Lecture 9 Flashcards
Describe the steps that are taken in the passing of electrons in complex 1.
NADH will drop off electrons 2 at a time to FMN. Then FMNH will pass e-s (1 at a time). Each electron will go up the Fe-S chain until the e- reaches Q. Q will become fully reduced with a pair of e-s from FMNH and then take the e-s on the complex III.
What brings electrons to complex II?
Succinate.
p.s. Complex II is the enezyme in step 6 of the citric acid (the only citric acid cycle enzyme that is not a soluble enzyme found on the matrix of the mitochondria – it is embedded in the inner membrane of the mitochondria.
Describe the passing of electrons through complex 1I.
Succinate will be oxidized to fumarate and those electrons that enter into complex II will reduce CoQ.
1) Succinate will give e-s to FAD and become Fumurate. FAD will become FADH2. The protons come from the succinate.
2) One at a time FADH2 will give electrons to iron-sulfur centers. And the Iron-sulfur centers will transition from Fe3+ (oxidized form) to Fe2+ (reduced form).
3) Electrons from Iron-sulfur chain will reduce CoQ one at a time making Qh2. And Qh2 will go to complex III. —CoQ needs to get 2 protons from the inside to do this.
KEY NOTE– NO protons pumped.
What is the final electron acceptor for glycerol-3-phosphate?
CoQ. Electons are transferred from NADH through glycerol-3-phosphate shuttle and give electrons to FAD. FADH2 gives electrons to Q making Qh2. Then those electons go on to Complex III and then Complex IV making it’s way to Oxygen.
True or False: Electrons from many different oxidative processes in the cell will all funnel into Ubiquinol (Qh2).
True. From Qh2 electrons will be passed to complex III.
What happens in Complex III?
It’s going to take the electrons off of Ubiquinol (Qh2) oxidizing it and pass electrons on to 2 moles of Cyt C (Cyt C is reduced).
Cyt C is a peripheral membrane protein.
When it receives the electrons from Qh2 there has to be 2 Cyt C since there is a heme group in Cyt C which can only except 1 e- at a time.
There are 4 protons, all of which end up on the P side.
What is the Q-cycle, where does it run and what is the result of it? What are the two stages of this cycle and how do they differ?
The Q cycle is the term we use to describe how electrons get through the subunits of complex III. It will take 2 rounds to get 2 electrons to Cytochrome C. Ubiquinol (QH2) coming from many oxidative processes will bind at the Qp site in Complex III. Qh2 will be oxidized in a single step with 1 electron going up to the Fe-S center and then this electron will move to C1, and then to Cytochrome C and then to Complex IV.
The other electron will reduce Heme BL and then reduce Heme Bh. This electron will go into the Qn site where ubiquinone and this electron will end up on the semiquinone radical. This will complete then one round of the Q cycle. One electron goes up and gets on cytochrome C and the other goes down and ends back up on a CoQ molecule.
The 2 protons on ubiquinol will diffuse by proximity into the P side and are considered “pumped” into the P side. These protons came from complex 1 or 2. All of this is round 1 of the Q cycle.
What is the difference between Cycle 1 and 2 in the Q cycle?
The biggest difference is the type of Q in the Qn site.
In cycle one it is Q and in cycle 2 it is *QH (semiquinoal radical) which will then make QH2 (ubiquinol) –using one proton from the N side.
In the Qp site an additional Qh2, ubiquinol comes into the Qp site, and electrons are split just like in cycle 1 and 2 protons are “pumped” into the P side.
How do electrons get to complex IV?
Electrons have found their way from either succinate or NADH and are now sitting on Cytochrome C that will now dock on to Complex IV and pass electrons onward to oxygen.
How many protons does Complex IV pump?
Complex IV pumps 2 protons to the positive side.
How many protons does complex 1 and III move?
4 each. Complex IV moves 2 protons.
Describe the Oxygen binding site in Complex IV.
Oxygen will covalently bind to the iron atom (Cu b) of Heme a3.
Describe the flow of electrons from beginning to end in Complex IV
First Cyt C (Fe2+) will pass it’s electrons to copper B. Then electrons will jump to Heme a. Eventually, Oxygen will accept electrons from Heme a and will become water.
As this happens, two protons will be pumped to the P side.
What is the result of Oxidative Stress?
R.O.S. Reactive oxidative species. If you have a cell under oxidative stress it is because there is an accumulation of reactive oxygen species.
Oxidative stress can lead to things such as cancer.
It can lead to cardivascular disease, cell death, diabetes, and overall death.
What can lead to oxidative stress?
metabolism, UV radiation, Ionizing radiation, Smoking, Everything else.
Disrupts lipids in the membrane, damage to DNA and RNA which can lead to cancer, proteion funtion. etc.
How do you reduce oxidative stress?
Antioxidants like glutathione and NADPH. These reduce the lethal amount of .2 to 2% of oxidative stress.
What is an example of something that leads to R.O.S?
The slow down of the ETC such as when there are low levels of ADP in the matrix of the mitochondria. This then leads to a reduced rate of ATP synthase and this leads to a decrease of protons across the membrane thus reducing the flux of the ETC.
What is the problem if there is a slow down of FLUX in the ETC?
If electrons keep feeding into the ETC and there’s nowhere for the electrons to go, then the electrons accumulate in the chain and then the oxygen that’s made it into the cell is exposed to free radicals such as the free radicals produced in Ubiquinone, Flavin Nucleotides.
So there are a lot of place Oxygen can pick up an electron and become a radical form a then become a Reactive Oxidative Species.
How can oxygen become Reactive Oxidative Species?
Oxygen in the matrix of the mitochondria can interact with ubiqitonal or a semi quinol and pick up the electron and become a superoxide radical.
The superoxide radical can then become a hydroxy free radical.
These two are the primary components of ROS.
How do our cells deal with these ROS’s?
The first line of defense is superoxide dismutase in both the cytosol and the mitochondrial matrix. It will take 2 protons and make Hydrogen Peroxide from the Superoxide radical.
Hydrogen Peroxide can then be converted to H2O by glutathione peroxidase using GSH (reduced glutathione) and oxidizing it to GSSG using the two hydrogen and elctrons that come with it to create 2 waters.
Glutathione reductase will then regenerate reduced glutathione using NADPH.
Nicotinamide will regenerate NADPH from NADH.
How flux of NADH affect the bodies process to reduce oxidative stress?
ETC flux has gone down which means that the rate of NADH [] if the citric acid cycle keeps going are going to go up which supplies electrons to Nicotinamide nucleotide transhydrogenase.
This allows the production of NADPH to keep glutathione reductase working which allows gluthione perosidase to keep working.
Describe the oxidative phase of the pentose phosphate pathway.
1) It starts at G6P. G6P will be oxidized to 6-phosphogluconate in 2 steps. step 1 by glucose 6 phosphate dehydrogenase. It will use 1mole NADP+ to make NADPH. *Rate limiting step.
3) 6-phosphogluconate undergoes decarboxylation to ribulose 5-phosphate by enzyme 6-phosphogluconate dehydrogenase. This enzyme also uses 1mole NADP+ to make NADPH.
2 moles of NADPH are created. this is important for reoxidizing glutathione by glutathione reductase.
A family on vacation in Greece sit down to an authentic Greak meal complete with falafel. A day later one of their children’s skin has a yellowish tint and is clearly in need of medical attention. In the ER the physicians determine that the child’s RBC count is low and their kidneys are beginning to fail. Upon learning of the childs diet in the past 24 hours the head physician orders a genetic screen. What gene do you think is the target of this screen?
G6P dehydrogenase. These symptoms are indicitive of hemolysis. Jaundice is due to hemoglobin being metabolized as heme and billirubin causing yellowish tint.
Divicine is found in flafa. There is no mitochondria in red blood cells so the pentose phosphate pathway is really the only resource to NADPH. But if the pentose phosphate pathway is not working due to a genetic mutation in G6P dehydrogenase then oxidative damage to lipids, proteins, and DNA is the outcome. The lipids will undergo damage and red blood cells will begin to lyse.
True or False: Increased levels of HIF will lead to increased levels of PDH kinase (Pyruvate Dehydrogenase Kinase).
True. This is the kinase that will phoshorylate E1 of the PDC complex and inhibit it.
The increased levels of PDH kinase will increase inhibition of Pyruvate Dehydrogenase –thus preventing conversion of Pyruvate to Acetyl-CoA.
This will slow down Citric Acid Cycle Flux which will reduce NADH which will reduce the number of electrons entering the ETC preventing the accumulation of electrons in the ETC—By doing this you will reduce oxidative stress.
Why does an increase in HIF-1 lead to an increase in Glucose Transporter, Glycolytic enzymes, and lactate dehydrogenase.
many tumors are hypoxic (low oxygen supply) because they do not have much vascular capillaries. And an increase in HIF will increase glucose to get into the cell, things that use glucose, and conversion of pyruvate to lactate.
What are some of the mechanism with anti-cancer therapeutics?
inhibit G6P Dehydrogenase and hexokinase with the goal to decrease levels of glucose 6 phosphate so that G6P can’t feed into the pentose phosphate pathway.
So if you are not feeding into the pentose phosphate pathway then you are not getting NADPH and you’re also not getting ribose (which means these cells would not be able to produce as much nucleotides).
How many ATP can be made from NADH and FADH2 respectively?
Depends on the size of the ring. If the c ring is 9 subunits, it takes 4 protons to make 1 ATP. so one NADH will make 2.5 ATPs.
Succinate will make 1.5 ATP
*****Working backwards, how would cyanide back up the ETC?
Mention the effects on Cyt c, complex II, Q cycle, and Complex 1.
Cytochrome C giving up it’s electrons to complex IV will be inhibited.
Thus, this means that the Q cycle will be inhibited.
Complex 1 will be inhibited and complex II will be inhibited. This will lead to oxidative stress–> oxidative stress will lead to tissue damage.
NADH since it can no longer feed into the ETC, NADH will also start to rise. This will feedback and inhibit the citric acid cycle. NADH will also active pyruvate dehydrogenase complex kinase which will inhibit dehydrogenase complex. This means that pyruvate will no longer be converted to Acteyl-CoA.
Pyruvate will then be converted to lactate. –> Thus the reason why lactate levels increase.
ATP levels will also go down.
How many protons need to be pumped to the inner mitochondrial space for each NADH electrons to make ATP?
10.
*However, if it take 4 protons to produce 1 ATP then, NADH can lead to the production of 2.5 ATP.
Describe the natural uncoupler process in brown fat.
First the hypothalumus recongizes body temperature. When the body temperature cools down–> This will then release to the release of norepinephrine.
Norepinephrine will bind to a G protein coupled receptor. -> activated G protein-> activates Adenylyl Cyclae-> activates cAMP—> activate Protein Kinase A.
Protein Kinase A activate beta oxidation–> fatty acids are broken down and a bunch of FADH and NADH are made driving the ETC.
The UCP-1 channel will, however, pump protons back into the matrix. Thus, the body is producing a lot of heat.
TRUE or FALSE: DNP and FCCP are examples of unnatural uncouplers?
True
What dictates which of the 3 confirmation forms the alpha/beta pairs can have in Fo?
It’s the alpha and beta pair’s interaction with the stick (y shaft).
If it goes from T to O, ATP will come out. It rotates in the clockwise direction.
