Concept explainers
ΦC31 is a type of bacteriophage that infects Streptomyces bacteria. One gene in the bacteriophage genome specifies a recombinase called ΦC31 integrase that works through a mechanism slightly different from that of the recombinase shown in Fig. 6.30. Most importantly, the two target DNA sequences are different from each other. One called attP is 39 base pairs and is found on the circular bacteriophage chromosome, while the other—attB—is 34 base pairs long and is located on the much larger circular bacterial chromosome. Excepting two base pairs roughly in the middle of both targets that are identical and at which recombination takes place, the DNA sequences of attP and attB are completely different from each other.
a. | Diagram the reaction that ΦC31 integrase performs. How could this reaction be important for the life cycle of the bacteriophage? |
b. | Using the diagram you just drew, explain why ΦC31 integrase cannot reverse the reaction. |
c. | Now consider how you might exploit this site-specific recombination to place genes from another species (a transgene) into the genome of an experimental organism like Drosophila. Assume you can make any DNA sequences you want and that you can introduce these DNA sequences into fruit fly germ-line cells by injection. Why is the irreversibility of the ΦC31 integrase–mediated reaction valuable for placing the transgene into the Drosophila genome? |
d. | Bacteriophage ΦC31 must eventually reverse this reaction. Why? How do you think the bacteriophage can achieve this reversal? |
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Genetics: From Genes to Genomes
- F ′strains in E. coli are derived from Hfr strains. In some cases, these F ′strains show a high rate of integration back into the bacterial chromosome of a second strain. Furthermore, the site of integration is often the site occupied by the sex factor in the original Hfr strain (before production of the F ′strains). Explain these results.arrow_forwardDifferent Hfr strains have the F factor DNA integrated into their chromosome at different locations due to Different Hfr strains have the F factor DNA integrated into their chromosome at different locations due to homologous recombination between the F factor's origin of replication and the bacterial chromosome's origin of replication. homologous recombination between an IS element within the F factor and an IS element that may be located at different chromosomal locations in different E. coli strains. random breaks that occur within the bacterial chromosome. recombination between homologous chromosomal regions of donor and recipient cells during conjugation.arrow_forwardThe DNA of a deletion mutant of λ bacteriophage has a length of 15.4383 μm instead of 19.6356 μm. How many base pairs are missing from this mutant? *arrow_forward
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- Consider the following plasmid (size 8000 bp), with restriction sites at the positions indicated: (see image) a) This plasmid is digested with the enzymes listed below. Indicate how many fragments will begenerated in each case, and give the sizes of the fragments.PstIXhoICombination of PstI + XhoI + EcoRI (triple digest) b) Draw the banding pattern you would expect to observe if each of these digestions is loaded into a separate well of an agarose gel, and the fragments separated by electrophoresis. In the first well you load a DNA marker (M) containing fragments with sizes of 1000 bp, 2000 bp, 4000 bp and 8000 bp. c) This gel is transferred to a membrane in a Southern blot experiment, and hybridised to a radioactively labelled 200 bp probe, which anneals to the plasmid DNA at the position indicated on the diagram above. Draw the autoradiographic profile you would expect to observe for the membrane.arrow_forwardSTR sites are the basis of the FBI Laboratory's Combined DNA Index System (CODIS). One STR site that is used in CODIS is TPOX, named for its location within an intron of the thyroid peroxidase gene located on human chromosome 2. Different versions of the TPOX site are known, due to varying number of repeats of the short sequence "AATG". For TPOX, the number of repeats found on different chromosomes varies from 6 to 13. Hence the "alleles" of this site are called "6", "7", "8", "9", "io", "11", "12" and "13". Any given individual can be homozygous for any one of these eight different alleles or heterozygous for two different alleles of this set. The frequency of any given genotype, however, depends upon the individual frequency of each of these eight alleles in the population. For example, allele "6" occurs at a frequency of less than 5%, allele "11" at a frequency of 20% and allele 8" at a frequency of about 46%. See below for the double stranded DNA sequence of the *11" allele and a…arrow_forwardSupercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?arrow_forward
- DNA is extracted from the blood cells of two different humans, individuals 1 and 2. In separate experiments, the DNA from each individual is cleaved by restriction endonucleases A, B, and C, and the fragments are separated by electrophoresis. A hypothetical map of a 10,000 bp (base pair) segment of a human chromosome is shown (1 kbp= 1,000 bp). Individual 2 has point mutations that eliminate restriction recognition sites B* and C*. After the 10kbp segment is digested with each restriction endonuclease A, B, and C one by one, samples are loaded onto an agarose gel for electrophoresis for analysis. I would like you to draw the result of gel electrophoresis. To answer the question, first, draw the whole gel image (on the right) on your answer sheet, then indicate where you expect to see the bands on the gel. (Hint1: Same-size bands will appear at the same position. Hint 2: On the gel bands from individuals 1 and 2 might be at different positions.) The left lane (with an M label on the…arrow_forwarda) A plasmid DNA in bacteria has a length of 14,000 bp and an Lk of 1300. Calculate the superhelical density o for this plasmid. Show your work for partial credit, round to one digit after the decimal point. b) You use a Type II topoisomerase to change the linking number of this plasmid to 1310. How many turnovers must the topoisomerase perform? Is this resulting plasmid underwound or overwound?arrow_forwardIn a clinical context, a scientist is working with a viral DNA which is about 24000bps long. There are two known variants of the virus that share almost the same DNA but for a final fragment; with reference to Figure Q2b, the regions A and B are conserved in both variants, while the region C differs and is either 320bps (variant 1) or 380bps (variant 2). The scientist wants to set up a procedure to identify the variant they are dealing with. Viral dsDNA (i) (ii) (iii) Stable region (A) Variable region (C) Figure Q2b Known sequence (B) 5-GACCTCAATGTCCAGCGGTACGCTCATAAA-3' 3'-CTGGAGTTACAGGTCGCCATGCGAGTATTT-5' The scientists want to design a primer to amplify the variable region and to do so, they sequence a small fragment (sequence B) the conserved region close to the variable region C. Why is the scientist targeting a region outside of the fragment of interest? [3] The sequence of the fragment B is reported in Figure Q2b. Suggest a primer that can efficiently target this region and…arrow_forward
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