PCR Amplification of The Human Dimorphic Alu PV92 Site 3/17 Honors Biomedical Science 2 Redwood High School Name: [ETRLMBR] Background T he human genome (the total sum of our genetic makeup) is made up of approximately 6 billion base pairs distributed on 46 chromosomes. All cells in your body, except red blood cells, sperm, and eggs, contain 46 chromosomes (sperm and egg cells contain only 23 chromosomes). Approximately 3 percent of this DNA is actually used to directly code for the proteins required for supporting cellular metabolism, growth, and reproduction. The protein-encoding regions are scattered throughout the genome. Genes are often separated by many thousands of base pairs. Furthermore, most genes in the human organism are themselves broken into smaller protein-encoding segments called exons with hundreds or thousands of base pairs intervening between them. These intervening regions are called introns. Although it is true that intron function is not well understood, whatever their function, examination of these intervening DNA regions has revealed the presence of unique genetic elements that can be found in a number of different locations within the genome. One of the first such repeating elements identified is Alu. Alu repeats are approximately 300 base pairs in length. These repeating sections of DNA are so named because they include the base sequence AGCT, the recognition site for the Alu I restriction endonuclease. There are over 500,000 Alu repeats scattered throughout the human genome. How they arose is still a matter of speculation but evidence suggests that the first one may be have appeared in the genome of higher primates about 60 million years ago. Approximately every 100 years since then, a new Alu repeat has inserted itself in an additional location in the human genome. Alu repeats are autosomal and therefore inherited in a stable manner; they come intact in the DNA your mother and father contributed to your own genome at the time you were conceived. Some Alu repeats are fixed in a population, meaning all humans have that particular Alu repeat. Others are said to be dimorphic; different individuals may or may not carry a particular Alu sequence at a particular chromosomal location. Moreover, they are stable markers that reflect unique evolutionary events, namely the insertion of an Alu element into a new chromosomal location. Another advantage of these markers is that there is no parallel gain or loss of Alu elements at a particular chromosomal location, and thus all chromosomes that carry a polymorphic Alu element are identical by descent. Furthermore, the ancestral state of these polymorphisms is known to be the absence of the Alu insert, and this information can be used in analyses of human population relationships. Studies have thoroughly examined Alu patterns world-wide, and compared this data to other chromosomal markers used to locate the evolutionary homeland of modern humans. Data from these Alu studies further confirm Africa as the ancestral homeland of modern humans. Possible Role in Mitosis During mitosis a class of proteins called cohesion protein complexes hold sister chromosomes together. These proteins bind together newly replicated chromosomes prior to nuclear division, which leads to cytokinesis and the advent of daughter cells. While studying cohesion protein activity researchers discovered that the attachment sites had a common sequence. This made sense, as multiple cohesion sites are required on each chromosome to keep the newly replicated pairs together. There would be no advantage for nature to have to have a sequence other than a common sequence act as the cohesion-binding site. It is believed evolution has favored the rise of Alu sites as a reliable binding site for cohesion proteins during one of life s must-get-it-right every time processes.
In the following laboratory exercise, you will use PCR to amplify a dimorphic Alu repeat (designated PV92) found on your number 16 chromosome. You will use your own DNA as template for this experiment. DNA is easily obtained from the human body. A simple saltwater mouthwash will release cheek cells, from which you will extract the DNA. After you amplify the Alu repeat region, you will determine whether or not you carry this particular Alu sequence on one or both of your number 16 chromosomes. This will be accomplished by electrophoresing your PCR sample on a polyacrylamide gel. Protocol Day 1: Chelex Assisted DNA Extraction of Buccal Cells 1. Working individually, initial your Chelex tube (200µl PCR Tube), RNASE/DNASE FREE 1.5ml tube, and your 0.5ml Skirted tube. Label another 1.5 tube with your initials for step 6. 2. Secure a cup with 10ml of 0.9% NaCl. 3. With vigor, swirl the entire 10ml of NaCl in your mouth for 30 seconds making sure to repeatedly slam the solution up against the inside of your cheeks. 4. Expel the NaCl solution back into the cup and swirl to mix the cells instantly begin step 5. 5. Using a sterile disposable 1ml transfer pipet, transfer 1.5ml of the expelled NaCl solution into the RNASE/DNASE FREE tube (will almost fill the tube). Store your sterile pipet in its sleeve for step 6. In a balanced centrifuge spin sample at 14,000 rpm for 4 minutes. 6. Observe the off-white colored cell pellet at the bottom of the tube. Using your 1ml transfer pipet, carefully remove almost all of the supernatant to your initialed 1.5ml tube. Do not to disturb pellet. Using a p-200 add 40µl of the supernatant to the top of your cell pellet. Do not to disturb pellet. 7. Using a motorized pellet pestle and RNASE/DNASE FREE pestle, re-suspend your cell pellet for 5 seconds. *need to really work the pellet (12,000 rpm for 4min really mashes the cells to the bottom of the tube) 8. Immediately: Using a p-200, flush and then withdraw 30µl of the cell suspension and add it to the PCR tube containing 200µl of 5% Chelex. Note: Do not pipet up and down at this step or else you will clog the tip with Chelex beads. See the note below The Chelex attracts certain cations such as Mg +2. These ions are co-factors for DNA degrading enzymes called DNases. These ions and the DNA degrading enzymes are found in the cytoplasm of your cells (and are therefore normally kept separate from DNA). During the next set of procedures you will be lysing your cheek cells to release DNA. If certain enzymatic co-factors are not removed from the supernatant as the cells are lysed, the ions in question, the DNases, and the DNA will co-mingle and your DNA will be degraded. 9. Firmly close your Chelex tube and place the tube in the thermal cycler holding at 99 C for 10 minutes. Record your initials on the grid provided. 10. Vortex your tube for 7 seconds and then place the tube in a balanced microcentrifuge. Spin for 1 minute. [2,000 rpm] 11. From your Chelex tube, withdraw 60µl of supernatant VERY SLOWLY. Keep the tip away from the Chelex beads. Transfer the supernatant to your skirted tube. 2
12. Store the skirted tube containing your genomic DNA at 4 C. Day 2: Polymerase Chain Reaction Use a fresh tip for every transfer even if it is the same reagent. Add reagents to the bottom of the reaction tube, not to its side. Add each reagent directly into the existing pool of reagents. Flush after each transfer 1. Secure a 200µl PCR tube. Label the tube with your initials. 2. Transfer 20µl of Master Mix (TUBE labeled MM) into your PCR tube. 3. Add 20µl of Primer Mix (TUBE labeled PM) into your PCR tube. FLUSH!!! 4. Add 10µl of your purified DNA into your PCR tube. FLUSH!!! 5. Place your PCR tube into the thermal cycler which is set at 4 C. Record the location of your tube on the grid provided. 6. The cycling protocol for amplification of this Alu region is: 95 C 2 minutes 94 C 30 seconds 60 C 30 seconds 72 C 2 minutes 72 C 10 minutes 4 C hold 27 cycles *been 27 since 2014 (30 prior to that) Product to be stored at -20 C until day 3 Controls (these reactions will be prepared for you) Positive (+) Control: 20µl Master Mix, 20µl Primer Mix, 10µl Control DNA Negative (-) Control: 20µl Master Mix, 20µl Primer Mix, 10µl Sterile dh 2 O Day 3: Electrophoresis of Amplified DNA 1. Prepare your 5% TBE PAGE gels. Load your gel into your vertical gel box. Add the 1X TBE running buffer to the inner and outer chambers. 2. Retrieve your PCR tube and micro centrifuge for 5 seconds to pool your PCR products. Do the same to the control tubes. 3. Using a filtered p-20 tip, add 5µL of loading dye set the micropipet to 20µL and thoroughly flush with your loading dye tip. Do the same to the control tubes (if you have them). 4. Using a PAGE capillary tip load the lanes of the PAGE gels according to the map provided. 20µL - your DNA sample 20µL - control DNA sample 5µL - 100 bp ladder 5. Run your gel at 100V until the xylene cyanol (turquoise dye) gets close to bottom of the gel *should be apx. 50 minutes 3
Day 3: Staining and Photographing 1. Make the following changes to the standard Et-Br staining protocol: a) leave the gel in the Et-Br for 15 minutes instead of 10 minutes b.) de-stain for 5 minutes instead of 10 minutes 2. View and photograph your gel. Results By examining the photograph of your agarose gel, you will determine whether or not you carry the Alu repeat on one, both, or neither of your number 16 chromosomes. PCR amplification of this Alu site will generate a 415 bp fragment if the repeat is not present. If the 300 bp Alu insert is present, a 715 bp fragment will be made. The ladder is a 100bp ladder and should feature a brighter band for both the 1000bp and 500bp bands. When you examine your results and others, it should be apparent that there are differences between individuals of the same species at the level of their DNA. Even though you amplified only one site, a site that every one has in their DNA, you will notice that not all individuals have the same pattern of bands. Some individuals will have only one band, while others will have two. The term allele is used to describe different variants of a given gene. For those who have the Alu repeat (they have at least one 715 bp band), we can say that they are positive for the insertion and denote that allele configuration with a + sign. If the Alu repeat is absent (a 415 bp band is generated in the PCR), we assign a - allele designation. If an individual has a single band, whether it is a single 415 bp band or a single 715 bp band, then both their number 16 chromosomes must be the same in regards to the Alu insertion. They are said to be homozygous and can be designated with the symbols -/- or +/+, respectively. If a person s DNA generates a 415 bp band and an 715 bp band during PCR, the person is said to be heterozygous at this site and the designation +/- is assigned. A person s particular combination of alleles is called their genotype. The following table illustrates the possible genotypes possible for the PV92 locus. One Allele The Other Allele Genotype 715 bp 715 bp Homozygous for the insert (B+/B+) 715 bp 415 bp Heterozygous for the insert (B+/B-) 415 bp 415 bp Homozygous for no insert (B-/B-) What is your genotype for the Alu PV92 insert? 4
Day 4: Calculating Allele and Genotype Frequencies * Objectives: 1. calculate allele frequencies. 2. calculate genotype frequencies. A.) Allele Frequencies Within your class, how unique is your particular combination of Alu alleles? By calculating allele frequency, you can begin to answer this question. An allele frequency is the percentage of a particular allele within a population of alleles. It is expressed as a decimal. You can calculate the allele frequency for the Alu PV92 insertion in your class by combining all the data. For example, imagine that there are 100 students in this class and the genotype distribution within the class is as follows: Genotype Number of Students having that Genotype +/+ 20 +/- 50 -/- 30 Since each person in your class has two number 16 chromosomes (they are diploid for chromosome 16), there must be twice a many total alleles as there are people: 2 alleles 100 students= 200 alleles student To calculate allele frequencies for the class, therefore, 200 would be used as the denominator value. To calculate the + allele frequency, we must look at all those students who have a + in their genotype. There are 20 students who are +/+ ; they are homozygous for the insertion. Since these 20 students have two copies of the Alu insert on their chromosome, they contribute 40 + alleles to the overall frequency. There are 50 students heterozygous ( +/- ) for the Alu insertion. Each heterozygous individual, therefore, contributes one + allele to the overall frequency, or 50 + alleles. Adding all + alleles together gives us: The frequency of the + allele in this example is: 90 "+" alleles 200 total alleles = 0.45 The frequency for the PV92 - allele is calculated in a similar manner. There are 30 students homozygous for the - allele. This group, then, contributes 60 - alleles to the frequency. There are 50 students heterozygous for the Alu insertion. They contribute 50 - alleles to the frequency. Adding all - alleles together gives us 110 - alleles. The frequency for the - allele in this example is: 110 "-" alleles 200 total alleles = 0.55 Notice that the sum of the frequency for the + and - alleles is 1.0. If the allele frequencies do not add up to 1.0, then you have made an error in the math. 5
Use the spaces below to calculate the + and - allele frequencies for your class. Number of + and - alleles: B.) Allele frequencies: Genotype Number of Students Number of + Alleles Number of - Alleles +/+ 0 +/- -/- 0 Total "+" allele frequency = "-" allele frequency = total "+" alleles total alleles total "-" alleles total alleles = = Do these allele frequencies add up to 1.00? C.) Genotype Frequencies How does the distribution of Alu genotypes in your class compare with the distribution in other populations? For this analysis, you need to calculate a genotype frequency, the percentage of individuals within a population having a particular genotype. Remember that the term allele refers to one of several different forms of a particular genetic site whereas the term genotype refers to the specific alleles that an organism carries. You can calculate the frequency of each genotype in this class by counting how many students have a particular genotype and dividing that number by the total number of students. For example, in a class of 100 students, let s say that there are 20 students who have the +/+ genotype. The genotype frequency for +/+, then, is 20/100=0.2. Given the ethnic makeup of your class, might you expect something different? How can you estimate what the expected frequency should be? If within an infinitely large population no mutations are acquired, no genotypes are lost or gained, mating is random, and all genotypes are equally viable, then that population is said to be in Hardy-Weinberg equilibrium. In such populations, the allele frequencies will remain constant generation after generation. Genotype frequencies within this population can then be calculated from allele frequencies by using the equation: p 2 + 2pq + q 2 = 1.0 where p and q are the allele frequencies for two alternative forms of a genetic site. The genotype frequency of the homozygous condition is either p 2 or q 2 (depending on which allele you assign to p and which to q). The heterozygous genotype frequency is 2pq. The convention is to use let p 2 represent the +/+ condition. 6
Using the calculated allele frequencies found in your class, determine the expected class genotype frequencies. (Let p represent the + allele and q the - allele.) Expected +/+ genotype frequency: p 2 = Expected +/- genotype frequency: 2pq = Expected -/- genotype frequency: q 2 = Use the table below to calculate how many students in your class should have each genotype. genotype +/+ expected genotype frequency total number of students in class expected number of students with specific genotype +/- -/- Now, calculate the actual genotype frequencies for this class genotype +/+ number of students with this genotype total number of students in class frequency of specific genotype in this class +/- -/- Discussion [type and print your answers] *Attach a copy of your gel picture 1. Compare your calculated expected genotypic frequencies with the actual data. Discuss the variation you discovered. What could explain this variation? 2. Discuss your individual PCR results. Include information related to the amplification process. Were your results optimal or less than optimal? If less than optimal, what may explain your results? 7
AluPV92 Troubleshooting Guide The following gel has all possible allele combinations. Agarose gel of homozygous and heterozygous individuals for the Alu PV92 insertion. A 100 base pair ladder is loaded in the first lane and is used as a size marker, where these bands differ by 100 bp in length. The 500 bp band and the 1,000 bp band are purposely spiked to be more intense than are the other bands of the ladder when stained with ethidium bromide. The next 5 lanes contain the results of homozygous and heterozygous individuals. A negative control (-C) does not contain any template DNA and should therefore contain no bands. The positive control (+C) is heterozygous for the Alu insertion; it contains both the 415 bp and 715 bp bands. 8
Alu Gel Analysis Desired Alu results. The bands of the 100 bp ladder are sharp and well resolved. Samples should contain either one band (for homozygous -/- or +/+) or two bands (for heterozygous +/- for the Alu insertion). For heterozygous genotypes, it is typical that the larger 715 bp band will be less intense than the smaller 415 bp band. This is called preferential amplification and results from the fact that shorter fragments are amplified by PCR more efficiently than larger fragments. Artifact bands. This gel shows a number of the artifact bands that might be generated during the Alu PCR amplification. In the +/- lane, we see the desired 415 bp and 715 bp products but also a number of other bands, primer dimer (at the bottom of the gel) forms by an interaction between the primers. We also see a number of large fragments that we believe are heteroduplex molecules generated by annealing between a 415 base single strand and an 715 base single strand. Such a heteroduplex will migrate down the gel at a slower rate than might be expected for its actual length. Shown in the -/- lane is a band that migrates at about 300 bp. This may appear as a single band, or, if the gel is run longer, two bands in this area might appear. These probably result from nonspecific amplification; primers annealing elsewhere on the template and generating a PCR fragment. In the +/+ lane, a band at 415 bp appears. Since this band is less intense than the 715 bp band (the opposite of what we should expect from preferential amplification), it is an artifact. It can result from spill-over from the lane next to it or from contamination. 9
100 bp ladder shows bands but no sample bands are present. Reactions failed to produce amplification. This might result from: 1. Inadequate template amounts added to the reaction. 2. Some component of the reaction was not added. 3. The reagents were inactive due to improper storage. Reagents should be stored frozen or in the refrigerator until use. 4. Master Mix was vortexed violently resulting in loss of AmpliTaq DNA Polymerase activity. No bands. If no bands are present on the gel not even for the 100 bp marker lane, then the possibilities are: 1. Samples were not properly loaded into the wells. 2. The staining solution does not have adequate ethidium bromide stain. Staining solution should contain ethidium bromide at a concentration of 0.5 µg/ml and staining should be allowed to proceed for at least 15 minutes at room temperature. 3. Gel was stored too long in buffer prior to staining and the bands diffused out of the gel. 10