Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 11, Problem 19P
A survey of organisms living deep in the ocean revels two new species whose DNA is isolated for analysis. DNA samples from both species are treated to remove nonhistone proteins. Each DNA sample is then treated with DNase I that cuts DNA not protected by histone proteins but is unable to cut DNA bound by histone proteins. Following DNase I treatment, DNA samples are subjected to gel electrophoresis, and the gels are stained to visualize all DNA bands in a gel. The staining patterns of DNA bands from each species are shown in the figure. The number of base pairs in small DNA fragments is shown at the left of the gel. Interpret the gel results in terms of chromatin organization and the spacing of nucleosomes in the chromatin of each species.
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Extreme UV exposure leads to the SOS response in bacteria. By what mechanism does the SOS response function?
Answer choices
induction of photolyase and the addition of white light to remove the thymine dimer
destruction of lexA, which leads to expression of an alternate, error-prone DNA polymerase
homologous recombination repair
non-homologous end joining
exinuclease removal of a segment of DNA including a thymine dimer, followed by the replacement of DNA using the complementary strand of DNA
The technique of fluorescence in situ hybridization (FISH) is described. This is another method for examining sequence complexity within a genome. In this method, a DNA sequence, such as a particular gene sequence, can be detected within an intact chromosome by using a DNA probe that is complementary to the sequence.For example, let’s consider the β-globin gene, which isfound on human chromosome 11. A probe complementary to theβ-globin gene binds to that gene and shows up as a brightly colored spot on human chromosome 11. In this way, researchers can detectwhere the β-globin gene is located within a set of chromosomes. Becausethe β-globin gene is unique and because human cells are diploid(i.e., have two copies of each chromosome), a FISH experimentshows two bright spots per cell; the probe binds to each copy ofchromosome 11. What would you expect to see if you used thefollowing types of probes?A. A probe complementary to the Alu sequenceB. A probe complementary to a tandem array near…
As described in Table, what is the difference between a rapidstop and a slow-stop mutant? What are different roles of the proteins that are defective in rapid-stop and slow-stop mutants?
Chapter 11 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Bacterial DNA is compacted by two principal...Ch. 11 - 10.2 The human genome contains contains base...Ch. 11 - 10.1 Give descriptions for the following...Ch. 11 - 10.4 Describe the importance of light and dark G...Ch. 11 - In eukaryotic DNA, Where are you most likely to...Ch. 11 - Prob. 8PCh. 11 - Human late prophase karyotypes have about 2000...Ch. 11 - 10. What are the two or three most essential...
Ch. 11 - Prob. 11PCh. 11 - Prob. 12PCh. 11 - A researcher interested in studying a human gene...Ch. 11 - Prob. 14PCh. 11 - 10.11 In what way does position effect variegation...Ch. 11 - 16. What are chromosome territories, and what...Ch. 11 - Prob. 17PCh. 11 - Prob. 18PCh. 11 - 10.18 A survey of organisms living deep in the...Ch. 11 - A eukaryote with a diploid number of 2n=6 carries...Ch. 11 - The accompanying chromosome diagram represents a...Ch. 11 - Suppose the genome of a bacterium contains a...Ch. 11 - DNaseI cuts DNA that is not directly associated...Ch. 11 - 10.17 Histone protein isolated from pea plants...Ch. 11 - 25. The molecular probes used in FISH can detect...Ch. 11 - Experimental evidence demonstrates that the...Ch. 11 - Prob. 27PCh. 11 - Genomic DNA from the nematode worm...Ch. 11 - What function do histone proteins perform in...Ch. 11 - Based on discussions of specific proteins and...
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- You have discovered a new species of plant. You isolate chromatin from the plant and examines it with the electron microscope and see characteristic beads on a string structure. After addition of a small amount of nuclease, which cleaves the string into individual beads that each contain 280 bp of DNA. With more digestion a 120-bp fragment of DNA remains attached to a core of histone protein. Analysis of the histone core reveals histones in the following proportions: H2A 33.3% H2B 33.3% H3 0% [no histone H3 found] H4 33.3% Based on these observations, what conclusions can you make about the probable structure of the nucleosome in the chromatin of this plant? Be specific in describing the nature of the nucleosome: which histones form the core (), how many of each are present () and size in bp of core-DNA that envelopes it ().arrow_forwardThe regulation of replication is essential to genomic stability. Normally, the DNA is replicated just once in every eukaryotic cell cycle (in the S phase). Normal cells produce protein A, which increases in concentration in the S phase. In cells that have a mutated copy of the gene encoding protein A, the protein is not functional, and replication takes place continuously throughout the cell cycle, with the result that cells may have 50 times the normal amount of DNA. Protein B is normally present in G1, but disappears from the cell nucleus during the S phase. In cells with a mutated copy of the gene encoding protein A, the levels of protein B fail to drop in the S phase and, instead, remain high throughout the cell cycle. When the gene for protein B is mutated, noreplication takes place. Propose a mechanism for how protein A and protein B might normally regulate replication so that each cell gets the proper amount of DNA. Explain how mutation of these genes produces the effects just…arrow_forwardIn chromosomes, doubly-stranded DNA wraps tightly around histone proteins. Think about the structure of DNA, especially the "backbone," and determine what category of amino acids make up the majority of histones: Histone Octamer tetramer dimer HZA dimer dimer dimer Histone "fastener" Here the blue represents histones, and the red is the DNA strand. To help the DNA fold up in an aqueous environment, the histones are nonpolar to exclude water, so many of the amino acid side chains are nonpolar. The backbone of DNA is the alternating phosphate-sugar part, and the phosphate is negatively charged. Positive charges on the histone residues will bind tightly here, so the acidic amino acids are used here. The backbone of DNA is the alternating phosphate-deoxyribose part, and the phosphate is negatively charged. Positive charges on the histone residues will bind tightly here, so the basic amino acids are used here. The histones are strongly attracted to the hydrogen-bonding of the base pairs, so…arrow_forward
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