Wednesday 10 5 05

Lecture 10/5/05
• The heme ligand binds reversible to proteins. Heme binds so tightly and irreversible.
• Iron is the core that does the binding. It is Iron 2+, so it is in ionic form.
• Iron 2+ is soluble, which we see as rust, so globin is used to keep Iron 2+ in aqueous solution. Helps prevent hydrogen-peroxide from binding to the iron and oxygen.
• Oxygen + Iron + Oxygen is an oxidation reduction reaction. That is why in heme; only one iron and one oxygen are bound together.
• Figure 5-2 Histidine takes one site that lies deep in the pocket so that no more hemes can enter.
• Table 5-1 Reaction between proteins and ligand has a dissociation constant based on binding differences or biological conditions.
• Overhead
Kd = [P] x [L] for [PL] -> [P] +[L]
[PL] <-

[PL] + [P] =[Pt] t=total

Kd = ( [Pt] – [PL] ) x [L]
PL

Define θ = [PL] = fxn. Ligand bound to protein
[Pt]

Kd= ( {[PL]/θ }- [PL] ) x [L]
[PL]

Kd = ([PL]x[L])/θ - [PL]x[L]
[PL] [PL]

multiply both sides by θ
Kd θ = [L] – [L] θ

Kd θ + [L]θ = [L]

θ = [L]
(Kd + [L])

• This represents what fraction of protein is bound to legend.
• This math can also be found on page 160. He don’s not like the use of Ka because we association that with acid/base things, do he used Kd, d is for dissociation.
• Overhead take the double reciprocal
1 = Kd + [L]
θ [L]
1 = Kd (1/[L]) + 1
θ
of the form
1/y proportional 1/x => rectangular hyperbola
• Figure 5-4 keeping pressure constant, P50 partial pressure to bind to 50%
• Special case when [L]=Kd gives ½
• Kd has units of molarity
• Figure 5-5 a and b Free heme is more likely to bind to CO (25,000 more times likely) than O2. Also more tightly because they can line up collinear.
• Figure 5-5 c Oxygen bound to y-ligand position. Heme is asymmetrical. In all proteins the x is occupied. The y position is closed but do not form with Fe.
• O and Fe(iron) are at an angel because histidine is in the way making it so it can not be collinear. If they could line up collinear than the bond would be too strong, which would not be a good thing.
• Figure 5-7 comparison of myoglobin and alpha and beta hemoglobin. 27 conserved (meaning there are 27 amino acids in the same position) between the two of them. These must be important to binding heme.
• Biochem. is taught structure -> function. Function is more important.
• Overhead
[Mb x O2] = θ Fraction of Mb in a complex
[ Mb]t

[Mb]f = 1 – θ Fraction of Mb free
[Mb]t

and
θ = [O2]/ (Kd + [CO2])
1-θ Kd/( Kd + [O2])

θ = [O2]
1-θ Kd
Equation holds for 1 to 1 complexes, like myoglobin

Hemoglobin
Hb + nO2 <=> [Hb x nO2]

Hb x nO2 = [O2]^n
[Hb]t Kp+[O2]^n
n is the number of equivalence that interact with that protein

[Hb]f = Kp
[Hb]t Kp[O2]^n
θ = [O2]^n
1-θ Kd

Taking the log of both sides

Log {θ/(1-θ)} = log[O2]^n –log Kd

= n log [O2] – log Kd
• Figure 5-14 Myoglobin is a straight line while Hemoglobin is not. This is because there is another variable; Hemoglobin was changing while picking up oxygen.
• Figure 5-12 the s shape of the graph confirms that something is changing in the hemoglobin.