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Concept explainers
a.
To determine:
The appearance of chromosome 1 after FISH having genotypes Del1/Del2.
Introduction:
Human chromosome 1 is a large metacentric chromosome. Some of the telomeric regions on the chromosome were barcoded by multicolor banding. Fluorescence In situ Hybridization technique (FISH) is used to label the DNA probes.
b.
To determine:
The appearance of chromosome 1 after FISH having genotypes Del1/+.
Introduction:
Multicolor banding is done towards the telomeric end of chromosome 1 using different fluorescent dyes. This technique of banding is called FISH. The banding pattern is unique for different genotypes.
c.
To determine:
The appearance of chromosome 1 after FISH having genotypes Inv1/+.
Introduction:
Inversion refers to a type of chromosomal mutation in which a segment of a chromosome gets reversed from one end to the other. Inversion is of two types, paracentric and pericentric inversion. Pericentric inversion includes the centromere, whereas; paracentric inversion does not involve centromere.
d.
To determine:
The appearance of chromosome 1 after FISH having genotypes Inv2/+.
Introduction:
Inversion can be classified into two types, paracentric and pericentric inversion. Pericentric inversion includes the centromere, whereas; paracentric inversion does not involve centromere.
e.
To determine:
The appearance of chromosome 1 after FISH having genotypes Inv2/Inv2.
Introduction:
The technique of banding the chromosomal region by using different fluorescent dyes is called FISH. The banding pattern is unique for different genotypes.
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Chapter 12 Solutions
Genetics: From Genes to Genomes, 5th edition
- A 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_forwardIn each of the illustrations below, a segment of a chromosome has two copies of a transposable element. In panel a, they are oriented in the same direction, whereas in panel b they are in opposite directions. A double strand break occurs in element A and is repaired by homologous recombination using element B as a repair template. For each case, what will the chromosome look like after homologous recombination occurs? Choose one of the five options below, 1-5.arrow_forwardThe yeast genome has class 1 elements (Ty1, Ty2, and so forth) but no class 2 elements. What is a possible reason why DNA elements have not been successful in the yeast genome?arrow_forward
- A molecular biologist is investigating homologous recombination. One aim of this study is to reconstitute stages of the process in vitro. Draw diagrams to show how the four synthetic oligonucleotides below could base-pair to form a stable model Holliday junction. W 5’ GATCGCATTGTAGCCGTAGGTCCACTGTAA 3’ X 5’ GTCCCATACGTAGCCGTAGGACATGTACCG 3’ Y 5’ CGGTACATGTCCTACGGCTACAATGCGATC 3’ Z 5’ TTACAGTGGACCTACGGCTACGTATGGGAC 3’arrow_forwardThe presence (+) or absence (−) of six sequences in each of five bacterial artificial chromosome (BAC) clones (A–E) is indicated in the following table. Using these markers, put the BAC clones in their correct order and indicate the locations of the numbered sequences within them. Sequences BAC clone 1 2 3 4 5 6 A + − − − + − B − − − + − + C − + + − − − D − − + − + − E + − − + − −arrow_forwardArabidopsis thaliana has among the smallest genomes in higher plants, with a haploid genome size of about 100 Mb. If this genome is digested with BbvCl, a restriction enzyme which cuts at the sequence CCTCAGC GGAGTCG 1. approximately how many DNA fragments would be produced? Assume the DNA has a random sequence with equal amounts of each base.arrow_forward
- Based on the attached image, if we are using the Holliday junction model of recombination, where exactly would be the positions where DNA is cut? Would it be to the right because of branch migration?arrow_forwardE. coli chromosomes in which every nitrogen atom is labeled (that is, every nitrogen atom is the heavy isotope 15N instead of the normal isotope 14N) are allowed to replicate in an environment in which all the nitrogen is 14N. Using a solid line to represent a heavy polynucleotide chain and a dashed line for a light chain, sketch each of the following descriptions:a. The heavy parental chromosome and the products of the first replication after transfer to a 14N medium, assuming that the chromosome is one DNA double helix and that replication is semiconservative.b. Repeat part a, but now assume that replication is conservative.c. If the daughter chromosomes from the first division in 14N are spun in a cesium chloride density gradient and a single band is obtained, which of the possibilities in parts a and b can be ruled out? Reconsider the Meselson and Stahl experiment: What does it prove?arrow_forwardYou have isolated 8 mutants in yeast that fail to grow on minimal media plates but do grow when they are supplemented with Arginine. You know that Arginine is synthesized in a biochemical pathway within wild-type yeast, but you do not know how many gene products it takes for the pathway. You have all of the lines as both a and a cells and mate each strain to each other in pairwise crosses and plate them on minimal media to see if they grow. You obtain the following results with (+) representing growth, and (-) indicating no growth: a 1 5 1 a 4 5 6 7 8 How many genes are represented? O 1 3 7 O Cannot tell from the data a + + + + + • + + i 0 +, + + + • + + 7 + + + + + , . + + + + + m + + + + + + + 2 + + + + + i + -I + + . . + + +arrow_forward
- Explain how DNA probes with different fluorescence emissionwavelengths can be used in a single FISH experiment to map thelocations of two or more genes. This method is called chromosomepainting. Explain why this is an appropriate term.arrow_forwardE. coli strains diploid for the lac region were constructed by introducing a plasmid carrying the lac genes. The plasmid carries one copy of the lac region, and the chromosome carries the other copy. The two copies of the lac region have different genotypes, as shown in the chart below. Indicate whether the products of the lacy gene (permease) and the lacZ gene (B-galactosidase) will be inducible, uninducible, or constitutive in each strain (assuming glucose is absent). lac region on plasmid lac region on chromosome permease B-galactosidase I-o+Z+Y- I+o+ Z-Y+ I+o+Z+Y- I+o° Z=Y+ I- oº Z+Y- I+o+ Z-Y+ Is o+ Z+Y- I+o+ Z-Y+ I+ oc Z+Y- IS O+ Z-Y+arrow_forwardYou have two tubes, each with a pair of DNA fragments inside them. Tube #1 has fragments that are 500bp and 1000 bp in length. Tube #2 has fragments that are 7500bp and 8000bp in size. If you were to perform agarose gel electrophoresis and run the contents of each tube in two separate lanes on the same gel, what would you expect to see? O That the difference between the distances migrated by the two fragments in Tube #1 was much greater than the difference between the distances migrated by the two fragments in Tube #2. O That the difference between the distances migrated by the two fragments in Tube #1 was the same as difference between the distances migrated by the two fragments in Tube #2. O That the difference between the distances migrated by the two fragments in Tube #1 was much less than the difference between the distances migrated by the two fragments in Tube #2. O It is not possible to estimate what we would expect to see.arrow_forward
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