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 18.10, Problem 1MQ
- How could NanoSIMS be used to identify a nitrogen-fixing bacterium?
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Chapter 18 Solutions
Brock Biology of Microorganisms (14th Edition)
Ch. 18.1 - Describe the enrichment strategy behind...Ch. 18.1 - Why is sulfate (So42) added to a Winogradsky...Ch. 18.1 - What is enrichment bias? How does dilution reduce...Ch. 18.2 - How does the agar dilution method differ from...Ch. 18.2 - How might you isolate a morphologically unique...Ch. 18.2 - What is meant by high-throughput in culturing...Ch. 18.3 - How does viability staining differ from stains...Ch. 18.3 - Prob. 2MQCh. 18.3 - Why is it incorrect to say that the GFP is a...Ch. 18.4 - What structure in the cell is the target for...
Ch. 18.4 - FISH and CARD-FISH can be used to reveal different...Ch. 18.5 - What could you conclude from PCR/DGGE analysis of...Ch. 18.5 - What surprising finding has come out of many...Ch. 18.6 - MINIQUIZ
• What is a phylochip and what can it...Ch. 18.6 - What are the advantages and disadvantages of...Ch. 18.6 - Prob. 3MQCh. 18.7 - Prob. 1MQCh. 18.7 - How do environmental genomic approaches differ...Ch. 18.7 - Prob. 3MQCh. 18.8 - Prob. 1MQCh. 18.8 - If a large pulse of organic matter entered the...Ch. 18.9 - Prob. 1MQCh. 18.9 - What is the simplest explanation for why lunar...Ch. 18.9 - What is the expected isotopic composition of...Ch. 18.10 - How could NanoSIMS be used to identify a...Ch. 18.10 - Prob. 2MQCh. 18.10 - How does MAR-FISH link microbial diversity and...Ch. 18.11 - How can stable isotope probing reveal the identity...Ch. 18.11 - What key method is required to do genomics on a...Ch. 18.11 - Prob. 3MQCh. 18 - Prob. 1RQCh. 18 - Prob. 2RQCh. 18 - Prob. 3RQCh. 18 - Prob. 4RQCh. 18 - Prob. 5RQCh. 18 - Prob. 6RQCh. 18 - Prob. 7RQCh. 18 - Prob. 8RQCh. 18 - Prob. 9RQCh. 18 - REVIEW QUESTIONS
10. Why is a microarray not...Ch. 18 - Prob. 11RQCh. 18 - Prob. 12RQCh. 18 - Q What are the major advantages of radioisotopic...Ch. 18 - Prob. 14RQCh. 18 - Prob. 15RQCh. 18 - REVIEW QUESTIONS
16. What is the advantage of...Ch. 18 - Prob. 17RQCh. 18 - Design an experiment for measuring the activity of...Ch. 18 - You wish to know whether Archaea exist in a lake...Ch. 18 - Design an experiment to solve the following...Ch. 18 - Design a SIP experiment that would allow you to...
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- You implant an FGF10-coated bead into the anterior flank of a chicken embryo, directly below the level of the wing bud. What is the phenotype of the resulting ectopic limb? Briefly describe the expected expression domains of 1) Shh, 2) Tbx4, and 3) Tbx5 in the resulting ectopic limb bud.arrow_forwardDesign 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_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_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_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
- 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
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