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 29.6, Problem 1MQ
How do the genomes of the measles virus and the German measles virus differ?
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You aim to test the hypothesis that the Tbx4 and Tbx5 genes inhibit each other's expression during limb development. With access to chicken embryos and viruses capable of overexpressing Tbx4 and Tbx5, describe an experiment to investigate whether these genes suppress each other's expression in the limb buds. What results would you expect if they do repress each other? What results would you expect if they do not repress each other?
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Briefly describe the expected expression domains of 1) Shh, 2) Tbx4, and 3) Tbx5 in the resulting ectopic limb bud.
Chapter 29 Solutions
Brock Biology of Microorganisms (14th Edition)
Ch. 29.1 - Why can it be said that respiratory pathogens have...Ch. 29.1 - Prob. 2MQCh. 29.2 - Prob. 1MQCh. 29.2 - What is the primary virulence factor for...Ch. 29.3 - Prob. 1MQCh. 29.3 - What measures can be taken to decrease the current...Ch. 29.4 - Why is Mycobacterium tuberculosis a widespread...Ch. 29.4 - Prob. 2MQCh. 29.5 - Identify the symptoms and causes of meningitis.Ch. 29.5 - Describe the infection by Neisseria meningitidis...
Ch. 29.6 - How do the genomes of the measles virus and the...Ch. 29.6 - Describe the potential serious outcomes of...Ch. 29.6 - Identify the effects of immunization on the...Ch. 29.7 - Prob. 1MQCh. 29.7 - Discuss the possibilities for effective treatment...Ch. 29.8 - Distinguish between antigenic drift and antigenic...Ch. 29.8 - Prob. 2MQCh. 29.9 - What is the normal habitat of Staphylococcus...Ch. 29.9 - Prob. 2MQCh. 29.10 - Prob. 1MQCh. 29.10 - How can gastric ulcers be diagnosed? How can they...Ch. 29.11 - Prob. 1MQCh. 29.11 - Prob. 2MQCh. 29.12 - MINIQUIZ
• Explain at least one potential reason...Ch. 29.12 - Prob. 2MQCh. 29.13 - Prob. 1MQCh. 29.13 - Prob. 2MQCh. 29.14 - Prob. 1MQCh. 29.14 - Prob. 2MQCh. 29.14 - Prob. 3MQCh. 29 - Prob. 1RQCh. 29 - Prob. 2RQCh. 29 - Prob. 3RQCh. 29 - QDescribe the process of infection by...Ch. 29 - QDescribe the symptoms of meningococcemia and...Ch. 29 - Prob. 6RQCh. 29 - QWhy are colds such common respiratory diseases,...Ch. 29 - Prob. 8RQCh. 29 - QDistinguish between pathogenic staphylococci and...Ch. 29 - QDescribe the evidence linking Helicobacter pylori...Ch. 29 - Prob. 11RQCh. 29 - REVIEW QUESTIONS
12. Why did the incidence of...Ch. 29 - REVIEW QUESTIONS
13. For the sexually transmitted...Ch. 29 - REVIEW QUESTIONS
14. Describe how human...Ch. 29 - Prob. 1AQCh. 29 - Application Questions
Why does active tuberculosis...Ch. 29 - APPLICATION QUESTIONS
3. Your college roommate...Ch. 29 - APPLICATION QUESTIONS
4. Contrast an HIV infection...Ch. 29 - Prob. 5AQCh. 29 - APPLICATION QUESTIONS
6. As the director of your...
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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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