In Chapter 23, 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
A. A probe complementary to the Alu sequence
B. A probe complementary to a tandem array near the centromere of the X chromosome
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Genetics: Analysis and Principles
- Nucleosomes can be assembled onto defined DNA segments. When a particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests DNA that is not located within the nucleosome, uniform fragments 147 bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 bp and 110 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes? How would you interpret this result?arrow_forwardA person with a rare genetic disease has a sample of her chromosomessubjected to in situ hybridization using a probe that is known to recognize band p11 on chromosome 7. Even though her chromosomes look cytologically normal, the probe does not bind to this person’s chromosomes. How would you explain these results? How would you use this information to positionally clone the gene that is related to this disease?arrow_forwardIt is possible to take the DNA of a gene from any source and place it on a chromosome in the nucleus of a yeast cell. When you take DNA of a human gene and put it into a yeast cell chromosome, the yeast cell can synthesize the human protein. However, when you remove the DNA for a gene normally present on yeast mitochondrial chromosomes and put it on a yeast chromosome in the nucleus, the yeast cell cannot synthesize the correct protein, even though the gene comes from the same organism. Explain. What would you need to do to ensure that such a yeast cell could make the correct protein?arrow_forward
- The period gene of Drosophila melanogaster encodes for a stretch of Thr-Gly repeated in tandem. In natural populations, the three most common alleles encode for 17, 20 and 23 Thr-Gly repeats. The amplification by PCR of the allele encoding for 20 Thr-Gly repeats produces a fragment of 320 bp. Using the same set of primers, what is the size expected when amplifying the 17 Thr-Gly allele? 317 303 314 302 In a certain species of plant loci A, B and C have an additive effect on the colour of the flower. Alleles A, B, and C are dominant and alleles a, b and c are recessive. Knowing that a plant with genotype AAbbCc has a pink flower, which genotype, among the ones listed below, will produce the same phenotype? Aabbcc. aabbcc. AaBbCc. AABBCc. In pea plants, tall (T) is dominant to dwarf (t) and yellow (Y) is dominant to green (y). In a cross of true-breeding tall yellow peas x dwarf green…arrow_forwardThe 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…arrow_forwardA new technique for rapidly determining the nuclesome structure for regions of chromatin is called Mnase-Seq. Briefly, in MNase-Seq experiments, the chromatin is digested with Micrococcal Nuclease (MNase) and then the resulting digest is subjecting to high throughput sequencing to identify sequences digested by enzyme. See this short article for complete explanation. In the experiment below, researchers conducted an Mnase Seq experiment on a region of a chromosome in the absence (OHT-) and presence (OHT+) of a Protein X. This protein has a dramatic effect on the expression of the Igll1 gene, but no effect on Top3b nor Vpreb1 genes. Based on this information, explain what Protein X is doing to influence the Igll1 gene (limit 4-5 sentences).arrow_forward
- Below is a portion of an exon from a gene that encodes protein X in the genome of the plant Arabidopsis. Wildtype DNA3’ TTC AAT GCT CCG AAT ACC 5’ template strand5’ AAG TTA CGA GGC TTA TGG 3’ non-template strand A new strain (Strain B) of Arabidopsis is identified with the same region of the gene coding for protein X: 3’ TTC AAT GCT CCC AAT ACC 5’ template strand5’ AAG TTA CGA GGG TTA TGG 3’ non-template strand Compare the two DNA sequences and look for any differences. Based on what you find a. There is no mutation in Strain B compared to Strain A. b. After the point of the mutation, all the amino acids encoded by the Strain B template will be different than the Strain A protein X. c. Protein X made from the Strain B template will be much shorter than protein X made from the Strain A template d. Protein X from Strain B will have one amino acid difference that would not affect protein function. e. There is a mutation but there will not be any difference in the…arrow_forwardBelow is a portion of an exon from a gene that encodes protein Y in the genome of the plant Brassica. Wildtype DNA3’ CTT AAT GCT CCG AAT CCA 5’ template strand5’ GAA TTA CGA GGC TTA GGT 3’ non-template strand A new strain (Strain X) of Brassica is identified with the same region of the gene coding for protein Y:3’ CTT AAT GCT GCG AAT CCA 5’ template strand5’ GAA TTA CGA CGC TTA GGT 3’ non-template strand Compare the sequence of Wildtype with Strain X DNA, and note the following: Whether there is a mutation. If there is a mutation, what is the type of mutation (be as specific as possible) and explain the rationale for your decision. Assuming this is the only difference between the Wildtype and Strain X, describe the potential impact of the mutation on the structure and function of the protein.arrow_forwardYou are using the restriction enzyme HAEIII to digest different samples of the taster gene isolated from cheek cells of different people and amplified by PCR. When viewing the bands on the electrophoresis gel, one would expect that a taster (homozygote) would have---------band(s), whereas a carrier (heterozygote) would show--------band(s), and a non-taster would show------band(s).arrow_forward
- The genome of Drosophila melanogaster, a fruit fly, was sequenced in 2000. However, this “completed” sequence did not include most heterochromatin regions. The heterochromatin was not sequenced until 2007 . Most completed genome sequences do not include heterochromatin. Why is heterochromatin usually not sequenced in genome-sequencing projects?arrow_forwardReferring to Figure 7-20, answer the following questions:a. What is the DNA polymerase I enzyme doing?b. What other proteins are required for the DNApolymerase III on the left to continue synthesizingDNA?c. What other proteins are required for the DNApolymerase III on the right to continue synthesizingDNA?arrow_forwardSee the attachment and answer the following parts of the question: A) If the binturong genome is 2.87 x 109 base pairs, and the "highly repetitive DNA" fraction is composed entirely of copies of sequence 5'ATGGTCC3' and its complement, how many copies of this sequence are present in the binturong genome? B) Briefly explain, in your own words, why the fraction of the binturong DNA fragments that reannealed relatively slowly took so much longer to renature than the other DNA fragments. C) If you took more of the same randomly generated 1000 bp fragments of binturong DNA (the same sample that you used in the equilibrium density gradient centrifugation experiment described in part a and the C0t curve described in part b of this question) and used them as a sample in agarose gel electrophoresis, how many bands would you expect to find in the gel when you turned off the current and stained the gel with ethidium bromide? Briefly explain why you would predict that number of bands.arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning