Name:
2 points
Multiple choice (4 points apiece):
1. Which of the following statements about energy conservation in the mitochondrion is false?
A) Drug that inhibits the ATP synthase will also inhibit the flow of electrons down the chain of carriers.
B) For oxidative phosphorylation to occur, it is essential to have a closed membranous structure with an inside and an outside.
C) The yield of ATP per mole of oxidizable substrate depends on the substrate.
D) Uncouplers (such as dinitrophenol) have exactly the same effect on electron transfer as inhibitors such as cyanide; both block further electron transfer to oxygen.
E) Uncouplers "short circuit" the proton gradient, thereby dissipating the proton motive force as heat.
2. The oxidation of a particular hydroxy substrate to a keto product by mitochondria has a P/O ratio of less than 2. The initial oxidation step is very likely directly coupled to the:
A) oxidation of a flavoprotein.
B) oxidation of a pyridine nucleotide.
C) reduction of a flavoprotein.
D) reduction of a pyridine nucleotide.
E) reduction of cytochrome a3.
3. DNA in a closed-circular, double-stranded molecule with no net bending of the DNA axis on itself is:
A) a left-handed helix.
B) a mixed right- and left-handed helix.
C) relaxed.
D) supercoiled.
E) underwound.
4. If the structure of a fully relaxed, closed-circular DNA molecule is changed so that the specific linking difference ( ó ) is -0.05, the number of:
A) bases is decreased by 5%.
B) bases is increased by 5%.
C) helical turns is decreased by 5%.
D) helical turns is increased by 5%.
E) helical turns is unchanged.
5. Bacterial chromosomes:
A) are highly compacted into structures called nucleoids.
B) are seen in electron microscopy as "beads on a string".
C) are surrounded by a nuclear membrane.
D) contain large numbers of nucleosomes.
E) when fully extended are as long as the bacterial cell.
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6. When a DNA molecule is described as replicating bidirectionally, that means that it has two:
A) chains.
B) independently replicating segment.
C) origins.
D) replication forks.
E) termination points.
7. The 5' 6 3' exonuclease activity of E. coli DNA polymerase I is involved in:
A) formation of a nick at the DNA replication origin.
B) formation of Okazaki fragments.
C) proofreading of the replication process.
D) removal of RNA primers by nick translation.
E) sealing of nicks by ligase action.
Longer questions. 14 points each. Do five: transport. What were these three other electron carriers, give me a brief description of their structure, and how do they fit in the overall electron transport pathway.
Ubiquinone. -Lipid soluble benzoquinone carries both electrons and protons in the michondria membrane. Can do both 1 and 2 electron reaction
Cytochromes - protein with a metal bound to a heme group. Three classes:a,b,&c based on spectral properties. A,b and some c’s are integral membrane proteins, but cytochrome c of mitochondria is associated with the outer surface of the inner mitochondiral membrane via ionic interactions
Iron-sulfur proteins - Proteins is which iron is complexed with cys residues and inorganic sulfur, can have between 1 and 4 Fe’s in a complex
Overall flow of electrons is not nice an linear
Starting with NADH 6 complex I 6 Q 6 complex III( cytochrome bc ) 6 Complex IV
(cytochrome oxidase)
Starting with Succinate 6 complex II 6 FADH
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6 Q 6 complex III 6 Complex IV
Within Complex I NADH 6 FMN 6 Iron Sulfur 6 Q
Within Complex II Succinate 6 FAD 6 FeS 6 Q
Within Complex III Q 6 FeS 6 cytochrome c
Within Complex IV Cytochrome c 6 Copper 6 heme a 6 O
2
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1.5 ATP.
A. (5 points) Show me the math. For each complex involved in electron transport, tell me how many protons are pumped out of the mitochondrial matrix. Also, for the ATPsynthase, how many protons must be let into the matrix to generate an ATP.
Do the numbers match or is something missing? If something is missing tell me what it is, show me where the 2.5 and 1.5 numbers come from.
NADH
Complex I: NADH + 5H +Q 6 NAD +QH +4H +
2 c
+
N c
2 P
Complex III: QH +2cyt (Ox)+2H 6 Q+2cyt (Red)+4H
+
+
P
+ Complex IV: 2cyt (Red) + 4H + ½ O
+
N 2
6
2
6
+
2cyt (Ox)+2H +H O
P c
10H +H O
FADH
2
Complex II: FADH +Q 6 QH +FAD
2 c c
+
2
+
N
Net: FADH + 6H +1/2O
+
2
N
6
6 Q+2cyt (Red)+4H
+
2
6 c
+
P
2cyt (Ox)+2H +H O
6H +H O
ATP synthase
ADP + P +3H +
P
6 ATP + 3H +
N
At this point NADH should make 10/3 = 3.333 ATP
What is missing is that for each P transported into the cell, 1 H +
P
6 1H +
N
So the net ATP synthase, including proton transport is:
ADP+P +4H =ATP + 4H +
Trying again
N
NADH =10/4 = 2.5
The numbers match
B. (9 points) I have found several mutant mitochondria, for each mutant, predict the number of ATP’s generated for NADH and FADH
2
Mutant 1. The Complex I reaction for this mutant is:
NADH + 3H + Q 6 NAD + 3H + QH
2
In this case the net for NADH will be
Net: NADH + 9H +1/2O
2
6 8H +H O
8/4=2
Mutant 2. The complex IV reaction in this mutant is:
4 Cyt c (reduced) + 4H + O
(2 +2 +1/2O
Net: NADH + 9H +1/2O
2
6 +
6
2
4 Cyt c (oxidized) + 2H O
6 2 +1)
8H +H O
8/4=2
Net: FADH + 4H +1/2O
2
6 4H +H O
4/4=1
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Mutant 3: The ATPsynthase reaction is:
4 H + ADP + P 6 4H +
N
+ ATP
Adding in the H for transport of H +
5 H + ADP + P
Net: NADH + 11H +1/2O
2
6
6
+
5H
P
+
N
+ ATP
10H +H O
10/5=2
Net: FADH + 6H +1/2O
2
6 6H +H O
6/5=1.2
10. I have a 1000 base pair piece of ccc DNA. To this I will bind 5 nucleosomes, and then treat the DNA with a topoisomerase to relax any supercoils induced by the binding of the nucleosomes. If I then remove the nucleosomes, what is the superhelical density
( ó ) of the DNA?
The starting linking # is the number of base pairs/10.5 bo/turn
=1000/10.5 = 95.24
But the linking number must be an integer, so I will round to 95
Each nucleosome binds about 200 bp of DNA so you can bind 5 nucleosome
Binding a nucleosome, and then relaxing the DNA introduces a Ä Lk of -1, therefor the overall Ä Lk=-5
Superhelical density( ó ) = Ä Lk/Lk =-5/95
=-.05
11. Tell me about the steps involved in placing nucleosomes on a piece of bare DNA in a eukaryotic cell.
In general 2 H3’s and 2 H4's bind first. After that H2A and H2B dimers join in to complete the histone core. Assembly after chromosomal replication starts with the binding of RCAF(Replication-coupling assembly factor). This factor includes acetylated
H3 and H4 histones, a three-subunit protein called chromatin assembly factor 1 (CAF-
1), and the anti-silencing factor 1 (ASF-1). Additional details are not known at this time.
12. Tell me about the 5 different DNA polymerases found in E. Coli
DNA Polymerase I Molar mass 103,000. Composed of a single polypeptide. Has both 3' 6 5' and 5' 6 3' exonuclease activity. Relatively low polymerization rate and high
Processivity (Falls off the DNA after only a few bases are added). Thought to remove
RNA primers and replace the RNA with DNA on the lagging strand of the DNA.
DNA Pol II 7 subunits, of which the actual polymerase subunits has a molar mass of 88,000. Had a 3' 6 5' exonuclease proofreading activity, but does not have 5' 6 3' nuclease activity. Slightly faster than Pol I and has slightly lower processivity. Thought to be involved in DNA repair
DNA pol III very complicated 17 different polypeptides consisting of 10 different subunits with a total molar mass of about 800,000. It has only 3' 6 5' exonuclease proofreading capabilities. It is extremely fast, polymerizing up to 1000 nucleotides/second, and very low processivity, staying on the DNA for up to 500,000 bp or more before falling off. This is the polymerase that is actully used to replicate the chromosome
Polymerase IV and V only recently identified, are involved in specialized DNA repair. In section 14.2 you will see that they are involved in SOS repair, a highly error prone repari mechanism that is only used when the cell is highly damaged.
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13. Describe the steps involved in the initiation of DNA replication in E. Coli.
First 8 DnaA proteins (with bound ATP) bind at the 5R and 3 I sites on the oriC
DNA sequence. These proteins form a right handed helical structure that is thought to denature the AT rich DNA at the DUE site. The formation of this complex is aided by the proteins HU, IHF, and FIS, but we don’t know their exact role at this point. On each strand of the denatured DNA a hexamer of DnaC protein (with bound ATP) load loads a hexamer DnaB protein. When he ATP bound to Dna C is hydrolyzed, the protein falls off and leaves the DnaB bound to the DNA. The PolII enzyme is then assembled on this core of two DnaB hexamers, and each of the hexamers acts as a helicase to begin unwinding each strand further in the 5' 6 3' direction. When the complete PolIII is assembled and the â clamps attached to the DNA, protein Hda bind to both the â clamp and to the DnaA to stimulate its hydrolysis of its bound DNA., once this hydrolysis is complete the DnaA complex falls of the oriC region of the DNA.
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1. D 2. C 3. C 4. C 5A 6D 7D
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