Principles of Biology
Principles of Biology
2nd Edition
ISBN: 9781259875120
Author: Robert Brooker, Eric P. Widmaier Dr., Linda Graham Dr. Ph.D., Peter Stiling Dr. Ph.D.
Publisher: McGraw-Hill Education
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Chapter 36.9, Problem 1CC
Summary Introduction

To discuss:

The change in the oxygen-hemoglobin dissociation curve when the person is infused with HCO3- at resting position.

Introduction:

Hemoglobin brings oxygen from the lungs to the tissues of the body in red blood cells and returns carbon dioxide to the lungs from the tissues. Hemoglobin consists of four connected protein molecules (globulin chains).

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Blood pressure is usually given as the ratio of the maximum pressure (systolic pressure) to the minimum pressure (diastolic pressure). For example a typical value for this ratio for a human would be 120/70, where the pressures are in mm Hg (millimeters of mercury, SG = 13.6). What would these pressures be in kPa (kilo Pascals)?: (a) systolic pressure (kPa), (b) diastolic pressure (kPa).
Using the normal arterial blood gas values (pH = 7.35-7.45; Paco2 = 35-45 mm Hg; HCO3- = 22-26 mEq/L), identify both the primary disturbance and the degree of compensation for these three arterial blood gas readings: (a) pH = 7.20 (b) pH = 7.20 (c) pH = 7.36 Paco2 = 35 mm Hg Paco2 = 25 mm Hg Paco2 = 20 mm Hg [HCO3-] = 17 mEq/L [HCO3-] = 17 mEq/L [HCO3-] = 17 mEq/L
Systemic blood pressure is defined as the ratio of two pressures—systolic and diastolic—both expressed in millimeters of mercury. Normal blood pressure is about 120mm/80mm which is usually just stated as 120/80. What would normal systemic blood pressure be if, instead of millimeters of mercury, we expressed pressure in each of the following units, but continued to use the same ratio format? Part A) atmosheres = a)120/ 80 b) 2.32/ 1.54 c) 0.158/ 0.105 d) 1.60x10^4/ 1.06x10^4   Part B) torr = a) 120/ 80 b) 2.32/ 1.54 c) 0.158/ 0.105 d) 1.60x10^4/ 1.06x10^4   Part C) Pa = a)120/ 80 b) 2.32/ 1.54 c) 0.158/ 0.105 d) 1.60x10^4/ 1.06x10^4   Part D) N/ m^2 = a)120/ 80 b) 2.32/ 1.54 c) 0.158/ 0.105 d) 1.60x10^4/ 1.06x10^4   Part E) psi = a)120/ 80 b) 2.32/ 1.54 c) 0.158/ 0.105 d) 1.60x10^4/ 1.06x10^4
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