Biochem protein function ch 5

statement true about Hb and Mb

- Fe+ heme prosthetic grp bound at five and six sites to nitrogen atoms

humoral immune sys

- bact. infection -extracellular viruses -recognition proteins immuoglobulins -major cell type B

-Binding of O2 is inversely related -at low pH [H+] is high and [CO] high --at high pH [H+] is low and [CO] low

- binding of H+and CO2 -Hb affinity for O2 low (T-state) --Hb affinity for O2 high (R-state)

chemical transformations occur, ligand-binding site is called

- catalytic site or active site. -ligands =substrate

When a vertebrate dies, its muscles stiffen as they are deprived of ATP, a state called rigor mortis. -Explain the molecular basis of the rigor state

- the absence of ATP, actin and myosin bind tightly to ea. other

-H+binds to aa residues and -- CO2 binds to the

-(His 146) --amino terminal end of Hb polypeptide chain.

Hb bind to

-4 O2 , -multi subunint -Two achains (141 residues each) -two bchains (146 residues each) -Hb (64,500 kDa)

rxn of CO2 with H2O contribute to Bohr's effect

-H+ decreases pH stabilizing deoxy Hb -H+ lowers affinityfo Hb for O2,releases more O2 -CO2 dissolved forms carbaminohemoglobin ,release O2

Hb transports

-H+and CO2 lungs and kidneys excreted -both end products cellular respiration

help transport and store O2 ,classic examples.

-Myoglobin (153 aa; ~ 17 kDa; Mb) - Hemoglobin (Hb) Hb (64,500 kDa)

Protein A has a binding site for ligand X with a Kd of 10-6M ,Protein B has a binding site for ligand X with a Ka of 10^-9M. Which protein has a hlgher affinity for ligand X? Explain your reasoning Convert the Kd to Ka. for both proteins

-Protein B has higher affinity for ligand X. -the lower Kd indicates that protein B will be half saturated with bond ligand X at much lower [ ] of X than will protein A. -Ka=1/Kd , protein A has Ka =10^6M-1 , protein B 10^9M-1.

Hb structure

-Quaternary structure -globing family -has symmetry axis -identity only 27 positons

structural adjustments between binding partners referred

-as induced fit -adaptation that occurs between protein and Iigand

deoxy forms of Hb is stabilze by

-binding of CO2 -BPG

binding of O2 to Hb subunit in T state triggers

-change in conformation to R state -likewise R to T is favored when O2 leaves R

Induced fit allows for

-complementary groups to interact fully -Antibody-antigen interactions strong (Kd~ 10-10M) -Extraordinary specificity and affinity

What is the dissociation constant, Kd, for each protein? Which protein (X or Y) has a greater affinity for ligand A?

-constant.For X, Kd : 2 ; for Y ,Kd = 6 --Because X is half-saturated at a lower [A], it has a higher affinity for the ligand.

cellular immune system

-destroy host cell -destroy parasites -major cell type Tc and Th(helpers)

Hb binds O2 cooperatively

-first O2binds to T state weakly, the second one binds more tightly, followed by third and fourth with higher affinity

Hb S (Hb from sickle cells) is

-insoluble and forms polymers aggregate into tubular fibers. -O2 transport hamper

-T state -- R state

-less affinity for O2 you release the ball from your hand --you retain the ball from your hand , high affinity for O2

-the bigger the Kd --the smaller the Kd

-less affinity, --more affinity and tighter the binding

ligandis

-molecule bound reversibly to a protein. - Ligand can be a water molecule, prosthetic group or another protein.

Mb is found primarily in

-muscle tissue mammals. -Similar proteins distributed including single-celled organisms. -belongs globins family similar primary and tertiary structures -single polypeptide chain w prosthetic grp=heme

Sickle-cell anemia

-mutation in gene encodes Hb -found homozygous individuals -

-first order reactioninvolves --second orderreaction involves

-only one molecule (kd; s-1) , reverse rxn -- two molecules (ka; M-1s-1) fwd rxn

Binding affinity of ligands to heme varies from its

-protein-bound state to protein-unbound state. --Conformational changes --Steric clashes ,not favorable for linear CO/forced at angle Optimal geometry, felicitated by H-bond explain for the difference

-protoporphyrin --Heme grps

-proto are the ch2 substitutions -porphyrin is the ring --Fe2+ has six co-ordination bonds --His side chain and O2 fulfill the other bonds. --It can also bind to CO excluding O2. ---CO is highly toxic,CO compete with O2

Antibodies

-target specific chemical structure

statements true concerning structure of Mb an Hb

-teritiary structure Mb similar to Hb -Mb one binding site for O2 /molecule -Mb/Hb one binding site O2/heme

What is the effect of the following changes on the O2 affinity of Hb? (a) A drop in the pH of blood plasma (b) A decrease in the partial pressure of CO2 in the lungs (c) An increase in the BPG (d) An increase in CO

-the affinity of Hb for O2 is regulated by binding of ligands H+, CO2 and BPG. a=decrease affinity b=increase affinity c=decrease affinity d=decrease affinity

The isolated alpha subunit binds oxygen,but the O2-saturation curve is hyperbolic rather than sigmoid. binding oxygen to the isolated a subunit is not affected by the presence of H+, CO2,or BPG. What do these observation indicate about the source of the cooperative in hemoglobin?

-the sigmoid O2 binding curve and pos cooperative in ligand binding -of Hb arises form interaction btw subunits.

Kd is the concentration at

-which half of the available binding sites are occupied - reciprical of ka or 1/Ka -kd =theta

Immunoglobulin G

0(150 kDa) major class antibody -abundant blood serum

muscle contraction

1.atp binds to myosin 2.atp hydrolyzed head fo myosin 3.myosin head rebinds 4.power stroke , filament moves

regulation of muscle contraction

1.nerve impulse, Ca+ release 2.Ca+ binds troponin 3.formation complex btw Ca+ troponin 4. tropomyosin chngs pos. 5. head mysosin bind actins

Ig are about 20% of blood protein and are produced by

B cells

distal histidine bind oxygen with

H-bonds

2,3-BPG also regulates

O2 binding by Hb

ligand binds to protein at specific and complementary binding site what must be matched before binding ?

Shape, size, charge, and polar or non-polar characteristics

helper T cells(TH) coordinate in the proliferation of both

T and B cells fight the pathogen.

rate constant talk about the speed , so Ka is

affinity ligand for protein (association constant Pix)

Approximately 78% of the aa residues

are in a-helices.

HIV (human immunodeficiency virus) that causes AIDS

attacks TH cells.

Hb is an allosteric protein

binding ligand Affects binding of ligands other sites on same protein

release O2 promoted from Hb TO THE TISSUES needed by

binding of H+ and CO2

protein flexibility is necessary for

bio function

allosteric interaction affect

conformation and fnc of protein

Hb also binds

excess H+ and CO2 produce by respiratory tissues

heme binds oxygen atoms of CO

false, bind C tripple bonds

Rate constants refer to

fraction of pool reactants react in given amount time

How Oxygen is bound to Mb or Hb

iron bing to O2 attaches to prosthetic grp= heme

in enzymes, where chemical transformations occur,

ligands and the binding site are called substratesand

effect of BPG on O2

lowers O2 affinity , stabilizing T-state

Monoclonal antibodies-

made by identical B cells, recognize same epitope of antigen

Myosin and Actin(80% of protein mass of muscle) each other resulting in

muscular contraction

Hb is hold together by

non-covalent bonds, H-bond, H.phobic , ions

Polyclonal antibodies-

produced by different B cells respond antigen recognize antigen or protein

antigenic epitope

recognize by antigen

Hb bind O2 efficiently lungs and

release or exhibit low affinity for O2 in the tissues

Gaseous oxygen is sparingly soluble in aqueous solutions,Therefore, tissues will not be able to receive

sufficient supply O2 in blood serum. Molecular evolution designed proteins help transport and store O2.

protein and ligands undergo conformational changes resulting in

tighter binding

free heme binds CO with Fe+, C, O, in linear array

true

proximal histidine bind iron

true

An antibody binds to an antigen with a Kd of 5 x 10-8 M At what concentration of antigen will theta be (a) 0.2, ( b) 0.5, ( c) 0.6, (d) 0.8

use formula [L]=theta*Kd/(1-theta) divide first then multiply the theta a) (.2)(5 x 10-8) /(1-.2)= b) (.5)(5 x 10-8) /(1-.5)= c) (.6)(5 x 10-8) /(1-.6)=