1 SEM ACC W/RAVEN CARDED
12th Edition
ISBN: 9781265486297
Author: Raven
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 15.9, Problem 3LO
Summary Introduction
To list: The various chromosomal mutations and their effects.
Introduction: Each chromosome contains a single DNA molecule. All these chromosomes must be compacted in order to fit the nucleus of the cell. All eukaryotic organisms contain chromosomes, and the number of chromosomes varies in different organisms. In humans and several other species, diploid (2n) number of chromosomes is present. Therefore, for humans, the diploid and haploid numbers are 46 and 23, respectively. The diploid number of chromosomes indicates that each parent contributes equally to the offspring. The variant forms of a particular gene located at the same genetic locus or position on a chromosome are regarded as alleles.
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1) Look at the ideal results. Were your predictions accurate, and how did they compare with your results?
2) You used aseptic technique during this lab. Why is it important to work in a sterile manner when working with bacteria in the lab?
3) Why are the cells incubated at 42°C?
Overview of Transformation Protocol
-Prepare competent bacteria for transformation:
Treat starter E. coli bacteria with CaCl2and Competent Cell Solution (CCS). Store on ice until transformation procedure.
Competent cells are cells that are likely to take up foreign DNA and be transformed. This step increases the likelihood that the E. coli cells will take up the introduced vector and be transformed.
-Transformation procedure:
Obtain two microcentrifuge tubes containing your competent cells. Label one tube +DNA and one -DNA.
Add CaCl2 to both tubes.
Add the transformation mix containing the plasmid DNA to the tube labeled +DNA. Do not add any plasmid DNA to the -DNA tube.
Incubate both tubes on ice for 10 minutes. Then, place both tubes in a 42\deg C water bath for 45 seconds. Replace the tubes in an ice bucket for 2 minutes.
Add recovery broth to both tubes.
Incubate both tubes in a 37 C water bath for 5 minutes.
Questions:
1) What differences would you expect to see between the…
Overview of Transformation Protocol
-Prepare competent bacteria for transformation:
Treat starter E. coli bacteria with CaCl2and Competent Cell Solution (CCS). Store on ice until transformation procedure.
Competent cells are cells that are likely to take up foreign DNA and be transformed. This step increases the likelihood that the E. coli cells will take up the introduced vector and be transformed.
-Transformation procedure:
Obtain two microcentrifuge tubes containing your competent cells. Label one tube +DNA and one -DNA.
Add CaCl2 to both tubes.
Add the transformation mix containing the plasmid DNA to the tube labeled +DNA. Do not add any plasmid DNA to the -DNA tube.
Incubate both tubes on ice for 10 minutes. Then, place both tubes in a 42\deg C water bath for 45 seconds. Replace the tubes in an ice bucket for 2 minutes.
Add recovery broth to both tubes.
Incubate both tubes in a 37 C water bath for 5 minutes.
Questions:
1)What is the selectable marker in this experiment? How…
Chapter 15 Solutions
1 SEM ACC W/RAVEN CARDED
Ch. 15.1 - Prob. 1LOCh. 15.1 - Prob. 2LOCh. 15.1 - List the roles played by RNA in gene expression.Ch. 15.2 - Prob. 1LOCh. 15.2 - Describe the characteristics of the genetic code.Ch. 15.2 - Prob. 3LOCh. 15.3 - Prob. 1LOCh. 15.3 - Differentiate among initiation, elongation, and...Ch. 15.3 - Prob. 3LOCh. 15.4 - Prob. 1LO
Ch. 15.4 - Prob. 2LOCh. 15.4 - Explain the differences between bacterial and...Ch. 15.5 - Prob. 1LOCh. 15.5 - Prob. 2LOCh. 15.5 - Prob. 3LOCh. 15.6 - Explain why the tRNA charging reaction is critical...Ch. 15.6 - Prob. 2LOCh. 15.7 - Prob. 1LOCh. 15.7 - Prob. 2LOCh. 15.7 - Compare translation on the RER and in the...Ch. 15.9 - Prob. 1LOCh. 15.9 - Explain the nature of triplet repeat expansion.Ch. 15.9 - Prob. 3LOCh. 15 - Prob. 1DACh. 15 - Prob. 2DACh. 15 - Prob. 1IQCh. 15 - Prob. 2IQCh. 15 - Prob. 3IQCh. 15 - The experiments with nutritional mutants in...Ch. 15 - What is the central dogma of molecular biology? a....Ch. 15 - In the genetic code, one codon a. consists of...Ch. 15 - Eukaryotic transcription differs from prokaryotic...Ch. 15 - An anticodon would be found on which of the...Ch. 15 - RNA polymerase binds to a ________ to initiate...Ch. 15 - During translation, the codon in mRNA is actually...Ch. 15 - You have mutants that all affect the same...Ch. 15 - The splicing process a. occurs in prokaryotes. b....Ch. 15 - The enzyme that forms peptide bonds is called...Ch. 15 - In comparing gene expression in prokaryotes and...Ch. 15 - The codon CCA could be mutated to produce a. a...Ch. 15 - An inversion will a. necessarily cause a mutant...Ch. 15 - What is the relationship between mutations and...Ch. 15 - Prob. 1SCh. 15 - Frameshift mutations often result in truncated...Ch. 15 - Describe how each of the following mutations will...Ch. 15 - There are a number of features that are unique 10...
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