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The map below illustrates three alleles in a genome segment. The alleles differ in the number and location of restriction sequences. Restriction digestion and Southern blotting with the molecular probe whose hybridization location is indicated results in detection of a single DNA
band for each allele.
List the size (in kilobases) of DNA bands detected by Southern blotting of restriction digested DNA from organisms with the genotypes
Restriction
Organisms with the genotype
Why does this genotype produce a single detectable band, and why are the
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Genetic Analysis: An Integrated Approach (2nd Edition)
- Single and double digestion of plasmid pJC716 were performed using the restriction enzymes EcoRI and BamHI. DNA fragments were shown in an electrophoretogram below. Construct a restriction map of plasmid pJC716 for enzymes EcoRI and BamHI.arrow_forwardA restriction map lists the locations of DNA sequences that are cut by a particular restriction enzyme for a piece of DNA, such as a chromosome or a plasmid. Restriction maps are important when generating a construct for experimental use. Digesting the DNA sequence with the restriction enzymes will result in fragmented DNA of predictable sizes, based on the restriction map, that allow a researcher to analyze if his or her construct was generated correctly when visualized using gel electrophoresis. Use the linear restriction map to predict where bands would be expected on a gel if a digest is performed using the specified restriction enzymes. Assume that there is enough restriction enzyme that every possible restriction site on each molecule of DNA will be cut.arrow_forwardThe human gene for hemophilia is on the X chromosome. Below is a pedigree from a family afflicted with hemophilia. The man with blackened symbol (III-1) has hemophilia. To help with genetic diagnosis, a probe that detects an RFLP (restriction fragment length polymorphism) on the X chromosome is used. This probe detects either a 7 kb restriction enzyme fragment or 3 kb and 4 kb restriction enzyme fragments. The RFLP pattern for all the members of the pedigree is shown. The recombination distance between the RFLP and the hemophilia locus is 40%. What is the chance that their unborn child (III-7) is affected? 4 kb 3 kb Select one: a. 10% b. 20% C. 40% d. 60% e. 80% f. 90% g. None of these above II IIarrow_forward
- Describe the process for shotgun sequencing of a genome. Practice aligning the two sets of sequenced fragments below, to determine the order of the fragments and the complete sequence.arrow_forwardThe following DNA sequence has been identified and named ‘Gene Z’. It is thought that this sequence codes for a single protein. Design 5’ and 3’ PCR primers that will amplify ONLY the open reading frame of the genes. Give these sequences in a 5’ to 3’ orientation. Include EcoR1 and NotI restriction enzyme recognition sites on the 5’ ends of the 5’ and 3’ primers respectively.arrow_forwardDNA samples from four individuals were cleaved with the same MW restriction endonuclease. The DNA fragments were separated by gel clectrophoresis, transferred to a membrane, and hybridized with a 12 kb 10 kb DNA probe complementary to a region between sites C and D (see 8 kb hybridization line). The image of the southern blot shows the labeled DNA bands and 6 kb molecular weight (MW) markers. The lane labels I, II, III, and IV -5 kb correspond to individuals I, II, III, and IV. Assume that fragments such as C-D and C-E are clearly resolved in this gel system. Fragment sizes are as given: A-B is 4 kb, B-C is I kb, C-D is 5 kb, and D-E is 650 bp. Individual I has five cleavage sites (A, B, C, D, and E) for the restriction endonuclease. DNA homologous to probe Which individual has at least one point mutation that eliminates restriction site C only? O II IV cannot be determined II Which individual has at least one point mutation that climinates restriction sites B and C? III O IV cannot be…arrow_forward
- All of the following are performed during restriction fragment length polymorphism analysis. 1. splitting of double-stranded into single-stranded DNA 2. gel electrophoresis 3. autoradiography 4. immersion in radioactive probes 5. digestion of DNA with restriction endonucleases 6. use of a positive charge to transfer single-stranded DNA from a gel to a membrane. The correct sequence of these operations is whatarrow_forwardSickle cell anaemia is caused by a mutation in the HBB gene. The normal wild-type βA allele contains an MstII restriction site; in the mutated sickle-cell βS allele this restriction site has been lost.PCR amplification of part of the gene encompassing the mutation site from a normal unaffected individual results in a product of 110bp. Digestion of the PCR product with MstII and subsequent gel electrophoresis results in a single band of 55bp. How many bands and of what size would you expect to see in individuals who: i) has sickle-cell anaemia (has two βS alleles) ii) carry the disease (sickle cell trait) (has one βS and one βA allele) a. Individual Number of bands Size of bands (bp) Sickle-cell anaemia(has two βS alleles) 2 55 Sickle cell trait)(has one βS and one βA allele) 1 110 b. Individual Number of bands Size of bands (bp) Sickle-cell anaemia(has two βS alleles) 2 110 Sickle cell trait)(has one βS and one βA allele) 2 55 & 110…arrow_forwardThe map of plasmid pUC19 is shown below. The restriction site coordinate is the position of the 5’base on the top strand of each site sequence. The restriction enzyme sites are in bold type if there is only one site in pUC19. Please list the fragments in order of size, largest to smallest, which will result from a complete digestion by the restriction enzyme PvuII. Please list the fragments in order of size, largest to smallest, which will result from a complete digestion by the restriction enzyme DrdI.arrow_forward
- A gel pattern displaying PCR products shows four strong bands. The four pieces of DNA have lengths that are approximately in the ratio of 1 : 2 : 3 : 4. The largest band is cut out of the gel, and PCR is repeated with the same primers. Again, a ladder of four bands is evident in the gel. What does this result reveal about the structure of the encoded protein?arrow_forwardAn individual is heterozygous for an allele. Allele A has a SNP that results in disruption of an Not1 restriction enzyme site. Allele B has a functional Not1 restriction enzyme site. PCR is performed on the region of the gene containing the SNP to produce a 255bp PCR product. Restriction digestion was subsequently performed using EcoR1 restriction enzyme and the fragments were electrophoresed on an agarose gel. (EcoR1 cuts allele B at a site 115bp from the end of the sequence.) What are the sizes of the subsequent DNA fragments? 1) 115bp + 140bp 2) 205bp + 255bp 3) 140bp 4) 205bp + 255bp + 115bp 5) 140bp + 205bp + 255bp 6) 255bp + 115bp + 140bp 7) 115bp 8) 255bp 9) 115bp + 255bp 10) 255bp + 115bp + 140bp + 205bp 11) 140bp + 205bp 12) 205bp 13) 205bp + 115bp 14) 140bp + 255bparrow_forwardIn the following "gene library" cloning experiment Digested genomic DNA AmpR gene TCR gene TCR is tetracycline resistant marker, AmpR is ampicillin resistant marker and BamHI is the unique restriction enzyme on plasmid. A PhD student digests/cuts the plasmids with BamHI restriction enzyme and the genomic DNA with EcoRI restriction enzyme. After performing the cloning experiment and obtaining colonies on a selection plate, the obtained cells will be ..... (Hint: this question is even more challenging; the PhD student was later demoted to an MSc student). a) resistant to ampicillin and tetracycline b) sensitive to tetracycline and ampicillin c) resistant to tetracycline and sensitive to ampicillin d) resistant to ampicillin and sensitive to tetracycline e) sensitive to ampicillin and tetracycline BamHIarrow_forward
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