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 7.17, Problem 1MQ
What is feedback inhibition?
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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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Chapter 7 Solutions
Brock Biology of Microorganisms (14th Edition)
Ch. 7.1 - Prob. 1MQCh. 7.1 - Prob. 2MQCh. 7.1 - Prob. 3MQCh. 7.2 - What is protein domain?Ch. 7.2 - Why are most DMA-binding proteins specific to...Ch. 7.3 - Why is negative control so named?Ch. 7.3 - How does a repressor inhibit the synthesis of a...Ch. 7.4 - Prob. 1MQCh. 7.4 - Prob. 2MQCh. 7.5 - Prob. 1MQ
Ch. 7.5 - Prob. 2MQCh. 7.5 - Explain how the lac operon is both positively and...Ch. 7.6 - What is the major difference between...Ch. 7.6 - How do transcriptional activators in Archaea often...Ch. 7.7 - What are kinases and what is their role in...Ch. 7.7 - Prob. 2MQCh. 7.8 - What are the primary response regulator and the...Ch. 7.8 - Why is adaptation during chemotaxis important?Ch. 7.8 - How does the response of the chemortaxis system to...Ch. 7.9 - Prob. 1MQCh. 7.9 - Prob. 2MQCh. 7.10 - Prob. 1MQCh. 7.10 - Prob. 2MQCh. 7.10 - Prob. 3MQCh. 7.11 - How are different sets of genes expressed in the...Ch. 7.11 - Prob. 2MQCh. 7.12 - Why are the levels of DnaA protein controlled...Ch. 7.12 - Prob. 2MQCh. 7.14 - Prob. 1MQCh. 7.14 - Prob. 2MQCh. 7.15 - What happens when a riboswitch binds the small...Ch. 7.15 - What are the major differences between a repressor...Ch. 7.16 - Why does attenuation control not occur in...Ch. 7.16 - Prob. 2MQCh. 7.17 - What is feedback inhibition?Ch. 7.17 - Prob. 2MQCh. 7.18 - Prob. 1MQCh. 7.18 - Prob. 2MQCh. 7 - Prob. 1RQCh. 7 - Describe why a protein that binds to a specific...Ch. 7 - REVIEW QUESTIONS
3. Most biosynthetic operons need...Ch. 7 - What is the difference between an operon and a...Ch. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Adaptation allows the mechanism controlling...Ch. 7 - How can quorum sensing be considered a regulatory...Ch. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 16RQCh. 7 - Prob. 17RQCh. 7 - Application Questions
1. What would happen to...Ch. 7 - Prob. 2AQCh. 7 - Prob. 3AQCh. 7 - Most of the regulatory systems described in this...Ch. 7 - Many amino acid biosynthetic operons under...Ch. 7 - How would you design a regulatory system to make...
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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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