Exam 1 Flashcards

Question

True or False: The secondary structure of proteins includes α-helices and β-sheets.

Answer 1

Covert MW in kDa to Da and Divide the MW by 110 Da The average molecular weight per amino acid is approximately 110 Da.

Answer 2

Secondary structures are established by non-covalent interactions, including hydrogen bonds.

Answer 3

K = [PL] / [P][L] ## Footnote K represents the equilibrium constant for the formation of the complex.

Answer 4

k_a = k_on / k_off k_on=k_a k_off=k_d ## Footnote It measures how fast the protein and ligand come together.

Answer 5

k_d = k_off / k_on ## Footnote It measures how fast the complex PL breaks apart.

Answer 6

K_d = 1 / K_a

Answer 7

θ = 0.5 (50% of binding sites are occupied)

Answer 8

θ ≈ 1 (Almost all sites are occupied)

Answer 9

θ ≈ 0 (Almost no sites are occupied)

Answer 10

θ = [L] / ([L] + K_d)

Answer 11

P50 = K_d = [O2] that results in 50% (half saturation) of Mb

Answer 12

Myoglobin **accepts oxygen released from hemoglobin** and **delivers** it to the **mitochondria**.

Answer 13

pO2 high (100 mm Hg, 13.3 kPa)

Answer 14

pO2 low (30 mm Hg, 4.0 kPa)

Answer 15

The heme group consists of Fe2+ bound within heme, which is buried within the structure of a globin.

Answer 16

* T (tense) state without O2 * R (relaxed) state with O2

Answer 17

The Monod/Wyman/Changeux (MWC) model

Answer 18

IgG (immunoglobulin G)

Answer 19

* Fab (Fragment antigen binding) * Fc (binds to Fc receptors on macrophages) signal

Answer 20

To exploit antibody specificity for rapid screening and quantitation for antigen in samples.

Answer 21

SILMVKLASEG

Answer 22

* Covalent bonds between cysteine residues - disulfide bonds * Higher entropy for water molecules when they are not caged around hydrophobic groups * Ionic bonds between charged side chains * Hydrogen bonds between main chain atoms

Answer 23

Lower entropy for the polypeptide when it has a buried hydrophobic core

Answer 24

0 ## Footnote The average of the two flanking pKa’s (4.25 and 6.00) is closest to 5, making it the best answer.

Answer 25

* Arg: 12.48 * Asp: 3.65 * Glu: 4.25 * His: 6.00 * Lys: 10.53 ## Footnote These values are crucial for determining the behavior of the peptide in different pH environments.

Answer 26

[higher] affinity.

Answer 27

[lower] affinity.

Answer 28

The amount of ligand bound to each protein is a function of the ligand’s concentration in solution. Ligand binding increases with concentration, but the rate and pattern depend on cooperativity and affinity K_d Hemoglobin binds oxygen cooperatively → sigmoidal curve. Myoglobin binds oxygen non-cooperatively → hyperbolic curve.

Answer 29

Dissociation constant

Answer 30

Arg (12.48)

Answer 31

They determine the **ionization state** of amino acids at different pH levels.

Answer 32

* Purify complexes from native sources by isolating a protein directly from the tissue or organism rather than expressing it recombinantly in bacteria, yeast, or mammalian cells. * pro: native complexes present * con: yields often limiting, difficult to engineer

Answer 33

start: Lyse/homogenize/break cells then: centrifuge High speed centrifugation of lysed cells produces two components: the supernatant (cytosol) and the pellet (organelles and plasma membrane)

Answer 34

Generates one molecule of ATP

Answer 35

* based on protein’s affinity to ligand covalently attached to solid resin * highly selective — proteins are very selective with ligand they bind to * can allow for single step purification * generality greatly enhanced by the ability to generate fusion proteins of your target with a protein with known affinity

Answer 36

* based on protein’s charge at a particular pH * not so selective, since many proteins can have the same charge at a given pH — but very general * resin in figure has negative charge = binds positive proteins (cation exchange chromatography) * one can also use positively-charged resins to bind negatively-charged proteins (anion exchange chromatography)

Answer 37

* based on protein’s MW * not so selective, since many proteins have similar MW * porous resin with lots of different sized channels; smaller proteins enter resin particles while larger proteins cannot = longer path for smaller proteins than larger ones, takes more time to go through column

Answer 38

* proteins are coated with SDS **negative charge** * SDS mostly removes contributions due to conformation **denaturant** and inherent charge in the folded protein (instead, now there is equal charge per MW) * SDS **denatures proteins**, so it does not give information about native structure, folding, or activity. * You can infer if disulfide bonds exist by comparing reducing vs. non-reducing SDS-PAGE, but SDS-PAGE alone does not confirm their role in the native protein structure. * this effectively allows separation by MW * smaller things move faster through the gel matrix — so at the end, low MW at bottom and high MW at top * resolution ~1 kDa = ~1000 Da

Answer 39

* first, electrophoresis to separate proteins by MW * Then treat with a stain such as Coomassie blue to visualize * note calibration markers (“ladder”) at left * example shown here: recombinant protein expression (not a recombinant process) * start with a complex mixture, become simpler with purification

Answer 40

Few AA (like pentapeptide = 5mer below)

Answer 41

Many AA (Molecular Weight (MW) < 10,000 Daltons)

Answer 42

Big polypeptide (Molecular Weight (MW) > 10,000 Daltons)

Answer 43

The ideal α-helix formers are **Alanine (Ala/A), Leucine (Leu/L), Methionine (Met/M), and Glutamate (Glu/E)** because they have side chains that fit well into the helical structure and **favor hydrogen bonding** patterns. Conversely, **Proline, Glycine, bulky, charged, and strongly polar** amino acids tend to destabilize α-helices. “Breaking residues” of α-helix: * adjacent turns of **Glu(E)/Asp(D)** (**negative** charges repel) * adjacent turns of **Lys(K)/Arg(R)** (**positive** charges repel) * bulky R-groups (**W,Y,F**) * Pro - unable to H-bond and restricted dihedral angles A bigger number of destabilizing charges = less propensity to take up the α-helical conformation

Answer 44

No, b/c the same-charged residue separated by 4 residues (n,n+4): RxxxK; DxxxE; Y, P How to analyze: 1. Assess Each Amino Acid's Helix Propensity Different amino acids have varying tendencies to stabilize or destabilize α-helices. Below is a quick classification: Strong Helix Formers: Alanine (Ala, A) → Most favorable Leucine (Leu, L), Methionine (Met, M), Glutamate (Glu, E), Lysine (Lys, K) → Common in α-helices Neutral or Moderate Helix Formers: Aspartate (Asp, D), Arginine (Arg, R), Threonine (Thr, T) → Can be present but not always ideal Helix Breakers: Proline (Pro, P) → Major helix breaker Glycine (Gly, G) → Too flexible, usually destabilizing 2. Analyze the Given Peptide (RDFTKEYP) 3. Prediction: Is It Helical? The peptide has some strong helix formers (K, E), which help α-helix formation. However, Proline (P) at the end is a known helix breaker, meaning it can terminate or severely disrupt an α-helix. The presence of Asp (D) and Thr (T) can also destabilize the helix due to electrostatic repulsion or H-bond competition. 4. Conclusion ❌ This peptide is unlikely to form a stable α-helix because: Proline at the end disrupts the helix. Aspartate and Threonine can interfere with hydrogen bonding. The mix of bulky and charged residues may prevent a stable helix from forming.

Answer 45

* R-groups protrude outward * the repeating unit is a single turn of the helix extending ~ 5.4 Å (3.6 AA/turn, 1.5 Å rise per AA = 5.4 Å rise/turn) * 3.6 AA/turn * H-bonds between carbonyl of residue n and amide protons of residue n+4 * every peptide bond has potential to participate in H-bonding * Q: What is the exception? PROLIN : lack of free amide proton

Answer 46

Effect on Protein Function (Effect Z) Loss of function: If the mutation disrupts critical interactions. Gain of function: Rare, but could stabilize an active conformation or enhance binding. Neutral effect: If the mutation is conservative (similar size/charge) or in a flexible loop. Example Cases Aspartate (D) → Glutamate (E): Similar charge, minimal effect. Glycine (G) → Proline (P): Could disrupt secondary structure due to rigidity. Cysteine (C) → Serine (S): Might eliminate a disulfide bond, reducing stability. Leucine (L) → Arginine (R): Drastic—changes from hydrophobic to positively charged, likely destabilizing.

Answer 47

Describe the Distribution of Secondary Structure in a Protein * describes the dihedral angles phi (φ) and psi (ψ) of a polypeptide * Ramachandran plot is required to test the modeled geometry of a protein structure

Answer 48

Hemoglobin is a tetrameric protein (α2β2) that transports oxygen in blood. It must efficiently pick up oxygen in the lungs and release it in tissues. Cooperativity helps achieve this: T (Tense) State: Low O₂ affinity, stabilizing Hb in deoxygenated form. R (Relaxed) State: High O₂ affinity, forming upon oxygen binding. Oxygen Binding Switches T → R: Each O₂ bound increases affinity at remaining sites, making Hb an efficient oxygen transporter.

Answer 49

1. Hill Coefficient (nH) Measures the degree of cooperativity. **nH > 1 → Positive cooperativity** (e.g., Hb, nH ≈ 2.8). **nH = 1 → No cooperativity (e.g., myoglobin)**. nH < 1 → Negative cooperativity. Determined from the Hill plot, where the slope = nH. 2. Oxygen Binding Curve - **Sigmoidal (S-shaped)** Curve: Characteristic of cooperative binding (**Hb**). - **Hyperbolic Curve**: Seen in non-cooperative binding (e.g., **myoglobin**).

Answer 50

Cooperativity occurs when the **binding of a ligand** to **one site on a multi-subunit protein influences the binding of additional ligands at other sites**. **Positive** Cooperativity: Ligand binding **increases affinity** at other sites (e.g., **hemoglobin and oxygen**). **Negative** Cooperativity: Ligand binding **decreases affinity** at other sites (less common). **Non-Cooperative** Binding: **Binding at one site does not affect other sites (e.g., myoglobin)**.

Answer 51

- Explains **cooperative binding** (e.g., hemoglobin’s oxygen affinity). - Assumes only two states, **T and R** with all subunits switching together. - Used in **allosteric regulation of enzymes and receptors**.

Answer 52

IgG consists of two heavy chains and two light chains, with a Fab region for antigen binding and an Fc region for signaling.

Answer 53

It binds to antigens.

Answer 54

Kd as low as 10^-10 M indicates high affinity and specificity.

Answer 55

hydrophobic patch

Answer 56

A mutation replacing **Glu 6 with Val on the β subunits** of hemoglobin (HbS).

Answer 57

A **hydrophobic patch** that leads to interactions between different Hb complexes, forming **long fibers**.

Answer 58

**CO binds to Fe^2+** in hemoglobin with ~200x stronger affinity than O2, **blocking O2 transport**.

Answer 59

It helps **facilitate O_2 release by hemoglobin** through the **Bohr** effect.

Answer 60

~**67%** transported as **bicarbonate (HCO3-)**.

Answer 61

It **catalyzes** the conversion of **CO2 & H2O** to **H+ & HCO3-** in red blood cells.

Answer 62

**Lower pH increases [H+] bound to Hb, resulting in O2 release**

Answer 63

~**25%** carried by Hb **via reversible reactions** with amino groups.

Answer 64

The Bohr Effect describes how **lower pH in tissues leads to increased H+ binding to hemoglobin (Hb), reducing its affinity for O2 and facilitating O2 release**.