Prescott's Microbiology
10th Edition
ISBN: 9781259281594
Author: Joanne Willey, Linda Sherwood Adjunt Professor Lecturer, Christopher J. Woolverton Professor
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 27.3, Problem 2MI
Summary Introduction
Bacteriophage f1 is classified based on its structure as class I filamentous phage. Phage f1 in short Pf1 is a rod-shaped bacteriophage that contains single-stranded DNA. This phage is approximately 850nm in length and 4.3 to 6.3nm in width. The sheath of this phage is made of thousands of coat proteins that are set in a fishcale-like alpha helix.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The 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 A
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.
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.
Chapter 27 Solutions
Prescott's Microbiology
Ch. 27.1 - List some characteristics used in classifying...Ch. 27.1 - Consider these terms: Equine torovirus,...Ch. 27.2 - What enzyme found in the T4 baseplate facilitates...Ch. 27.2 - Why do you think T4 evolved to initiate DNA...Ch. 27.2 - What function does HMC glycosylation serve?Ch. 27.2 - How is the envelope of this virus formed? How does...Ch. 27.2 - Explain why the T4 genome is circularly permuted.Ch. 27.2 - Prob. 1.2RIACh. 27.2 - How is a prophage induced to become active again?Ch. 27.2 - Describe the roles of cII, CIII, repressor (CI),...
Ch. 27.2 - How do the temperate phages Mu and P1 differ from...Ch. 27.2 - The CRISPR/Cas system has been a boon to...Ch. 27.2 - Why do cold sores recur throughout the lifetime of...Ch. 27.2 - In what part of the host cell does a herpesvirus...Ch. 27.2 - Many small DNA viruses rely on host enzymes for...Ch. 27.3 - Why is the X174 genome considered plus stranded?Ch. 27.3 - Prob. 2MICh. 27.3 - Why is it necessary for some ssDNA viruses to...Ch. 27.3 - From the point of view of the virus, compare the...Ch. 27.3 - How do parvoviruses trick the host DNA polymerase...Ch. 27.4 - The rotavirus genome encodes 12 proteins. Suggest...Ch. 27.4 - Describe the life cycle of 6 phage. What makes...Ch. 27.4 - Prob. 3RIACh. 27.4 - In what ways are the life cycles of 6 and...Ch. 27.5 - Where in the host does the plus-strand RNA genome...Ch. 27.5 - How do some plus-strand viruses use polyproteins...Ch. 27.5 - What is an IRES? Why is it important?Ch. 27.5 - Prob. 3RIACh. 27.6 - How does that use of a segmented genome by...Ch. 27.6 - Prob. 2RIACh. 27.7 - Prob. 1MICh. 27.7 - Prob. 1RIACh. 27.7 - Prob. 2RIACh. 27.7 - What role does alternative splicing play in the...Ch. 27.8 - Prob. 1RIACh. 27.8 - Trace the HBV multiplication cycle, paying...Ch. 27 - No temperate RNA phages have yet been discovered....Ch. 27 - The choice between lysogeny and lysis is...Ch. 27 - Prob. 3CHICh. 27 - You are studying RNA viruses and have discovered a...Ch. 27 - Prob. 5CHICh. 27 - Upon infection of host epithelial cells,...Ch. 27 - Associated with the envelope of herpesviruses are...
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
- 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_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_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
- 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_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_forwardExplain what age of culture is most likely to produce an endospore?arrow_forward
- Explain 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_forwardReactunts C6H12O6 (Glucose) + 2NAD+ + 2ADP 2 Pyruvic acid + 2NADH + 2ATP a. Which of the above are the reactants? b. Which of the above are the products? c. Which reactant is the electron donor? GHz 06 (glucose) d. Which reactant is the electron acceptor? NAD e. Which of the products have been reduced? NADH f. Which of the products have been oxidized? g. Which process was used to produce the ATP? h. Where was the energy initially in this chemical reaction and where is it now that it is finished? i. Where was the carbon initially in this chemical reaction and where is it now that it is finished? j. Where were the electrons initially in this chemical reaction and where is it now that it is finished? 3arrow_forward
- There is ________ the concept of global warming. Very strong evidence to support Some strong evidence to support Evidence both supporting and against Evidence againstarrow_forwardHow many types of reactions can an enzyme perform?arrow_forwardYour goal is to produce black seeds resistant to mold. So you make the same cross again (between a homozygous black seeded, mold susceptible parent and a homozygous white seeded and mold resistant parent), and, again, advance progeny by SSD to create 100 F10 generation plants. Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold? Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer. Info from Question 4 a. P Locus (Seed Color): Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus. Observed Data: Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51 Total black seeds: 7 (resistant…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles Of Radiographic Imaging: An Art And A ...Health & NutritionISBN:9781337711067Author:Richard R. Carlton, Arlene M. Adler, Vesna BalacPublisher:Cengage Learning
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Principles Of Radiographic Imaging: An Art And A ...
Health & Nutrition
ISBN:9781337711067
Author:Richard R. Carlton, Arlene M. Adler, Vesna Balac
Publisher:Cengage Learning
Concepts of Biology
Biology
ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:OpenStax College