PLB161A Laboratory XI a Genome Mapping

Transcription

1 PLB161A Laboratory XI a Genome Mapping Restriction Digests and Agarose Gel Electrophoresis of Genomic DNA. A. Restriction Digests. Introduction Restriction enzymes are a class of DNA endonucleases, which occur in nature in various microorganisms. They have evolved as a defense mechanism in bacteria to degrade foreign DNA such as that introduced by a bacteriophage. Such exogenous DNA is rapidly cleaved into smaller fragments by the restriction enzyme and the bacterium is thus protected from viral infection. The bacterial DNA is protected from cleavage by methylation of sequences that would otherwise be cut. Type II restriction endonucleases are the type used most often in molecular biology. The proteins bind specifically to double stranded DNA and then cleave it, and after this has occurred, the enzyme dissociates from the DNA molecule. Some enzymes cleave exactly at the axis of symmetry, generating fragments of DNA that carry blunt ends. Other enzymes cleave each strand at similar locations, on opposite sides of the axis of symmetry, creating fragments of DNA that carry protruding single-stranded termini also referred to as sticky ends. The vast majority of type II restriction endonucleases recognize specific sequences that are four, five, six or eight nucleotides in length and display two fold symmetry: Eco RI (from Escherichia coli) cleaves GAATTC Pst I (from Providencia stuartii) cleaves CTGCAG Taq I (from Thermus aquaticus) cleaves TCGA Some restriction enzymes (rare-cutters) cut the DNA very infrequently generating a small number of very large fragments (several thousand to a million bp). These enzymes are generally used for physical mapping. For genetic studies using RFLPs restriction enzymes that cut DNA more frequently are generally used. These restriction enzymes generate a large number of small fragments. Restriction enzymes with 4-base recognition sites will yield pieces of 256 bases long on average and those with 6-base recognition sites will yield pieces of 4,096 bases long on average. These numbers are affected by the differential abundance of nucleotides in different organisms. Since hundreds of different restriction enzymes have been characterized and are available commercially, DNA can be cut into many different small fragments. Methylation effect on restriction enzymes Plant DNA has a high content of 5-methylcytosine. Approximately 80% of the CG, and CNG sequences are methylated in plants. Some restriction enzymes will not cut 5m CG sequences (i.e. HhaI, HpaII, SalI, XhoI) and others will not cut 5m CNG (i.e. EcoRII, PstI, PvuII). I some cases there are isoschizomers (restriction enzymes from different sources but with the same recognition site) that differ in their sensitivity to -17-

2 methylation. BstNI is an isoschizomer of EcoRII but BstNI is not blocked by 5m CNG methylation. Sometimes, the 5m CG site is not within the restriction site, but can be formed with the adjacent bp in the DNA sequence. For example, Apa I (GGGCCC) does not have a CG site, but if the next bp is a G or the previous base pair is a C, there will be a CG pair that if methylated ( 5m CG) will not be cut by Apa I. This is called overlapping methylation. Methylation and overlapping methylation affect the average size fragments produced by a restriction enzyme. In the case of internal methylation, if we assume random distribution of the 4 base pairs (=0.25 probability of occurrence for each one), the expected average size of a six-cutter restriction enzyme sensitive to 5m CG methylation (e.g. SnaBI TACGAT) the probability of finding an un-methylated C is 0.25 * 0.20=0.05 (because 80% of the CG are methylated). Therefore the expected size will be 1/[(0.25) 6 *0.2]= 20,480-bp. Therefore, the average fragment size will be five times larger larger than in a restriction enzyme without sensitivity to methylation (1/(0.25) 6 = 4,096- bp. In the case of overlapping methylation, you need to include in the calculation the probability that the adjacent bases will generate a 5m CG or 5m CNG. For example if your enzyme recognition sequence is GATC, and the enzyme is sensitive only to 5m CG methylation, the GATC site will have adjacent C or G in 7/16 of the combinations indicated in red below. In 80% of the 7/16 the CG will be methylated and the enzyme will not cut (7/16 * 0.80= 0.35). agatca tgatca cgatca ggatca agatct tgatct cgatct ggatct agatcc tgatcc cgatcc ggatcc agatcg tgatcg cgatcg ggatcg The probability of a GATC site cut, based on the adjacent bases is 9/16 + 7/16*0.2 = Therefore the expected size will be 1/[(0.25) 4 *0.65]= 394-bp. If the enzyme is sensitive to both 5m CG and 5m CNG overlapping methylation the denominator needs to be multiplied again by 0.65: 1/[(0.25) 4 * ]= 605-bp Once a plant DNA segment is cloned and introduced into a bacteria, for example in a BAC clone, the methylation pattern of the plant disappears and is replaced by the methylation of the bacteria. Therefore, when digesting a BAC clone, you need to consider the effect of Dam (G m ATC) and Dcm (C m C(A/T)GG). For example if a BAC clone is cut with a restriction enzyme sensitive to overlapping Dam methylation as Xba I (TCTAGA), the sites followed by TC (TCTA G m ATC) will not be cut in the bacteria but will be cut in the plant genomic DNA producing differences in the observed restriction fragments. -18-

3 Technical information: Homogenization: One of the primary problems encountered with digestions of high molecular weight DNA is that the DNA is sometimes not evenly distributed in solution, leading to localized regions of high DNA concentrations. These clumps of DNA are relatively inaccessible to restriction enzymes, and result in only the partial digestion of the DNA. If clumps of DNA are seen, it is necessary to homogenize the solution before the digestion. Mixing high molecular weight genomic or nuclear DNA requires gentle handling in order to minimize mechanical shearing of the DNA during extraction. Use wide-bore pipet tips prepared by removing the tip of a regular pipet tip with a sterile razor blade. Solutions of high molecular weight DNA are quite viscous and are therefore difficult to pipet. Time and number RE units: By definition one unit of restriction endonuclease will completely digest 1 µg of substrate DNA (lambda for example) in a 50 µl reaction in one hour. Restriction digestion of high molecular weight genomic DNA requires longer digestion times (2 to 3 hours) and 3 units of restriction enzyme per µg of genomic DNA. If after digesting the DNA for three hours the restriction is not complete; a second aliquot of enzyme can be added to continue to digest the DNA. DNA amount and concentration: Wheat has a very large genome with huge amounts of repetitive DNA. Consequently, to obtain an adequate detection of low copy number RFLP sequences it is necessary to load more DNA in each well than in RFLP studies of species with small genomes like rice or tomato. For wheat RFLPs, µg of DNA are loaded in each well. In our class we will load only 5 µg because we will use a high copy number probe. The maximum volume that can be loaded in one well is generally lower than 80 µl. To load this amount of DNA in this volume the DNA concentration of the samples should be higher than 150 ng/µl (10µg /150 ng/µl= 67µl). Although DNA solutions of high molecular weight DNA are viscous, due to the length of the molecules, the concentration of genomic DNA preparations can be quite low. If the concentration is lower than 150-ng/µl the restriction digest can be performed in a larger volume and then concentrated in a vacuum concentrator before loading. Procedures: Restriction Digests of Wheat Genomic DNA 1. Determine the total µl of each DNA sample to be digested with an enzyme in a single tube as follows: Total µl DNA = (Total µg DNA) / (DNA concentration, µg/µl). We will digest 5 µg of DNA with restriction enzyme Taq I (65 C). Example: if a DNA concentrations is 200-ng/µl add 25 µl of DNA. -19-

4 Some enzymes as TaqI work better with BSA (Bovine Serum Albumin). Use 1/100 of the total volume 2. Determine the units (U) and volume of enzyme necessary to digest each DNA sample. In general, it is best to use 3.0 U/µg DNA to prevent partial digestions. Total U Enzyme = (Total µg DNA) x 3.0 Total µl Enzyme = (Total U enzyme) / (enzyme concentration, U/µl) For 5 µg of DNA use 15 U. TaqI is at 20 U/µl so use 0.75 µl of enzyme per sample. Note: Do NOT add enzyme in excess of 10% of the reaction volume. Many enzymes exhibit star activity and cut the DNA at nonspecific sites in the presence of glycerol > 5%. The restriction endonucleases are supplied in 50% glycerol. 3. Based on the DNA and enzyme volumes, determine the total reaction volume. In this example total volume is 50 µl. 4. Determine the volume of ddh 2 0 per tube as follows: µl ddh 2 0 = 50µl 5 µl buffer - µl DNA - µl enzyme -µl BSA For example if DNA concentration is 200 ng/µl, and the restriction enzyme is 10U/µl add = µl of ddh Use permanent black marker to label 20 sterile PCR tubes. Set up a reaction for each of the 14 nulli-tetrasomic lines. Add the proper amount of water first to each tube (you can use the same tip) and then the DNA samples (using new sterile tips for each sample). If all DNA samples were adjusted to the same concentration, the water can be included in the digestion mix with the buffer and the enzyme. 6. Calculate a bulk digestion mix containing the buffer, and enzyme needed for the total number of different DNA samples to be digested by the same enzyme. Prepare bulk mix on ice, adding enzyme last; mix well. To allow for pipetting errors prepare extra bulk mix (usually one extra every 20 samples). 5 µl buffer * 15= 75 µl 0.75 µl enzyme * 15= µl 0.5 µl BSA * 15= 7.5 µl 7. Add 6.25 µl of mix to each tube. An extremely important, yet often overlooked, element of a successful restriction digest is mixing. The reaction must be thoroughly mixed to achieve complete digestion. Gently pipet the reaction mixture up and down (avoid bubbles). Close tube tops and pool reagents by pulsing in a microfuge or by sharply tapping the tube bottom on lab bench. 8. Place reaction tubes in a water bath or PCR machine with the optimum temperature for the enzyme. Note: The recommended incubation temperature for most restriction -20-

5 endonucleases is 37 C. restriction endonucleases isolated from thermophilic bacteria require higher incubation temperatures ranging from 50 C to 65 C. Always check catalog for optimum incubation temperature. 9. Following incubation, add 5 µl of loading dye and freeze reactions at 20 C until ready to continue. 10 X Loading Dye: [FINAL] 50 ml 100 ml Ficoll type % 7.5 g 15.0 g Bromophenol blue 0.25% g 0.25 g Xylene cyanol 0.25% g 0.25 g 50 X TAE buffer 10 X Tae 10 ml 20 ml ddh 2 0 To 50 ml To 100 ml B. AGAROSE GEL ELECTROPHORESIS Introduction Electrophoresis through agarose or polyacrilamide gels is the standard method to separate, identify and purify DNA fragments. Although agarose gels have a lower resolving power than polyacrilamide gels, they have a greater range of separation and are much simpler and easier to handle than the polyacrilamide ones, and have been more widely used in DNA electrophoresis. Agarose, a purified form of agar isolated from seaweed, is a linear polymer. Various grades and forms of agarose are available commercially. Agarose gels are cast by melting the agarose in the presence of the desired buffer until a clear, transparent solution is achieved. The melted solution is then poured into a mold and allowed to harden. Upon hardening, the agarose forms a matrix, the density of which is determined by the concentration of the agarose. The gel is then submerged in running buffer and the DNA samples are loaded into wells that were cast into the gel using a plastic comb at the time of preparation. An electrical field is then applied across the length of the gel and the negatively charged DNA molecules migrate toward the anode of the electrophoretic system. The rate at which linear DNA molecules migrate within the gel depends on their length: larger molecules migrate more slowly than smaller ones because it is more difficult for them to pass through the agarose pores. Thus, the DNA molecules become separated according to their size. At high gel concentrations the resolution of smaller fragments is favored, while at low agarose concentrations that of larger fragments is favored. DNA is visualized by soaking the gel in a solution of ethidium bromide; this fluorescent dye intercalates between the two strands of DNA and makes them visible under ultraviolet light. When plant or animal DNA is digested with restriction endonucleases and the DNA fragments are separated by gel electrophoresis, the result is a smear of millions of -21-

6 bands. In the DNA hybridization laboratory we will learn how to visualize specific sequences within this smear. It has been shown empirically that individuals are not identical with respect to the distance (in bases of DNA) between restriction endonucleases cleavage sites. We will use these differences in sizes as molecular markers for genetic studies. Optimum agarose concentrations for different DNA sizes. % Gel Optimum Resolution for Linear DNA (kb) to to to to to 0.2 Procedure 1. Prepare a 1% agarose gel cast in 1X TAE using a gel rig and a 2-mm comb with 15-wells. For 100-ml: i) Weigh 1g of SeaKem LE agarose, transfer to a 250-ml flask ii) Add 1XTAE to 100 ml iii) Melt agarose in microwave oven iv) Pour molten agarose (cooled to 60 o C) into rig with comb; let gel cool 20 minutes to harden. Be sure that the comb is at least 1mm above the surface of the rig. vi) Cover gel with 1X TAE buffer 2. Heat the λ marker at 60 o C for 3-5 minutes, cool on ice, and briefly microcentrifuge to collect sample in bottom of tube. If the restriction digests have been stored at 4 o C, heat the digests at o C for 2-3 minutes to disrupt any base pairing between protruding cohesive termini before loading the gel. 3. Load the molecular weight standard and the samples slowly. 4. Run the gel slowly overnight at constant voltage (35 Volts). This maximizes resolution of restriction fragments, prevents smearing of bands, and reduces distortion or smiling of the DNA bands. 5. After electrophoresis is completed, stain the gel for 10 minutes in an ethidium bromide staining solution (0.25 µg/ml) if the ethidium bromide was not added to the gel. Rinse the gel briefly in ddh 2 O and photograph the gel using the still-image video system. Be sure to place a fluorescent ruler along the side of the gel before recording the image. 6. Using permanent waterproof ink (i.e. India ink) and an 18-gauge needle, mark the molecular weight markers. Take precautions to prevent nicking the UV filter! Place -22-

7 a sheet of UV-transparent Plexiglas under the gel before marking the lanes. Restriction Enzymes Exercises 1) What is the optimal digestion temperature? Taq I: Bsa I: Bst EII: 2) Which of the following enzymes need BSA? ClaI: XbaI: SpeI: 3) Which is the best New England BioLabs buffer? For NdeI: Double digestion EcoRV PstI: EcoRI HindIII: 4) What is the average size fragment generated by RE assuming random distribution of base pairs? DdeI: AscI: AccI: AvaII: 5) Calculate the expected average size fragment generated by Dra I assuming a) random distribution of base pairs and b) using the known base pair composition of the wheat genome: A=27.2%, T=27.1%, C=22.7%, and G=23%. 6) What of the following enzymes will have problems to digest plant DNA because of methylation and at which base methylation will occur? SalI: PvuII: TaqI: 7) Calculate the expected average size fragment generated by BglII (AGATCT) considering that this enzyme is sensitive to overlapping methylation at the beginning and the end of the recognition site. Assume random distribution of base pairs and that 80% of the CG pairs are methylated. 8) Calculate the expected average size fragment generated by BstZI (CGGCCG) considering that this enzyme is sensitive to internal and overlapping 5m CNG methylation. Assume random distribution of base pairs and that 80% of the CG and CNG are methylated. 9) Explain why you can get different fragment sizes for a segment form total plant genome DNA and from a BAC from the same species and from the same region when digested with Taq I. -23-

8 10) Find the isoschizomers of the following enzymes and indicate the differences in methylation sensitivity between them BstNI: MspI: 11) Calculate the number of copies of a single copy gene present in a restriction digest of 10 µg of diploid wheat DNA and in 5 µg of rice DNA. Assume that there is no restriction site within the gene. The size of the wheat genome is 5.6 pg per haploid genome and that of the rice genome 0.5 pg per haploid genome. 12) How do you know if a RE is methylation sensitive by looking at a digestion of genomic DNA in the gel? 13) Indicate if there are higher possibilities to find polymorphisms with a four- cutter restriction enzyme or a six-cutter RE in the following two scenarios: a) All the polymorphisms originated by insertion deletions (no point mutations) b) All point mutations (no insertions or deletions). How does the size of your probe affect your answer? There are 6 additional questions for this Lab report after the Southern chapter. -24-

9 Exercise 1. Indicate restriction maps compatibles with the observed polymorphisms. Use insertion/ deletions or base pair mutations -25-