Gentics Lab Results：遗传实验结果.docx

Sordaria Pictures, compliments of Macie Schick and the cool LSRI cameraClass results for 10/9/07- mostly lab 2- lab 1 can use this data; if you did not find any hybrid asci, then analyze the pictures below for your data; add it to the class data Lab 2 class dataNone112+:2g:2+:2g112+:4g:2+ 42g:4+:2g6Weird2 Lab 2872603310Right sideLeft sideLab 1Instructions for PCR Lab of 10/16-10/23Here are some useful web sites:/bio/ubl/Alu.doc ; /Manual/cr_D1S80_pcr.doc Here are the results from your gels. It appears that 10 of 17 had DNA that amplified in lab 1, and lab 2 had 7 of 18 that amplified. As we discussed in class, the D1S80 locus is one of the places with VNTRs. There are many different possible alleles. Your assignment is to produce a report on this PCR exercise. Include the following in your report:Introduction: briefly describe the purpose of this lab, the principle of PCR, and the principle of identifying a person based on PCR.Procedure: Indicate that DNA was obtained from saliva samples from the class, and amplified by primers for D1S80 and separated by agarose gel electrophoresis. Summarize the sampling procedure, and the PCR procedure. Include the following information about the gel: It was a 1.5% agarose gel, run for about 2 hours at 75 volts. The DNA was amplified with 30 cycles: Denaturing: 94o C, 1 minAnnealing: 68o C, 1 minElongation: 72o C, 1 min.Results: Give the the results of your PCR- if you dont remember which lane was yours, I can tell you. Indicate whether you are heterozygous or homozygous at the D1S80 locus, and the size of the bands that you produced. If you did not get bands, give results for Lane nine of lab 1, or the “crime scene” DNA for lab 2 (Lane 9 on both the left and right gel). How many different genotypes did you find for the two labs? Determining the size of DNA: We will use the size standards and Excel to produce a graph of the distance migrated from the well (in mm) (X axis) vs size of DNA (Y axis). The rate at which DNA migrates in a gel is inversely proportional to the log of its size (usually in base pairs or kilobase pairs of DNA); this means that small DNA fragments migrate faster than large DNA fragments, and if a fragment of 100 bp migrates at 40 mm/hour in an electric field, then a fragment of 1000 bp would migrate at 26 mm/hour in the same field- dont worry about the math. After plotting the standard curve, then determine the distance that the fragments migrated in the lane or lanes that you analyzed. Use your “standard curve to determine the size of the fragment produced. You really only have to plot four points- the 1078, 872 bp, 603 bp, and 310 bp standards.1078872603310Example: Lets determine the size of the two pieces of DNA in the box. These were fragments amplified from a student in a previous lab.The lane with all the bands had pieces of known size; four are given, and the largest and smallest are marked with an arrow. Our job is to determine the size of the bands in the box.To do this, we get an actual picture of this get (on paper- there are probably fancier ways of doing this!) and measure the distance that the four known fragments migrated. You also measure the distance that these two bands migrated. You get a table that looks like this:Distance migrated, mmSize, BpSample, Distance migrated, mm46107859.550872615760369310We then use Excel to graph our results, using a scatter graph. On Excel, go to “insert”, then pick “scatter” graph, and highlight the table; youll get a graph that looks like this:Next, we change the Y axis to a logarithmic scale. Click on the Y axis, right click to give you options, and set the axis options as shown above: Logarithmic scale, with the values shown. You should get a graph that looks like this:Then, add major and minor gridlines to both axes. Now click on one of the points, and add a trendline. Youll want an exponential trend, showing both the equation and the r2 value.Now that youve done that, you can determine the size of your fragment either graphically or by using the equation shown. The equation may be the easiest, although make sure it passes the sense test (you should get a size between 400 and 550 base pairs)If you substitute your value for x (distance migrated) in the equation, it will give you the size of the fragment in base pairs. Into one of the cells in Excel put the equation for Y, and plug in 59.5 for x:=13108*EXP(-0.054*59.5)This will then give you 527.4099. This is clearly false accuracy, but a value of 527 bp is reasonable. Plugging in the value for the second fragment gives you a size of 486 bp.Do this same exercise for determining the size of your DNA fragments, or of another one, if yours didnt turn out. Discussion: Comment on the success of your sample- why it may have worked or may not have worked. The “Crime Scene” DNA was lane 24 (last one on the right with DNA) for lab 1, and lanes 9 for both gels in lab 2. Who are possible suspects from the class (i.e., that have similar patterns)? If someone had DNA that is consistent with their being the person who left their DNA at the crime scene, what would be the next step?Mutation, Mutants, and Mutagen labs- Here are the numbers for the 2 labs.The lab 1 201 & 202 plates- with 1 g/ml tryptophan: total # of bacteria per plate201: 6.6 X 109/plate202: 8.8X 109/plateHere are the result from the 2 labs. The numbers are the # of Trp+ revertants that you recorded.NOTE: the diameter of the plate was 85 mm.RS201Control 10, 21, 24MMS-100%0-; 85 mm zone of inhibition (entire plate)MMS-10%134NaN38NQO21EMS-100%95, 207;17 mm zone of inhibitionEMS-10% 128;17X 20 mm zone of inhibitionRS202Control0MMS-100%0; 35 mm zone of inhibitionMMS-10%No dataNaN30; 43 mm zone of inhibitionNQO1?( possible contaminant)EMS-100%0EMS-10%No data The lab 2, 201 and 202 plates: with 5 g/ml tryptophan:201: 2.2X 1010/plate205: 2.4 X 1010/plate.RS201Control 75MMS-100%0-; 65 mm zone of inhibitionMMS-10%226; 26 mm zone of inhibitionNaN33NQO110EMS-100%1;22 mm zone of inhibitionEMS-10% 54;17X 20 mm zone of inhibitionRS2202Control0MMS-100%0MMS-10%1? ( possible contaminant)NaN30NQO0EMS-100%2EMS-10%0NOTE: I checked the four potential revertants from the two labs; one was a contaminant; two didnt grow in the absence of trp; and one may be a revertant that grows slowly in the absence of trp, which is a relevant observation.For both Labs: Use the estimate of the total # of bacteria/plate to estimate the frequency of reversion of RS201-2 and RS202-5, in the absence of mutagens. This is obtained by dividing the number of revertants by the total number of bacteria/plate. Which mutagen was the most toxic? Which increased the frequency of mutation the most? Did any of the mutagens actually have an “antimutagen” effect? How do you explain this result (note: dont go too deeply on this one)What is the effect of requiring two mutations to revert, instead of one?From the results with the suspected mutagens, indicate whether they actually increased the number of revertants with the strain tested.For your Discussion.Discuss the classresults- were you able to show the acquisition of the TrpA+ function? What effect did mutagenesis have on the rate at which this occurred?From the class results, what effect did a requirement for two independent mutations have on the ability of the cell to acquire a TrpA+ function? How did mutagenesis affect that rate?: RWSResults from gel of 11/20/07:The DNA size Standards: Each band is 25 ng, except that the 1 kb band is 60 ng. You ran 10 l of 24 l total of your reaction mixtures.Assignments: 1) Determine the size of the Pst I fragments of pUC119.kan, and the size of the repetitive DNA in the cow DNA. Good (i.e., complete) Pst I digestions of pUC119.kan are found in most of the lanes; in lab 1, lanes 7&8 show partial digestion, and lane 6 on lab 2 shows partial digestion. The uncut cow DNA is in lane 3(lab 1) and lane 1 (lab 2); the EcoR1-digested cow DNA is lane 2 for both labs. The only uncut plasmid DNA is lane 14 of lab 1.2) Determine the amount of DNA in the large PstI fragment of your digest, or that of another sample, if yours didnt cut well. Other questions, from the handout: Uncut DNA: How did the uncut chromosomal DNA compare with the uncut plasmid DNA?Cut DNA: What changes occurred in the plasmid and chromosomal DNA as a result of digestion with the restriction enzyme? Results from PCR Demonstration of 10/18/05. Heres the picture of the gel.Size Standard: size of bands in base pairs is on the leftINSTRUCTIONS FOR PCR DEMONSTRATION OF 10/18/05Your assignment is to produce a report on this demonstration. Include in your report the following:Introduction: briefly describe the purpose of this lab, the principle of PCR, and the principle of identifying a person based on PCR.Procedure: Indicate that DNA was obtained from saliva samples from four volunteers, and amplified by primers “D1S80”, and separated by agarose gel electrophoresis. Summarize the sampling procedure, and the PCR procedure. Include the following information about the gel: It was a 1.4% agarose gel, run for 2 hours at 50 volts. Analyzing the results:We will pretend that there was DNA left at a crime scene (CS), and four suspects. We obtained DNA from the CS and the suspects, amplified it by PCR, and ran the samples on the gel, obtaining the results shown.We obtained reasonably good results from our PCR and gel, although it took 2 PCRs, three gel runs and four pictures to get them. We will only concern ourselves with “suspects” lanes 2,3&4, which gave the brightest bands; the “crime scene” lane (CS); and the standard lane (to the right of the CS lane. The other two lanes show an amplified trpA gene of E. coli (my favorite gene) and the Rubisco gene from onion DNA (rubisco, remember, is ribulose 1,5 bisphosphate carboxylase- the most abundant enzyme in the universe, and the enzyme that catalyzes the first step in the Calvin cycle, which is how CO2 is fixed in plants. We actually amplified the small gene for the small subunit of this enzyme, which is found in the onion nucleus; the gene for the large subunit is found in the chloroplast. None of the rubisco information is testable).Use the size standards and the semi-log graph paper to graph a “standard curve” of distance migrated from the well (in mm) (X axis) vs size of DNA (Y axis). The rate at which DNA migrates in a gel is inversely proportional to the log of its size (usually in base pairs or kilobase pairs of DNA); this means that small DNA fragments migrate faster than large DNA fragments, and if a fragment of 100 bp migrates at 40 mm/hour in an electric field, then a fragment of 1000 bp would migrate at 26 mm/hour in the same field- dont worry about the math. After plotting the standard curve, then determine the distance that the fragments migrated in “suspect” lanes 2,3& 4, and use this information and the standard curve to determine the size of these fragments. You can consider that the largest fragment in lanes 2& 3 are the same size, and the single band in lane 2 is the same size as the small fragment in lane 4.After determining the size of these fragments, then indicate which of the suspects has DNA similar to that found at the crime scene.Discussion: Give me your impressions of this demonstration, and whether you think the evidence gathered would be enough to convict a suspect of the crime. Keep in mind that, due to time and material constraints, we were unable to have each of you perform this exercise. Also, you will have the chance to handle DNA later in the course.12/02/05: Plating results for sewage sample and phage sample- combined data for labs 1 & 2 # colonies dilutionUnselected plates- nutrient agar122, 6610-38,8,8,1110-4MacConkey agar14total/3 Lac+17 total/4 Lac+11 total/4 Lac+10-32 total/ 0 Lac+2 total/0 Lac+2 total/1 Lac+3 total/1 Lac+10-4Tetracycline plates3216476611210-1066310-2Plaques from phage assay268, 26610-8Where there are several numbers reported, give the average and standard deviation for the result.The tetracycline results are a bit odd, in that the numbers varied so much. This is partly due to the number of very small colonies that developed- whether or not those are counted influences the results.12/05/05: Link to the Linkage Homework FOR FALL 2005: IGNORE ANYTHING BELOW THIS POINT!Lab results for gel electrophoresis lab.For your analysis, youll be using both the gel from lab 1, and the gel from lab 2.In lab 1, I made a BAD mistake- I had the electrodes reversed! Your DNA ran off the gel. However, I ran the samples again.After the well, the next band on many lanes is DNA from your run, which shouldnt be there; ignore it. I ran seven samples, and was able to identify the symbols for six of them.For lab 2, the camera cut off the left-most well (lane 1) and most of the right-most well (lane 13)Size of lambda DNA standards (Lane 5, lab 1; lane 9, lab 2), from top to bottom, in kb: 23.1, 9.4, 6.6, 4.4, 2.3, 2.0Amount of DNA in each band: 23.1: 72 ng; 9.4: 29 ng; 6.6: 21 ng; 4.4: 14 ng; 2.3: 7 ng; 2.0: 6 ng. These bands are labeled in the Lab 2 gel.Questions:How did the uncut plasmid DNA compare to the uncut cow and uncut onion DNA? Use the data from lab 1, lanes 1, 10, and 12. You can also use the data from lab 2, lanes 3 and 11.What changes occurred in the plasmid DNA as a result of digestion with the restriction enzyme? Use the data from lab 1, lanes 1 (uncut) with lane 6.What changes occurred in the chromosomal DNA as a result of digestion with the restriction enzyme? Use the data from lab 1, lanes 10 and 11Use the Lab 2 gel to determine the size of the fragments marked by a ? , and the amount of DNA in the larger of the two fragments.Determining the size of a fragment: DNA migrates through the gel according to its size in kb; the smaller it is, the faster it runs. To determine the size of your fragments, use the lab 2 gel. Measure the distance from the top of the gel to the marks on the side of the photo corresponding to each fragment. Use the semi-log graph paper to plot the size in kb vs the distance migrated for the known points, to produce a standard curve. Then measure the same distance for the unknown points. Use the standard curve and the distance traveled to determine the size of the size of the unknown fragments.Determining the amount of a fragment: The amount is how much of the DNA you have, in ng. For this, we use the brightness of the bands, which correlates to the amount of DNA in a band. The amounts of the DNA in the lambda standard are given above. Ignore this bandLab 2: the first lane is lane 2, the last full lane, lane 12 Determine the size of the DNA in this bandDetermine the size and amount of the DNA in this band Results from 2001 labsGentics of Bacteria Lab ResultsLab 1; the numbers represent averages, when 2 plates were countedSample #Colonies DilutionSewage: 226 10 -5TotalAssaying for Phage Plaques# Plaques Dilution52 10 -7Sewage: r- 5 10 -1Selective w: 6TA1538 73 10 -7Total:TA1538 328 10 -1Selective- no mutagenTA1538 117 10 -1Selective- with NQOLab 2 results the numbers represent averages, when 2 plates were countedSample #Colonies DilutionSewage: 189 10 -5TotalAssaying for Phage Plaques# Plaques Dilution52 10 -7Sewage: r- 40 10 -1Selective w: 58TA1537 310 10 -6Total:TA1538 3 10 -1Selective- no mutagenTA1538 7 10 -1Selective- with NQOGel of 12/04/01Size of lambda DNA standards (Lane 5), from top to bottom, in kb: 23.1, 9.4, 6.6, 4.4, 2.3, 2.0Amount of DNA in each band: 23.1: 72 ng; 9.4: 29 ng; 6.6: 21 ng; 4.4: 14 ng; 2.3: 7 ng; 2.0: 6 ng. NOTES: Youll focus on lane 3 (from the left), pUC.kan cut with Pst, and lane 4, pUC.kan cut with Bam HI. The smallest band on lane four is 1.4 Kb in size.Your job: Determine how big the large band in lane three, and the large band in lane four is.Also answer the other two questions on the cow DNA, cut and uncut- how do they differ? The pUC.Kan, cut and uncut- how do they differ? How does uncut Cow differ from uncut plasmid?Two extra credit brain teasers: The band on lane two and the band on lane three are supposed to be the same size, but clearly are not! One thing that can affect a band is the amount of DNA; sometimes larger amounts run