Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 10, Problem 18P
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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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…
In one, simple sentence define the function of the following
1. Helicase =
2. Alpha subunit of DNA polymerase III =
Chapter 10 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 10 - 10.1 Give descriptions for the following...Ch. 10 - 10.2 The human genome contains contains base...Ch. 10 - In eukaryotic DNA, Where are you most likely to...Ch. 10 - 10.4 Describe the importance of light and dark G...Ch. 10 - Human late prophase karyotypes have about 2000...Ch. 10 - Prob. 6PCh. 10 - From the following list, identify the types...Ch. 10 - 10.8 If the haploid number for a plant species is...Ch. 10 - Mating between a male donkey (2n=64) and a female...Ch. 10 - A researcher interested in studying a human gene...
Ch. 10 - 10.11 In what way does position effect variegation...Ch. 10 - 10.12 A pair of homologous chromosomes in...Ch. 10 - 10.13 An animal heterozygous for a reciprocal...Ch. 10 - Dr. Ara B. Dopsis has an idea he thinks will be a...Ch. 10 - A normal chromosome and its homolog carrying a...Ch. 10 - The accompanying chromosome diagram represents a...Ch. 10 - 10.17 Histone protein isolated from pea plants...Ch. 10 - 10.18 A survey of organisms living deep in the...Ch. 10 - In humans that XX/XO mosaics, the phenotype is...Ch. 10 - 10.20 A plant breeder would like to develop the...Ch. 10 - In Drosophilia, seven partial deletion (1to7)...Ch. 10 - Two experimental varieties of strawberry are...Ch. 10 - 10.23 In the tomato, Solanum esculentum, tall ()...Ch. 10 - A boy with Down syndrome (trisomy 21) has 46...Ch. 10 - Experimental evidence demonstrates that the...Ch. 10 - Prob. 26PCh. 10 - Genomic DNA from the nematode worm...Ch. 10 - 10.28 A small population of deer living on an...Ch. 10 - A eukaryote with a diploid number of 2n=6 carries...Ch. 10 - Prob. 30PCh. 10 - For the following crosses, determine as accurately...Ch. 10 - A healthy couple with a history of three previous...
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- The 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_forwardUsing the figure below identify: What is the role of histones and nucleosomes? How the process of chromatin condensation is performed? What is a function of introns and exons? What is a role of mobile DNA elements? What is a meaning of simple-sequence DNA?arrow_forwardWhat is/are the attributes that make nucleotide excision repair (NER) and base excision repair (BER) similar and/or different from each other? Select the correct response: The NER pathway is the only one that can remove DNA lesions in the strand regardless of their size which is followed by attaching the correct strand, then sealed by a DNA ligase. They both use the enzyme DNA glycosylases that recognizes the damaged DNA segments and proceed with repairing the faulty base in the strand. They differ NER only repairs purine bases while BER repairs pyrimidine bases. They both remove the damaged parts of the DNA where the BER pathway corrects only the identified damaged bases which are usually non-bulky lesions. The NER pathway, on the other hand, repairs the damage by removal of bulky DNA adducts which is a short-single stranded DNA segment. They both utilize the enzyme photolyase to reverse the damages created by the faulty section of the DNA. They both remove the damaged parts of the…arrow_forward
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