Isolation of Human DNA and PCR amplification of an Alu insertion site and PCR amplification of a VNTR site

Transcription

1 BIO440 Genetics Laboratory Humboldt State University Isolation of Human DNA and PCR amplification of an Alu insertion site and PCR amplification of a VNTR site DNA fingerprinting is an efficient and highly accurate means of determining identities and relationships. It has revolutionized the field of forensics, especially concerning violent crimes such as rape cases. It has exonerated hundreds of falsely convicted people, and removed from suspicion countless more falsely accused people, while also implicating guilty parties where there were no other indications of their involvement. It allows scientists to identify the remains of victims of highly disfiguring accidents or soldiers lost in battle. It is routinely used (even anonymously, via companies advertising on the internet) to settle child custody cases by identifying the child's true biological parents. This amazing technology is the most powerful and accurate form of human identification ever used. In this laboratory exercise, you will first isolate your own DNA using a crude but effective protocol that will yield relatively impure DNA. The DNA is pure enough, however, for use in PCR reactions. A simple saltwater mouthwash will release cheek cells, which you will then lyse by boiling. You will boil the cells in a mixture that contains Chelex beads. Chelex will bind divalent cations required by nucleases. You will then spin the solution down, and the Chelex beads and cell debris will pellet, leaving your crude DNA prep in the supernatant. In the first experiment, you will use the polymerase chain reaction (PCR) to amplify a region of chromosome16 called the pv92 Alu locus. Some versions of chromosome 16 have a 300 nucleotide insert in this region (called an Alu insert), while other versions don't. After you amplify the Alu repeat region, you will determine whether or not you carry this particular Alu sequence on one, both, or none of your number 16 chromosomes i.e., you will directly determine your genotype. This will be accomplished by electrophoresing your PCR sample on an agarose gel and observing the size of the bands that have been created. In the second experiment, we are going to use PCR to amplify a portion of chromosome 8. The portion that we are going to amplify is in an intron in the gene for the tissue plasminogen activation (TPA) factor. In some copies of this gene there is an insertion of a 300 bp Alu I sequence. Since we each have two copies of chromosome 8, the possible genotypes are +/+ (both copies have the insertion), +/- ( one copy has the insertion), or (-/-) both copies lack the insertion. The different alleles of these non-coding regions are inherited in Mendelian fashion. Introduction: pv92 Alu The human genome 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 these 46 chromosomes (sperm and egg cells contain only 23 chromosomes, while RBC's

2 don't have nuclei). Only 3 to 10 percent of this enormous amount of DNA, however, 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 may be 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, in many cases, hundreds or thousands of base pairs intervening between them. These intervening regions are called introns. Examination of introns and other non-protein-coding 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 was Alu. Alu repeats are approximately 300 base pairs in length. They carry within them the base sequence recognition site for the Alu I restriction endonuclease. There are over 500,000 Alu repeats scattered throughout the human genome (thus making up ~5% of the genome--equal to the protein coding portion!). On average, one can be found every 4,000 base pairs along a human DNA molecule. How they arose is still a matter of speculation but evidence suggests that the first one may have appeared in the genome of higher primates about 60 million years ago. Alu repeats are 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. Some versions of chromosome 16 have a PV92 Alu insert; some don't. The Polymerase Chain Reaction PCR is a method for the in vitro exponential amplification of a specific DNA region (called target region), that lies between two regions (called primers) of known DNA sequence, resulting in µg quantities of DNA from fg or zg starting masses. PCR has a wide variety of uses. The diversity of applications derive from the ability of PCR to: a) distinguish a specific target sequence from a large excess of background, non-target sequences, and b) produce a large number of copies of a specific sequence from a very small initial amount (as low as a single molecule of the target sequence). The extreme specificity and sensitivity of PCR have allowed it to be used to amplify and clone DNA from such sources as mummified human tissues and samples of extinct plants

3 and animals. To accomplish this task, PCR utilizes DNA polymerase, an enzyme that catalyzes the synthesis of a complementary strand of DNA from a template strand. This synthesis always takes place in a 5 -->3 direction, and DNA polymerases require a 3 - OH on the growing daughter strand in order to catalyze the formation of a new phosphodiester bond. This requirement for a 3 -OH is exploited to give the process the high degree of specificty for which it is known. A target DNA sample is mixed with two different primers, which are short (20-30 bases), single-stranded, synthetically-made polynucleotides that are complementary to the sequences flanking the target sequence. This mixture is heated to denature the double-stranded target DNA. When the mixture is then cooled, the primers can anneal to the sequences that they are complementary to, that is, they can form a double-stranded DNA molecule consisting of the primer hydrogen-bonded to a complementary target DNA sequence. The primer thus provides a 3 -OH group for the DNA polymerase, and a copy of the region adjacent to the primerbinding-site is synthesized. After the complementary strand is synthesized, the mixture is then heated again, and the double-stranded molecule consisting of the target sequence and the newlysynthesized molecule is denatured to single strands. When this mixture is cooled, the newly synthesized strands and the original target strands can serve as templates for the synthesis of more product sequences; thus, the process is exponential. In theory, the process can double the amount of the specific target with each round of denaturating, annealing, and synthesis. Thus, after thirty rounds, 2 29 copies of each original target molecule would be present. In practice, three temperatures are typically included in a PCR cycle, corresponding to three reactions: denaturation, annealing, and extension. 94 C for 1 minute is usually sufficient for denaturation, and 72 C for minute is usually sufficient for extension (the DNA polymerase used is very active at this temperature). The annealing temperature is typically set at 5 C lower than the Tm of the primers, and thus will be dependent on the actual primer sequence (i.e, the higher the G+C content, or the longer the primers, the higher the annealing temperature). The optimum annealing temperature is usually determined empirically. A typical annealing temperature is 55 C. Clearly, at these elevated temperatures, a human DNA polymerase would rapidly denature and cease to catalyze synthesis of complementary DNA. Therefore, a variety of DNA polymerases that have been cloned from extreme thermophiles, microorganisms (Bacteria and Archea) which thrive at elevated temperatures. Taq is commonly used, a DNA polymerase isolated from Thermus aquaticus, a bacterium isolated from a hot spring in Yellowstone National Park. The yield of a PCR reaction is defined as the number of target molecules produced by the PCR, and can be described by the equation: Y = N o * (1+E) a-1 where Y = yield N o = number of starting target sequence molecules E = efficiency a = number of cycles

4 The efficiency of a PCR reaction is the fraction of target molecules that actually serve as a template for the synthesis of new product molecules. Therefore E can have values between 0 and 1. There are many factors which contribute to the efficiency of a PCR reaction. Another important concept is the specificity of a PCR reaction. Specificity refers to the number of target molecules that are bound by primers relative to the number of non-target molecules that are bound by primers. For example, decreasing the annealing temperature of a PCR reaction typically decreases the specificity of that reaction, because the temperature decrease allows the primer to form duplexes with molecules that are lessthan-perfect complements. Typically, decreases in specifity are accompanied by increases in efficiency. In designing a PCR reaction, you need to balance out the desire for a large amount of product with the desire to amplify only the desired target sequence. Some of the factors which affect specificity are listed in the table below. Conditions Favoring Enhanced Specificity decrease Mg++ decrease [dntps] decrease [Taq polymerase] decrease cycle times decrease number of cycles decrease primer concentration decrease primer degeneracy increase annealing temperature optimized primer design use of hot start use of Touchdown PCR use of enhancing agents to decrease 2 structure of template Components of a typical PCR reaction 10 ng of starting template DNA 1 U of Taq polymerase 0.1mM dntps 1.5 mm Mg++ 1 µm each primer buffer (Tris buffer ph 8.5/KCl/ BSA) There is a good animation of PCR at

5 Protocol: DNA isolation from Cheek cells. 1.Use permanent marker to label your name on a 15-ml culture tube containing saline solution. (this is commercial drinking water with table salt added to 0.8%). 2.Pour all of the saline solution into your mouth, and vigorously swish for 10 seconds. Save the empty 15-ml tube for reuse in the next step. 3.Expel saline mouthwash into a paper cup. Then carefully pour saline mouthwash from paper cup back into 15-ml tube from Step Transfer 1 ml of the cell solution to a 1.5 ml tube labeled with your name. Securely close cap, and place mouthwash tube in a balanced configuration with other tubes in rotor of microcentrifuge. Centrifuge on high setting for 1 minute to pellet cells. 5.Being careful not to disturb cell pellet, pour off as much supernatant as possible into sink or paper cup. Place tube with mouthwash cell pellet on ice. 6.Use micropipettor to add 30 µl of saline solution to cell pellet. Resuspend cell pellet by vortexing and by pipetting up and down. It is important to resuspend the cell pellet entirely - no clumps. 7. Add 30 µl of your resuspended cell pellet to a 0.5 ml tube containing 200µl 10% Chelex solution. Label the tube with your initials 8. Incubate the 0.5 ml sample tube at 99 C in the thermalcycler for 10 minutes. 9. Following incubation, remove sample tube from thermalcycler, briefly vortex the sample, and cool tube on ice for approximately one minute. 10. Place sample tube in a balanced configuration in a microfuge rotor, and spin for 30 seconds to pellet Chelex beads at bottom of tube. 11. Transfer 50 microliters of the supernatant to a fresh 1.5 ml tube labeled with your name, and place tube in freezer. Avoid transferring any of the Chelex pellet.

6 BIO440 Genetics Laboratory Analysis and Interpretation TPA-25 Alu Insertion / Pv92 Human Genotyping Name: Tape a picture of your gels below, indicating which lanes represent your samples. Interpret the results of your amplifications, including expected results and possible explanations of unexpected results:

7 Analysis of Alu TPA-25 results For this calculation, include the values determined for your class, and with previous classes as depicted on the example results gel at the end of this handout. We will symbolize the presence of an Alu insertion with a + and the absence with a -. For Hardy- Weinberg calculations, we will have + =p and - = q. 1. How many class members were homozygous for the insertion (+/+)? 2. How many class members were heterozygous for the insertion (+/-)? 3. How many class members were homozygous for no insertion (-/-)? 4. What is the frequency of p? of q? 5. Given the frequency of p and q, and assuming that we are in Hardy-Weinberg equilibrium, how many students would we expect to be +/+? +/-? -/-? 6. Presumably, there will be some differences between the actual distribution of genotypes that we observe and the distribution predicted by Hardy-Weinberg equilibrium. To determine whether this difference appears due to chance, use chisquared analysis to analyze the deviation. A table of p-values appears later in the handout - use 1 degree of freedom.

8 Analysis of Alu Pv92 results 1. How many class members were homozygous for the insertion (+/+)? 2. How many class members were heterozygous for the insertion (+/-)? 3. How many class members were homozygous for no insertion (-/-)? 4. What is the frequency of p? of q? 5. Given the frequency of p and q, and assuming that we are in Hardy-Weinberg equilibrium, how many students would we expect to be +/+? +/-? -/-? 6. Presumably, there will be some differences between the actual distribution of genotypes that we observe and the distribution predicted by Hardy-Weinberg equilibrium. To determine whether this difference appears due to chance, use chisquared analysis to analyze the deviation. 7. For the Pv92 insertion, what is the p-value associated with the chi-squared value that you calculated? In your own words, describe what this p-value means.

9 Students in a genetics class at the University of Costa Rica performed the Pv92 amplification. Of 56 students, they had allele frequencies of 0.22 =+ and 0.78 = -, with a genotype distribution of +/+ = 0.02, +/- = 0.40, and -/- = Students in a genetics class at the University Texas performed the Pv92 amplification. Of 31 students, they had allele frequencies of 0.42 =+ and 0.58 = -, with a genotype distribution of +/+ = 0.26, +/- = 0.32, and -/- = Using a chi-squared approach, compare your results with their results to see if our class can be distinguished as an interbreeding population from either of these groups. Table of p-values for Chi-squared calculations P value(%) *** 1 Chi 2-1 DOF Chi 2-2 DOF Chi 2-3 DOF

10 9. Some populations are not in Hardy-Weinberg equilibrium. Why? What are the possible causes of a non-h-w distribution of genotypes? 10. The following problem refers to the gel pictured below. Forensic scientists from time to time must reconstruct the DNA profile for a missing person from analysis of DNA profiles of close relatives. In this case, a mother of four children is missing. All children have the same biological father. Results from a single locus probe DNA fingerprint analysis for the four children and their father are shown in the figure. Unfortunately, the forensic scientist forgot to label the lane with the father's DNA. Which lane is the father's DNA? Which alleles (A-D) does the mother have? Explain.

11 11. A scandal has broken. It's revealed that Ashley Olsen has four children from her early teen years, and she is claiming that Rollin Richmond is the father. He claims that she's not his type (he's more a Mary Kate fan), that he rebuffed her advances, and that she has been stalking him for years. Results from a single locus probe DNA fingerprint analysis for Ashley, Rollin, and the four children are shown in the illustration below. Which child, if any, can be excluded as being the biological offspring of Rollin? Explain. 11. It is determined that the father of the children, and Dr. Richmond, have the following genotype: the same as you for the TPA 25 and the pv92 locus B/D for the locus from question 9 (frequency in population = 0.15) A/D for the locus from question 10 (frequency in population = 0.05) C/E at another locus (frequency in population = 0.05) Assuming the frequencies for the TPA 25 genotype and the pv92 genotype that you determined as part of this lab, what fraction of individuals are expected to have the same genotype as the father and Dr. Richmond? How many people at Humboldt State, assuming these genotype frequencies, would have that genotype? How many people in America, assuming these genotype frequencies, would have that genotype?

12 BIO440 Genetics Laboratory Example results TPA-25 Alu Insertion BIO440 Genetics Laboratory Example results Pv92 Alu Insertion