what we're going to see sometimes in the electron transport chain
there are molecules that can carry only electrons. Only electrons but it's been given a hydrogen
molecule. So if it can only carry electron, something
has to happen to the proton. What happens to that proton is it gets pumped to some location.
In other situations, there are components and electron transport chain that can carry
only a hydrogen, but it's been given an electron therefore it has to grab a proton.
So what we'll see is by alternating hydrogen care yeses, electron only carriers and electron
transport chain we pump protons from one face to the membrane to the other, from the stoma
to the thylakoid space. The analogy is this, if this is the thylakoid membrane here, so
phospholipid by layer here, and then I'm donating hydrogen, grab it, move it on, oh, whoop,
problem she is only carry electrons but she's been given a hydrogen so grab the hydrogen,
and toss the proton behind you. Now we've moved the proton to this face of the membrane,
we've made a gradient, right there's now more protons here than before now go ahead and
donate your electron. But now, you can only grab hydrogen, you were
given an electron so what do you have to do? Grab a proton.
So now, from this face not that face because we're going keep that gradient going. All
right. All right. So you'll see that's what's happening
it's basically we have iron, sulfur proteins that can go from + 3 to + 2, in electrons
only. So let's follow this.
Now photosystem 1 is oxidized, because it lost its electron, right?
So it can't continue the process until it gets reduced it gets reduced by the electron
transport chain, so electrons are flowing from here they come from here so lets run
through photosystem how it works. Photosystem 2, the antenna gets excited, transfer the
energy to the reaction center, transfer electrons to the primary electron acceptor, which donates
to plastoquinone, but plastoquinone is a hydrogen carrier. So here's an example, where we have
a carrier like plastoquinone, that carries this, but it was given this, so it has to
grab protons. Okay.
It was given electrons. From that primary electron in this case, but
it can carry only hydrogen so it grabs a proton from out on this face of the membrane, out
in this stroma, it grabs protons. Grabs two protons, gets to electrons, becomes
PQHT, it's reduced. It goes through the cytochrome complex but it can only carry electrons. So
it's begun given hydrogen, but prove the protons across that membrane into the thylakoid space
so this label's wrong here this is from cam Pell we're not making ATP here, directly,
what are we doing? We're making a proton gradient, and then later we'll make ATP through the
ATP synthase. Okay.
We're not moving protons, sorry we're not moving ATP we're moving protons. Again, we
have a hydrogen carrier, electron only carry it and a hydrogen carry so it gets electrons,
so it can grab proteins so we funnel electrons here to this reaction center that was oxidized
now rite deuced and it can continue the process, right.
But what's the problem? What happened to photosystem 2 now? It's oxidized so where does it get
reduced from? Water. So what we'll see is that water will donate
electrons here, so we take two water molecules we'll get 4 electrons, 4 protons and oxygen
so what we're doing is moving electrons from water through this complexes, to NADPH +,
to form NADPH. It's not the same electron. Okay.
There are many electrons around here that are moving around its not the same electron
but effectively we're moving electrons from water to here, okay.
Question? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: So the question was: Does it did this one electron at a time.
Sort of. It's actually quite complicated it takes 4 electrons to do this, so what happens
is the oxygen evolving complex and the photosystem 2 it takes 4 photons nor to happen. And it
would be nice the system doesn't set so if we butt it in the dark, and then we say let's
hit with 4 photons and we get the water in that, oxidized, it doesn't work that way.
It actually rests at a state where you hit 3 photons, it does the reaction four photon
it's back to where it stops at state 1 it's not critical for you to know but there's 4
steps. Step 1, 2, 3, and 4. It will stop at the end
of this, so when you hit it with 3 photons it will do that oxidation of water, but the
4th photon will put it back to this state and just repeat that.
Okay. Back to this state.
So in the dark, it's at this state, which was made it a little complicated at first,
but it's curves a cycle of 4 photons. Okay. Detail you don't really need to know in terms
of the level. Other questions?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: All right. So let's look at
this. Yes?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: So the question is does it take
energy to move the protons from here to here? Absolutely it does P. Where is the energy
coming from? The capturing of light. As you go through here, these things follow a gradient,
a redox gradient. So with each transfer, you lose a little bit of energy.
Where did the energy all come from? The capturing of light it takes it to that higher redox
state. Did that answer your question?
Okay. So, this is just look at this a little bit.
This is a balanced reaction I like balance reaction it is other thing is, a lot of times
I will only draw one end of an arrow, and I assume you can draw the other end of the
arrow. Okay.
So let's just run through this. Photosystem 2, 1.
Let's start with photosystem 1, we excite it. We take two electrons, two protons we
reduce *** + to NADPH + that proton. Balance reaction just happens to be the NADPH ***
+ can grab two electrons and one proton, but balance to make it NADP + the proton is that
okay? That's the balance. We have this problem here that photosystem
is oxidized it gets electrons from electron transport chain that allow it is process to
continue but the problem is, photosystem is oxidized gets electrons from water, two water
molecules, 4 protons it's oxygen. So notice the gradient here, as we go through the electron
transport chain here we move protons from the stroma into the thylakoid space, and we
also increase the bucket of protons from the lysis of water so we're increasing the proton
gradient that represents potential energy, that potential energy is used to make ATP,
the phosphorylase in ADP to ATP. So now if I two solutions, and remember, there's a gradient,
so if I have a 0 molar protons here and 0.4 molar here, that's state A and state B, 0
molar here, and 30.8 molar there, which represents more potential energy? State A or state B?
B. So realize that to create state B is going
to take more energy. Okay.
And that we'll see important for one of the experiments you'll be doing. So the way you
do this experiment this a little bit more balance showing you various things, page 700,
P 680, so the excitation, show it is NADPH, now this is all in vibrio, in vibrio, and
in act in a plant, it would be in a chloroplast in a bacterium?
No. Because they don't have chloroplasts. Okay.
In our experiment when we do the hill reactions and when we're measuring rates of photosynthesis
we will measure not the reduction of NADP + anymore, or sugar production we will measure
electron transport chain because we'll have a chemical that we will, DCPIP which is blue
and as it gets reduced to DCPIPH 2 it become colorless so we'll see a change in the absorbents
from blue to colorless as the reaction proceed. The faster reaction the proceeds the faster
we lose the blue color. So for this to happen, for us to be able to measure this, we have
to have the chloroplasts broken open. Okay.
Normally, in vibrio, this is producing a lot of sugars so there's also ore [Indiscernible]
and in this chloroplast so it has a lot of osmotic [Indiscernible] so if you put that
in pure water the water would come in, rush in, and basically blow this up, question over
here? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: No. Okay.
So, when we do the experiment, we will not have sucrose present. Which will cause the
envelope membranes to rupture, but the thylakoid membranes remain in tact.
Thylakoid membranes remain in tact, but the envelope membranes will rupture if no sucrose
is present, if the sucrose is present the chloroplasts will remain in tact.
I saw a lot of puzzled looks. Okay.
Okay I'm not going to draw the phospholipid by layers just to be fast.
Let's label stroma. The lumen.
The envelope membranes. Lots of sugars are being produced.
Normally. Cell let's go back to the terms tonicity.
Isotonic and hypotonic. These terms are meaningless unless you realize
they have to have a reference. I'm taller than some people but I'm shorter
than others there's always a reference I'm comparing to be able to say whether I'm taller
or shorter. So if I I have two solutions, if A is hypertonic to B, I can also say B
is hypotonic to A, so there's always a comparison. Okay.
So in this case, there's lots of sugars there so it's pretty a high tonicity, a lot of
colligative properties there in terms of the ions the sugars. If I put this in pure water.
The water is hypo or hypertonic to this? Hypo. So the water would rushing in.
It is like a balloon. You keep blowing in air in the balloon what will happen? It bursts.
Okay. So we can have in tact chloroplasts, we can maintain the intact, if we have sucrose
in our phosphate buffer because it acts as osmotic to prevent lysis.
So when I say the chloroplasts and I'm putting them on ice, we want them on sucrose but when
he we do the reactions to measure the reduction of DCPIP we have to get the DCPIP adjacent
to those thylakoid membranes because that's where that reaction is occurring, DCPIP, to
colorless, and there is no white chalk, there's one small piece and that's where we're measuring
this reaction to DCPIP H 2, so when we do the hill reactions there's no sucrose.
We need to break those chloroplasts open. Okay.
But when we eyes late the chloroplasts we want them in tact, so be familiar with the
concept of as osmotica, so in our actual experiments for the hill reactions they're broken open.
Okay. Here's the experiment you'll do, this is techniques
aspect of this, we're going to take some spin itch leaves we get them from Costco, nothing
special about them we put them in sucrose phosphate buffer, we blend them those sheer
forces are dramatic they break up the cell walls things like that, things get released,
we then spin it in a centrifuge, please balance the two tubes make sure there's about the
same volume in each, put them opposite from one not the rotor, spin them you'll end up
with a pallet and a [Indiscernible], the pellet should have mostly chloroplasts there might
be a few mitochondria, but the pellets is where the chloroplasts should be. They're
levy, so it is short spin. Not really high G forces there's a supernatant.
ER, things like that. You will take the supernatant and decant it away.
So, basically, pour it off, get rid of the sucrose. So you have a pellet.
Okay. We're going have two tubes.
One tube you're going to add acetone to, acetone is it hydrophilic or hydrophobic? Trick question
it's both. Hydrophobic, most students say it's hydrophilic
because that's what they emphasize in O chem, so it will abstract the pig transmit the membranes
because it's so hydrophobic, the pigments will go into that, denature proteins. So the
acetone is hydrophobic it extracts the pigments goes into solution.
Okay. Then now that we have that acetone extract
we put it on a piece of paper, [Indiscernible], take a pencil, mark the line here, don't use
ink or excuse me use chromatography on it as well. This is a hand in case you didn't
notice. I'm a lousy with drawings. Come on you didn't know the fingernail polish?
All right. Now, there are a little tricks with everything.
And this seems like a minor point, so I'll just go over it.
We had the paper like this, there's your pencil line, here's a microcapillary tube it's got
a very narrow diameter, when you put in the acetone extract lit be whipped up due to capillary
reaction, there's the pigment solution, when you put that in the paper, that paper will
suck that solution right out and you will get a huge blob. So to overcome that put your
finger over the end here, to seal that off and when you put your tube here, this vacuum
will counter act the capillary action from the paper and you'll looking at a nice small
dot. Move it over here you to take your finger off to release the vacuum and you want to
blow on it to dry it we want a nice green line here.
It's green because all of the chlorophyll but there's chlorophyll A and Bing and there's
carotenoids present so make that green line and we'll do the chromatography on it. And
what will happen is the developing solvent here gets whipped up the paper by the capillary
reaction the moment it interacts with the band there and a line is better than a circle,
get that line, the moment you're [Indiscernible] and we start partitioning between the mobile
faze, the developing solvent and the paper, is the paper hydrophobic or hydrophilic? Hydrophilic.
Bounty the quicker picker upper. It picks up water. Why does it love water so much?
What is it made up? Silius(?) and what it's made up of? Not starch. Beta 14 linkages of
glucose. All of the hydroxyl groups so it will partition back and forth.
And then in the end we'll see the pigments partition.
We're going to two green bands and three yellow bands, the yellow bands correspond to the
carotenoids, the green bands correspond to the chlorophyll, one is chlorophyll A and
one is B let's think about this, we can predict which one is which.
Okay. So, chlorophyll B has the aldehyde so it is
more hydrophilic or hydrophobic? Hydrophilic so should be retained by the paper more? More
so the [Indiscernible] should be lower, smaller? So this is the chlorophyll B and this is A.
So you can cut these out, just put them in a test tube with acetone you extract out the
pigment from the paper in the acetone remove the paper, you could then do absorption spectra
on purified pigment. So that's that part there. Okay.
Later on, we're going to have you use the same acetone solution to do an absorption
spectrum what should be the blank be in that case? Acetone should be in the blank so take
a tube with acetone put it in the speck, calibrate it. These specks will great you will calibrate
it at every wavelength, it goes to the entire wavelength of visible light, and then when
you put your sample and shutter, and it will give an absorption spectrum so it takes no
more than 3 minutes to do this. Entire spectrum. Okay. So remember, you have to use acetone
as the blank because the pigment is dissolved in acetone.
Okay. Now, there's a slight technical tissue here.
Sense this was a sucrose solution, and you decant it by tipping it upside down, the tube's
upside down like this, round bottom there, your pellet here, you decant this like this,
there's going to be some residual liquid right here, due the adhesion here to this.
So take the chem wipe and wipe this off with a chem wipe the reason why that's necessary
is because the residual sucrose here, when you add the acetone it will come out of solution,
it won't precipitate it just stays in solution so when you do the absorption spectrum all
that the sucrose in solution will hit the light and you can't get an absorption spectrum
so one of the technicals, so make sure you wipe it there. Because it comes out of solution,
its solubility in acetone is very low. Any questions about that?
So you're going to do the chromatography it takes about an hour to develop. So while this
is happening we're going to do the hill reactions. The light reactions in terms of electron transport
chain. So, you work through this, add the acetone
the second tube you put on ice, okay. People always ask why it's on ice because
you've broken it open from the normal environment so you want to keep it as active as possible
because you've released protease and things like that. So for the hill reactions we want
to have a concentration when we do the experiment of about 0.1 migs of chlorophyll per millimeters
so we've chosen that so it doesn't happen in 10 seconds, you don't get any data, nor
does it happen in 5 hours, you take a reading every 15 seconds, maybe 12, 14 readings. The
concentration of 0.1 migs of chlorophyll per mill will give you that. Any more the reaction
goes to too fast the other he takes too long. We have this pallet what we're going do with
the pellet we're going to rhesus spend it in sucrose phosphate buffer, so 2.5 mils
we have paintbrushes you want to gently paint that pellet back in suspension you don't want
clumps because we're working with the micropipetters and that he have fine openings in the tips
so if you get clumps the tip will get plug as you release the plunger it create a vacuum
and shoot the solution in and you won't get the proper volumes so we don't want have clumps
so with the paintbrushes, generally resuspend the pellet they're camel hair they're the
finer brushes we can find, it's a standard use bid research labs, okay. So paint those
back into suspense it's very concentrated so I'm going call that enriched so we have
to figure out how many chloroplasts are there to make the solution that's 0.1 migs so we
have a concentrated solution, what we're going to do conceptually is take a subset of it
out, measure the OD of the subset, and from that value determine what we need to dilute
this to. And it's the No. 1 mistake made in lash, by 5 x more people than a mistake
in this lab than any other mistake in the lab. So to avoid I'm going to write on the
board. You have this pellet, you've resuspended in 2.5 mils with sucrose.
And it's cold. And this is very concentrated. So what I'm going to do is take out 0.1 mils
of this, I'm going to add it to 3.5 mils of phosphate buffer.
I'm going to measure the OD, and solve a formula. And that formula would tell me that let's
say its 46 mils just made up that up. That would tell me I need to dilute this 2.5 up
to a final volume of 46 and all 46 mils would be 0.1 migs of chlorophyll per mill.
So the theory is, concentrated subset, solve the formula figure out what to dilute this
to. Is that clear? But we don't need 46 mils and it's really
difficult to make 46 mils because the tubes only hold about 13.
So we found a work around, so here's the work around.
You have this concentrated one with you resuspended in 3.5 mils you take 0.1mil out, and measure
the OD because you use the speck what do you have to do? Have a blank so you have a blank
of phosphate buffer, first. Blank it. And then take your sample measure the OD gives
you some OD, multiply this by 46 comes out with a final volume in this case I would have
the said the OD was 1 to make my math easier to that OD of 1 I need to take this to 1.25
and I get 46 mils of the proper concentration. But I can't really make 46 mils it's just
too time consuming and you don't have the right tubes and stuff so we're going to make
5 mils + X, so we're going the figure out what the X volume is. The X volume is going
to be a value that I add too 5 Miles of cold sucrose, the 5 from here and that will give
the 5 + X so here's the formula you saw. Here's the final volume from that OD, the
X value is going to be 12 divided by VF 2.4 so this would be 46 in this case, 46 2.4,
so it would be 0.3 mils or something. Approximately math done in my head.
What that means is, if I take 0.3 mils of the concentrated one, this is the X value
now, if I take 0.3 mils of that, and I add it to 5 mils of the cold sucrose I get in
this case, 5.3 mils. And that 5.3 mils is of this concentration.
So the No. 1 mistake made by far is students take this tube, which is very already 36 fold
solution and they use this as a source of X.
That is not the source of X, because that's the really dilute one.
Right. The source of X is the concentrated one. Because
the concept was, subset of concentrated, measure solve formula that's what we add to the 5.
No. 1 mistake made by far. When you have the solution and it's proper concentration
it should be fairly light green you did it wrong.
The problem is, if you use this one, the reaction still foresees but it's about 15 hours for
each experiment and you're doing 5 of them so about 75 hours you don't want to be we
wont don't what to be there. So concentrated one. So in the end we end
up with 5point + X of a proper concentrated and we're going to do 5 tubes.
The 5 tubes are as follows: We're going to have to have a blank. We have a blank because
we're using the spec. The chloroplast will absorb, contribute to
the background, the phosphate buffer will absorb contributes so we have a blank that's
pretty straightforward we have the phosphate. We have a we're going to do some conditions
we don't have DCPIP here, no DCPIP. We have the chloroplasts so that's our blank and we're
going do four sorts of experiments we're doing do a dark, [Indiscernible].
So let's just run through these. The dark will have DCPIP and they will have chloroplasts
and water, etc. For the phosphate buffer, excuse me. Now that
5 mils how much photosynthesis should be occurring in the dark?
None. So, ideally we'll see no change in the OD,
now blank did not have DCPIP, the dark had half a mil so the dark, the initial value,
in terms of OD, the initial value will be a high value.
And since I said the reactions usually take about 3 minutes. We'll wrap that tube after
we take the initial value, wrap it in aluminum foil, check the OD 3 minutes later. So this
is going to be time here. Approximately 3 minutes later, so you should
have an initial value like this, and it shouldn't change after 3 minutes let's say. Okay. Shouldn't
change. And if it changes that means when we get to the light data we have to take that
into account. And the reason it can potentially change is mitochondria also does reduction
reactions and it can reduce the DCPIP we did a purification, but it's not elaborate so
there might be mitochondria contaminating. So let's do light reactions we should expect
some change in OD. The way you're going to do this experiment
is we have some lamps in the lab, goose neck lamps.
Light bulbs inside here. Here's the bench top. You want to take the ruler and measure
15 centimeters down like this, with a ruler, and when you hold the tube, and expose it
to the light, hold it a at an angle like this and rock it because those chloroplasts are
heavy and if I hold the tube like that, for 15 seconds, the chloroplasts are already started
selling and what I would see sit would be clear at the bottom where the chloroplasts
have settled and blue at the top. Because [Indiscernible] so by gently rocking
it you keep the chloroplasts in suspense so you will change that in blue through out the
tube. So rock for it for 15 seconds your lab partner will say time, take the spec, take
the OD value, back in light. So every 15 seconds of light exposure.
So 15 seconds in the light, OD, take it out. 15 seconds, OD, 15 make sense?
15 seconds of light exposure. It will probably stop after 3 minutes you'll
probably be down to 0 or no change anymore. Any questions about the light? What happens
in that light reaction, we can go back to some slides here.
What we're doing here in the light we're broken up the chloroplasts because it's a phosphate
buffer no sucrose so we have DCPI outside here.
So as those electrons gets transferred, we're reducing the blue color DCPIP to the colorless
DCPIP 2 so you'll see some decrease you'll see some values does that make sense? We'll
do some other experimental conditions here. All right. One of the other experiments will
be light + DCMU. Now I have to explain to you what it does, because I haven't told you
what it does, stands for die chloral methyl urea I would not ask you on the lab exam what
it stands for, but uncouples electron transport so we have photosystem 2 here, photosystem
1 here, we have the electron transport chain here we have plastoquinone here.
DCMU mimics this, and it blocks electron transport chain.
So no electron transport chain should be occurring, so in that case, photosystem 1, will potentially
reduce the DCPIP that's blue to a colorless, but it can't do that for very long, because
all of the photosystem 1 will be oxidized and there aren't many or any chloroplasts
so you probably won't see change in the OD if you're quick and good, you might see a
slight degrees but probably not. The expectation is, that with DCMU, there should be no change
in absorb answer the. That's the expectation, okay.
Because we've blocked the electron transport chain.
Is this clear? In fact you can do these experiments and see how big a pool size you have, so the
other experiment we'll do, so we've done light + DCMU, the light I introduced first so then
I can make sense with the DCMU and then we have this other reaction. Light + methylalanine,
it's a pH buffer and remember, this is all in the background of a thylakoid membrane,
so we're no DCMU so he would be doing the electron transport chain, but now we have
methylalanine, and it's small and it's free to cross into the thylakoid space.
But notice, it has this amine group and when we pump these protons in and when we make
the protons from the oxidation of water, this methylalanine and grab it is protons but effective
it will removes the proton gradient so it will abolish the protein gradient. So remember
here how you said it took more disrupts into more energy? So for me to make more of a gradient
it takes more and more work, correct? So let's think about this, no more process
no [Indiscernible] I pump protons I start to make a gradient, I pump more protons, I
make more gradient, more energy, working against the back pressure, now with methylalanine
present I pump the protein it's neutralized so there's no gradient so the next proton
gets pumped, easier and faster. So we actually expect the data here, with
light + methylalanine, to be faster. Now, notice we're measuring photosynthesis
as the electron transport chain and the reduction of DCPIP, if we measure ATP production, we
would be getting very different results. So let's just think about what we would see
if we did ATP production. We have a different graph, label it differently.
And we'll just do, not rate but again the amount of ATP.
So the dark should and we'll say, here's the 0point how much we're making additional.
How much should we make in the dark? Lots. None?
None. Because we're capturing light so these are
dark. The light will be some value.
What would light + DCMU be, we block an electron transport chain so what should it be? 0, like
dark. Since we're not having electron transport
chain operate, we can't make the proton gradient. We can't make the proton gradient we can't
make ATP, how about light + methylalanine, should it go higher? Lower? Or like this?
Let's think about it. In the presence of methylalanine, do I make
a proton gradient? No.
Do I make ATP? No. So here the light + methylalanine looks just
like the dark in DCMU. So depending upon what parameter you're measuring the results can
be dramatically different so you always have to think what parameter am I measuring when
I'm talking about some reaction? Okay.
And in this case we're measuring the reaction of DCPIP. So we do the controls that is the
dark, because we want to know how much that reduction is actually due to the presence
of light, i.e. photosynthesis, we do the DCMU and the methylalanine to see how the other
parameters affect reduction, now I'm going to show you a couple more slides.
Okay. The next few slides will not be on the bio1A
lab exam, but they should help tremendously with the bio1A lab exam. So you have been
studying the Calvin cycle yet, but you've studied glycolysis, so in glycolysis, if you
remember, we start off with glucose 6 carbon, cleave it to 2 or 3 electrons, one of which
is 3 phosphoglyceraldehyde, aldehyde. And we take and add an inorganic phosphate and
we spend we get off an NADH we spend it NADH +, and we get NADH, I reduce the ***
+ so what do I do to this? I oxidize it and I corporate that inorganic phosphate group
this is acid and aldehyde. So this is oxidation of that, reduction of the *** +.
Okay. So again, we go from an aldehyde, we oxidize
that, we spend a PI here get an [Indiscernible] and what we will see for the Calvin cycle
we will see a [Indiscernible] it is called bis because the phosphates are in different
carbon numbers, if it was di, it would be on the same carbon number. That's the difference
between bis, and di. We incorporate CO2 we end it with a two molecules
of PGA so the acid, so what we're going to do is reverse this to get the aldehyde.
So in glycolysis, if we get out a reduced molecule what do we have to spend to do the
reverse? We have to spend a reduced molecule that's
the NADPH and the light reactions and here we got out ATP so what do we have to do here?
Spend it. So we're getting ATP and NADPH from the light
reactions and we're spending it to reduce the sugar here, to eventually the two aldehydes.
Okay. So this is the Calvin cycle, pretty similar to glycolysis because it's pretty
much the reverse of it, instead spending NADH, you spend NADPH. So that's just a I was
showing you what I showed earlier, this is talking about where do you capture the carbon.
So the initial capturing of carbon, is how you take CO2 from the environment, later on,
all whether it's CAM or C 4 plants they all have to do the C 2, the Calvin cycle they
have to do [Indiscernible], CAM and plants is how do we capture the initial CO2, C 4
plants will capture by having two different locations so it's a spatial solution, and
CAM plants capture at night, and so it's temporal solution. what they will go in both cases
we'll make a 4 carbon acid, in both cases we make a 4 carbon acid, eventually this 4
carbon acid will be decarboxylated to give CO2, but that CO2 will then enter the Calvin
cycle. But the initial capturing is different.
Spatial temporal, in this spatial we have two different cell compartments in the temporal
we have night and day. And what they're trying to do there is creating an acid that can deliver
the carbon, you can get very concentrated solutions of acids, and as a result you can
increase the concentration of CO2, and the enzyme that does the Calvin cycle is called
rubisco it reacts with oxygen, so what this is doing is increasing the CO2 levels relative
to the oxygen, because remember, oxygen's increasing due to the light reactions, so
you minimize photorespiration so good luck on the exam on Monday and enjoy lab.
And that's not for the lab class that's just for the 1A. Students tend to do very poorly
on questions about the Calvin cycle.