ESC102/BME FINAL LAB REPORT Genetic Testing of Alpha-Thalassemia by PCR and Restriction Digest Name: Maria Yancheva Lab Section: PRA 0103 Student Number: 996742173 TA Name: Elizabeth Berndl Partner Name: Laila Hulbert Date 13 April 2009 1 ESC102 FINAL LAB REPORT–GENETIC TESTING OF ALPHA-THALASSEMIA ABSTRACT Experiment conducted consisted of collecting DNA samples from four individuals and testing them for Alpha-Thalassemia. Thalassemia is an inherited autosomal (not sex-determining chromosome) recessive blood disease. The genetic mutation tested is the His50Arg in the HBA1 gene. The DNA samples collected were cheek cells whose gene-encoding region was amplified using PCR and cleaved by restriction digest using the SacII restriction enzyme. Finally, the digested samples were run through gel electrophoresis, stained and imaged. The imaged results were analyzed based on band sizes expected for wild-type and mutant samples. None of the individuals tested were found to have Alpha-Thalassemia. INTRODUCTION The objective of the experiment was to determine whether the individuals tested had Thalassemia, based on analysis of the DNA samples provided. The methods used in the experiment utilize PCR and restriction digest for medical diagnostics – this novel way of determining genetic differences between individuals is significant in its inexpensive and feasible nature. The use of PCR techniques for information-gathering and molecular analysis of clinical samples is an innovation in the area of genomics and allows for much smaller amounts of DNA as input material [1]. Thalassemia is a recessive inherited blood disorder. The presence of Thalassemia causes the decreased production of healthy red blood cells, as well as decreased levels of hemoglobin production. Specifically, Alpha-Thalassemia causes defects in the two alpha globin chains of the hemoglobin A gene. As a result of the insufficient amount of hemoglobin in red blood cells, people with thalassemia suffer from mild to severe anemia [2]. The disorder is treated with blood transfusions in combination with iron chelators, and other means of treating the iron overload caused by transfusions such as pump-infusion therapy [3]. There are a number of genetic mutations causing Thalassemia. In this experiment, the His50Arg substitution in the HBA1 gene was tested. In particular, this means that an amino acid at location 50 is changed from Adenine to Guanine, thereby coding for Arginine instead of Histidine [4]. 1. Difference between an intron and an exon: An intron is part of the primary transcript that is removed by splicing during DNA processing and is not included in the mature, functional mRNA, rRNA or tRNA. On the other hand, an exon is part of the primary transcript which is included in the mature mRNA, rRNA or tRNA molecule. 2. What is needed from the cells for PCR: A sample of DNA which contains the DNA sequence of interest; the sequence to be amplified must be present and intact. 3. What structures must be broken to release the DNA from the cell: The plasma membrane and nuclear membrane. 4. Why it is necessary to have a primer on each side of the DNA segment to be amplified: The primers are complementary to sequences of about 18 bases flanking the region of interest; they are required to identify an initiation site for replication to begin. In other words, DIVISION OF ENGINEERING SCIENCE, UNIVERSITY OF TORONTO 2 ESC102 FINAL LAB REPORT–GENETIC TESTING OF ALPHA-THALASSEMIA primers determine what segment of DNA will be amplified, because during each PCR cycle the DNA polymerase binds to, and extends each primer from its 3' end, generating newly synthesized strands. PROCEDURE Protocol Modifications [4] Part 1: Cheek Cell DNA Template Preparation As in Lab 2. Part 2: PCR Amplification As in Lab 2, except: At step 3, transfer 15 μL (instead of 20 μL) of DNA template from the supernatant to the PCR tube. After step 3, transfer the remaining InstaGene liquid, without any beads, to a 1.5mL tube with a removable label with your name on it. At step 4, transfer 15 μL (instead of 20 μL) of master mix into PCR tube. After step 4, handed back remainder of DNA template (1.5 mL tube) to TA. The TA removed student names and replaced them with a number, for anonymity of results. The TA then distributed two additional samples (#148 and #090) to test. The two additional samples were collected; then transferred 15 μL of each to a correspondingly labelled PCR tube and then added 15 μL (instead of 20 μL) of Thalassemia PCR master mix to the all PCR tubes. Added 40 μL of control to control PCR tube and added 40 μL of master mix to control. Proceed to step 6 in lab 2. Part 3: Restriction Digest and Electrophoresis As in Lab 3, except: Before step 1, took 15 μL PCR product (for each of the four samples labelled "LH", "MY", "148" and "090" test tubes) and transferred 15 μL of SacII enzyme into each, then incubated at 37oC for 1 hour. Proceed to step 1. In step 1, the gel prepared should be 2% [4]. Microwaved gel for about 2.5 minutes instead of 3 min. Proceed to steps 3-5. After the incubation, added 5 μL of dye to 30 μL of each of the four digested samples. Obtained 10 μL of undigested sample, and added 2 μL of dye to the undigested sample. Proceed to step 6. In step 6, loading placement as follows: 1 – Ladder (5 μL), 2 – Digested Sample "MY" (10 μL), 3 – Digested Sample "LH" (10 μL), 4 – Digested Sample "#148" (10 μL), 5 – Digested Sample "#090" (10 μL), 6 – Undigested Sample "Control" (10 μL). Part 4: SybrGreen Staining As in Lab 3. 1. Why are there nucleotides in the master mix? What are the other component s of the master mix and what are their functions? There are nucleotides in the master mix because they are DIVISION OF ENGINEERING SCIENCE, UNIVERSITY OF TORONTO 3 ESC102 FINAL LAB REPORT–GENETIC TESTING OF ALPHA-THALASSEMIA used form the strands complementary to the existing strands of DNA during PCR (since DNA is made up of nucleotides). The other components of the master mix are the forward and reverse primers which identify the DNA replication initiation site; the DNA polymerase (typically Taq polymerase) which extends the primers and assembles the nucleotides to the existing DNA strand; a salt buffer which maintains a neutral pH for the PCR reaction to occur; and magnesium which is a cofactor required by DNA polymerase. 2. The three main steps of each cycle of PCR amplification and the reactions and temperatures at which they occur: Step one, denaturation of DNA strands – occurs at 94oC for 1 min. Heating the strands to this temperature allows for the breaking of the hydrogen bonds which stabilize the complementary base pairs. As a result, the two DNA strands separate. Step two, Annealing of forward and reverse primers to corresponding strand – occurs at 60oC for 1 min. The strands are cooled to this temperate to allow the primers to anneal to the complementary regions on each strand, thereby flanking the region to be amplified. Step three, DNA polymerase extends the primers – occurs at 72oC for 2 min. Typically, Taq polymerase is used to attach to the primers and extend each primer from its 3' end, generating newly synthesized strands. As a result of these three steps, two double-stranded DNA molecules are generated equal in length to the sequence of the region to be amplified [6]. 3. Temperature at which restriction enzyme was incubated and why: The Thalassemia restriction enzyme was incubated at 37oC because that is the optimal incubation temperature for SacII. Typically, this is the recommended incubation temperature for the majority of restriction enzymes, except for those which are isolated from psychrophilic or thermophilic microorganisms [5]. Since that is not the case with SacII, its incubation temperature is 37oC. 4. To which electrode pole of the electrophoresis field does DNA migrate and why? DNA would be expected to migrate towards the positive electrode pole because DNA molecules are negatively charged and opposite charges attract. RESULTS AND DISCUSSION The image obtained after electrophoresis and staining is presented on the following page (Fig. 1). To analyze the band sizes produced, the simulated gel data provided was used. A graph of band size versus distance is also presented on the following page (Fig. 2). Using Fig. 2, the following analysis can be made about the experimental results (Samples 3-7 as labelled in Fig. 1 are analyzed): Sample 3 – distance of 3.31 cm corresponds to band size of 3908. Sample 4 – distance of 3.29 cm corresponds to band size of 3920. Sample 5 – distance of 3.23 cm corresponds to band size of 3954. Sample 6 – distance of 3.22 cm corresponds to band size of 3960. Sample 7 – distance of 1.92 cm corresponds to band size of 4926. DIVISION OF ENGINEERING SCIENCE, UNIVERSITY OF TORONTO 4 ESC102 FINAL LAB REPORT–GENETIC TESTING OF ALPHA-THALASSEMIA Fig. 1. Legend: 1 – Ladder; 2 – Failed Attempt at loading Digested Sample "MY"; 3 – Digested Sample "MY"; 4 – Digested Sample "LH"; 5 – Digested Sample "#148"; 6 – Digested Sample "#090"; 7 – Undigested Sample "Control”. Fig. 2. Simulated Gel Data. DIVISION OF ENGINEERING SCIENCE, UNIVERSITY OF TORONTO 5 ESC102 FINAL LAB REPORT–GENETIC TESTING OF ALPHA-THALASSEMIA Analysis of Subject Genotypes based on Band Size: Samples 3-6 which are the four student samples tested indicate a single band of comparable size (approximately 3936 on average). This shows that all four students tested had the same genotype for the disease. The Alpha-Thalassemia protocol does not specify the exact band sizes that would be expected for each genotype [4], but examining the cutting patterns of SacII for the wild-type and mutant gene, it can be inferred that a homozygous wild-type will have a single large band; a homozygous mutant type would have two smaller bands; and a heterozygous type would have one large band and two smaller bands. Therefore, the results show that all four students have a homozygous wild-type genotype for Alpha-Thalassemia. Deviation from Expectations and Sources of Error: The results obtained are expected because it is not likely that students who have Alpha-Thalassemia mutant genotype would be unaware of it. However, the undigested sample has a band size which appears to be larger than expected. This can be due to fact that the undigested sample bands are very unclear on the image. The fact that the undigested sample (Sample 7) was barely distinguishable on the image is due to the fact that the amount of undigested sample loaded into the well was less than the expected 10 μL due to insufficient amount in the provided test tube. Furthermore, the gel was taken out of the electrophoresis apparatus earlier than the prescribed 45 min, due to insufficient time at the end of the lab period. As a result, the electrophoresis process may not have been completed properly. Further Discussion: The experimental procedure did work in that it provided the expected results. The results show that none of the students tested have Alpha-Thalassemia. This completes the experimental objective. The strength of the experimental design lies in its inexpensive and easy methodology, which allows testing for a genetic disease simply by processing a small amount of cheek cells (easy to obtain). The limitations of the experimental design lie in the possibility of obtaining unclear results due to incomplete process of electrophoresis (e.g. bands not showing up on image). 1. Which of the simulated bands did not fragment? Samples 3-6 did not fragment. Sample 7 (Control) is too unclear to be able to determine its band size. 2. Which of the DNA samples were fragmented? What would your gel look like if the DNA were not fragmented? None of the student samples were fragmented. You would expect the samples to have the same band size as the undigested control (Sample 7), however the control band is not clearly visible on the image. 3. What kind of controls are run in this experiment? Why are they important? Could others be used? The control used in this experiment is Sample 7 (the undigested sample) which shows what the band size should be for a sample of DNA which was not cut by the restriction enzyme. Since for Alpha-Thalassemia, the homozygous wild-type is not cut by SacII, the wild-type genotype should therefore look the same as the control. This is important in order to be able to analyze the samples' genotype based on band sizes produced. Other samples which could have been used are samples of people with knows genotypes, so as it make the comparison simpler (for example, look at the heterozygous control and compare the number and size of bands with the samples). Additionally, the ladder is also used as a control to determine band sizes based on distance from the wells. DIVISION OF ENGINEERING SCIENCE, UNIVERSITY OF TORONTO 6 ESC102 FINAL LAB REPORT–GENETIC TESTING OF ALPHA-THALASSEMIA 4. Why is it necessary to chelate the metal ions from solution during the boiling/ lysis step at 100°C? What would happen if you did not use a chelating agent lysis step at 100°C? What would happen if you did not use a chelating agent? It is necessary to chelate the ions because that would grab enzyme cofactors which disable enzymes that degrade DNA. The degrading enzymes are released along with the DNA at the lysis step. If the cofactors were not present, the DNA would be degraded. 5. After DNA samples are loaded into the sample wells, they are “forced” to move through the gel matrix. What size fragments (large vs. small) would you expect to most quickly? Explain. You would expect the smallest size fragments to move most quickly through the gel because they can fit through smaller physical spaces. Therefore, the smallest fragments would be at the bottom of the gel and the largest at the top. CONCLUSIONS The four student samples tested show that the students don't have Alpha-Thalassemia. The band sizes produced by the student samples indicate that all four students have a homozygous wild-type genotype. FINAL QUESTIONS Explain why the precise length target DNA sequence doesn’t get amplified until the third cycle. The precise length target DNA does not get amplified until the third cycle because it is at the third cycle when the primers serve as the beginning and end of the region to be amplified. A restriction endonuclease “cuts” two DNA molecules at the same location. What can you assume is identical about the molecules at that location? The two DNA molecules have the same sequence at that location, and that sequence is one which the restriction enzyme recognizes as a cut site ("recognition site"). REFERENCES [1] Clinical Laboratory International. "21st-century Diagnostics: Recent Innovations in the Clinical Laboratory." Retrieved April 12, 2009, from http://www.cli-online.com/index.php?id=2102. [2] National Heart, Lung and Blood Institute: Diseases and Conditions Index. "What Are Thalassemias." Retrieved April 12, 2009, from http://www.nhlbi.nih.gov/health/dci/ Diseases/Thalassemia/Thalassemia_WhatIs.html. [3] Thalassemia Foundation of Canada. Retrieved April 12, 2009, from http://www.thalassemia.ca/ viewarticle.asp?aID=31&searchQ=F.A.Q. [4] "Genetic Testing of Alpha-Thalassemia by PCR and Restriction-Digest". Final Proposal in ESC102. University of Toronto. [5] Colorado State University. "Factors that influence Restriction Enzyme Activity." Retrieved April 13, 2009, from http://www.vivo.colostate.edu/hbooks/genetics/biotech/enzymes/ cuteffects.html. [6] Lodish, Harvey. Molecular Cell Biology. W.H. Freeman and Company: New York, 2008. DIVISION OF ENGINEERING SCIENCE, UNIVERSITY OF TORONTO