Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 10, Problem 3NST
German measles results from an infection of the rubella virus, which can cause a multitude of health problems in newborns. What conclusions can you reach from a viral
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German measles results from an infection of the rubella virus, which can cause a multitude of health problems in newborns. What conclusions can you reach from a nucleic acid analysis of the virus that reveals an A + G/U + C ratio of 1.13?
Heinz Shuster collected the following data on the base composition of ribgrass mosaic virus (H. Shuster, in The Nucleic Acids: Chemistry and Biology, vol. 3, E. Chargaff and J. N. Davidson, Eds. New York: Academic Press, 1955). On the basis of this information, is the hereditary information of the ribgrass mosaic virus RNA or DNA? Is it likely to be single stranded or double stranded? Percentage A G C T U Ribgrass mosaic virus 29.3 25.8 18.0 0.0 27.0
A two mL stock of Marburg virus (Filamentous, non-segmented, negative strand RNA virusin Filoviridae family) contains 5.6 x 1017 virions (individual viruses)/mL. The sample mustbe diluted serially to the point in which 2.4 x 106 virions exist in every 100 µL of sample foran infection experiment which requires at least 7.2 mL (you will NEED more than 6 mL).Using dilutions of no greater than 1/1000 in volumes no greater than 10 mL and includingone dilution of 1/8.8 show below how you would set up this dilution series. Remember smallvolumes carry inherently larger error.
Chapter 10 Solutions
Concepts of Genetics (11th Edition)
Ch. 10 - Would an experiment similar to that performed by...Ch. 10 - In sea urchin DNA, which is double stranded, 17.5...Ch. 10 - German measles results from an infection of the...Ch. 10 - Smallpox, a once highly lethal contagious disease,...Ch. 10 - Prob. 2CSCh. 10 - Prob. 3CSCh. 10 - Prob. 4CSCh. 10 - HOW DO WE KNOW? In this chapter, we first focused...Ch. 10 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 10 - Discuss the reasons proteins were generally...
Ch. 10 - Prob. 4PDQCh. 10 - When Avery and his colleagues had obtained what...Ch. 10 - Why were 32P and 35S chosen for use in the...Ch. 10 - Does the design of the HersheyChase experiment...Ch. 10 - What observations are consistent with the...Ch. 10 - What are the exceptions to the general rule that...Ch. 10 - Draw the chemical structure of the three...Ch. 10 - How are the carbon and nitrogen atoms of the...Ch. 10 - Adenine may also be named 6-amino purine. How...Ch. 10 - Draw the chemical structure of a dinucleotide...Ch. 10 - Describe the various characteristics of the...Ch. 10 - What evidence did Watson and Crick have at their...Ch. 10 - What might Watson and Crick have concluded had...Ch. 10 - How do covalent bonds differ from hydrogen bonds?...Ch. 10 - List three main differences between DNA and RNA.Ch. 10 - What are the three major types of RNA molecules?...Ch. 10 - Prob. 20PDQCh. 10 - What is the physical state of DNA after it is...Ch. 10 - What is the hyperchromic effect? How is it...Ch. 10 - Why is Tm related to base composition?Ch. 10 - What is the chemical basis of molecular...Ch. 10 - What did the WatsonCrick model suggest about the...Ch. 10 - A genetics student was asked to draw the chemical...Ch. 10 - Considering the information in this chapter on B-...Ch. 10 - One of the most common spontaneous lesions that...Ch. 10 - In some organisms, cytosine is methylated at...Ch. 10 - Because of its rapid turnaround time, fluorescent...Ch. 10 - Prob. 31PDQCh. 10 - Prob. 32ESPCh. 10 - Newsdate: March 1, 2030. A unique creature has...Ch. 10 - Prob. 34PDQCh. 10 - During gel electrophoresis, DNA molecules can...Ch. 10 - Electrophoresis is an extremely useful procedure...Ch. 10 - Following is a table (modified from Kropinski,...
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- Heinz Shuster collected the following data on the base composition of ribgrass mosaic virus (H. Shuster, in The Nucleic Acids: Chemistry andBiology, vol. 3, E. Chargaff and J. N. Davidson, Eds. New York: Academic Press, 1955). On the basis of this information, is the hereditary information of the ribgrass mosaic virus RNA or DNA? Is it likely to be single stranded or double stranded?arrow_forwardSeveral cell culture lines of epithelial cells, called “Cell Line A, B, or C,” are incubated with 104 infectious particles of influenza virus, and viral titers in the culture media are measured 2 days later. Looking at the results of this experiment, it is apparent that the three lines do not all show the same response to the virus. To investigate these differences, mixing experiments are performed, where cells from two different cell lines are mixed together at a 1:1 ratio before the Influenza infection. Based on the results shown in the figure below, propose an explanation for these data.arrow_forwarda variant of the SARS-CoV-2 virus has been found in brazil. this variant contains the mutation K417N, where the amino acid K has been mutated to the amino acid N. K and N refers to the one letter codes of the amino acids. a) state the names of the amino acids K and N. b) state one difference between the properties of these 2 aminoacidsarrow_forward
- Imagine that you are a student in Alfred Hershey and Martha Chase’s lab in the late 1940s. You are given five test tubes containing E. Coli bacteria infected with T2 bacteriophages that have been labeled with either 32P or 35S. Unfortunately, you forget to mark the tubes and are now uncertain about which tubes is which. You performed their blender experiment and got the following results. Which tube out of these 5 contains E. Coli infected with 32P-labeled phage? Explain your answer.arrow_forwardEscherichia coli dsDNA contains 51% GC on a molar basis. What are the mole percentages of A, T, G and C respectively?arrow_forwardYou are given two samples of DNA, one from Clostridium perfringens and the other from Mycobacterium tuberculosis. Unfortunately, the labels on the tubes of DNA were accidentally removed, and you do not know which DNA sample belongs to which organism. As you were doing some work with Escherichia coli in the lab, you also have some of its DNA available as well. The %G + %C values for each of the organisms is known (C. perfringens, 27%; M. tuberculosis, 67%; E. coli, 50%). Based upon this information, how could you quickly determine the source of the DNA for your two samples?arrow_forward
- The base composition of an RNA virus was analyzed and found to be 14.1% A, 14.0% U, 36.2% G, and 35.7% C. Would you conclude that the viral genetic material is single-stranded RNA or doublestranded RNA?arrow_forwardChoose all of the statements that correctly describe the base pairs drawn below. A C H H-N -H-N N-H- -N B H D موعة Rita N -H---- 2 NHN O- -H-N H -H- N- -H-N The non-Watson-Crick base pair shown in A is much less stable than the base pairs shown in B and C, because the smaller size of the two pyrimidine bases induces a distortion in the structure of the double helix that decreases the stability of the helix when compared to helices with the normal Watson-Crick base pairs. The base pair shown in B is found in BOTH DNA and RNA The base pair shown in C is found ONLY in RNA and NOT DNA The base pair seen in B is more stable than the Watson-Crick base pair shown in C partly because of a larger number of hydrogen bonds and partly because of more favourable pi-stacking interactions with adjacent base pairs.arrow_forwardFrom the complete genome sequence of an organism, you have determined the following percentages of each nucleotide: Adenine (20); Cytosine (30) Guanine (30); Thymine (20). The mol% G + C for this organism isarrow_forward
- What type of nucleic acid genome does the human T-cell lymphotropic virus type-1 (HTLV-1) contain as defined by the Baltimore classification system?arrow_forward, The E. coli genome is 4.6 Mbp (mega base-pairs). What is the physical length of the E. coli genome?arrow_forwardTwo theoretical genetic strains of a virus (a-b-c- and a+b+c+) were used to simultaneously infect a culture of host bacteria. Of 10,000 plaques scored, the following genotypes were observed. Determine the genetic map of these three genes on the viral chromosome. Decide whether interference was positive or negative. a+ b+ c+ 4100 a- b+ c- 160 a- b- c- 3990 a+ b- c+ 140 a+ b- c- 740 a- b- c+ 90 a- b+ c+ 670 a+ b+ c- 110arrow_forward
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