BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
9th Edition
ISBN: 9781319425784
Author: BERG
Publisher: Macmillan Higher Education
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Chapter 7, Problem 14P
Interpretation Introduction

(a)

Interpretation:

An oxygen-binding curve for a hypothetical two subunit hemoglobin with n=1.8 and P50=10 torr should be plotted.

Concept introduction:

Hill equation is represented as follows:

log(Y1Y)=nlog(pO2)nlog(P50)

Here, Y- fractional saturation

n − a measure of the degree of cooperativity in ligand binding.

P50 − partial pressure of oxygen at which hemoglobin is half saturated.

The Hill plot of log(Y/1-Y) versus log(P50) should be a linear graph with slope of n.

Expert Solution
Check Mark

Answer to Problem 14P

BIOCHEMISTRY (LOOSELEAF)-W/ACCESS      , Chapter 7, Problem 14P , additional homework tip  1

Explanation of Solution

The values of n=1.8 and P50=10 torr are substituted to the Hill equation and the partial pressure of oxygen at different fractional saturation values is calculated. A plot of fractional saturation versus oxygen partial pressure is drawn.

Y log(Y/(1-Y) log(Y/(1-Y)-n log (P50) log (pO2) pO2
0.1-0.954242510.8457574910.4698652.950294
0.2-0.602059991.1979400090.6655224.629374
0.3-0.367976791.4320232150.7955686.245518
0.4-0.176091261.6239087410.9021727.983099
0.501.8110
0.60.1760912591.9760912591.09782812.52646
0.70.3679767852.1679767851.20443216.01148
0.80.6020599912.4020599911.33447821.60119
0.90.9542425092.7542425091.53013533.89493

BIOCHEMISTRY (LOOSELEAF)-W/ACCESS      , Chapter 7, Problem 14P , additional homework tip  2

Interpretation Introduction

(b)

Interpretation:

An oxygen-binding curve for shypothetical two subunit hemoglobin with n=2, L = 1000, c = 0.01 and KR=1 torr should be plotted using concerted model

Concept introduction:

Concerted model equation is written as follows:

Y=α( 1+α)n1+Lcα( 1+cα)n1( 1+α)n+L( 1+cα)n

Here, Y − fractional saturation

a − ratio between the substrate concentration and the dissociation constant for a ligand binding to a single site in R state.

L − The ratio of the concentrations of the T and R states with no ligands bound

c − the ratio between the dissociation constant for a ligand binding to a single site in R state and that of T state.

n − number of binding sites

Also, the ratio can be calculated as follows:

α=pO2KR

Here,

pO2 − partial pressure of oxygen

KR - the dissociation constant for a ligand binding to a single site in R state.

Expert Solution
Check Mark

Answer to Problem 14P

BIOCHEMISTRY (LOOSELEAF)-W/ACCESS      , Chapter 7, Problem 14P , additional homework tip  3

Explanation of Solution

Y=α ( 1+α ) 21+1000×0.01α ( 1+0.01α ) 21 ( 1+α )2+1000 ( 1+0.01α )2Y=α+α2+10α+0.1α21+2α+α2+1000+20α+0.1α2Y=11α+1.1α21001+22α+1.1α2

When Y=0.1

0.1=11α+1.1α21001+22α+1.1α211α+1.1α2=100.1+2.2α+0.11α20.99α2+8.8α100.1=0α=6.55

α=pO2KR6.55=pO21pO2=6.55 torr

Likewise, partial pressure of oxygen for several fractional saturation values are calculated and a plot of fractional saturation versus oxygen partial pressure is drawn.

Y pO2
0.16.55
0.317.09
0.530.15
0.753.21
0.9138.91

BIOCHEMISTRY (LOOSELEAF)-W/ACCESS      , Chapter 7, Problem 14P , additional homework tip  4

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