Ashwin Suresh 18 December 2012 Period 7 Wolbachia Lab Report Purpose_________________________________________________________________ The purpose of this lab Introduction______________________________________________________ Wolbachia is a parasitic bacterium that infects 20% of known insect species. Wolbachia interferes with a variety of tissues but mainly the sex organs. An infected sperm cell cannot properly produce offspring with an uninfected egg cell. However, when an infected sperm cell and an infected egg cell merge, they result in perfectly fine offspring. Therefore, infected females produce more offspring. In many insect species, unfertilized eggs result in males and fertilized eggs result in females. The unfertilized eggs of an infected female always duplicate their chromosomes always resulting in female offspring. The primary effect of Wolbachia is insect populations with a high percentage of females. (Timmer Arstechnica.com) The shape of a DNA molecule is a double helix consisting of nucleotides. Each nucleotide has a sugar, deoxyribose, and a phosphate group which make up the backbone of the DNA molecule. Each nucleotide also has a nitrogenous base, which could be adenine, guanine, cytosine, or thymine. The nitrogenous bases are key to how the DNA molecule stores information. Each DNA molecule is arranged so that adenine (A) is opposite thymine (T) and guanine (G) is opposite cytosine (C). During the replication (S) phase of the cell cycle the enzyme helicase separates the opposite nucleotides resulting in two single helix strands. Each side is used as guide for a new complementary strand to be attached to it. DNA polymerase matches up corresponding nitrogenous bases; A to T and G to C. DNA ligase joins the nucleotides into one strand opposite of the original. A similar process is used when the cell reads the DNA. RNA polymerase separates the DNA strands and uses one strand as a template to create mRNA. mRNA diffuses into the cytoplasm where it is read by transfer RNA. Each three base sequence on the mRNA, called a codon, code for a different amino acid. The amino acids are then assembled to form a protein. The structure of DNA allows it form many different proteins, thus creating very complex organisms. In a eukaryotic cell, the double helix is wound around proteins called histones and then further compiled into nucleosomes, then coils, then supercoils, and finally chromosomes which are visible during cell division. Prokaryotic DNA is not wrapped around any histones therefore it is not formed into chromosomes. Eukaryotic DNA is linear and is contained in a number of chromosomes. Prokaryotic consists of only a single circular DNA molecule and a variety of smaller plasmids. DNA in a eukaryotic cell is enclosed within a nucleus, while prokaryotic cells do not have a nucleus. (Fancher) PCR (Polymerase Chain Reaction) is a process that amplifies a segment of DNA into billions of identical copies. This enables scientists to study certain genes in an entire genome. The materials involved in PCR are nucleotides, primers, and Taq polymerase. Primers are usually 20 nucleotides long and are complementary to the ends of the DNA that is wished to be amplified. A PCR reaction takes up several hours and has twenty to thirty-five cycles. The first step of a cycle is to heat the DNA to 95o Celsius, separating the hydrogen bonds between the two strands. The second step is too cool down the mixture to 60oC, allowing the primers to bond to their complementary sequence on each DNA strand. The third step consists of raising the temperature to 72oC so the Taq polymerase can add nucleotides to the 3’ end of each strand. After four cycles, half the DNA strands contain only the target DNA. The number of target DNA strands increases exponentially after more cycles. Ashwin Suresh 18 December 2012 Period 7 During gel electrophoresis, one side of the gel has positive charge and one side of the gel has a negative charge. DNA that has reacted with restriction enzymes is placed at the negative end of the gel in wells. The electric current is then turned on. Because DNA has a negative charge from the phosphate group, it is repelled from the negative side and attracted to the positive side. The restriction enzymes cut up the DNA where ever it reads the sequence “GAATTC”. The shorter pieces of the DNA move faster through the porous gel and end up closer to the positive side. The longer strands take longer to navigate through the gel and end up closer to the negative side. DNA from different individuals have this sequence at different points in their DNA. Therefore, each individual has a unique pattern of DNA segment sizes resulting in a unique pattern of bands on the gel. Hypothesis____________________________________________________________ If the sample arthropods are infected with Wolbachia, then we expect to find additional bands on the gel other than the bands for mitochondrial DNA. If the sample arthropods are not infected with Wolbachia, then we only expect to see the bands from the mitochondrial DNA. Materials and Method______________________________________________________ The first part of the lab is DNA extraction. Get two microcentrifuge tubes and number them 1 or 2 in addition to your period and group number. Use P200 pippettor to pipet 180 microliters of buffer ATL into each tube. ATL is a detergent that breaks down the insect. If the insect was stored in ethanol, blot away the ethanol with a Kimwipe. Do the same for the + or – Nasonia. Using tweezers place the sample insect into tube 1 and the nasonia into tube 2. Using a microtube pestle, macerate the insect in tube 1. It is important to do this step quickly, because macerating the insect frees DNases that eat DNA. Immediately after the previous step, add 20 microliters of proteinase K. Proteinase K is an enzyme that breaks down histones to unravel the DNA. Proteinase K also breaks down DNases to prevent the DNA from being harmed. Now, add 200 microliters of buffer AL to the mixture. Buffer AL is a detergent that breaks down the phospholipid membranes, which are made of fatty acids. Vortex the mixture for ten seconds. Repeat for the nasonia in tube 2. Incubate both tubes at 70oC for at least ten minutes. After the tubes are done in the incubator, add 200 microliters of 95% ethanol to each tube. Ethanol is polar and extracts the DNA from the other content in the tube. Vortex for ten seconds. Store the tubes at 4oC overnight. Get two spin columns fitted in 2.0 milliliter collection tubes. Using a Sharpie marker, label the lid of each spin column 1 or 2 along with your period and group number. Using a P1000 pippetor, pipet 600 microliters of liquid from yesterday’s “1” tube into spin column 1. Centrifuge the spin column at 8,000 rpm for one minute. The DNA is now stuck to the filter of the spin column. Take out the tube and dispose of the waste in the collection tube. Place the spin column back into the same collection tube that you just emptied. Pipet 500 microliters of buffer AW1 into the spin column. AW1 denatures the proteins still mixed with the DNA. Centrifuge for one minute at 8,000 rpm. The DNA is still caught on the filter of the spin column. Take out the tube and empty the waste from the collection tube. Place the spin column back into the same collection tube that you just emptied. Pipet 500 microliters of buffer AW2 into the spin column. AW2 is another wash buffer that helps purify the DNA. Centrifuge for three minutes at 14,000 rpm. Take out the spin column and place them into a labeled 1.5 milliliter microcentrifuge tube. Pipet 100 microliters of buffer AE into the spin column. The AE is an elution buffer and unsticks the DNA from the filter and into the microcentrifuge tube. Incubate the tube at room temperature for one minute. Centrifuge at 8,000 rpm for one minute. Discard the spin column but keep the labeled microcentrifuge tube that contains the DNA. Store at -20oC until ready for PCR. Pipet Ashwin Suresh 18 December 2012 Period 7 600 microliters of liquid from yesterday’s “2” tube into spin column 2. Repeat the same steps for spin column 1. The second part of the lab is the polymerase chain reaction or PCR. Get two PCR ready tubes and label like all the other tubes. Each PCR tube contains a pellet, that is preformulated with Taq polymerase, MgCl2, Buffer, and dNTPs. Pipet 23 microliters of master mix into each PCR tube. The master mix contains 12 microliters (5 micrometers) of the primers, Wspec-F, Wspec-G, CO1-F, CO1-R. The master mix also contains 90 microliters of sterile water. The Wspec primers correspond to a DNA sequence unique to Wolbachia. The CO1 primers correspond to a region in mitochondrial arthropod DNA that codes for the protein, cytochrome oxidase I. Pipet, using a P20 pippettor, 2 microliters of DNA from microfuge tube 1 into PCR tube 1. Do the same for the nasonia DNA in microfuge tube 2. Close and tap the tubes gently to mix the components in the tube. Place the PCR tubes in the thermal cycler. The first cycle should be of temperature 94oC and last two minutes. The heat is strong enough to break the hydrogen bonds between the nucleotides, thus separating the strands. The second cycle should last two minutes and fifteen seconds. The first thirty seconds should be 94oC again to break the hydrogen bonds between the nucleotides, thus separating the strands. The next 45 seconds should be at 55oC. This lower temperature allows for the primers to attach to the DNA strands. The next minute should be at 72oC, which is the best temperature for Taq polymerase to attach nucleotides to the existing strand. Each cycle of your next 29 cycles should look like this. The last (32nd) cycle should last ten minutes at 72oC. Finally after the thermal cycler process is completed, about two hours, store the PCR tubes at -20oC. The Polymerase chain Reaction is now over. The third and final part of the lab is gel electrophoresis. First, pipet 2 microliters of All the results from our Period 7 class failed. Ashwin Suresh 18 December 2012 Period 7 Results________________________________________________________________ Class & Arthropod Results Results Prepared by notes (+/-) Group Order (+/-) COI Wolbachia Thomas Yan, Ricky Yang, Jan Clasen, Theo DeFuria, Luke 1-1 arachnida - - Zheng, Erik Viggh Michelle Yao, Rose Zhao, Shawnae Ma, Brianna Hoff and Judie 1-2 diptera - - Wang Mark Benati, Tharun Sankar, Ben Fanucci-Kiss, Robert 1-3 hymenoptera - - Tabachnik, Siddharth Limaye, and Jeffrey Shao 1-4 lepidoptera - - Madhuri Jois, Michelle Wu, Pelin Ozturk, Jennifer Ma, Kaitlin 1-5 lepidoptera - - Hoang Talia Quaadgras, Hannah Hoffman, Sarah Mamlet, Stephanie 1-6 hymenoptera - - Margolis, and Brooke Blackshaw 2-1 hymenoptera - - 2-2 hymenoptera - - Emily Liu, Alli Sedler, Jack Hoggard, Matt Bullock, Sam Toulmin, 2-3 orthoptera - - Josh Linker 2-4 hymenoptera - - 2-5 isopoda - - 2-6 lepidoptera - - Jack Billings, Marshall Grant, Gabe Kline, Nick Duncan, and 3-1 isopoda - - Melody Mao 3-2 Grace Wu, Julie Yao, Carol Zhang, Fiona Koshy diptera - - 3-3 hymenoptera - - 3-4 orthoptera - - 3-5 hymenoptera - - Teja Pallikonda, Nina Prakash, Roshni Sahoo, and Maggie 3-6 isopoda - - Sannicandro 4-1 coleoptera - - 4-2 Akshay Karthik, William Hu, Ajay Suresh, Alex Mateev hymenoptera - - 4-3 orthoptera - - Michelle Chen, Rebecca Kane, Avani Khatri, Vanessa Low, Julia 4-4 hymenoptera - - Ma 4-5 hemiptera - - 4-6 hymenoptera - - Emily Ropiak, Selena Gonzales, Charles Watt, Ellen 7-1 Isopoda - - McCormick,Bekah Pierce 7-2 Heteroptera - - Kevin Hu, Akash Purohit, Shirley Xu, Ezra Stein, and Sam 7-3 Isopoda - - Haufler Srinivas Setty, Saahil Claypool, Dylan Caldwell, David Chen, 7-4 Hymenoptera - - Darren Yang, Ashwin Suresh 7-5 Arachnid - - Jamie Young, Julian Waugh, Abishek Kumar, Varan 7-6 Lepidoptera - - Culanathan, Kevin Leung , and Bobby Smieszny 8-1 Ankit Datta, Matt Gilligan, Richard Tang, Valerie Han, Sarah Gao, Anuradha Hymenoptera - - 8-2 Lepidoptera - + Julianne Higgins, Tara Jawahar, Katya Muzin, Vennela 8-3 Pandaraboyina, Peter Rakauskas, Leanne Quinn and Ben Coleoptera - - Wurman 8-4 Diptera - - Ashwin Suresh 18 December 2012 Period 7 8-5 Diptera - - 8-6 Artem Petrov, James McClung, Alex Kim, Sri Nevvula, Joie Liba Isopoda - - Ashwin Suresh 18 December 2012 Period 7 Conclusion____________ _____ Literature Cited__________________________________________________________________ Ashwin Suresh 18 December 2012 Period 7 Here are a few ways enzymes play a role in the human body. You have probably heard about people that are lactose intolerant. These people cannot consume dairy products because their body cannot digest it. In order for the body to digest it, the disaccharide lactose must be broken down into the monosaccharaides glucose and galactose. This is done with the help of the enzyme lactase, something that lactose intolerant people lack. This shows how important enzymes are because not having lactase prevents you from eating a whole entire category of foods.