Pearson eText for Evolutionary Analysis -- Instant Access (Pearson+)
5th Edition
ISBN: 9780137521029
Author: Jon Herron, Scott Freeman
Publisher: PEARSON+
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Chapter 1, Problem 11Q
Summary Introduction
To determine: Whether the given statement is consistent with the evolution of HIV
Introduction: Acquired immunodeficiency syndrome (AIDS) is a life-threatening disorder that is caused by a retrovirus called human immunodeficiency virus (HIV). This is a life-threatening syndrome because the virus interferes with the body’s ability to fight against diseases by destroying T cells of the immune system.
Summary Introduction
To determine: The reason why the given hypothesis is not consistent with HIV.
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A sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation?
The cell membranes are more resistant than normal.
An isotonic solution had been added instead of water.
A solution of 0.1 M NaCl had been added instead of water.
Not enough water had been added to the red blood cell pellet.
The man had sickle-cell anaemia.
A sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation?
The cell membranes are more resistant than normal.
An isotonic solution had been added instead of water.
A solution of 0.1 M NaCl had been added instead of water.
Not enough water had been added to the red blood cell pellet.
The man had sickle-cell anaemia.
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-haemoglobin
Chapter 1 Solutions
Pearson eText for Evolutionary Analysis -- Instant Access (Pearson+)
Ch. 1 - Prob. 1QCh. 1 - Prob. 2QCh. 1 - Prob. 3QCh. 1 - Given the risk of evolution of resistance, why do...Ch. 1 - Prob. 5QCh. 1 - Prob. 6QCh. 1 - In a monograph published in 1883, Alexander Graham...Ch. 1 - Design a study to test our prediction that human...Ch. 1 - Prob. 9QCh. 1 - Suppose that HIV were the ancestor of the SIVs,...
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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_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_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
- During a routine medical check up of a healthy man it was found that his haematocrit value was highly unusual – value of 60%. What one of the options below is the most likely reason? He will have a diet high in iron. He is likely to be suffering from anaemia. He lives at high altitude. He has recently recovered from an accident where he lost a lot of blood. He has a very large body size.arrow_forwardExplain what age of culture is most likely to produce an endospore?arrow_forwardExplain why hot temperatures greater than 45 degrees celsius would not initiate the sporulation process in endospores?arrow_forward
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- How many types of reactions can an enzyme perform?arrow_forwardYour goal is to produce black seeds resistant to mold. So you make the same cross again (between a homozygous black seeded, mold susceptible parent and a homozygous white seeded and mold resistant parent), and, again, advance progeny by SSD to create 100 F10 generation plants. Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold? Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer. Info from Question 4 a. P Locus (Seed Color): Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus. Observed Data: Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51 Total black seeds: 7 (resistant…arrow_forward10. Consider the following enzyme and its substrate where the "+" and "-" indicate cations and anions, respectively. Explain which of the following inhibitors could inhibit this enzyme? Which type of inhibitor would it be and why? (Video 5-2) Substrate Enzyme Potential inhibitorsarrow_forward
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