Brock Biology of Microorganisms (14th Edition)
14th Edition
ISBN: 9780321897398
Author: Michael T. Madigan, John M. Martinko, Kelly S. Bender, Daniel H. Buckley, David A. Stahl, Thomas Brock
Publisher: PEARSON
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Chapter 3.5, Problem 3MQ
Summary Introduction
Energy is defined as the ability to do work and is measured in kilojoules (KJ) unit. All
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Chapter 3 Solutions
Brock Biology of Microorganisms (14th Edition)
Ch. 3.1 - Which four chemical elements make up the bulk of a...Ch. 3.1 - Which two classes of macromolecules contain most...Ch. 3.1 - Prob. 3MQCh. 3.2 - Why would a complex culture medium for Leuconostoc...Ch. 3.2 - In which medium shown in Table 5.1, defined or...Ch. 3.2 - What is meant by the word sterile? Why is aseptic...Ch. 3.3 - Prob. 1MQCh. 3.3 - Prob. 2MQCh. 3.4 - What is free energy?Ch. 3.4 - Prob. 2MQ
Ch. 3.4 - Using Table 3.2, calculate G0 for the reaction...Ch. 3.5 - Prob. 1MQCh. 3.5 - Prob. 2MQCh. 3.5 - Prob. 3MQCh. 3.6 - Prob. 1MQCh. 3.6 - Prob. 2MQCh. 3.6 - Prob. 3MQCh. 3.7 - How much free energy is released when ATP is...Ch. 3.7 - Prob. 2MQCh. 3.7 - Prob. 3MQCh. 3.8 - Which reactions in glycolysis are redox steps?Ch. 3.8 - Prob. 2MQCh. 3.8 - Prob. 3MQCh. 3.9 - Prob. 1MQCh. 3.9 - Prob. 2MQCh. 3.10 - Prob. 1MQCh. 3.10 - Which electron carriers described in this section...Ch. 3.11 - How do electron transport reactions generate the...Ch. 3.11 - Prob. 2MQCh. 3.11 - What structure in the cell links the proton motive...Ch. 3.12 - Prob. 1MQCh. 3.12 - What two major roles do the citric acid cycle and...Ch. 3.12 - Why is the glyoxylate cycle necessary for growth...Ch. 3.13 - Prob. 1MQCh. 3.13 - Prob. 2MQCh. 3.13 - Prob. 3MQCh. 3.14 - What form of activated glucose is used in the...Ch. 3.14 - Prob. 2MQCh. 3.14 - What functions does the pentose phosphate pathway...Ch. 3.15 - Prob. 1MQCh. 3.15 - List the steps required for the cell to...Ch. 3.15 - Which nitrogen bases are purines and which are...Ch. 3.16 - Prob. 1MQCh. 3.16 - Prob. 2MQCh. 3.17 - Prob. 1MQCh. 3.17 - Prob. 2MQCh. 3.17 - Prob. 3MQCh. 3 - Prob. 1RQCh. 3 - Prob. 2RQCh. 3 - What is aseptic technique and why is it necessary?...Ch. 3 - Prob. 4RQCh. 3 - REVIEW QUESTIONS
5. Describe how you would...Ch. 3 - Prob. 6RQCh. 3 - Why are enzymes needed by the cell? (Section 3.5)
Ch. 3 - Prob. 8RQCh. 3 - 9. What is the reduction potential of the...Ch. 3 - Prob. 10RQCh. 3 - How is ATP made in fermentation and in...Ch. 3 - Prob. 12RQCh. 3 - MINIQUIZ
13. Besides lactic acid and ethanol, list...Ch. 3 - List some of the key electron carriers found in...Ch. 3 - What is meant by proton motive force and how is it...Ch. 3 - Prob. 16RQCh. 3 - MINIQUIZ
17. How much more ATP is possible in...Ch. 3 - Prob. 18RQCh. 3 - MINIQUIZ
19. What is the major difference between...Ch. 3 - Prob. 20RQCh. 3 - Describe the process by which a fatty acid such as...Ch. 3 - Prob. 22RQCh. 3 - Prob. 1AQCh. 3 - Application Questions
2. Desulfovibrio can grow...Ch. 3 - Using the data of Figure 3.10, predict the...Ch. 3 - Prob. 4AQ
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- Design a grafting experiment to determine if limb mesoderm determines forelimb / hindlimb identity. Include the experiment, a control, and an interpretation in your answer.arrow_forwardThe Snapdragon is a popular garden flower that comes in a variety of colours, including red, yellow, and orange. The genotypes and associated phenotypes for some of these flowers are as follows: aabb: yellow AABB, AABb, AaBb, and AaBB: red AAbb and Aabb: orange aaBB: yellow aaBb: ? Based on this information, what would the phenotype of a Snapdragon with the genotype aaBb be and why? Question 21 options: orange because A is epistatic to B yellow because A is epistatic to B red because B is epistatic to A orange because B is epistatic to A red because A is epistatic to B yellow because B is epistatic to Aarrow_forwardA 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.arrow_forward
- 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.arrow_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_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_forward
- Which 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_forwardDuring 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_forward
- Explain 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_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_forward
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