1Solutions to 7.014 Problem Set 7Question 1The plasmid vector pet83 contains an ampicillin resistance gene (Amp) and a kanamycinresistance gene (Kan). The plasmid is cut at the BamHI site shown below, and successfulyligated with a compatible human BamHI fragment. The recombinant plasmid is thentransformed into E. Coli. The transformation mixture is then spread on petri plates andallowed to grow.a) What type of E. Coli cells would you choose as the host strain for your transformation?You would begin with a strain that is ampSkan.b) What antibiotic would you include in the medium to ensure that only bacteria containingthe pet83 plasmid will grow?Ampicillinc) What two antibiotic resistance phenotypes will be found among the transformed bacteria?ampRkanR and ampRkanSd) Of the antibiotic resistance phenotypes listed above, which will contain the human BamHIfragment?ampRkanSSpring 2002 7.0142e) The human BamHI fragment that you cloned into pet83 encodes the enzyme helicase. Youdecide to clone the mouse version of the helicase gene.i) Briefly explain how you would make a mouse genomic library in E. Coli.To make a mouse genomic library in bacteria:1) Digest mouse genomic DNA with a restriction enzyme and digest a plasmid vector with thesame restriction enzyme.2) Ligate the mouse DNA into the vector.3) Transform ampicillin and kanamycin sensitive bacteria with this ligation mixture.4) Plate the transformed bacteria on media containing ampicillin. This selects for cells that arecarrying a plasmid.ii) Once you have the mouse genomic library in E. Coli, how would you find the cellsthat contain the mouse helicase gene?When thinking about this question, recognize that a mouse gene of similar structure andfunction as a human gene will probably have a similar DNA sequence. We can take advantage ofthis by assuming that DNA molecules with similar sequences will hybridize to each other.1) Plate the E. Coli cells that compose the library on petri plates with ampicillin containingmedia and make a copy or replica of the colonies on each plate using a piece of solid matrix(nitrocellulose).2) Lyse (break open) the bacteria on the nitrocellulose in such a way that single-stranded plasmidDNA is accessible.3) Wash the nitrocellulose with a solution containing radioactively labeled human helicase geneas a hybridization probe to identify colonies containing DNA similar to this gene.f) The restriction endonuclease sites for several enzymes are given below.BamHI: BglII: EcoRI t5'...G G A T C C...3'3'...C C T A G G...5' s t5'...A G A T C T...3'3'...T C T A G A...5' s t5'...G A A T T C...3'3'...C T T A A G...5' sPvuII PvuI EcoRV t5'...C A G C T G...3'3'...G T C G A C...5' s t5'...C G A T C G...3'3'...G C T A G C...5' s t5'...G A T A T C...3'3'...C T A T A G...5' si) DNA that has been cut with PvuII can be ligated to DNA that has been cut withEcoRV. Indicate what would be the DNA sequence of the ligated DNA by filling in theblanks below.5'...C A G A T C...3'3'...G T C T A G...5'ii) EcoRI fragments can be ligated together because the ends have compatibleoverhangs.• Which two enzymes above create compatible ends? BamHI and BglII• If a DNA fragment cut with one of these enzymes was ligated to DNA cut with theother, what would be the DNA sequence of the ligated DNA? Label 5' and 3' ends.5'...G G A T C T...3'3'...C C T A G A...5'• Can this new DNA be cut with either of the enzymes originally used? No.Spring 2002 7.0143Question 1, continuedg) You successfully clone a DNA fragment (2.1 kb in length) carrying the mouse helicase geneinto a unique EcoRI restriction site of the plasmid pQE1 (4.1 kb in length). You digest this newplasmid with the restriction enzymes EcoRI, BamHI, and PstI alone and in pairs. After runningout the digests on an agarose gel you get the following results. Lengths are in kb.Draw a restriction map of the plasmid; include the fragment lengths, restriction sites, and labelthe vector and the insert.Note that the PstI and EcoRI site withinthe insert are two separate sites. Theyare close enough, however, to look likeone site in the restriction digest.EcoRI BamHI PstIEcoRI & BamHIEcoRI & PstIBamHI& PstI5.54.11.01.0 1.00.7 0.73.72.5 2.52.40.60.60.50.23.91.1 1.11.52.6Spring 2002 7.0144Question 2a) Complete the following life table for the gray squirrel.Gray Squirrel Life TablexnxlxdxmxLxTxex0-1 530 1.00 0.747 0.747 332 536 1.011-2 134 0.253 0.147 0.581 95 204 1.522-3 56 0.106 0.032 0.302 47.5 109 1.953-4 39 0.074 0.030 0.405 31 61.5 1.584-5 23 0.043 0.021 0.488 17.5 30.5 1.335-6 12 0.023 0.013 0.565 8.5 13 1.086-7 5 0.009 0.006 0.667 3.5 4.5 0.907-8 2 0.004 0.004 1.0 1.0 1.0 0.508-9 0 0 0 0 0 0 0b) Complete the Fecundity table for the gray squirrel.xlxbxlxbx0-1 1.00 0.0 0.001-2 0.253 2.56 0.6482-3 0.106 4.56 0.4833-4 0.074 6.48 0.4804-5 0.043 6.48 0.2795-6 0.023 4.96 0.1146-7 0.009 4.56 0.0417-8 0.004 4.56 0.0188-9 0 0 0c) Is the population of squirrels increasing, decreasing, or remaining constant?Ro = 2.06. Therefore the population is increasing.Spring 2002 7.0145Question 2a) Draw three different types of survivorship curves (label your axes) and briefly explain thesurvival strategy associated with each.Type IType IIType III0%50%100%Age of Individuals in CohortType I: Low mortality throughout life until an age threshold is reached, after which time individuals die at a relatively highrate. This survival strategy usually entails parent(s) investing a lot of energy in raising their offspring, which keepsmortality among young individuals low.Type II: Constant death rate throughout life. Intermediate between types I and III, this survival strategy entails a moderateinput of energy from parent(s) in caring for their young. Populations which are subject to a constant rate of predation mayexhibit type II survivorship curves.Type III: High mortality among the young; individuals which survive past young age tend to live for long time. Thisstrategy usually entails parents generating a lot of offspring without investing much energy in caring for any one individual.The strategy is successful when at least a few of the many offspring survive their youth and reproduce.b) For each of the following cohorts (group of individuals born at the same time), decidewhich type of survivorship curve best applies. Explain briefly.i) Cohort of 20 sugar maple seedlings growing in the shadow of the parent, a tall sugarmaple in mature forest.Type III curve best applies, since the parent has generated
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