Exam 2 Flashcards

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1
Q

L8. Sketch the organization of a mitochondrion of a plant cell. Label the inner and outer membranes, the matrix, and the intermembrane space.

A

Diagram

B-oxidation and TCA occur in matrix

Electron carriers for OxPhos are in the inner membrane (cyt C is in the IMS)

ATP synthase is in the inner membrane, with catalytic heads facing the matrix.

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2
Q

L8. Where do the reactions of B-oxidation and the TCA cycle take place? Where are the electron carriers of oxidative phosphorylation? The ATP synthase?

A

B-oxidation and TCA occur in matrix

Electron carriers for OxPhos are in the inner membrane (cyt C is in the IMS)

ATP synthase is in the inner membrane, with catalytic heads facing the matrix.

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3
Q

L8. Diagram the electron transport reactions of oxidative phosphorylation on a redox/energy diagram. Indicate at which stages and in what form the energy released during electron transport is “harvested” and “stored”?

A

Diagram (!!!!!!!)

Energy is harvested by using the flow of electrons to pump protons across the inner membrane. Energy is then stored in a proton gradient.

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4
Q

L8. Diagram the electron transport chain of oxidative phosphorylation in the mitochondrial membrane. Indicate at which stages and in what form the energy released during electron transport is “harvested” and “stored”?

A

Diagram(!!!!)

The major electron
transport complexes (NADH
deHyd , the b-c1 complex,
cytochrome oxidase) act as a
proton pumps. In addition,
protons are pumped as CoQ transfers a pair of electrons from NADH dehydrogenase to Cyt b-c1.
Protons are pumped across the inner membrane into the IMS/cytoplasm (the outer
membrane is permeable to protons), creating a H +
gradient across the IM, high in the cytoplasm,
low in the matrix.

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5
Q

L8. The inner mitochondrial membrane in brown fat cells, which are abundant in infants, contain a protein that allows protons to flow across the membrane without generating ATP. Considering the first law of thermodynamics, what do you think happens to the potential energy of the proton gradient in brown fat cells? Why might that be beneficial to young babies?

A

1st law: Energy is neither created or destroyed, merely conserved.

Protons flow through the membrane carriers in the inner membrane of the mitochondria in brown fat cells, the PE of the proton gradient is dissipated as heat, which helps keep the baby warm.

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6
Q

L8. If radiolabeled Oxygen was used to trace the fate of the oxygen during the complete oxidative metabolism of stearate to Carbon Dioxide and Water, what would be the first chemical compound into which radiolabeled oxygen atoms were incorporated? explain

A

H2O

The only place molecular O is used in fatty acid metabolsim is as an electron acceptor during oxPhos, and it ends up in water.

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7
Q

L8. Sketch a chloroplast, identifying all of its important features and compartments.

A

Sketch

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8
Q

L8. Given a sample of chloroplasts isolated from leaf mesophyl cells, how might you physically separate ribulose bisphosphate carboxylase (rubisco) from cF0 cF1 ATP synthase?

A

Gently break open the chloroplasts and spin in a centrifuge to separate the membrane fragments from soluble components. The ATP synthase will be in the membrane fraction, Rubisco will be found in the soluble fraction.

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9
Q

L8. Diagram the electron transport chain of non-cyclic photosynthesis. Indicate at which stages and in what form the energy released during electron transport is “harvested” and “stored”?

A

Diagram

Energy is harvested by pumping/using/releasing protons, generating a proton gradient across the thylakoid membrane.

Energy stored in this gradient will be used by ATP synthase to synth. ATP

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10
Q

L8. Compare the roles of chlorophyll in the antenna complexes with those in the reactoin centers of PSII and PSI during photosynthesis

A

Chlorphyls in the antenna complex absorb photons, and pass energy to other chlorophyll molecules by a quantum effect known as resonance energy transfer. in RET, energy is passed form electron to electron, but the electrons are not themselves passed from one molecule to the next.

The reaction center chlorophyll molecules accept energy from the antenna complex, boosting electrons to a higher energy state P680* in PSII and p700* in PSI. These electrons are then stripped from the reaction center chlorophylls and passed along the ETC

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11
Q

L8. Diagram the “Calvin-Benson” cycle by which plants fix Carbon Dioxide into carbohydrates, indicating important reactants and intermediates and where/what form energy is required.

A

Diagram

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12
Q

What is the energy cost (In ATP equivalents) required to synthesize a molecule of glyceraldehyde-3-phosphate in the “Calvin-Benson” cycle? Assume NADPH=NADH=2.5ATP

A

Each G3P made in CAlvin cycle requires:
6 NADPH+9 ATP

Total: 24 ATP equivalents.

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13
Q

What is the yield (in ATP equivalents) of the complete aerobic metabolsim of one molecule of glyceraldehyde-3-phosphate?

A

Complete aerobic metabolism of one G3P yields:
3 ATP (2 in glycolysis and 1 in CAC)
4 NADH
2 FADH2

Total: 16 ATP equivalents.

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14
Q

Why do plants expend so much enery synthesizing G3P from Carbon Dioxide?

A

The G3P made in the dark RXN’s of photosynthesis provide the precursors for EVERY biomolecule in plant cells. Thus for most biomolecules on Earth

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15
Q

L8. Considering photosynthesis, Which reactant is the source of the oxygen atoms liberated as O2 during photosynthesis? B) IN which product would you find the oxygen atoms derived from CO2 during photosynthesis

A

A) H20

B) CH2O

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16
Q

L8. Make simple sketches the show the location of the electron transport carriers, direction of proton pumping, ATP synthase, and direction of proton flow through ATP synthase in
A) a respiring prokayote
B) mitochondrion
C) Chloroplast
D) can you think of an evolutionary pathway that connects the topology of proton flow and the ATP synthase in these examples

A
DIAGRAM(!!!!!!!)
A
B
C
D: Inner membrane of mitos is topologically identical to the plasma membrane of respiring proks with infoldings to increase surface area. The thylakoid membrane is also similar in orientation, if we consider it to be "budded off" from the inner membrane. Both mito IM and chloroplast TM may be evolutionarily derived from the plasma membrane of endosymbiotic bacteria.
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17
Q

L8. Although chloroplasts and mitochondria both generate and use proton gradients to power ATP synthesis, the mitochondrial proton gradient has a larger electrical component, while the chloroplast’ gradient has a large chemical component. Why might this be so?

A

Mitochondria are pumping protons to the IMS, which is the same as the cytoplasm. Although
they can raise the matrix pH to ~8, they cannot greatly acidify the cytoplasm without affecting
other chemical reactions. Thus, the cytoplasm stays at pH ~7… and the concentration gradient is
only ~10-fold. However, there is a significant electrical component to the proton motive force

Chloroplasts, on the other hand, pump protons into a small, distinct compartment… the
thylakoid lumen (or space). There are less strict constraints on the pH of the thylakoid… which is
acidified to a pH of ~5. The Stromal pH is ~8, so the chemical gradient is ~1000-fold. However,
other ions balance much of the electrical component of the thylakoid proton gradient, so that the
proton motive force is mostly chemical.

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18
Q

L8. According to the first law of thermodynamics, energy is neither created nor destroyed, but can be converted from one form to another. What are three “forms” of stored energy used by cells. How are these forms interconverted?

A
  1. Chemical energy of high energy phosphate bonds (ATP
  2. Chemical energy and reducing power of electrons in NADH and FADH2
  3. The potential energy of electrochemical ion gradients
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19
Q

L10. Draw the structure of the tri-deoxyribonucleotide ATG (remember 5’-3’) with 5’ triphosphate and free 3’ hydroxyl. circle a phosphoester bond

A

Diagram

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20
Q

L10. What causes melting temp. of DNA to be higher?

A

More GC bonds, because they have 3 H-bonds between them

Larger double helices make more H bonds, therefore they have higher melting points.

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21
Q

L10. Briefly outline or diagram the packaging of DNA into chromatin in a euk. What are the major structural proteins found in chromatin?

A

Diagram

Double Helix
-Core Histone H2A, H2B, H3, H4

Beads on a string(nucleosomes)

30 nm fiber (solenoid)
-non-histone chromosomal proteins

Chromatin loops
“condensins

Mitotic Chromatin.

Two copies each of histones H2A,
H2B, H3, and H4 (the “core”
histones) constitute the histone
core, around which are wrapped ~146
bp of DNA.
The “linker” histone H1 binds the
nucleosome linker DNA and coils the
nucleosomes into the 30 nm fiber or
“solenoid.”
Additional proteins (“non-histone chromosomal proteins,” incl “condensins”) organize the chromatin
into loops and higher order structures.
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22
Q

L10. What provides energy for adding nucleotides to a growing DNA or RNA strand? How much energy (ATP) equivalents is required to add each nucleotide?

A

Beta and Gamma phosphates are cleaved off and released as PPi, which is then cleaved to 2Pi. These reactions, which are equivalent to using 2 ATP, provide the energy for adding the nucleotide to the growing strand.

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23
Q

L10. Sketch a replication fork that is moving from left to right, corrctly indicating the polarity of each DNA strand
A)which strand is lagging/leading?
B)an Okazaki fragment with its RNA primer.

A

Diagram

24
Q

L10. Sketch the two replication forks of a prelication bubble, correctly indicating the polarity of each DNA strand, the leading, lagging of each fork and the most recent Okazaki fragment with its RNA primer

A

Diagram

25
Q

L10. Briefly describe the steps in synthesis of an Okazaki fragment, from priming to sealing into the growing lagging strand. For the following factors/proteins
A) indicate the order in which they act during synth of Okazaki fragments and
B) Briefly indicate the role they play in synth/integration of an Okazaki fragment into the lagging strand
(PROTEINS:DNA polymerase, Helicase, DNA Ligase, Primase, Repair exonuclease, Repair Polymerase, Single-strand binding protein, Topoismoerase)

A

A)

  1. Helicase-unzips DNA helix
  2. SSBB-binds to and stabilizes DNA
  3. Primase-synthesizes a short RNA primer
  4. DNA Polymerase-elongates primer
  5. Repair exonuclease- removes previous primer
  6. Repair polymerase- fills the gap resulting from removal of primer
  7. DNA Ligase- seals the nick by making a single phosphodiester bond to link the okazaki fragment to growing DNA strand.
  8. Topoisomerase I- creates ss nick ahead of replication to allow unwinding.
  9. Topoisomerase II- cuts ds DNA to pass one DNA through another.

B)

26
Q

L10. Why does DNA replication require an ample supply of all four ribonuclotide triphosphates?

A

To make primers for each Okazaki Fragment

27
Q

L10. Explain based on the mechanism for replicating/elongating the ends of euk. chromosomes

A

ing/elongating the ends of eukaryotic chromosomes.
Telomeres at the ends of all the chromosomes (human and tetrahymena) would shorten with
each round of DNA replication and cell division. However, cultured human cancer cells have
telomerase to re-extend them. The sequence of the repeats added is determined by the template
used, which is a component of the telomerase. Since human cells have human telomerase, which
includes the template for human repeats, it will add human repeats (TTAGGG) onto the ends of the
tetrahymena chromosomes.

28
Q

L10. What role does telomerase play in the replication of DNA in euks? Is telomerase needed by proks?

A

Telomerase elongates the telomere sequences at the end of eukaryotic chromsomes,
ensuring that the ends can be properly replicated w/o losing needed sequence information. It is not
needed by proks (why?)

29
Q

L10. What other essential roles do telomeres play in euks?

A

They protect the ends of the

chromosome from nuclease and sequester them from inappropriate repair/recombination.

30
Q

L10. Replication of the circular bacterial chromosomes results in formation of two interlocked DNA circles. How are these chromosomes “unlinked”?

A

Topoisomerase II makes a transient double-strand cut and passes one DNA
molecule through the other, before resealing the cut DNA. Despite not being closed circles, the
extreme length of eukaryotic chromosomes requires topo II activity to allow them to separate and
segregate during mitosis and meiosis.

31
Q

L10. How does a human cell repliate its genome w/o introducing mutations due to replication errors?

A

DNA polymerase includes an intrinsic “proof reading” function to correct mismatches. DNA
polymerase includes a “3’>5’ exonuclease” activity that can remove mismatched bases, and then
polymerization continues with the addition of the correct base (the fact that each nucleotide brings
in the energy for its own addition, in the form of its terminal 5’ phosphates, is a crucial element of
the mechanism). Proof-reading has an error frequency of ~10-2 . When combined with later mismatch
repair (error rate ~10-2 ), the overall error frequency is 10-5 x 10-2 x 10-2 = 10-9

32
Q

L12. Draw the RNA trinuclotide AUG, labeling or otherwise indicating the 5’ and 3’ ends, the sugar phosphate backbone, and a phosphodiester bond. What distingusihes this structure from that of a single stranded DNA?

A

Diagram

Has U’s in place of Ts, and ribose in place of 2-deoxyribose

33
Q

L12. Briefly compare the overall processes of DNA replicatoin and transcription. How do these processes differ? How are they similar?

A

Replication:
ds DNA to ds DNA + ds DNA (2 identical daughter DNA molecules)
Requires DNA polymerase, helicase, primase, repair exonculease, repair polymerase, ligase and dNTPs and NTPs

Transcription:
ds DNA to ds DNA +ss RNA
Requires RNA polymerase and s-factor (proks) or general transcription factors (euks) plus NTPs

In both, addition of each nucleotide to the growing chain requires the equiv of 2 high energy bonds.
Both polymerases elongate chain in 5’63’ direction (read templates 3’65’). DNA pol can only add to
pre-existing 3’ hydroxyl (thus the need for primer and primase). RNA pol does not have this
constraint

34
Q

L12. Why is fidelity in transcription not as important to the cell as fidelity of replication?

A

RNA doesn’t serve as a permanent repository for the genetic info that must be passed on. mRNA is transient intermediate in the process from DNA to protein. It is likely that an mRNA with a few base changes will still encode a functional protein as the genetic code is redundant.

35
Q

L12. Briefly outline and compare the steps leading up to transcription of a typical protein coding gene in a prokaryote and a euk. Which eukaryotic factors have a function “analogous” to that of bacterial sigma factor? What DNA sequences direct the initiation and termination of transcription in proks and euks?

A

? Lecture 12 Notes. Transcription.
Prok:
-Promoter starts everything: gives template and denotes direction.
-RNA polymerase transcribes DNA sequence to RNA
-sigma factors help identify promoters
-internal helicase unzips DNA
-Terminator ends transcription

Euk: (occurs in nuc.)

  • pol II is what transcribes mRNA, need general transcription factors
  • BASAL PROMOTER: TATA box(30 bp upstream) BRE (40 bp up stream) CAT (70 bp upstream)
  • TFIID binds to promoter
  • other transcription factors bind and kink DNA
  • Finally TFIIH completes intiation complex, acts as helicase and kinase.
  • It phosphorylates RNA pol II and begins transcription.
36
Q

L13. What are restriction endonucleases? (aka restriction enzymes) Why are they so useful to biologists?

A

They cut DNA at certain (target) sequences.

They help bactria defend themselves against foreign double-stranded DNA.
They allow biolgists to cut DNA at specific sequences, and paste the pieces back together. Allow for genetic engineering

37
Q

Which restriction enzyme would give the smallest fragments?
HaeIII “4 cutter”
EcoR1 “6-cutter”
Not1 “8-cutter”

Why might sticky ends not be useful?

A

4 cutter gives smallest average fragment size, cutting on average 4^4 base pairs, while th 8 cutter cuts every 4^8 basepairs and would thus give the largest.

Sticky ends can anneal and be ligated back together, making it easy to specifically piece DNA fragments together

38
Q

L13. Humane genome totals 3 E 9 basepairs, Estimate the number of fragments that would result by cutting human DNA with a 4,6 and 8 cutter

A

4 cutter. divide by 256
6 cutter divide by 4096
8 cutter divide by 65536

divide by 4^#cutter

39
Q

L13. What is a plasmid? What are 3 essential features or elements found or engineered into a plasmid that is used as a “cloning vector”? What role does each of these elements play when the plasmid vector is used to “clone” or amplify DNA created using recombinant DNA?

A

Plasmids are small circular DNA found in bacteria. they have an origin of replication allowing them to be replicated by the bacterial replication machinery and often carry a gene conferring resistance, which is used to select for bacterial colonies carrying the plasmid.

Through engineering we’e inserted multiple cloning sites, multiple cut sites. Inducible promoters, terminators, polyA signals, GFP coding regions and other features. Plasmids can be purchased

40
Q

L13. Explain how PCR (polymerase chain reaction) can be used to amplify DNA

A

L13 slides 15& 16 figure 10-15
1 Heat DNA to denature it (90-95C)
2 Anneal forward and reverse primers (50-70C) (oligonucleotide)
3 Elongate primers with Taq polymerase (72C)

Repeat repeat repeat

41
Q

L13. What variant of PCR is used to amplify DNA from RNA transcripts? Briefly explain how it differs from PCR

A

RT-PCR (reverse transcriptase)
Adds a first step in which reverse transcriptase is used to make a single-stranded cDNA from the RNA molecule, which can then be amplified by PCR

1 Isolate mRNA
Use forward primer and reverse transcriptase to make cDNA…
3 & 4. Use forward and reverse primers to amplify by RT-PCR…

42
Q

L13. Explain how to create genomic and cDNA libraries from developing frog embyos

A

ECB4 Fig 10-10 10-15 10-13

Genomic libraries are made from sheared genomic DNA, which is inserted into a cloning vector.

cDNA libraries are made from cDNAs reverse transcribed from mRNA transcripts in the cell, which are then inserted into a vector

43
Q

L13. How does a genomic library differ from a cDNA library in terms of information?

A

Genomic DNA libraries contain all DNA sequences from the genome, with equal representation including non-transcribed regions, introns and exons.

cDNA libraries contain copies of all the RNA transcripts from the particular cell type or tissue. Non-transcribd sequences and introns would not be present. Abundance of individual sequences would reflect the abundances of the transcripts from which they were copied

44
Q

L13. Which kind of library would you choose to identify/compare/clone RNA transcirts present at different stages of development or from different tissues?

A

cDNA libraries from the various stages of development would allow you to clone/compare transcripts from the different stages

45
Q

L14. What color will these types of E coli.s cells be based on how the react with Xgal

a) wild type
b) mutations in the lac promoter that prevent polymerase binding
c) mutations in the lack operator that prevent repressor binding
d) mutations in the CAP binding site that prevents CAP binding
e) mutations in promoter for I gene that prevent polymerase from binding

A

a) white plae blue
b) white white white
c) blue plae blue
d) white pale pale
e) blue pale blue

WT: when lactose is present, lac is transcribed

46
Q

L14. Transcription factor (1) binds to the (2), a regulatory region found ~30 bp upstream of the transcription start that includes a consensus sequence known as the (3) box. After binding of additional transcription factors, RNA polymerase (4) binds (in association with more transcription factors) to form an “initiation complex”. Transcription factor (5) initiates transcription, by using its (6) activity to unwind the DNA and its (7) activity to phosphorylate the tail of the polymerase, releasing it from the initiation complex to begin transcription. Additional regulatory sequences commonly referred to as (8)

A
1 IID
2 promoter
3 TATA
4 II
5 IIH
6 helicase
7 kinase
8 enhancers
47
Q

Diagram mRNA splicing and explain

A

DIAGRAM (series)

Many pol II transcripts must have intron sequences removed. Requires Branch-binding protein (BBP). U2-activiating factor, and snRNP’s U1,U2, U4, U5 and U6.
BBP and U2AF recruit the U2 to the branch point A. U1 binds to the 5’ splice site. U4, u5 and u6 bind, forming the splicesome, which holds the transcript in a configuration allowing the 2’OH of the branchpoint A to attack the 5’ splice site, freeing the 3’ and 5’ of the exon. The free exon 3’ attacks the 3’ splice site, joining the exons and releasing the intron sequence as a lariat.

48
Q

Briefly describe the mechanism of Fo F1 ATP synthase

A

As protons flow through the channel formed by the ATP synthase F0 component (the rotor), they turn the rotor and stalk (gamma subunit). The rotating gamma subunit induces sequential conformational changes (open> loose>tight) in the alpha/beta catalytic subunits (F1 component), which catalyses ADP+Pi into ATP

49
Q

Hybridization to an affinity column consisting of oligo-dT coupled to microscopic agarose beads can be used to rapidly isolate

A

euk mRNA

Binds to poly-A tracts in RNA. only euk pol II transcripts are polyadenylated

50
Q

Explain 5 prime cap

A

The 5’ end of pol II transcripts is capped with 7mGppp5’(base). Phosphatase removes a 5’ phosphate from the transcript, and guanyl transerfase adds GMP in a 5’-5’ linkage. Methyl transferase methylates the 7 position of the G ring (and occasionally the 2’OH of first transcribed nucleotide)

Diagram

51
Q

Explain Poly-A tail

A

The 3’ end of pol II transcripts is polyadenylated. Poly-A factors recognize the transcribed poly-A signal (AAUAAA) in the nascent RNA, recruiting nuclease to cut the transript (30 nucs downstream) and poly-A polymerase, which adds 100-200 A’s in a template independent process

52
Q

In vitro replication in the prescene od ddNTP’s is the basis for a DNA sequencing method. How are ddNTP’s used to sequence DNA?

A

By setting a series of four in vitro replication reactions, each with a limited amount of a different ddNTP, sets of nested fragments corresponding to the different bases are generated, with each rxn set ending at a distinct base. Running these rxns of a polyacrylimide gel gives a sequencing ladder, with each band corresponding to one base in the sequence of the DNA. Since shorter fragments run faster, gels are read from the bottom to the top (5’-3’)

53
Q

Lac Promoter

A

If it’s off, lac is always off

54
Q

lac operator

A

if it’s mutated, lac always on.

REPRESSOR BINDS TO OPERATOR

55
Q

Mutations in CAP

A

CAP means high transcription
CAP binds when Glc is low, if Glc is present, CAP doesn’t bind
ENHANCER

56
Q

Mutations in I gene

A

prevent polymerase from transcribing I= no repressor

lac is always transcribed