Exam 1 Flashcards

Question

What are the three types of energy?

Answer 1

* Free energy - energy that can do work at constant T and P ( we use free energy) * Heat energy - energy that can only do work through changes in T or P * Entropy - "disordered" energy that cannot be used to do work

Answer 2

* Calorie (cal): energy to raise the temperature of 1.0 g of water from 15 to 16 degrees C, at 1 atm P * Joule (J): energy required to apply 1 newton of force over a distance of 1 meter * 1 kilocalorie (kcal) = 1000 cal * 1 kcal = 4.184 J

Answer 3

* T = 298 degrees K (25 degrees C) * P = 1 atm * [Reactants] and [Products] start at 1 M concentration * In biological systems, pH = 7.4. [H+] = 10^-7 M

Answer 4

1. Delta G = negative * Spontaneous, forward, exergonic 2. Delta G = positive * Non spontaneous, energonic 3. Delta G = 0 * System is at equilibrium, no work being done

Answer 5

* delta G^o: the delta G for a reaction at standard state * delta G^o': the delta G for a biological reaction at standard state. 1. delta G^o = negative * Spontaneous, forward * Keq > 1.0 * Products < reactants 2. delta G^o = positive * Non spontaneous, backward * Keq < 1.0 * Products > reactants 3. delta G^o = 0 * Reaction is at equilibrium * Keq = 1.0

Answer 6

* delta G^o is at standard state, but real systems are rarely at standard state. * Actual delta G is for the real system conditions. * delta G's of summed reactions are additive.

Answer 7

* ATP: adenosine triphosphate * Carries 4 negative charges (ADP has 3). * Sometimes forms MgATP^2- with Mg²⁺ in cellular fluid * Acts as an energy shuttle * ATP --> ADP + Pi, delta G^o' = -7.3

Answer 8

* Due to its structural features and those of its hydrolysis products 1. Ionization: the PO₄ groups of ATP are ionized at cellular pH and will generate H+ during hydrolysis 2. Electrostatic forces: the O- of the PO₄ groups in ATP want to repel each other 3. Resonance structures

Answer 9

* A guanidinium phosphate that is similar to ATP * PC has limited resonance itself, but its hydrolysis products have high resonance stabilization * Acts as a reserve of PO₄ groups to donate to AMP/ADP to make ATP. * Transfer catalyzed by creatine kinase

Answer 10

* ATP hydrolysis supplies energy to move molecules against the gradient * The amount of energy to move a compound from 1 region to another: delta G = 2.303 RT log( C_in / C_out) * C_out > C_in = negative delta G, spontaneous * C_out < C_in = positive delta G, requires energy, active transport

Answer 11

* Some reactions proceed with the hydrolysis of two PO₄ groups, such as the activation of a fatty acid for lipid biosynthesis. * ATP + RCOOH + CoA-SH --> AMP + PPi + RCO-S-CoA * This reaction is energically unfavorable. * PPi + H₂O --> 2 Pi * This second reaction couples with the first, and, due to the additive property of delta G's, makes the entire reaction favorable.

Answer 12

* Oxidation: loss of electrons (or gain of oxygen) * Reduction: gain of electrons (or loss of oxygen) * The electron **donor** is **oxidized**. * The electron **acceptor** is **reduced**. * Oxidations are often accompanied by dehydrogenations due to the loss of hydrogen (catalyzed by dehydrogenases).

Answer 13

* **Direct transfer of electrons**: Fe²⁺ + Cu²⁺ <---> Fe³⁺ + Cu⁺ * **Transfer as hydrogen atoms**: AH₂ <---> A + 2e^- + 2H⁺ * **Transfer as hydride ion (H^-)**: H^- has two electrons. NAD^- linked dehydrogenases transfer hydride ions. * **Direct combination with oxygen**: R-CH₃ + 1/2 O₂ ---> R-CH₂-OH

Answer 14

* delta G = -nF (delta) E * This equations allows you to calculate the delta G for any redox reaction, if the reduction potential is known. Steps to solve for standard conditions: 1. Find the E^o' of both half reactions, when written in the forward direction 2. delta E^o' = E^o' (acceptor) - E^o' (donor) 3. Fill in the formula and solve.

Answer 15

1. Find the E for both half reactions using this formula: 2. E = E^o' + (RT/nF)(ln [reactant]/[product]) 2. Solve delta E = E (acceptor) - E (donor)

Answer 16

G, A, V, L, I, M, P

Answer 17

S, T, C, N, Q

Answer 18

* W and Y are strong chromopores * F is a weaker chromopore * W, Y and F absorb UV light at 270-280 nm

Answer 19

N-terminal end to C-terminal end

Answer 20

* D and L amino acids * Mainly L-amino acids are found in proteins

Answer 21

* Used to find the concentration of a solution with an amino acid * A = eLc * A - absorbance, e - Absorbance constant (specific absorptitivity), L - length of solution the light passes through, c - solution concentration

Answer 22

* At low pHs, amino acids are in their positively charged form * At high pHs, amino acids are in their negatively charged forms * The zwitterion form has both a positive and negative charge * This form exists between the pKa's of the carboxy and amino groups

Answer 23

* The carboxyl group will be pronated at pH < 2.3 and unprotonated at higher pHs * The amino group will be unprotonated at pH > 9.6 and protonated at lower pHs

Answer 24

* The isoelectric point (pI) is the specific pH where the net charge of an amino acid is 0 (pH = pI = net charge 0) * pH > pI - net negative charge * pH < pI = net positive charge * At the pI, the amino acid is the least soluble in water and does not migrate in electric fields. * For amino acids without ionizable side chains: pI = (pK1 + pK2) / 2 * For amino acids with ionizable side chains, the pI is the average of pK's on either side of isoionic species

Answer 25

* By comparing the pKa to the pH of the solution * pH < pKa = protonated * pH > pKa = deprotonated * pH ~ pKa = mixture of two forms

Answer 26

* Selenocysteine * Polar, uncharged * R group is similar to Cysteine, but with Selenium instead of Sulfur

Answer 27

* Amino acids are linked in a chain with peptide bonds connecting the carboxyl group and amino group. (Sometimes called planar amide bonds) * H₂O is removed when linking two amino acids together. * Each peptide bond has some double-bond character due to resonance and thus, cannot rotate

Answer 28

* Oligopeptides: a few amino acids * Polypeptides: many amino acids, molecular weight < 10 kDa * Protein: thousands of amino acids, molecular weight > 10 kDa

Answer 29

* Formed by the oxidation of two cysteine residues to form a covalent sulphur-sulphur bond * These can be intra- or inter- molecular bridges in peptides and proteins

Answer 30

* Protein purification is needed to isolate a single protein to be analyzed, to determine its structure and function. * Purification can be done with column chromotagraphy techniques * Ion exchange chromatography - charge * Gel filtration (size exclusion) chromatography - size * Affinity chromatography - specificity * Adsorption, reverse phase, hydrophobic chromatography - polarity

Answer 31

* Purification factor = final specific activity / starting specific activity * Percent recovery (yield) = total activity in purified fraction / total activity in crude extract * 100%

Answer 32

* Primary - amino acids linked in a chain * Secondardy - recurring structural patterns between nearby amino acid residues * Tertiary - 3D folding of the polypeptide with bonds between amino acids far away from one another * Quaternary - two or more polypeptide subunits linked together

Answer 33

* Sanger sequencing (classical method, not used much) * Edman Degradation * Mass Spectrometry

Answer 34

* Fragmenting proteins into smaller peptides through successive rounds of N-terminal modification, cleavage and identification * **Trypsin (bovine pancreas)**: cleaves after K and R residues * **Cyanogen bromide**: cleaves after M residues

Answer 35

* Identifies the mass of a peptide (after fragmentation), and thus, can determine the amino acid sequence * Peptides can be volatilized for MS analysis by **electrospray ionization (ESI)** or **matrix-assisted laser desorption (MALDI)**

Answer 36

* Tandem MS (MS/MS) has two mass filters in tandem * The first MS sorts cleaved peptides * The second MS measures mass : charge ratios (m/z) of charged fragments

Answer 37

* Phi (o): angle around the alpha carbon-amide nitrogen bond * Psi (w): angle around the alpha carbon-carbonyl carbon bond * In a fully extended polypeptide, both phi and psi are 180

Answer 38

* A Ramachandran plot shows the distribution of phi and psi dihedral angles that are found in a protein. * Shows the common secondary structure elements * Reveals regions with unusual backbone structure * Some combinations of phi and psi angles are unfavorable because of steric crowding * Some combinations are more favorable because of the chance to form H-bonding interactions

Answer 39

* Regular, common arrangements: alpha helix and B sheet * Irregular: B turn

Answer 40

* Right handed helix shape with R groups protruding away from the backbone (roughly perpendicular with the helical axis) * Peptide bons are aligned roughly parallel with the helical axis * 3.6 residues per turn of the helix * Helical backbone is held together by hydrogen bonds between backbone amides of n and n+4 amino acids * The inner diameter is too small for anything to fit "inside" * The outer diameter fits well into the major groove of dsDNA * Amino acids #1 and #8 align nicely on top of each other

Answer 41

* Proline: rotation around the N-C_a bond is impossible * Glycine: the tiny R group supports other conformations

Answer 42

Alanine, Leucine and other small, hydrophobic residues

Answer 43

* Tetrahedral geometry of the alpha-carbon * Hydrogen bonds between the backbone amides in different strands * Side chains protude from the sheet alternating in up and down directions

Answer 44

* Parallel: hydrogen bonded strands run in the same direction * Results in bent H-bonds that are weaker * Antiparallel: hydrogen bonded strands run in opposite directions * Results in linear H-bonds that are stronger

Answer 45

* B-turns are 180^o turns in B-sheets * The turn is accomplished over four amino acids * Stabilized by a hydrogen bond from a carbonyl oxygen to amide proton 3 residues down in the sequence

Answer 46

* Proline: usually in position 2 and forms Type I turns * Glycine: usually in position 3 and forms Type II turns

Answer 47

* Circular dichroism (CD) spectroscopy * Measures differences in the molar absorption of left-handed vs right-handed circularly polarized light * Chromopores (peptide bonds) in the chiral environment produce characteristic signals

Answer 48

* 3D folding of a polypeptide * Stabilized by numerous weak interactions between amino acid side chains (mostly hydrophic and polar interactions) * Can also be stabilized by disulfide bonds * Two major classes: Fibrous and globular

Answer 49

* Formed by the combination of two or more tertiary units * Stabilized by the same interactions found in tertiary structures (hydrophobic-polar interactions, hydrogen bonds, disulfide bonds, etc.)

Answer 50

* A protein is purified and crystallized * Diffraction data is collected and used to calculate the electron density, and thus determine the amino acid residues * Pros: no size limit, well-established * Cons: difficult for membrane proteins, cannot see hydrogens

Answer 51

* After a protein is purified, it's dissolved in a solution. * NMR measures nuclear spin angular momentum of H, C, N, F and P * The NMR signals are then analyzed and the structure of the protein is determined * Pros: no crystallization, can see hydrogens * Cons: works best with small proteins, difficult for insoluble proteins

Answer 52

* Protein is quick-frozen in vitreous ice and kept frozen while being observed in 2D with an electron microsope (EM) * This greatly reduces damage to the specimen * Useful for determining structure of large, dynamic, macromolecule complexes and integral membrane proteins

Answer 53

* Storage of ions and molecules - myoglobin, ferritin * Transport of ions and molecules - hemoglobin, serotonin transporter * Defense against pathogens - antibodies, cytokines * Muscle contractions - actin, myosin * Biological catalysis - chymotrypsin, lysozyme

Answer 54

* Hemoglobin, myoglobin, various enzymes * Tend to be soluble in aqeuous media, with hydrophobic amino acids pointing toward the core of the protein * Lots of hydrogen bonds, somewhat neutralizes the polar N-H and C=O bonds * The core consists almost entirely of fixed secondary structures (a-helix, B-conformations) * Folded in spherical or globular shape

Answer 55

* Heme: Protoporphyrin IX containing a bound iron atom * A porphyrin is composed of 4 pyrrole rings linked by methane bridges, with two open sites for binding to iron * Oxygen will only bind to the iron atoms of heme in the ferrous (2+) oxidation state * 2 perpendicular coordination bonds: one is binding site for O₂; the other is occupied by a side chain nitrogen of a highly conserved proximal His residue

Answer 56

* Protein side chains lack affinity for O * Some transitional metals bind O₂ but would generate free radicals * Fe²⁺ in free heme could be oxidized to Fe³⁺ and no longer bind to O₂

Answer 57

* Alters conformation * Pulls heme closer into the porphyrin ring * His residue is pulled along, distorting the shape of the a-helix

Answer 58

* Myoglobin is in muscle cells * Hemoglobin is in the blood * Both transport/store oxygen * Myoglobin has a higher affinity for oxygen * Hemoglobin can bind up to 4 oxygen molecules, whereas myoglobin can only bind 1

Answer 59

The hydrogen bond between the imidazole side chain of histidine E7 and bound O₂ electrostatically stabilizes the Fe-O₂ polar complex

Answer 60

* CO has similar size and shape as O₂, so it can fit to the same binding site * Binds over 20,000x better because the carbon in CO has a filled lone electron pair that can be donated to the vacant d-orbitals on the Fe^{2+^{* CO competes with oxygen to bind to myoglobin, hemoglobin and mitochondrial cytochromes, and can block their function}}

Answer 61

* Preferred configuration is linear * The His residue at position E7 sterically hinders its binding to heme, reducing its affinity

Answer 62

* T state: tense state, more interactions, more stabilized, lower affinity for O₂ * R state: relaxed state, fewer interactions, more flexible, higher affinity for O₂ * T state is stabilized by a greater number of ion pairs

Answer 63

* Positive cooperativity: first binding event increases the affinity at remaining binding sites * Negative cooperativity: first binding event reduces affinity at remaining sites

Answer 64

* H+ binds to Hb and stabilizes the T state, which leads to the release of O₂ in the tissues * The pH difference between lungs and metabolic tissues increases efficiency of the O₂ transport.

Answer 65

* H+ promotes the release of O₂ from hemoglobin * The pH difference between lungs and metabolic tissues increases efficiency of the O₂ transport.

Answer 66

* 2,3-BPG is a negative heterotropic regulator of Hb function * Small negatively charged molecule that binds to the positively charged central cavity of Hb * Stabilizes the T state * Release of 2,3-BPG is increased at high altitudes to release more O₂ to tissues

Answer 67

* CO₂ is exported in the form of a carbamate on the amino terminal residues of each of the polypeptide subunits. * The formation of a carbamate yields a proton that can contribute to the Bohr effect. * The carbamate also forms additional salt bridges, stabilizing the T state.

Answer 68

* Due to a mutation in hemoglobin * Glu6 is mutated to Val6 in the B chain of Hb * The new valine side chain can bind to a different Hb molecule to form Hb aggregates * Heterozygous individuals exhibit a resistance to malaria

Answer 69

* Hb with Fe^{3+^{instead of normal Fe^{2+^{results in a reduced ability to bind and deliver oxygen to tissues, causing hypoxia
* Newborns have half the level of NADH-methemoglobin reductase enzymes. Therefore, they are more sensitive to oxidative stress.}}}}

Answer 70

* Collagen superstructures are formed by the cross-linking of collagen triple-helices to form collagen fibrils * Lysyl oxidase is responsible for further cross-linking of collagen molecules. * Lysyl oxidase is a Cu²⁺ and vitamin C dependent enzyme.

Answer 71

* Collagen is an important constituent of connective tissue in tendons, cartilage, bones, cornea of the eye, etc. * Each collagen chain is a long glycine and proline rich left-handed helix * Three of these collagen chains twist together into a right-handed superhelical triple helix * Many triple helices assemble into a collagen fibril.

Answer 72

* Forces the proline ring into a favorable folding confirmation * Offers more hydrogen bonds between the three strands of collagen

Answer 73

* Catalyzed by prolyl and lysyl hydroxylase * Requires alpha-ketoglutarate, molecular oxygen, and ascorbate (vitamin C)

Answer 74

A vitamin C deficiency can cause scurvy, bleeding gums, swollen joints, poor wound healing, etc.