What are Nucleotides?
Nucleotides are the monomer unit for nucleic acids and are composed of nitrogenous base, 5-carbon sugar and a phosphate group.
Nucleic acids are biopolymers of: nucleotide monomers
● DNA (MORE STABLE): carrier of genetic information double-stranded
● RNA (LESS STABLE) : intermediate in expression of genetic information. RNA can form secondary structure (as in tRNA) by hydrogen bonding with itself. Not all forms of RNA are intermediates in expression of genetic information; single-stranded (usually).
○ Types of RNA:
○ tRNA: transfer RNA, holds amino acids used in protein synthesis, remember WOBBLE POSITION(Inosine)!
○ rRNA: ribosomal RNA, part of the ribosome
○ siRNA: small interfering RNA,short term silencing of pleiotrophin coding genes, targets mRNA for degradation
○ miRNA: micro RNA, silencing and post-transcriptional regulation
of gene expression.
etc. serve a wide array of different purposes (esp. Compared to DNA). Can also catalyze reactions (Ribozyme).
Use resonance to help to differentiate between all four: If needed nitrogen will donate!
● Purines (2 rings:pyrimidine and imidazole) - Adenine, Guanine are purines derivatives in
● Pyrimidines (one ring) - Cytosine, Thymine, and Uracil (in RNA only) are pyrimidine derivatives in nucleic acid
● A-T/U bond has 2 H-bonds
● G- C bond has 3 H-bonds
● Tm: temperature at which 50% of DNA is annealed
○ NOTE: Melting point of GC-rich DNA is higher than AT rich DNA ( due to the number of G-C hydrogen bonds(3))
Ways to remember Nucleotides:
● Cytosine and Thymine have Y's in them, and are pyrimidines, which also has a Y in it
● Pure As Gold (Purines-Adenine and Guanine)
● C and T cUt pie….pyrimidines or just remember TCU university
○ Purine (small word, 2 bases): lady GAGA (G and A) and the rest (3 more) is pyrimidine (long word has more bases) :)
Know Characteristics of the different forms of DNA (A/B/Z) and what affects their structures
● B DNA - 0.34 nm between bp, 3.4 nm per turn, about 10 bp per turn. right handed helix. the bases are stacked parallel and bases are facing each other at the same level making it have a high H-bonding.
● A DNA - Cylindrical CORE. 11bp per turn. forms when DNA is dehydrated, more tightly wound than B DNA (right handed helix). The bases are not parallel but at an angle (this angle gives it a less strong H-bonding making A conformation less stable than B DNA)
● Z DNA - 12 bp per turn. No grooves, left handed helix, occurs in G/C rich regions
Manipulation (REN, sequencing, PCR)
REN - Restriction Endonuclease Recombination
Sequencing This is specific to Sanger sequencing or dideoxy (which is what the dd in ddNT
stands for), other sequencing techniques work differently
● Uses ddNTP: ddNTP does not have OH (deoxy) on C2/C3 (dideoxy) , DNA polymerase will stop when ddNTPs are incorporated onto the DNA strand. ddNTPs are color labeled for different bases. (ex: ddGTP is yellow). Determines the nucleotide sequence (A,T,G,C) in one strand.
● Separates strands by denaturing (95 deg C)
● Anneal primers (65 deg C)
● Extend primers w/ Taq (can withstand higher temperatures) polymerase (72 deg C)
● Generates thousands→ millions of copies of the DNA
Helicase/DNAB: protein that catalyzes the ATP-dependent unwinding of DNA double helices (DnaB is
the helicase in E. coli)
Single Stranded Binding Protein: binds to the unwound strands to prevent reannealing
DNA polymerase: enzyme responsible for DNA replication, uses ssDNA as the template and makes a
complementary strand by polymerizing deoxynucleotides; synthesis in 5’ to 3’ direction
● In E. coli
DNA pol I: removes the RNA primers and fills the gaps, proofreads; DNA pol II: DNA repair enzyme expressed in response to DNA damage; DNA pol III: holoenzyme responsible for DNA replication
● In Eukaryotes
DNA polymerase �: nuclear, initiation of nuclear DNA replication;
DNA polymerase �: nuclear, principal polymerase in leading - and lagging- strand synthesis, highly processive;
DNA polymerase �: nuclear, leading- and lagging- strand synthesis, sensor of DNA
damage, checkpoint control;
DNA polymerase �: mitochondria, mitochondrial DNA replication; DNA polymerase �: DNA repair
RNAse H: a nuclease that specifically degrades RNA chains in DNA in eukaryotes (DNA Poly I in
DNA Ligase: seals nicks in double stranded DNA where a 3’-OH and a 5’-phosphate are
Telomerase: adds telomeres to the ends of chromosomal DNA, an RNA-dependent DNA
polymerase (has its own RNA primer, does need external RNA primers)
● Semi-conservative: One parent strand and one newly synthesized strand
● Discontinuous: In DNA replication, lagging strand makes Okazaki fragments that need to be glued together by ligase
Central Dogma : Replication, transcription, translation (DNA → RNA → Protein) DNA can self replicate via replication. DNA is coded into RNA via transcription and then translated into protein.
RNA is synthesized in the 5'-3' direction, while the DNA template is read in the 3'-5' direction
Thermodynamics related equations and side notes:
is used to solve for Keq and Q. Keq is using the concentrations at equilibrium and Q can be under any conditions.
; -n = #of e- transferred, F = Faraday constant (96.5 kJ/V), and E = Eacceptor - Edonor. When asked to solve for Keq/Q and not given concentrations, use the following equation:
A hydrogen bond is one of two things:
A hydrogen covalently bonded to an electronegative atom in one molecule, and can form a noncovalent bond with electronegative atoms of another molecule
N, O, F
Hydrogen bonds add to the solubility of a molecule in water
Ex: This is a correct Hydrogen Bond
--N - - - - -H---O----H (N is delta negative while H is delta positive)
When in the presence of Carbon, Hydrogen will not make a H-bond with Carbon, there isn’t enough electronegativity to have a net-dipole moment.
Acid | Base Chemistry
A Bronsted Lowry acid is a proton donor. HA --> H+ + A-
A Lewis acid is an electron acceptor.
The stronger the acid the larger the Ka.
The stronger the acid the smaller the pKa.
Kw = k [H2O] = [H+][OH-] = (10-7M) (10-7M) = 10-14
pH = -log[H+]
pKa = -log(Ka)
Henderson Hasselbalch : pH = pKa + log[A-]/[HA]
● Adding weak acid and conjugate base results in a buffer system, which resists large changes in pH.
● Buffer capacity is dependent on the buffer composition
The greater the amounts of conjugate-acid base pair, the greater the buffer capacity (can resist addition of base or acid)
EX. phosphate buffer system, imidazole group of histidine, bicarbonate buffer system
Chapter 4: Amino Acids
Structures and classification (solubility)
Amino acids are chiral (except Glycine which has 2H), Proteins have AA in L-configuration
D- configuration a.a. can be found in bacterial cell walls
The “CORN” rule for determining configuration of amino acids: The groups:
COOH, R, NH2 and H (where R is an unnamed carbon chain)
are arranged around the chiral center carbon atom. If these groups are arranged clockwise around the carbon atom, then it is the D-form. If counter-clockwise, it is the L-form. The convention that is used to designate the configurations of chiral carbons of naturally occurring compounds is called the D and L convention or system.
Three types of Amino Acids:
1.) Non-polar (hydrophobic)
2.) Polar (uncharged)
pKa and ionization states
● Amino acids have different charges at different pH
● the pKa of functional groups affects the overall charge of the molecule since pKa for N and C terminus are similar in all amino acids:
○ N-terminus - pKa 9.6
○ C-terminus - pKa 2.3
(If in a chain, N-terminus pKa is 8.5, C-terminus pKa is 3.5)
Example: Dipeptide Ala-His This type of question is on the 2003 ACS biochem test
pH = 2 → +2 pH = 6 → +0.5 pH = 4.5 → +1 pH = 9 → -0.5
● PI (isoelectric point) is where the amino acid residue has a net charge of zero, and is therefore insoluble.
This point occurs directly between two transition states, take the example of Ala-His, at
pH of 6 the charge is +0.5, at pH of 9 the charge is -0.5 so PI of the amino acid chain is at a pH of 7.5.
Protein purification using ion exchange column (net charge and charge on column)
● Cation exchange column (positive protein) elute negatively charged molecules first.
● Anion exchange columns (negative protein) elute positively charged molecules first.
● Can also further elute with changing pH: for example in a cation exchange column if you have a positively charged protein, after eluting all the negatively charged protein, you increase the pH past the pIs so the protein becomes negatively charged and be eluted to collect your protein
● The 3 letter amino acid abbreviations will be given but the structure and the one letter abbreviation will NOT be given so you need to memorize those. There are nice apps to help you do that :)
Primary, secondary, and tertiary structures and their driving forces.
Primary structure: (ALL PROTEINS HAVE)
acids in a polypeptide chain
Held together by: Peptide bonds, or amide bonds note their planar configuration due to resonance
Secondary structure: amino acid chains with bonding between a.a.
ex: Alpha helix | Beta sheet
Held together by: Hydrogen bonds beta:
● parallel (longer loop) tilted H-bonding (less stable)
● antiparallel (smaller loop) straight H-bonding (more stable) Tertiary structure (ALL PROTEINS HAVE: Ex. MYOGLOBIN) Many secondary structures into domains
ex: Fibrous or globular proteins
Driving factor in creating tertiary structure is - hydrophobic effect
Held together by: Hydrophobic effect, Ion pairs, Disulfide bonds and
Hydrophobic effect: The tendency of water to minimize its contact with hydrophobic molecules. As such, hydrophobic molecules, when placed in a solution (of water) will begin to aggregate. This is energetically favorable: there is less organization; entropy will increase. However, it is important to note that, upon immersion of the hydrophobic substance, there is a reduction in entropy.’
Quaternary structure: (NOT ALL PROTEINS HAVE : Ex. HEMOGLOBIN)
- Highest structure a protein can have proteins that have multiple subunits which makes up the whole protein such as hemoglobin.
Hemoglobin as a model quaternary structure --- binding curves of hemo and myoglobin.
Myoglobin is more easily saturated with oxygen, which explains its location in muscle tissues, hemoglobin carries oxygen through the bloodstream where it saturates myoglobin and allows muscle to carry oxygen for cellular processes. Allosteric effect of O2 binding to Hb and Mb causes conformational change in the molecule.
oxygen binds with higher affinity to myoglobin vs hemoglobin (P50 of O2) but hemoglobin displays cooperative/concerted binding and binds/releases O2 more readily than myoglobin.
Myoglobin = non cooperative binding = HYPERBOLIC CURVE Hemoglobin = cooperative binding = SIGMOIDAL CURVE
All enzymes use proximity and orientation (of substrate(s)) effects to increase catalysis
Catalytic Mechanisms: Acid/Base, covalent and metal
Acid → Proton donors Base → Proton acceptors pH dependent
Acid base catalysis
. Electron sharing: nucleophilic attack by base pairs on amino acids
Usually requires a nucleophile
Metal Ion catalysis
Help oxidation-reduction (OIL RIG = oxidation is losing electrons |
reduction is gaining electrons) Provides electrostatic effects
The intermediate is relatively stable and requires another larger energy barrier to proceed to the eventual product which is ultimately more stable than the reactants.
Serine proteases: class of proteolytic enzymes whose catalytic mechanism is based on an active- site
Chymotrypsin: digestive enzyme, cleavage Acid/base and covalent catalysis Catalytic triad of serine proteases:
- tetrahedral intermediate #1
-tetrahedral intermediate #2
Transition state stabilization:
-Oxyanion hole: of negatively charged O of first tetrahedral intermediate
-Low barrier H bond: of substrate NH/His 57/Asp 102
Rate equation, Km and Vmax on graphs
-1st, 2nd, and zero order
- k = rate constant, has different units for each order
1/Vo = (Km/Vmax)(1/[S]) + (1/Vmax)
Km is an indicator of substrate binding affinity. High Km = Low Affinity
Km is the substrate concentration required to reach ½ Vmax
Vmax is the maximum enzymatic velocity
How to determine the Vmax and Km on a double reciprocal plot (Lineweaver-Burk plot)
-Units of Km = x axis
-Units of Vmax = y axis
Inhibition graph identification
-Inhibitors cause a decrease in enzyme activity, therefore with a competitive inhibitor there is a larger amount of substrate needed to reach half of Vmax
-Competitive inhibition - inhibitor binds in active site. ES or EI only, as a competitive
inhibitor in the active site means both cannot be bound at once (therefore NO ESI).
-Vmax unchanged: can overwhelm inhibitor with excess substrate
-Km increased: need more substrate to reach ½ Vmax
There is lesser formation of E-S complex. High [S] solves problem.
-Noncompetitive inhibition (or mixed) - inhibitor binds most distant from active site. EI
-Km may raise or lower
-Both the enzyme and enzyme-substrate complex may bind the inhibitor
- If enzyme and E-S bind inhibitor with equal affinity, alpha and alpha prime are equal and Kapp max is unchanged- this is known as pure noncompetitive inhibition.
-Uncompetitive inhibition - inhibitor binds near active site. Only ESI.
- The inhibitor binds directly to the enzyme-substrate complex but not to the free enzyme.
Using a linear regression to solve for Vmax and Km
Intro to Metabolism
High energy compounds
ATP (and its triphosphate derivatives) Acetyl-CoA
Reduced cofactors: NADH, NADPH, QH2
Oxidation | Reduction
Many metabolic pathways include oxidation-reduction reactions. CoQ is an
Chapter 8: Carbohydrates
Identification of reducing sugars:
reducing sugars are those that contain an aldehyde group or have an anomeric carbon that does not participate in glycosidic bond
Notation for glycosidic bonds (see picture above)
If the anomeric carbon and sixth carbon are opposite → alpha
If the anomeric carbon and the sixth carbon are on the same side
Reactions often favor the formation of alpha bonds.
How to tell if a sugar molecule is D or L? If the Chiral carbon (C5) furthest
from Carboxyl group (C1) has it’s OH group on the right side it is D (as shown below) if it is on the left side then it is L configuration. D is naturally occurring. L for Amino Acids, but D for Carbohydrates
Chapter 9 & 10
Fatty Acids (saturated vs unsaturated and their melting points)
-saturated fats are completely saturated with hydrogen atoms (most reduced form)
-unsaturated fats have double bonds and take up more space due to those bonds
is down 3 C
-The melting point of unsaturated fatty acids is lower than saturated fatty acids (hence why unsaturated are liquid at room temp) Think butter vs. olive oil, which is healthier? What state does each occupy at room temp?
-longer acyl chain aka more C’s = higher melting point (mp)
-fatty acid nomenclature
-18:2 Δ9,12 ← from carboxyl end
-18:2n6 ← from methyl/omega end, assume next saturation
-more saturation = higher mp
Plants grown in hotter regions (C4) will have a greater amount of saturated fat than plants grown in cooler regions (C3), this is due to the accessibility of these fats depend on the fluidity and unsaturated fats are more fluid at lower temperatures. Colder climate plants need easier access to plant fats.
Membrane Lipids (classification: Phospholipids (PL), sphingolipids, glycolipids)
Make up the cellular lipid bilayer
Sphingolipids: Parent compound → Ceramide
3 classes: Sphingomyelins, cerebrosides, and ganglioside
Chapter 15: Glucose
Glycolysis (starting compound) enzyme [coenzyme, or ATP if needed]
1. (d-glucose) hexokinase [use ATP, need Mg]
2. (glucose-6-phosphate) phosphoglucoisomerase
a. phosphorylation ensures that the glucose remains in the cell due to (-) charge b. reduces the intracellular level of glucose (diffusion, etc)
c. necessary to maintain thermodynamics
3. (fructose-6-phosphate) phosphofructokinase [use ATP, need Mg]
a. RATE determining step
b. helps in facilitating the C-C bond cleavage c. allosteric inhibitors
i. ATP, citrate inhibits
ii. AMP, fructose-2,6-bisphosphate activates
4. (fructose-1,6-bisphosphate) fructose bisphosphate aldolase a. Makes GAP and DHAP
5. (dihydroxyacetone phosphate) triose phosphate isomerase a. Converts all DHAP to GAP to be used by the cell
6. 2x (glyceraldehyde-3-phosphate) glyceraldehyde-3-phosphate dehydrogenase [use 2 Pi and 2 NAD+, produce 2 NADH + H+]
a. Makes 1,3-bisphosphoglycerate
7. 2x (1,3-bisphosphoglycerate) phosphoglycerate kinase [use 2 ADP, produce 2 ATP, need
a. break even step
b. 2,3-bisphosphoglycerate an intermediate
8. 2x (3-phosphoglycerate) phosphoglycerate mutase
9. 2x (2-phosphoglycerate) enolase [produce 2 H2O, need Mg and K]
10. 2x (phosphoenolpyruvate) pyruvate kinase [use 2 ADP, produce 2 ATP, need Mg and K]
a. AMP, fructose-1,6-bisphosphate activate b. ATP, acetyl CoA and alanine inhibit
PRODUCES 2 pyruvate, 2 ATP (2 ATP used + 4 ATP produced = 2 net ATP, and 2 NADH)
Controlling points and controlling mechanisms
Sites of regulation: hexokinase, phosphofructokinase, pyruvate kinase
Step 3 in glycolysis: Phosphofructokinase--rate-determining step
Aldolase +G, but low [G3P] allows for -G in vivo
Fate of Pyruvate: 5 Different possible outcomes
Pyruvate → Lactic Acid (lactate dehydrogenase)
Pyruvate → Acetaldehyde → Ethanol (Pyruvate decarboxylase TPP
cofactor | Alcohol dehydrogenase, at pH=7 ethanol favored)
Pyruvate → Glucose (through gluconeogenesis)
Pyruvate → Oxaloacetate (pyruvate carboxylase, Biotin cofactor) Pyruvate → Acetyl-CoA (pyruvate dehydrogenase, E1/TPP cofactor | E2/lipoamide | E3/FAD)
Other substrates that can enter glycolysis
a. fructose b. mannose c. galactose d. glycerol
Chapter 16: Glycogen
Glycogen breakdown and synthesis
Glycogen- strands of glucose molecules
3 Enzymes in breakdown of glycogen (glycogenolysis)
Glycogen phosphorylase → PLP cofactor assists w/
Synthesizing glycogen requires configuration change of glucose molecules:
Glucose-6-phosphate → Glucose-1-phosphate
Glucose-1-phosphate + UTP → UDP-Glucose
UDP-Glucose + glycogen chain (n) → Glycogen chain (n+1)
-UDP glucose phosphorylase
Mainly occurs in liver, occurs during times of low blood sugar, to provide brain with glucose. Unlike other tissues, the brain can only metabolize glucose to gain energy.
Steps 10, 3, and 1 of glycolysis require different enzymes in gluconeogenesis
(sections of phosphorylation/dephosphorylation)
1. Pyruvate carboxylase - biotin dependent enzyme a. uses ATP and bicarbonate
b. allosterically activated by acetyl-CoA
c. compartmentalized to the mitochondrial matrix (pyruvate crosses the membrane of the mitochondria converts to oxaloacetate and is reduced to malate to cross the membrane once more)
2. PEP carboxykinase a. uses GTP
3. Fructose-1,6-bisphosphatase a. needs H2O
b. allosterically regulated
i. activated = citrate
ii. inhibitor = AMP, fructose-2,6-bisphosphate
a. requires transport proteins
Final step in gluconeogenesis, requires 2 step process. B. Requires G-6-Pase to return to glucose
Chapter 17: Citric Acid Cycle
Pyruvate dehydrogenase: 3 enzyme mechanism; pyruvate → Acetyl-CoA
Enzyme 1: pyruvate dehydrogenase
Enzyme 2: dihydrolipoamide acetyltransferase
Enzyme 3: dihydrolipoamide dehydrogenase
Keep in mind all of these are included in the PDH Complex.
2 NADH produced from this step
Enzymes in citric acid cycle: (starting material) enzyme [substrates and ATP]
1. (oxaloacetate) citrate synthase [acetyl CoA, H2O]
a. inhibitor: NADH
b. condensation reaction
2. (citrate) aconitase reaction is an isomerization
3. (isocitrate) isocitrate dehydrogenase [use NAD+, produce NADH + H+ and CO2]
4. (�-ketoglutarate) �-ketoglutarate dehydrogenase [use NAD+, produce NADH + H+ and
5. (succinyl-CoA) succinyl-CoA synthetase [use CoASH, Pi, GDP]
a. substrate-level phosphorylation
b. condensation reaction that produces ATP
6. (succinate) succinate dehydrogenase [use FAD, produce FADH2]
7. (fumarate) fumarase [use H2O]
8. (malate) malate dehydrogenase [use NAD+, produce NADH + H+]
Location in Cell
Citric acid cycle takes place in the mitochondrial matrix, glycolysis takes place in the cytosol.
To oxidize E3 NAD+ → NADH for two molecules
Net 2 molecules NADH → 5ATP for 2.5 P:O ratio
1 glucose (or 2 pyruvates) in the citric acid cycle produces
Chapter 18: Electron Transfer and Oxidative phosphorylation
What is happening?
Electrons are moved across the complexes until they are ultimately accepted by oxygen to form water.
Net products from 1 NADH: 10H+ | 2e- | ½ O2 | 1 H20 | ~2.5ATP
Net ATP from FADH2 = ~1.5 ATP
What does Beta oxidation do?
oxidize fatty acid 2 carbons at a time. Makes acetyl CoA during every oxidation step. Each round produces 1 acetyl CoA, 1 NADH, and 1QH2. Each round requires 4 enzymatic steps. Beta carbon is oxidized first
4 enzymatic steps: Acyl-CoA dehydrogenase, Enoyl CoA hydratase, 3-L-hydroxyacyl
dehydrogenase, Beta-ketoacyl-CoA thiolase.
- Even number unsaturation → 2.5 less ATP made because have to use a NADPH to move the double bond to the correct spot
-Odd number unsaturation → 1.5 less ATP made because skip first dehydrogenase step which means QH2 is not made
Fatty acid synthesis overview
-Occurs in cytoplasm so need to move acetyl coAs from mitochondria to cytosol, use tricarboxylic transport system
-Acetyl coA in cytoplasm
- Activated 3C unit malonyl coA, made using ACCase w/ Biotin cofactor. ACCase requires ATP.
-Acetyl ACP + Malonyl ACP
-Reverse of oxidation
-Uses NADPH → NADP+. Each round of fatty acid synthesis uses 2
Cholesterol: Chylomicrons, VLDL, LDL, and HDLs
Chylomicrons- largest of the transport molecules: transfer cholesterol from the intestine to the liver.
VLDL and LDL- second largest: transfer cholesterol from liver to other tissues HDL- smallest of the proteins: Transfer cholesterol from tissues to liver *good cholesterol*
Protein % composition from greatest to least by: HDL>LDL>IDL>VLDL>Chylomicrons
Tangier disease: defect in transporter → high cholesterol in
tissues, higher risk of cardiovascular disease & heart attack. Familial hypercholesterolemia: defect in receptor of liver cell (LDLR) → high cholesterol in plasma
General Topics for review
My SUGGESTIONS for Stryer Chapter reviews.
I will give you what my guess are the key chapters and concepts to study. This does not mean they are the ONLY concepts to study.
Chapter 2, 3 and 4. Pay close attention to biochemical lab methods and how they could be employed to solve a problem. Standard exams always ask about methods. Know basic chemical properties of amino acids, side chains and of course, acids and bases and pH.
Chapter 4,5. READ THESE. There will be some DNA,RNA, Molecular Biology types of stuff. The questions would probably be at a much lower level than all your MoBi and Cell courses, so this chapter should cover it. Be sure to freshen up on translation, transcription, protein synthesis, replication, operons.
Chapter 8. Review enzyme kinetics. Understand the basics of Km, Vmax, v, kcat, [S], assumptions of M-M kinetics and general mechanisms of inhibition. There
are always questions about relations between these quantities and enzyme activities.
Chapter 9. Be sure you understand basic organic strategies for catalysis, esp. the trypsin/ chymotrypsin model.
Chapter 7,10. Read these chapters. There are almost always hemoglobin and allosteric enzyme regulation type questions.
Chapters 3,6,11,12. Know fundamental chemical and physical properties of amino acids, lipids, carbohydrates and nucleic acids. Don’t memorize structures, but ,
e.g. be able to tell a ketose from an aldose, how sugars cyclize, anomers, etc.
Chapter 14 Read this for review. Chances are you already know as much as you will need about signal transduction from other classes.
Chapter 15. Energetics. There are always questions about basic thermodynamic properties for biochemical reactions. They are not likely to be any deeper than this chapter though. Understand the relations between DG, DH, DS, T, Keq, etc.
CHAPTER 16,17,18 !!!!!! These will probably cover the essential core knowledge about metabolism that you will need to know. You should know
Glycolysis and TCA intermediates and key enzymes and COFACTORS for different chemical transformations. Review the ETS and chemiosmosis.