GEN COMBO LOOSELEAF MICROBIOLOGY:A SYSTEMS APPROACH; CONNECT ACCESS CARD
5th Edition
ISBN: 9781260149364
Author: Marjorie Kelly Cowan Professor
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 18.6, Problem 13AYP
Summary Introduction
To enlist:
The possible causative agents for each of the infectious eye diseases: conjunctivitis, trachoma, keratitis, and river blindness.
Concept introduction:
Eye gives an individual the ability to visualize the things. The eye consists of different types of tissues and has the ability to process the visual details. The eye is the most crucial and complex organ of the body system. The interaction of microbiome to the eye can cause serious infections and disease conditions.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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?
You decide to delete Fgf4 and Fgf8 specifically in the limb bud. Explain why you would not knock out these genes in the entire embryo instead.
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.
Chapter 18 Solutions
GEN COMBO LOOSELEAF MICROBIOLOGY:A SYSTEMS APPROACH; CONNECT ACCESS CARD
Ch. 18.1 - Describe the important anatomical features of the...Ch. 18.1 - Prob. 2AYPCh. 18.2 - Prob. 3AYPCh. 18.3 - Prob. 4AYPCh. 18.3 - Prob. 5AYPCh. 18.3 - Prob. 6AYPCh. 18.3 - Prob. 7AYPCh. 18.3 - Prob. 8AYPCh. 18.3 - Prob. 9AYPCh. 18.4 - Prob. 10AYP
Ch. 18.4 - Prob. 11AYPCh. 18.5 - List the types of normal biota presently known to...Ch. 18.6 - Prob. 13AYPCh. 18.6 - Prob. 14AYPCh. 18 - Prob. 1MCQCh. 18 - Prob. 2MCQCh. 18 - Which of the following is probably the most...Ch. 18 - Prob. 4MCQCh. 18 - Prob. 5MCQCh. 18 - Prob. 6MCQCh. 18 - Prob. 7MCQCh. 18 - Prob. 8MCQCh. 18 - Prob. 9MCQCh. 18 - Prob. 10MCQCh. 18 - Prob. 11TFCh. 18 - Prob. 12TFCh. 18 - Prob. 13TFCh. 18 - Prob. 14TFCh. 18 - Prob. 15TFCh. 18 - Prob. 1CTQCh. 18 - Prob. 2CTQCh. 18 - a. Conduct additional research, and discuss...Ch. 18 - Prob. 4CTQCh. 18 - Prob. 5CTQCh. 18 - Prob. 1VCCh. 18 - Prob. 1CM
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- 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
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Anatomy & PhysiologyBiologyISBN:9781938168130Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark WomblePublisher:OpenStax College
Understanding Health Insurance: A Guide to Billin...Health & NutritionISBN:9781337679480Author:GREENPublisher:Cengage
Medical Terminology for Health Professions, Spira...Health & NutritionISBN:9781305634350Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. SchroederPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Anatomy & Physiology
Biology
ISBN:9781938168130
Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Publisher:OpenStax College
Understanding Health Insurance: A Guide to Billin...
Health & Nutrition
ISBN:9781337679480
Author:GREEN
Publisher:Cengage
Medical Terminology for Health Professions, Spira...
Health & Nutrition
ISBN:9781305634350
Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. Schroeder
Publisher:Cengage Learning
Cancer Types SIMPLY explained! MEMORIZE them QUICKLY and EASILY!; Author: CancerEdInstitute;https://www.youtube.com/watch?v=dEBi-yvSWmQ;License: Standard Youtube License