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 11.5, Problem 1MQ
Summary Introduction
The synthesis of oligonucleotides of DNA or RNA is referred to as the chemical synthesis of
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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 11 Solutions
Brock Biology of Microorganisms (14th Edition)
Ch. 11.1 - Prob. 1MQCh. 11.1 - Prob. 2MQCh. 11.2 - Prob. 1MQCh. 11.2 - Prob. 2MQCh. 11.3 - Why is a primer needed at each end of the DNA...Ch. 11.3 - MINIQUIZ
• From which organisms are thermostable...Ch. 11.3 - How does RT-PCR differ from traditional PCR?Ch. 11.4 - What is the purpose of molecular cloning?Ch. 11.4 - Prob. 2MQCh. 11.4 - Prob. 3MQ
Ch. 11.5 - Prob. 1MQCh. 11.5 - How can site-directed mutagenesis be useful to...Ch. 11.5 - What are knockout mutations?Ch. 11.6 - Prob. 1MQCh. 11.6 - Prob. 2MQCh. 11.7 - Prob. 1MQCh. 11.7 - Prob. 2MQCh. 11.7 - Prob. 3MQCh. 11.8 - Prob. 1MQCh. 11.8 - Prob. 2MQCh. 11.9 - MINIQUIZ
• Describe the components needed for an...Ch. 11.9 - Prob. 2MQCh. 11.10 - Prob. 1MQCh. 11.10 - Prob. 2MQCh. 11.10 - Prob. 3MQCh. 11.11 - What major advantage does cloning mammalian genes...Ch. 11.11 - Prob. 2MQCh. 11.12 - Prob. 1MQCh. 11.12 - Prob. 2MQCh. 11.12 - Prob. 3MQCh. 11.13 - Prob. 1MQCh. 11.13 - Give an example of a genetically modified plant...Ch. 11.13 - How have transgenic salmon been engineered to...Ch. 11.14 - Explain why recombinant vaccines might be safer...Ch. 11.14 - Prob. 2MQCh. 11.15 - Explain why metagenomic cloning gives large...Ch. 11.15 - Prob. 2MQCh. 11.16 - Prob. 1MQCh. 11.16 - Prob. 2MQCh. 11.17 - What are biobricks?Ch. 11.17 - How was Escherichia coli modified to produce a...Ch. 11 - Prob. 1RQCh. 11 - Prob. 2RQCh. 11 - Describe the basic principles of gene...Ch. 11 - Prob. 4RQCh. 11 - Prob. 5RQCh. 11 - Prob. 6RQCh. 11 - Prob. 7RQCh. 11 - REVIEW QUESTIONS
8. What is a reporter gene?...Ch. 11 - Prob. 9RQCh. 11 - Prob. 10RQCh. 11 - Prob. 11RQCh. 11 - Prob. 12RQCh. 11 - Prob. 13RQCh. 11 - Prob. 14RQCh. 11 - Prob. 15RQCh. 11 - Prob. 16RQCh. 11 - Prob. 17RQCh. 11 - What is the Ti plasmid and how has it been of use...Ch. 11 - What is a subunit vaccine and why are subunit...Ch. 11 - How has metagenomics been used to find novel...Ch. 11 - Prob. 21RQCh. 11 - Prob. 22RQCh. 11 - Prob. 1AQCh. 11 - Suppose you have just determined the DNA base...Ch. 11 - Prob. 3AQCh. 11 - 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
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