Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 11, Problem 3PDQ
Summary Introduction
To review:
The comparison between the sizes of the chromosome of bacteriophage lambda and T2 with those of Escherichia coli(E. coli) alsocompare the appearance and size of mitochondrial and chloroplast DNA (deoxyribonucleic acid).
Introduction:
The compact form of DNA is arranged into chromosomes. These chromosomes vary in size from organism to organism. The genetic diversity in an organism arises due to recombination of chromosomes, which occurs during meiosis. The mitochondrial and chloroplast DNA follows maternal inheritance pattern.
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Explain DNA replication in eukaryotes with the help of a neat diagram.
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Chapter 11 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 11 - CASE STUDY | Art inspires learning A genetics...Ch. 11 - Prob. 2CSCh. 11 - Prob. 3CSCh. 11 -
HOW DO WE KNOW?
1. In this chapter, we focused on...Ch. 11 - Review the Chapter Concepts list on p. 199. These...Ch. 11 - Prob. 3PDQCh. 11 - Describe how giant polytene chromosomes are...Ch. 11 - What genetic process is occurring in a puff of a...Ch. 11 - Prob. 6PDQCh. 11 - Why might we predict that the organization of...
Ch. 11 -
8. Describe the sequence of research findings...Ch. 11 - Prob. 9PDQCh. 11 - Prob. 10PDQCh. 11 - Provide a comprehensive definition of...Ch. 11 - Prob. 12PDQCh. 11 - Define satellite DNA. Describe where it is found...Ch. 11 - Prob. 14PDQCh. 11 -
15. Mammals contain a diploid genome consisting...Ch. 11 - Prob. 16PDQCh. 11 - Prob. 17PDQCh. 11 - Prob. 18PDQCh. 11 - Prob. 19PDQCh. 11 - The human genome contains approximately 106 copies...Ch. 11 - Prob. 21PDQ
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- A small section of Saccharomyces cerevisiae gene has the amino acid sequence valine, histidine, cysteine, and lysine. A mutation in the above section of the amino acid sequence resulted in the substitution of amino acid histidine with amino acid glutamine. The mutation in the antisense strand DNA of Saccharomyces cerevisiae gene described above involves a. the substitution of thymine base from GTG b. the deletion of second cytosine base from CAC c. the deletion of second guanine base from GTG d. the substitution of second cytosine base from CACarrow_forwardBelow is a diagram of DNA replication as currently believed to occur in E. coli. Arrows start from numbers and end at specific points. Answer the questions relating to the locations specified by the numbers (1) Which end (5' or 3') of the molecule is here? (2) Which enzyme is probably functioning here to deal with supercoils in the DNA? (3) Which enzyme is probably functioning here to unwind the DNA?arrow_forwardList five enzymes known to be involved in the replication of DNA in bacteria.arrow_forward
- A small section of Saccharomyces cerevisiae gene has the amino acid sequence valine, histidine, cysteine, and lysine. A mutation in the above section of the amino acid sequence resulted in the substitution of amino acid lysine with amino acid asparagine. The mutation in the antisense strand DNA of Saccharomyces cerevisiae gene described above involves Select one: a. the substitution of second adenine base from AAG b. the substitution of second thymine base from TTC c. the substitution of guanine base from AAG d. the substitution of cytosine base from TTCarrow_forwardSummarize the process of chromosomal DNA replication in E. coli.arrow_forwardDefine the following terms: Genome Bacteriophage λ DNAarrow_forward
- A particular variant of the lambda bacteriophage has a DNA double-stranded genome of 51,365 base pairs. How long would this DNA be?arrow_forwardChoose the CORRECT order of compaction of DNA in eukaryotes. DNA → nucleosome → loops → fiber → "beads on a string" → chromosome DNA → nucleosome → "beads on a string" → fiber → loops → chromosome DNA → nucleosome → fiber → loops → "beads on a string" → chromosome DNA → "beads on a string" → fiber → nucleosome → loops → chromosome DNA → fiber → loops → nucleosome → "beads on a string" → chromosomearrow_forwardDescribe how plasmids differ from bacterial chromosomes.arrow_forward
- Look at the picture carefully below and imagine inside a cell nucleus. a) encircle and name the parts where DNA is most accessible and least accessible b) how nucleosome positioning or spacing can interfere with DNA accessiblityarrow_forwardAs shown, telomerase attaches additional DNA, six nucleotides at a time, to the ends of eukaryotic chromosomes. However, it makes only one DNA strand. Describe how the opposite strand is replicated.arrow_forwardThe following statements are correct EXCEPT: A) Information in the DNA is transcribed int mRNA and translated into proteins B) Information in the DNA is translated int mRNA and transcribed into proteins C) Information in the DNA is copied by DNA polymerase D) The lagging strand during DNA replication is synthesized continuously to form the okazaki fragments E) B and Darrow_forward
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