Microbiology: A Systems Approach
4th Edition
ISBN: 9780073402437
Author: Marjorie Kelly Cowan Professor
Publisher: McGraw-Hill Education
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Chapter 14.2, Problem 4AYP
Define marker, and discuss its importance in the second and third lines of defense.
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Chapter 14 Solutions
Microbiology: A Systems Approach
Ch. 14.1 - Summarize the three lines of host defenses.Ch. 14.1 - Prob. 2AYPCh. 14.1 - Discuss the role of normal biota as a first-line...Ch. 14.2 - Define marker, and discuss its importance in the...Ch. 14.3 - Prob. 5AYPCh. 14.3 - List the components of the mononuclear phagocyte...Ch. 14.3 - Prob. 7AYPCh. 14.3 - Prob. 8AYPCh. 14.3 - Name six types of blood cells that function in...Ch. 14.3 - Prob. 10AYP
Ch. 14.3 - Prob. 2CFCh. 14.4 - List the four major categories of nonspecific...Ch. 14.4 - Summarize the steps in phagocytosis, and describe...Ch. 14.4 - Outline the steps in inflammation.Ch. 14.4 - Discuss the mechanism of fever and its role in...Ch. 14.4 - Compare and contrast the three different...Ch. 14.4 - Prob. 16AYPCh. 14 - Prob. 1CFCh. 14 - Ail example of a nonspecific chemical barrier to...Ch. 14 - Which nonspecific host defense is associated with...Ch. 14 - Prob. 3MCQCh. 14 - Which of the following is not a lymphoid tissue?...Ch. 14 - Prob. 5MCQCh. 14 - Prob. 6MCQCh. 14 - Prob. 7MCQCh. 14 - Prob. 8MCQCh. 14 - Which of the following substances is/are not...Ch. 14 - Which of the following is the end product of the...Ch. 14 - The liquid component of clotted blood is called...Ch. 14 - Prob. 12TFCh. 14 - Prob. 13TFCh. 14 - Prob. 14TFCh. 14 - Prob. 15TFCh. 14 - Prob. 1CTQCh. 14 - Prob. 2CTQCh. 14 - Prob. 3CTQCh. 14 - Prob. 4CTQCh. 14 - Prob. 5CTQCh. 14 - Prob. 6CTQCh. 14 - Prob. 7CTQCh. 14 - a. Inflammation is characterized by heat, pain,...Ch. 14 - Prob. 9CTQCh. 14 - HIV predominantly infects T helper cells, cells...Ch. 14 - Prob. 1CCCh. 14 - Prob. 2CCCh. 14 - Prob. 3CCCh. 14 - Prob. 4CCCh. 14 - Prob. 1VCCh. 14 - Prob. 1CM
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- With reference to their absorption spectra of the oxy haemoglobin intact line) and deoxyhemoglobin (broken line) shown in Figure 2 below, how would you best explain the reason why there are differences in the major peaks of the spectra? Figure 2. SPECTRA OF OXYGENATED AND DEOXYGENATED HAEMOGLOBIN OBTAINED WITH THE RECORDING SPECTROPHOTOMETER 1.4 Abs < 0.8 06 0.4 400 420 440 460 480 500 520 540 560 580 600 nm 1. The difference in the spectra is due to a pH change in the deoxy-haemoglobin due to uptake of CO2- 2. There is more oxygen-carrying plasma in the oxy-haemoglobin sample. 3. The change in Mr due to oxygen binding causes the oxy haemoglobin to have a higher absorbance peak. 4. Oxy-haemoglobin is contaminated by carbaminohemoglobin, and therefore has a higher absorbance peak 5. Oxy-haemoglobin absorbs more light of blue wavelengths and less of red wavelengths than deoxy-haemoglobinarrow_forwardWith reference to their absorption spectra of the oxy haemoglobin intact line) and deoxyhemoglobin (broken line) shown in Figure 2 below, how would you best explain the reason why there are differences in the major peaks of the spectra? Figure 2. SPECTRA OF OXYGENATED AND DEOXYGENATED HAEMOGLOBIN OBTAINED WITH THE RECORDING SPECTROPHOTOMETER 1.4 Abs < 0.8 06 0.4 400 420 440 460 480 500 520 540 560 580 600 nm 1. The difference in the spectra is due to a pH change in the deoxy-haemoglobin due to uptake of CO2- 2. There is more oxygen-carrying plasma in the oxy-haemoglobin sample. 3. The change in Mr due to oxygen binding causes the oxy haemoglobin to have a higher absorbance peak. 4. Oxy-haemoglobin is contaminated by carbaminohemoglobin, and therefore has a higher absorbance peak 5. Oxy-haemoglobin absorbs more light of blue wavelengths and less of red wavelengths than deoxy-haemoglobinarrow_forwardWhich ONE of the following is FALSE regarding haemoglobin? It has two alpha subunits and two beta subunits. The subunits are joined by disulphide bonds. Each subunit covalently binds a haem group. Conformational change in one subunit can be transmitted to another. There are many variant ("mutant") forms of haemoglobin that are not harmful.arrow_forward
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- Explain why hot temperatures greater than 45 degrees celsius would not initiate the sporulation process in endospores?arrow_forwardEndospore stain: Consider tube 2 of the 7-day bacillus culture. After is was heated, it was incubated for 24 hours then refrigerated. Do you think the cloudiness in this tube is due mostly to vegetative cells or to endospores? Explain your reasoningarrow_forwardReactunts C6H12O6 (Glucose) + 2NAD+ + 2ADP 2 Pyruvic acid + 2NADH + 2ATP a. Which of the above are the reactants? b. Which of the above are the products? c. Which reactant is the electron donor? GHz 06 (glucose) d. Which reactant is the electron acceptor? NAD e. Which of the products have been reduced? NADH f. Which of the products have been oxidized? g. Which process was used to produce the ATP? h. Where was the energy initially in this chemical reaction and where is it now that it is finished? i. Where was the carbon initially in this chemical reaction and where is it now that it is finished? j. Where were the electrons initially in this chemical reaction and where is it now that it is finished? 3arrow_forward
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