2. Protein Function Flashcards

1
Q

ligand

A

A molecule bound reversibly by a protein is called a ligand.

A ligand binds at a site of the protein called the binding site which is complementary in size, shape, charge, and hydrophobic and hydrophilic character.

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2
Q

Describe the difference between conformational changes and induced fit?

A

Conformational change and induced fit are similar.

Ligand binding often involves reversible protein conformational changes that makes the binding site more complementary to the ligand, permitting tighter binding.

The structural adaptation that occurs between the protein and ligand is called induced fit.

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3
Q

Describe the function of hemoglobin and why it is necessary.

How many and what polypeptide chains are contained within hemoglobin?

A
  • Hemoglobin (Hb) transports oxygen from lungs to other parts of the body.
  • Oxygen is poorly soluble in water. No amino acid is capable of binding to oxygen reversibly. Hb uses the prosthetic group heme for this purpose.
  • Hb contains four polypeptide chains (2 identical a-chains + 2 identical b-chains). Each subunit has a heme which can bind to oxygen.

Hb + O2 ⇔ HbO2

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4
Q

Wyman Linkage relationship

A

The Wyman Linkage relationship states that if the ligand specifically binds to one of the two states, then equilibrium between those states will be shifted in favor of the state with bound ligand with the increase of ligand concentration.

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5
Q

Iron (Fe) has a strong tendency to bind to oxygen. However, oxygen binding to free Fe generates highly reactive oxygen which can damage DNA and other macromolecules.

How does the structure of hemoglobin resolve this issue?

A
  • Fe is incorporated into a cyclic organic ring structure, protoporphyrin, to generate the prosthetic group, heme. Fe is in the ferrous (Fe+2) state.
    • Fe+2 has six coordination bonds, four to nitrogen atoms that are part of the porphyrin and two perpendicular to the porphyrin ring.
  • The four nitrogen atoms prevent the conversion of ferrous iron to ferric (Fe+3) state, because Fe+3 does not bind to oxygen.
  • Heme is sequestered deep in the protein matrix to prevent the binding of oxygen (O2) to two heme molecules which can result in irreversible conversion of Fe+2 to Fe+3.
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6
Q

What are the axial ligands to Fe?

A
  • One of the axial ligands to Fe is a sidechain nitrogen of a His residue.
  • The other is the binding site for molecular oxygen (O2).
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7
Q

What are the two conformations of Hb?

A

T and R states which have low and high affinity towards O2 binding, respectively.

  • When O2 is absent, Hb exists in T state.
  • O2 binding triggers a conformational change from T to R state.
  • In R state, the structures of the individual subunits change little, but the αß subunit pairs slide past each other and rotate, narrowing the pocket between the ß subunits.
  • In this process, some of the ion pairs that stabilize the T state are broken and some new ones are formed.
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8
Q

A protein that binds to O2 with high affinity would bind it efficiently in the lungs but would not release much of it in the tissues. If the protein binds to O2 with low affinity to release it in the tissues, it would not pick up much oxygen in the lungs.

How does hemoglobin overcome this problem?

A
  • Hb must bind O2 effectively in the lungs, where pO2 is about 13.3 kPa.
  • Hb must release oxygen in the tissues, where the pO2 is about 4 kPa.
  • Hb solves this problem by undergoing a transition from a low-affinity T state to a high-affinity R state as more O2 molecules are bound.

pO2 represents the partial pressure of oxygen.

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9
Q

Write the equation that can best describe the binding relationship between hemoglobin and O2.

A
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10
Q

Explain how O2 binds cooperatively with hemoglobin.

A

O2 binds to heme in the low affinity T-state, leading to conformational changes that are communicated to adjacent subunits, making it easier for additional molecules of O2 to bind. In effect, the T to R transition occurs more readily in the second subunit once O2 is bound to the first subunit. That means, individual subunits function cooperatively.

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11
Q

allosteric protein

A

An allosteric protein is one in which the binding of a ligand to one site affects the binding properties of another site in the same protein. Hb is an example of allosteric protein.

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12
Q

Draw and describe a Hill plot of hemoglobin in its low- and high-affinity state.

A

Slope of the curve is known as the Hill coefficient nH.

  • nH > 1 indicates positive cooperativity, which means the ligand binding at one site promotes the ligand binding at other site.
  • nH = 1 means no cooperativity, which means ligand binding is not cooperative.
  • nH < 1 indicates negative cooperativity, in which the binding of one ligand impedes the binding of others.
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13
Q

Why is exposure to carbon monoxide dangerous?

A
  • Carbon monoxide (CO) is responsible for more than half of yearly deaths due to poisoning worldwide.
  • CO has 250-fold greater affinity for Hb than does O2. Because of this, Hb gets trapped in high-affinity state and hence cannot release the bound O2 in tissues.
  • 15% COHb result in headaches whereas 50% COHb leads to coma.
  • When CO poisoning is suspected, rapid evacuation away from the CO source is essential.
    • However, COHb levels drop slowly (2-6.5 hrs) because CO binds to high-affinity state of Hb.
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14
Q

Hb transports about 40% of the total H+ and 15% to 20% of the CO2 (two end products of cellular respiration) formed in the tissues to the lungs and kidneys.

Describe how Hb transports H+ and CO2.

A
  • The binding of H+ and CO2 to Hb is inversely related to the binding of oxygen.
  • At the relatively low pH (high H+) and high CO2 concentration of peripheral tissues, the affinity of Hb for oxygen decreases as H+ and CO2 are bound, and O2 is released to the tissues.
  • Conversely, in the capillaries of the lung, as CO2 is excreted and the blood pH consequently rises, the affinity of Hb for oxygen increases and the protein binds more O2 for transport to the peripheral tissues.
  • This effect of pH and CO2 concentration on the binding and release of oxygen by Hb is called the Bohr effect.
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15
Q

Explain the Bohr effect as it relates to Hb.

A

HHb+ + O2 (tissues) ⇔ HbO2 + H+ (lungs)

  • pH affects O2 binding to Hb
  • The pH of blood is 7.6 (low H+) in the lungs and 7.2 (high H+) in the tissues.
  • Oxygen and H+ are not bound at the same sites in hemoglobin.
    • Oxygen binds to the iron atoms of the hemes
    • H+ binds to any of several amino acid residues in the protein.
  • When protonated, these amino acids form new ion pair interactions that stabilize the T state of deoxyhemoglobin.
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16
Q

antibodies or immunoglobulins (Ig)

A

Antibodies or Immunoglobulins (Ig) bind to bacteria, viruses, or large foreign molecules (antigens) and target them for destruction.

Antibodies make up 20% of the blood protein. Humans are capable of producing more than 100 million antibodies with distinct binding specificities.

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17
Q

antigen

A

Any molecule or pathogen capable of eliciting an immune response is called an antigen.

18
Q

How many classes of antibodies are there?

A

There are five classes of antibodies:

  1. IgG - major class of antibody, which is an abundant potein in blood serum
  2. IgA - found in saliva, tears, and mother’s milk
  3. IgD
  4. IgE
  5. IgM
19
Q

monoclonal antibodies

A
  • Most newly approved innovator drugs are mAbs.
  • mAbs are generally very safe drugs because of their target selectivity, thus avoiding unnecessary exposure to and consequently activity in nontarget organs.
  • The reduction of the xenogenic portion of the mAb structure decreased the immunogenic potential of the murine mAbs, thus allowing their wider application.
20
Q

Describe the nomenclature of monoclonal antibodies.

A
21
Q

Describe the IgG structure.

A
  • IgG contains four polypeptide chains: two large ones called heavy chains and two light chains, linked by noncovalent and disulfide bonds.
  • The heavy chains of IgG interact at one end, and then branch to interact separately with the light chains, forming a Y-shaped molecule. Each branch has a single antigen-binding site.
  • Cleaving IgG with a protease papain result in three fragments: Fc because it crystallizes readily, and two Fab’s, the antigen-binding fragments.
  • Heavy and light chains are made up of identifiable domains: constant domains with sequence and structure constant between IgG’s, and variable domains with varying amino acid sequences.
22
Q

How do carbohydrates interact with IgG?

A

Carbohydrates do not affect the IgG structure and function, but stabilize the IgG structure.

23
Q

The variable domains of antibodies associate to create two antigen-binding sites.

Describe binding of antigen to these antigen-binding sites.

Why are these sites referred to as hypervariable?

A

Each antigen-binding site binds only a particular molecular structure within the antigen, called antigenic determinant or epitope. Upon binding, the binding site undergoes conformational changes.

Antibodies bind tightly and specifically to antigen. The binding specificity is determined by the amino acid sequences in the variable domains, which are hypervariable.

24
Q

Describe how IgG interacts with macrophages.

A
  • When IgG binds to an antigen, it activates macrophages to engulf and destroy the invader.
  • Receptors on the macrophage surface recognize and bind the Fc region of IgG.
25
Q

One serious problem with the folding of large proteins is protein aggregation.

What proteins prevent protein aggregation during folding?

A
  • Molecular chaperones or chaperonins to interact with partially folded or improperly folded polypeptides, facilitating correct folded pathways or providing microenvironments in which folding can occur.
  • These proteins bind to regions that are rich in hydrophobic residues, preventing inappropriate aggregation.
  • They do not actively promote the folding of the substrate protein, but instead prevent aggregation of unfolded peptides.
    • They facilitate correct folding pathways or provide microenvironments in which folding can occur.
26
Q

What is the difference between chaperones and chaperonins?

A

Difference between chaperones and chaperonins is that chaperonins provide a cavity for the folding of an isolated substrate.

27
Q

GroEL/GroES system

A
  • Both GroEL and GroES are examples for quaternary structures. More than one polypeptide come together to form the protein structure.
    • GroEL (bottle)
      • Total 14 polypeptides
      • Double donut structure with 7 + 7 polypeptide chains
    • GroES (cap)
      • 7 polypeptide chains
28
Q

Explain the mechanism of GroEL/GroES.

A

Step 1: 7 ATPs bind to the top GroEL ring.

Step 2: Unfolded protein enters the top ring—conformation change

Step 3: GroES departs off the bottom ring, followed by the folded protein and 7 ADPs. GroES binds to the top ring triggering a massive twist inside the GroEL cavity.

Step 4: Unfolded protein dissociates from the inner wall into the cavity. It then has a chance to fold during the 10 seconds in which the high-energy ATP bonds are breaking, converting ATP to ADP.

Step 5: ATP to ADP conversion loosens GroEL’s grip on GroES and primes the bottom GroEL ring to accept ATP.

Steps 6-8: ATP binding restarts the cycle, this time on the bottom ring, with a new unfolded protein.

29
Q

Why is the GroEL/ES system called the “two-stroke machine?”

A

During one cycle, GroEL/ES system is able to fold two substrate molecules, which is why it is called the two-stroke machine.

30
Q

What are the three domains of GroEL?

A
  • GroEL contains three domains: apical, intermediate, equatorial.
31
Q

Upon GroES binding, a large conformational change occurs.

A

Non-native peptide binds to the hydrophobic patches in the apical domain. Upon ATP and GroES binding, the large conformation transition occurs that frees the non-native peptide.

The hydrophobic sites where substrate (red) binds move up. Green colors represent the GroES binding sites.

32
Q

Describe the allosteric states of GroEL.

A
  • T state: low affinity for ATP
  • R state: high affinity for ATP
  • ATP hydrolysis for the concerted intra-ring T to R step:
    • Positive cooperativity with Hill coefficient of 1.86 – 2.75
  • ATP hydrolysis for the inter-ring step:
    • Negative cooperativity with Hill coefficient of 0.003
33
Q

Describe the allosteric domain movements in GroEL.

A

Negative cooperativity between the two rings is communicated by the interacting residues in the equatorial domains.

34
Q

What two proteins generate the contractile force of muscle?

A

Myosin and actin.

  • These proteins are arranged in filaments that undergo transient interactions and slide past each other to bring about contraction.
    • The chemical energy is derived from ATP hydrolysis.
  • Together, actin and myosin make up more than 80% of the protein mass of muscle.
35
Q

Describe the structure of myosin.

A
  • Myosin has six subunits: two heavy chains and four light chains.
  • Heavy chains are arranged at their C-terminus as extended a-helices, wrapped around each other in a fibrous, left-handed coiled coil. At its N-terminus, each chain has a large globular domain containing an ATP hydrolysis site.
  • The light chains are associated with the globular domains of heavy chains.
  • The globular domain or the myosin head group is the motor domain that makes muscle contraction possible.
36
Q

Describe thick filaments.

A
  • In muscle cells, myosin aggregates to form structures called thick filaments.
  • Within a thick element, several hundred myosin molecules are arranged with their fibrous tails associated to form a long bipolar structure.
  • The globular domains project from either end of this structure, in regular stacked arrays.
37
Q

Describe the two different domains of actin.

A
  • The second muscle protein, globular actin (G-actin), (Mr 42,000) associates to form filamentous actin (F-actin).
  • Every G-actin monomer binds ATP and hydrolyzes it to ADP, which helps in the assembly of the filaments.
    • This ATP hydrolysis does not contribute directly to the energy expended in muscle contraction.
  • F-actin together with other proteins form thin filaments.
38
Q

Describe how actin and myosin interact together, structurally.

A

Each actin monomer in the thin filament can bind tightly and specifically to one myosin head group.

39
Q

What are the four steps of myosin and actin interaction that creates muscle contractions?

A
  1. ATP binds to myosin head, causing dissociation from actin.
  2. As tightly bound ATP is hydrolyzed, a conformational change occurs. ADP and Pi remain associated with the myosin head.
  3. Myosin head attaches to actin filament, causing release of Pi.
  4. Pi release triggers a “power stroke,” a conformational change in the myosin head that moves actin and myosin filaments relative to one another. ADP is released in the process.
40
Q

Describe the A, I, and Z disk and their interactions during muscle contraction.

A
  • A and I bands arise from the thick and thin filaments.
  • Z disk, made up of other proteins, is the structure serving as an anchor to which the thin filaments are attached.
  • Muscle contraction occurs by the sliding of the thick and thin filaments past each other so that the Z disks in neighboring I bands draw closer together.
41
Q

Motor proteins convert ________ energy into ________ energy.

A

chemical; kinetic