Chapter 5 Biochemistry

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1

The interactions of ligands with proteins:

are usually transient.

2

A prosthetic group of a protein is a non-protein structure that is:

permanently associated with the protein.

3

When oxygen binds to a heme-containing protein, the two open coordination bonds of Fe2+ areoccupied by:

one O2 molecule and one amino acid atom.

4

In the binding of oxygen to myoglobin, the relationship between the concentration of oxygen and thefraction of binding sites occupied can best be described as:

hyperbolic.

5

Which of the following statements about protein-ligand binding is correct?

The larger the Ka, the smaller the Kd (dissociation constant).

6

Myoglobin and the subunits of hemoglobin have:

very similar tertiary structures, but different primary structures.

7

An allosteric interaction between a ligand and a protein is one in which:

binding of a molecule to a binding site affects binding properties of another site on the protein.

8

In hemoglobin, the transition from T state to R state (low to high affinity) is triggered by:

oxygen binding.

9

Which of the following is not correct concerning 2,3-bisphosphoglycerate (BPG)?

It increases the affinity of hemoglobin for oxygen.

10

Which of the following is not correct concerning cooperative binding of a ligand to a protein?

It rarely occurs in enzymes.

11

The amino acid substitution of Val for Glu in Hemoglobin S results in aggregation of the proteinbecause of ___________ interactions between molecules.

hydrophobic

12

The fundamental cause of sickle-cell disease is a change in the structure of:

hemoglobin.

13

An individual molecular structure within an antigen to which an individual antibody binds is as a(n):

epitope

14

The proteins of the Major Histocompatibility Complex (MHC) bind and display:

antigen fragments.

15

Which of the following parts of the IgG molecule are not involved in binding to an antigen?

Fc

16

A monoclonal antibody differs from a polyclonal antibody in that monoclonal antibodies:

are synthesized by a population of identical, or “cloned,” cells.

17

Which of the following generalizations concerning motor proteins is correct?

They convert chemical energy into kinetic energy.

18

The predominant structural feature in myosin molecules is:

an α helix

19

The energy that is released by the hydrolysis of ATP by actin is used for:

actin filament assembly.

20

During muscle contraction, hydrolysis of ATP results in a change in the:

conformation of myosin.

21

Describe the concept of “induced fit” in ligand-protein binding.

Induced fit refers to the structural adaptations that occur when a ligand binds to a protein. This often involves a conformational change in the protein that alters the binding site to make it more complementary to the ligand.

22

Explain why most multicellular organisms use an iron-containing protein for oxygen binding ratherthan free Fe2+. Your answer should include an explanation of (a) the role of heme and (b) the role ofthe protein itself.

(a) Binding of free Fe2+ to oxygen would result in the formation of reactive oxygen species thatcan damage biological structures. Heme-bound iron is less reactive in this regard.

(b) Binding ofoxygen to free heme can result in irreversible oxidation of the Fe2+ to Fe3+ that does not bind oxygen.The environment of the heme group in proteins helps to prevent this from occurring.

23

Why is carbon monoxide (CO) toxic to aerobic organisms?

It binds to heme with a higher affinity than oxygen, and thus prevents oxygen from binding tohemoglobin.

24

Describe how you would determine the Ka (association constant) for a ligand and a protein.

An experiment would be carried out in which a fixed amount of the protein is incubated withvarying amounts of ligand (long enough to reach equilibrium). The fraction of protein molecules thathave a molecule of ligand bound is then determined. A plot of this fraction (θ) vs. ligandconcentration [L] should yield a hyperbola. The value of [L] when θ = 0.5 is equal to 1/Ka.

25

For the binding of a ligand to a protein, what is the relationship between the Ka (association constant),the Kd (dissociation constant), and the affinity of the protein for the ligand?

Ka = 1/Kd. The larger the Ka (and hence the smaller the Kd), the higher the affinity of the proteinfor the ligand.

26

Explain briefly why the relative affinity of heme for oxygen and carbon monoxide is changed by thepresence of the myoglobin protein.

The geometry of binding O2 and CO to heme is slightly different. In myoglobin there is a histidine residue that does not interact with the heme iron, but can interact with a ligand that is boundto the heme. It does not affect O2 binding but because of steric hindrance, it may prevent CO binding.As a result the relative affinity of protein-bound heme for CO and O2 is only 200, compared to 20,000for free heme.

27

Explain why the structure of myoglobin makes it function well as an oxygen-storage protein whereasthe structure of hemoglobin makes it function well as an oxygen-transport protein.

The hyperbolic binding of oxygen to the single binding site of myoglobin results in a highaffinity even at the relatively low partial pressures of O2 that occur in tissues. In contrast, thecooperative (sigmoidal) binding of O2 to the multiple binding sites of hemoglobin results in highaffinity at high partial pressures such as occur in the lungs, but lower affinity in the tissues. Thispermits hemoglobin to bind O2 in the lungs and release it in the tissues.

28

Describe briefly the two principal models for the cooperative binding of ligands to proteins withmultiple binding sites

In the concerted model, binding of a ligand to one site on one subunit results in an allostericeffect that converts all of the remaining subunits to the high-affinity conformation. As a result, all ofthe subunits are either in the low- or high-affinity conformation. In the sequential model, each subunit is changed individually to the high affinity conformation. As a result, there are many possiblecombinations of low- and high-affinity subunits.

29

How does BPG binding to hemoglobin decrease its affinity for oxygen?

BPG binds to a cavity between the β subunits. It binds preferentially to molecules in the low-affinity T state, thereby stabilizing that conformation

30

a) What is the effect of pH on the binding of oxygen to hemoglobin (the Bohr Effect)?
(b) Brieflydescribe the mechanism of this effect.

(a) The affinity decreases with decreasing pH.

(b) At lower pH (i.e., higher H+ concentration)there is increasing protonation of protein residues such as histidine, which stabilizes the low affinityconformation of the protein subunits.

31

Explain how the effects of sickle cell disease demonstrate that hemoblobin undergoes aconformational change upon releasing oxygen.

In Hemoglobin S, the wild-type glutamate at residue 6 of the B-chain is replaced by valine.When oxygen is bound, both Hemoglobin A and Hemoglobin S are soluble, but in the deoxy- form.Hemoglobin S (but not Hemoglobin A) becomes very insoluble, due to exposure of the hydrophobicvaline residue. This exposed “patch” causes aggregation of deoxy-Hemoglobin S into long insolublefibrous aggregates, resulting in distorted shapes of the red blood cells (and leading to the symptoms ofthe disease). (See p. 173 and Fig. 5-20.)

32

Why is it likely that the immune system can produce a specific antibody that can recognize and bindto any specific chemical structure?

As a result of genetic recombination mechanisms, antibody-producing B cells are capable ofproducing millions of different antibodies with different binding specificities.

33

What is the role of the Major Histocompatibility Complex (MHC) in the immune response?

MHC proteins are present on the surface of specialized immune system cells. Antigenfragments that are derived from external proteins are bound to the MHC proteins and elicit an immune response. A selection process eliminates those cells with MHC complexes that might bindnormal cellular proteins, leaving only those that can bind foreign proteins. Thus the MHC plays arole in the ability of the immune system to discriminate between self and nonself.

34

Describe briefly the basic structure of an IgG protein molecule.

An IgG protein contains two copies of a large polypeptide (heavy chain) and two copies of asmall polypeptide (light chain). β structure contributes significantly to the tertiary structure of domains of both chains. Disulfide bonds link the heavy chains to one another and to the light chains.The chains are arranged in a Y-shaped structure where the two arms are linked to the base by aprotease sensitive (“hinge”) region.

35

Describe the cycle of actin-myosin association and disassociation that leads to muscle contraction.

First, ATP binds to myosin and a cleft in the myosin molecule opens, disrupting the actin-myosin interaction so that the bound actin is released. Second, ATP is hydrolyzed, causing aconformational change in the protein to a “high-energy” state that moves the myosin head and changes its orientation in relation to the actin thin filament. Myosin then binds weakly to an F-actinsubunit closer to the Z disk than the one just released. Third, as the phosphate product of ATP hydrolysis is released from myosin, another conformational change occurs in which the myosin cleftcloses, strengthening the myosin-actin binding. Fourth, this is followed quickly by a “power stroke”during which the conformation of the myosin head returns to the original resting state, its orientationrelative to the bound actin changing so as to pull the tail of the myosin toward the Z disk. ADP isthen released to complete the cycle. (See Fig. 5-33, p. 186.)

36

What is the role of ATP and ATP hydrolysis in the cycle of actin-myosin association anddisassociation that leads to muscle contraction?

ATP binding to myosin results in a conformational change that causes dissociation of actin fromthe myosin. ATP hydrolysis results in a change of orientation of the myosin relative to the actin filament, which allows movement to the next actin subunit. This is followed initially by release of thephosphate hydrolysis product and weak binding of the myosin to this actin subunit, and,subsequently, by tight binding and release of the ADP hydrolysis product.

37

What is the relationship between G-actin and F-actin?

G-actin is a monomeric protein that can polymerize to form a long polymeric filament known asF-actin.

38

What is the chemical basis for the specificity of binding of an immunoglobin antibody to a particularantigen?

Specific binding results from complementarity between the chemical properties (such as size,charge, and hydrophobicity) of the antigen and the antigen-binding site of the antibody.

39

What is the concept of “induced fit” as it applies to antigen-antibody binding?

The conformations of the antigen and antigen-binding site of the antibody are influenced byeach other and change as binding occurs. These conformational changes increase the chemicalcomplementarity of the sites and result in tighter binding.

40

Describe how immunoaffinity chromatography is performed.

The specific antibody is covalently attached to an inert supporting material, which is thenpacked into a chromatography column. The protein solution is passed through the column slowly;most proteins pass directly through, but those for which the antibody has specific affinity are adsorbed. They can subsequently be eluted by a buffer of low pH, a salt solution, or some other agentthat breaks the antibody-antigen association.

41

What properties of antibodies make them useful biochemical reagents? Describe one biochemicalapplication of antibodies (with more than just the name of the technique).

The important properties are the high specificity of protein recognition, and the high affinity ofthe antibody-antigen association. These make possible immunoaffinity chromatography, immunocytochemistry, enzyme-linked immunosorbent assay (ELISA), and immunoblotting, all ofwhich are described on pp. 181−182.

42

Describe briefly the structure of myosin.

Myosin contains two copies of a large polypeptide (heavy chain) and four copies of a smallpolypeptide (light chain). The α helix contributes significantly to the structure of the heavy chains.At their carboxyl termini, the heavy chains are wrapped around each other in a fibrous left-handedcoil. At their amino termini, they each have a globular domain with which the light chains areassociated.