Download PIECING TOGETHER AN IDENTITY LAB

 PIECING TOGETHER AN IDENTITY LAB
BACKROUND PART ONE: ANTIGENS AND ANTIBODIES BLOOD BANK ANTIGENS AND ANTIBODIES Antigens are defined as substances recognized by the body as foreign, causing the body to produce an antibody to react specifically with it. Antibodies are proteins produced by lymphocytes as a result of stimulation by an antigen which can then interact specifically with that particular antigen. In order a substance to be an antigen to you it must be foreign (not found in the host): 
Autologous antigens are your own antigens (not foreign to you) 
Homologous, or allogenic, antigens are antigens from someone else (within the same species) that are foreign to you Blood group antigens: 
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There are over 300 known blood group antigens Over 1,000,000 different antigen sites on each red blood cell. These antigens are attached to proteins or lipids on the red cell membrane and are usually complex sugar groups. Some stick out far on the red cell membrane and some are buried within crypts on the membrane surface. http://faculty.matcmadison.edu/mljensen/BloodBank/lectures/blood_bank_antigens_and_antibodi.htm
Rules of Thumb For in vivo Antigen‐Antibody Reactions 1. If a person's cell have the antigen, the antibody should NOT be present in that person's serum 2. If an antibody to a blood group antigen is present in the serum of a person, his or her cells should lack that antigen 3. The antigens are on the cells and the antibodies are in the serum 1 | P a g e Stages of Antigen‐Antibody Interaction The first stage is sensitization. Sensitization occurs when antibodies react with antigens on the cells and coat the cells The second stage of the reaction is agglutination. Agglutination occurs when antibodies on coated cells form cross‐linkages between cells resulting in visible clumping. BACKGROUND PART TWO: BLOOD TYPING BASICS Karl Landsteiner opened the doors for blood banking with his discovery of the first human blood grouping system ABO, now also called ABH. His discovery marked the beginning of the concept of a person’s unique identity based upon the antigens present on the membrane of the erythrocytes (RBC) membrane. In 1901 Landsteiner drew blood from himself and five associates, separated the cells from the serum, and then mixed each cell sample with each serum. He was inadvertently the first person to perform both the forward and reverse grouping. The ABO group contains four blood types: A, B, AB, and O. The letters A and B represent antigens on the red blood cell membranes. A person with type A has A antigen on the RBC's, and someone with type B blood has the B antigen. Type AB means that both A and B antigens are present, and type O means that neither the A nor the B antigen is present. http://pathmicro.med.sc.edu/mayer/rx‐7.jpg
The frequency of these blood groups is as follows: O positive 38% B positive 9% O negative 7% B negative 2% A positive 34% AB positive 3% A negative 6% AB negative 1%
ABO typing: If your blood cells stick together when mixed with: (Uses patient’s whole blood or RBC’s and antisera) 
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2 | P a g e Anti‐A serum, you have type A blood Anti‐B serum, you have type B blood Both anti‐A and anti‐B serums, you have type AB blood If your blood cells do not stick together when anti‐
A and anti‐B are added, you have type O blood. Reverse typing or back typing: (uses the patient’s serum and type A cells and/or type B Cells from whole blood. http://i.dailymail.co.uk/i/pix/2008/06/21/article‐1028274‐01A7982200000578‐632_468x344_popup.jpg
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If the blood clumps together only when B cells are added to your sample, you have type A blood. If the blood clumps together only when A cells are added to your sample, you have type B blood. If the blood clumps together when either types of cells are added to your sample, you have type O blood. Lack of blood cells sticking together when your sample is mixed with both types of blood indicates you have type AB blood. Source of Blood Typing antisera: Monoclonal antibodies react with very specific antigenic determinants. They are not produced in humans or animals, but harvested from cells in cells grown in tissue culture. The tissue culture cells made from fusion of a plasma cell, which is the antibody producer and the myeloma cell, which provides longevity and ability to make large amounts of antibody Lectins are sugar‐binding proteins that are highly specific for their sugar recognition affinities. They play a role in biological recognition phenomena involving cells and proteins. Purified lectins are important in a clinical setting because they are used for identifying some of the ABO(H) antigens. Some of the, (the prefix gyco refers to the presence of sugars) glycolipids and glycoproteins on an individual's red blood cells can be identified by lectins. Lectins (a.k.a. toxalbumins, phytotoxins, or phytohemagglutinins) were originally discovered in plants more than a century ago by their ability to agglutinate (clot together) human red blood cells (erthyrocytes). Some lectins are so specific in agglutinating certain blood types that they were used in the characterizations of the human blood groups A, B, and O. The lectin is composed of amino acids, and these proteins bind to sugar molecules of glycoproteins attached to the outside of blood cells, but they can also bind to glycoproteins on other cell types and to simple sugars alone, such as sucrose. 
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A lectin from Dolichos biflorus, (the Hyacinth bean), is used to identify cells that belong to the A1 blood group. A lectin from Ulex europaeus, (spiny evergreen shrubs) is used to identify the H blood group antigen. A lectin from Vicia graminea, (broad beans and vetch), is used to identify the N blood group antigen. 3
RH typing: 
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If your blood cells stick together when mixed with anti‐Rh serum, you have type Rh‐positive blood. If your blood does not clot when mixed with anti‐Rh serum, you have type Rh‐
negative blood. Blood types are based on specific antigens and antibodies related to RBC's. Specific blood type antigens called agglutinogens are found in the cell membrane of erythrocytes. Antibodies called agglutinins are in the plasma and are formed after birth. When agglutinins in the plasma combine with agglutinogens on the surface of the RBC the result is agglutination or clumping of the RBC's. The agglutinogens on the red blood cells are organized into groups. Although 30 groups are recognized, the ABO and the Rh are the most important. Blood type antigens on the surface of the RBC's are agglutinogens. The antibodies in the plasma are called agglutinins. The ABO grouping system is the only blood group system in which individuals will have antibodies in their serum to antigens that are absent from their RBC. This occurs without any exposure to RBC’s by transfusion or pregnancy. Due to the presence of these antibodies transfusion of an incompatible ABO type can result in almost immediate lysis, (disintegration), of the donor RBC’s. This produces a very severe, if not immediately fatal, transfusion reaction in the patient. A patient's red blood cells can be mixed with antibody to a blood group antigen to determine a person's blood type. In a second example, a patient's serum is mixed with red blood cells of a known blood type to assay for the presence of antibodies to that blood type in the patient's serum. INHERITANCE OF BLOOD ANTIGENS The theory for inheritance of the ABO blood groups was first described by Bernstein in 1924. He demonstrated that an individual inherits one ABO gene from each parent and these two genes determine which ABO antigens are present on the membrane of the red blood cells, (RBC’s). The ABO system like other blood systems is co‐dominant in expression. One position or locus on each http://www.kinghawk828.com/en/up_files/xuexing1.jpg
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chromosome 9 is occupied by an A, B, or O gene. In case of O allele, the exon 6, (the sixth piece of active DNA in that chromosome) contains a deletion that results in a loss of enzymatic activity. The deletion causes a frameshift and results in translation of an almost entirely different protein that lacks enzymatic activity. This results in “H” antigen remaining unchanged in case of O groups. So the O gene is considered an amorph (present but not functional) because no antigen is produced in response to the inheritance of this gene. The formation of the blood antigens results from the interaction of genes at three separate loci, (ABO, Hh and Se). These genes do not actually code for production of antigens, but for the production of specific enzymes which add sugars to a basic precursor substance that is common to all of the blood types. A, B, and H antigens develop as early as the 37th day of fetal life. The RBC’s of the newborn have 25% to 50% fewer antigenic sites that those present on adult RBC’s and the complete complement of these antigens will not develop until 2‐4 years of age. The H gene also located on chromosome 9 is inherited by persons of all blood types and codes for the production of an enzyme that catalyzes the attachment of sugars to the RBC’s cell membrane in response to the A and B genes inherited as a part of another locus on chromosome 9. So if a person lacks the H substance even if they have inherited the A and B genes it will appear that they have type O blood because the sugars that are synthesized and attached to the basic precursor substance will not be present and a blood test will indicate that the person has type O blood (Bombay Phenotype). The “H” gene is present in 99.9 % of the random population. The allelic combination of “Hh” is quite rare and “hh” is exceptionally rare with an inheritance rate of 1/8,000 in Taiwan, 1/10,000 in India and 1/1,000,000,000 in Europe and the U.S. A patient who inherits the hh combination of genes lacks the normal expression of any of the ABH antigens. ABH antigens are an integral part of the membranes of RBC’s, endothelial cells,(The endothelium is the thin layer of cells that line the interior surface of blood vessels, forming an interface between circulating blood in the lumen and the rest of the vessel wall. Endothelial cells line the entire circulatory system, from the heart to the smallest capillary), platelets, responsible for blood clotting), lymphocytes,(white blood cells), and epithelial cells, (cover and line body surfaces). ABH soluble antigens can also be found in all body secretions. Their presence is dependent on the ABO genes inherited as well as the inheritance of another set of genes on chromosome 9 called the secretor genes (Se or se). Eighty percent of the U.S. populations are known as secretors because they have inherited one of the secretor genes. The inheritance of the Se gene coded for the production of the A, B, and H antigens only in bodily secretions and does not affect the production of ABH antigens on the RBC”s themselves. 5
The term “secretor” refers to the presence of A, B, and H substances in the body fluids, and the presence of these substances does confirm the inheritance of an A, B, H, and Se gene in an individual. In order to be a non‐secretor the selected A, B, and H antigens are absent because a person has inherited two copies of the “se” gene If a person is a secretor soluble A , B, and H antigens will be present in sweat, tears, saliva, and serum. When a blood test is performed on a patient there are actually two separate test one uses antiserum to test for the presence of the antigens on the surface of the RBC’s and a separate test to determine the presence of the antigens in the serum. Not all blood type antigens are inherited as a part of chromosome number 9. The “Rh” antigen also called antigen “D” is inherited as a gene on chromosome number one. The inheritance of this gene is considered by hematologists to be the most statistically significant of the 30 blood antigens with the exception of the A, B, and H antigens. Antigen D is present if 85% of the random population while 15% of the population lacks the “D” antigen. In 1939 Philip Levine and Rufus E. Stetson published their findings about a family who had a stillborn baby who died of hemolytic disease of the newborn. The mother was aged 25 and it was her second pregnancy and she suffered blood loss at the delivery. Both parents were blood group O and the husband's blood was used to give the mother a blood transfusion, but the mother suffered a severe transfusion reaction. They investigated this transfusion reaction. Since the mother and the father were both blood group O, they concluded that there must be a previously undiscovered blood group antigen that was present on the husband's RBCs but was not present on the mother's RBCs and that the mother had formed antibodies against the new blood group antigen. This suggested for the first time that a mother could make blood group antibodies because of immune sensitization to her fetus's RBCs. They did not name this blood group antigen, but it was subsequently found to be the Rhesus factor. Additional genes near the “Rh” locus are the C, c, and E, e genes that are thought to help regulate the expression the “Rh” antigens on the surface of RBC’s, and all must be tested to determine the patients “Rh” phenotype. MEDICAL CONDITIONS RELATED TO SECRETED BLOOD TYPE ANTIGENS o If a person is a secretor, the presence of the blood antigens will influence the populations of bacteria capable of taking up residence in the gastro‐
intestinal tract. This occurs because some of the bacteria in the GI tract are capable of producing enzymes that allow them to degrade the terminal sugar of the A, B, and H antigens and consume them as a food supply. o ABH secretors have longer blood clotting times because they lack the necessary concentrations of clotting factors 6
o Significant variations in carbohydrate concentrations in breast milk are found between secretors and non‐secretors. o ABH non‐secretors have a significantly higher number of duodenal and peptic ulcers o ABH non‐secretors are at a higher risk for urinary tract infections o ABH non‐secretors are at higher risk for Neisseria meningitis o ABH non‐secretors are at higher risk for Candida albicans infections o ABH non‐secretors have a higher risk of dental caries SO BLOOD TYPE CAN BE DETERMINED BASED ON THE PRESENCE OR ABSENCE OF ANTIGENS AND ANTIBODIES IN WHOLE BLOOD, SERUM OR OTHER BODY FLUIDS AND SOME TISSUES. BACKROUND LAB PART THREE: THE INHERITANCE OF SEX CHROMOSOMES AND THE PRESENCE OF BARR BODIES: Establishing individual identity is an imperative aspect of any investigative procedure. Determination of gender helps in investigations by narrowing down http://www.bio.miami.edu/~cmallery/150/gene/p14x17barr.jpg
potential pools of victims of potential pools of suspects. Sex chromosomes are particular chromosomes that are involved in determining the sex of an organism. In the cells of humans and many other organisms the sex chromosomes consist of a pair of chromosomes called the X and Y chromosomes. The X and Y chromosomes were first discovered in beetles by Nettie Stevens in 1906. She noticed that cells of female beetles had identical looking pairs of each of their several chromosomes, but that male beetles had one pair in which the chromosomes were very different in appearance from each other. She called these two chromosomes the X and the Y, and found that female beetles differed from males in containing two X chromosomes. The same situation is also found in humans where females are XX and males are XY. The X and Y chromosomes in humans are also very different in appearance, with the X chromosome being considerably larger than the Y. With the exception of only about nine shared genes, the X and Y chromosomes do not contain the same genes, unlike the other twenty‐two pairs of human chromosomes in which members of a pair share all the same genes. The Y chromosome contains the genes for determining a male pattern of development, and in the absence of a Y chromosome an embryo will follow a female pattern of development. Since cells in a male contain a single X chromosome and cells in a female contain two X chromosomes, females contain twice as many copies of the genes on the X chromosome per cell as do males. To equalize the dosage of X chromosome 7
genes between the two sexes, one of the two X chromosomes in each cell of all female mammals is inactivated early in embryonic development by becoming very tightly wound up or condensed. Most of the genes on the condensed X chromosome cannot be expressed. Since males carry only one copy of each X‐linked gene, they are much more likely to suffer from disease if they inherit a defective gene. The inactivation of an X chromosome in the cells of a developing female embryo occurs randomly, so that about half of the cells express the genes in one X chromosome and half express the genes in the other X chromosome. Once a particular X chromosome has been inactivated in a cell, it will remain inactivated in all of the descendants of that cell. If a female mammal has different forms or alleles of a particular gene on each of her two X chromosomes, then about half of her cells will express one of the alleles and about half the other allele. Chromosomes are ordinarily visible under a microscope only when the cell is dividing. However, when non‐dividing cells are treated with stains that bind to chromosomes, a darkly staining body is visible in the nuclei of cells from females but not in cells from normal males. This body is actually the condensed X chromosome, and it is called a "Barr body" after its discoverer, Murray Barr. The inactive X usually lies along the edge of the interphase nucleus in a highly condensed state. It is always the last to replicate. In 1948, before the discoveries of Lyon, Barr and Bertram found that in the interphase nucleus of female cat neurons there were a significant number of cells that had one "darkly staining body" lying along the edge of the nucleus, but they never found a "darkly staining body" in the neurons of male cats. Similar "darkly staining bodies" are found in buccal epithelial cells of human females, although they can usually be found in only 30% to 40% of the cells. Normal males never express these "Barr bodies." In all cases, the number of Barr bodies is one less than the number of X chromosomes in an individual. One Barr body means the individual has two X chromosomes, two Barr bodies means the individual has three X chromosomes, etc. We now know that the "darkly staining" Barr body is the condensed, inactive X chromosome. In 1961 Mary Lyon proposed that the condensation of the X chromosome into a Barr body was a mechanism for inactivating the genes on the chromosome. This is called "The Lyon Hypothesis," in her honor. The presence or absence of a Barr body in cells is used in medical and criminal forensics to determine and legally define the sex of an individual. The Barr body is most easily observable in cells that are in interphase. However, Barr bodies can only be observed in 30% to 40%of cells in a field of view, even if they are present in all cell samples. This is because observation of Barr bodies depends on the angle at which the nuclei are observed. Barr bodies allow one to distinguish between 8
male and female mammals but will not allow one to determine if the female possessing the Barr body is human. ACCORDING TO THE FBI LAB, THE PRESENCE OR ABSENCE OF BARR BODIES IN ANY EPITHELIAL CELL WILL ALLOW YOU TO DETERMINE THE GENDER OF THE REMAINS. REMEMBER EPITHELIAL CELLS COVER AND LINE ALL BODY SURFACES. In addition: if the tissues are not too degraded, the amelogenin gene for tooth pulp is six bases shorter on the X chromosome than on the Y chromosome. When amplified and examined on an agarose gel, a female will have one band, (two X’s) and a male will have two bands one for the X and one for the Y. BACKGROUND PART FOUR: Detection of ABO(H) substances in body fluids: In some cases traditional extraction of body fluids is not possible due to the degradation of body fluids that have been exposed to harsh environmental conditions or due to the putrefaction of a corpse. In such cases alternative means of extraction of body fluids and tissues must be used to identify human remains. Tissues and body fluids can be examined for the presence of AB0(H) antigens. In both cases, AB0(H) blood group substances were obtained by the adsorption elution method. Adsorption involves the combination of antibodies and antigens in sufficient enough quantities that the complexes precipitate out of solution and are collected from the supernatant after centrifugation. In auto adsorption, the AB0(H) antibodies are washed from the patients cells to remove unbound antibody, then the remaining cells are incubated with the patient’s serum for an hour. This solution is then inspected for signs of agglutination. The Elution stage involves the use, release, and purification of the AB0(H) antibodies. Washing (elution) of the cells to remove the antibodies changes the attractive forces between the antigen and the antibody, or may even change the structure of the surface of the red blood cell. The antibody is then freed into the resulting supernatant (eludate) and the solution is tested with anti‐sera to determine the antigens and antibodies present. In one series of tests conducted in Poland in the 1970’s, blood group identification was performed on the tissues of cadavers. This study used the fluid from the human inner ear and tissues from the salivary glands. http://pathmicro.med.sc.edu/mayer/ab‐
In the case of the cells of the salivary gland AB0(H) antigens are widely distributed. The antigens are detected by means of using an immunohistochemical technique using antibodies against the ABO(H) antigens or the lectin plant based 9
substances that cause agglutination. In this study 2 grams of salivary tissue was added to 1.5 ml of saline and boiled for 15 minutes. One drop of extract was mixed with one drop of anti‐A and B serum was added. (The anti sera must be diluted to a concentration of 1:16 before it can be used.) Antibodies can be manufactured to react either with the native protein component of the antigen or with parts of the denatured protein Another study conducted in Scotland in 1993, used Eliza assays of two forms: Radioimmunoassays (RIA) are assays that are based on the measurement of radioactivity associated with immune complexes. In any particular test, the label may be on either the antigen or the antibody. Enzyme Linked Immunosorbent Assays (ELISA) are those that are based on the measurement of an enzymatic reaction associated with immune complexes. In any particular assay, the enzyme may be linked to either the antigen or the antibody. The addition of a second antibody directed against the first antibody can result in the precipitation of the immune complexes and thus the separation of the complexes from free antigen and the amount of precipitant can be quantified. In this case, swabs from the submandibular gland were taken from 179 autopsies and in another 101 autopsies swabs were taken from the oral cavity. In each case there had been considerable contamination of the blood had occurred and the reliability of traditional blood testing means was not considered reliable or valid. As a control 100 uL of saliva from known secretors and non‐secretors was tested. SO THE PRESENCE OF BLOOD ANTIGENS AND ANTIBODIES CAN BE DETERMINED FROM SALIVARY GLAND TISSUE OR FROM SALIVA 10
THE CASE OF THE MISSING COLLEGE STUDENTS: 11
THE CASE OF THE MISSING COLLEGE STUDENTS: Aaron and Rebekah decided late on Wednesday night to take a weekend trip to Crossville, TN to one of their favorite parks for a camping trip. They needed to relax before their final exams began. Little did they know this would be the longest trip they would ever take. Aaron and Rebekah were experience hikers and rafters in the Blue Ridge Mountains and had worked several summers in Cumberland Mountain State Park that was just a few miles east of Nashville. The pair left the Vanderbilt campus early on Thursday morning with plans to return late the following Sunday. They were up early and eager to get started. They stopped at the convenience store across from Rebekah’s apartment to pick up last minute supplies and were seen on the store’s security tapes at 4:35 a.m. smiling and holding hands. They had decided not to call friends to tell them about the last minute trip, it was early and they had all been up late studying. When neither Rebekah nor Aaron was in class on Monday their friends were not too concerned because it was not unusual for the pair to skip Monday classes. When neither of the two was in classes all week, Rebekah’s roommate decided to call the police to report the couple missing. The authorities began an immediate search. This is when the police discovered the convenience store video tape and they discovered a second video of the pair in Aaron’s SUV at a freeway onramp for I 40 East. The timestamp on this video was 5:45 a.m. and corresponded with the time that the couple had been seen at the store. The interviews with the couple’s friends painted a picture of a happy couple who enjoyed spending their free time together sharing their passion for hiking and camping. The police searched both of the student’s apartments and discovered that their camping gear was missing. In subsequent interview authorities discovered that the area where Rebekah and Aaron liked to camp was difficult mountainous terrain, with a thick dense canopy of trees, dense undergrowth, and numerous deep valleys and gorges. Aarons SUV was discovered by park rangers in a remote area behind Byrd Lake in the Cumberland Mountain State Park. After a lengthy and extensive search, the couple was never found. More than a year later a group of hikers discovered two sets of remains near an unused trail at the bottom of a deep gorge. The remains were in an advanced state of decomposition. There was some soft tissue still intact, but it was decomposed to the point where ethnicity and gender were not immediately identifiable. There were no clothes on the remains. The pelvis of each skeleton was absent, the long bones were scattered and had obviously been gnawed on by scavengers, and the marrow had been exposed to the elements. There were obvious signs of foul play because only a few fragments of each skull were located http://www.wikidoc
Because of the lack of evidence, determination of gender will be .org/images/thumb/
difficult. The long bones were used to estimate height of each of each b/b2/Parotid_gland
of the sets of remains. Set two is estimated to be approximately 6’2” and set number _en.png/250px‐
Parotid_gland_en.p
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one is estimated to be 5’6”. These height estimations are consistent with the heights of Rebekah and Aaron. The ethnicity of the remains was judged to be Caucasian based on the lack of curvature in each of the femoral shafts, which is also consistent with the missing couple. Based on this preliminary information, Police believed that the remains might be those of the two missing students. But you must complete the analysis of the tissue samples in order for the police to make a positive ID. Police are sure foul play is involved in the case because remains of clothing were not located and the condition of the skull fragments indicated blunt force trauma. There were also injuries on the femurs of both victims that had been made with a wide sharp blade and that were inconsistent with the activity of predators. Tissue samples were obtained and sent to you for analysis. Due to the advanced state of decomposition, cyto‐technicians have fixed some of the samples for you onto slides so that you can more quickly complete your portion of the investigation. Unknown Tissue Sample Number 8 9 10
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Description of Remains is a fixed slide of squamous epithelial tissue taken from the basement membrane of the parotid land from the first set of remains is a fixed slide of squamous epithelial tissue taken from the basement membrane of the parotid land from the second set of remains is a fixed slide showing the karyotype from squamous epithelial tissue taken from the basement membrane of the parotid gland from the first set of remains is a fixed slide showing the karyotype from squamous epithelial tissue taken from the basement membrane of the parotid gland from the second set of remains is a sample of the parotid gland from the first set of remains for blood typing analysis is a sample of the parotid gland from the second set of remains for blood typing analysis is a sample of the parotid gland from the first set of remains for DNA extraction is a sample of the parotid gland from the second set of remains for DNA extraction From these tissue samples you will examine #8 and #9 for the presence or absence of Barr Bodies to determine the gender of the individual sets of remains. You will use #10 and #11 to confirm the gender of each individual and to examine the 13
karyotypes for the presence of genetic anomalies. You will use #12 and #13 to determine the blood type, secretor status of each individual and you will use the remain samples to extract DNA for electrophoresis and genotyping. Procedure One: To prepare the fixed slides for analysis of Barr bodies, the remaining oral tissue was rinsed with sterile saline. A sterile metal spatula was then drawn down the buccal, (oral), surface of the parotid gland and remaining masseter muscle. The cellular material was quickly smeared on the slide and gently flattened with a cover slip. The slides were then fixed for 15 minutes with 95% ethyl alcohol; they were then stained, rinsed, and stained again before final mounting. 1. What is the gender of the individual for unknown sample #8? 2. What is the gender of the individual for unknown sample #9? Procedure Two: PROCEDURES FOR PREPARING KARYOTYPES Before beginning the lab: place a dry slide vertically at a 45°angle. 1) With a pipette, gently re‐suspend the cells in the tube provided. Remove a small sample of cell suspension with a pipette and 2) Hold the pipette 2 feet above the slide. Allow one drop of cell suspension to “splat” onto the slide about 3/4 inch from the upper end and tumble down the slide. Carefully apply 6‐8 more drops from various heights, One drop at a time, onto the same region of the slide. It is important to release the cell suspension one drop‐at‐a‐time. Do not expel all of your cell suspension in one squirt, or you will obtain poor results. Gently blow across the slide for 2‐3 seconds. The drying will help “spread” the chromosomes. 3) Allow the cells to AIR DRY COMPLETELY. 14
http://www.betsiproject.org/pdf/outreach_hs_activities_chrom
osomelab.pdf 4. Dip the slide into the tube containing STAIN #1 for 1 second only. Remove the slide and dip into STAIN #1 again for 1 second only. Remove the slide and dip into STAIN #1 again for second only. 5. Drain off stain and dip the slide into tube containing STAIN #2 for 1 second only. Remove the slide and dip into STAIN #2 again for 1 second only. Remove the slide and dip into STAIN #2 again for 1 second only. Caution should be taken to avoid carryover of stains (wipe the bottom of slide with a paper towel before transferring). 6. Remove slide from stain and thoroughly rinse with distilled water. http://www.betsiproject.org/pdf/outreach_hs_activities_chrom
osomelab.pdf 7. Allow slide to AIR DRY COMPLETELY. A stream of warm air or blowing may help speed up the drying process. Incomplete drying will result in very poor resolution. 8. View your slides on your microscope. Scan on low power for cells that have been ruptured and released their chromosomes. Shift to high power(400x) to examine the chromosome spread more carefully. If you find a good spread, try to count the number of chromosomes present. Try and identify three characteristic chromosomes based on the location of their centromere. http://www.betsiproject.org/pdf/outreach_hs_activities_chrom
osomelab.pdf Procedure Part Three: YOU WILL PERFORM THIS ANALYSIS ANALYSIS OF SALIVA TO DETERMINE BLOOD TYPE: 1. Use the salivary gland cell sample that you have been provided. It should be already labeled with both a #12 and another tube with a #13. 15
2. Stand the tube upright in the test tube rack in a boiling water bath for ten minutes. Boiling the salivary gland cells inactivate the enzymes which otherwise might destroy the blood group components. Carefully remove the test tubes and rack and allow the tubes to cool. 3. Once cool, centrifuge the tubes for several minutes to sediment any of the coarse precipitate. You will use only the supernatant solution for the remaining parts of the experiment. If necessary the supernatant could be refrigerated for several weeks and still be used as a viable source of blood group antigens. 4. Place twelve micro‐centrifuge tubes in a test tube rack. Label the tubes as follows CA, 2A,4A,6A,8A,16A,32A.and in a second row CB,2B,4B,6B,16B,32B The CA, and CB tubes indicates the Control tubes to which no saliva will be added. While the numbers on the tubes indicate the final dilution of the saliva that each of the tubes will contain: 1:2, 1:4, 1:8, 1:16, 1:32. The following steps must be followed exactly in order for you to achieve accurate results. Before you begin this portion of the lab use a clean pipette and practice expelling one drop of saline from a Pasteur pipette at a time without any quantitative errors. 5. Use a clean Pasteur pipette, place one drop of the saline into each of your labeled tubes. 6. Use a clean Pasteur pipette, and place one drop of the saliva from tube 12 into the tube labeled 2A and one drop of saliva from tube 13 into 2B. You will now dilute the saliva saline mixture in tube two by vortexing or by drawing the contents into the pipette and expelling it back into the tube several times. 7. Using a clean pipette, take one drop of the saliva saline mixture from tube 2A and drop it into tube 4A, then take one drop of the saliva saline mixture from tube 2B and drop it into 4B. Expel the remainder of the saliva BACK INTO THE SOURCE TUBE AND DISCARD THE PIPETTE. You will now dilute the saliva saline mixture in tube 4 by vortexing or by drawing the contents into the pipette and expelling it back into the tube several times. 8. Using a clean pipette, take one drop of the saliva saline mixture from tube 4A and drop it into tube 8A, then take one drop of the saliva saline mixture from tube 4B and drop it into 8B Expel the remainder of the saliva BACK INTO THE SOURCE TUBE AND DISCARD THE PIPETTE. You will now dilute the saliva saline mixture in tube 8 by vortexing or by drawing the contents into the pipette and expelling it back into the tube several times. 9. Using a clean pipette, take one drop of the saliva saline mixture from tube 8A and drop it into tube 16A then take one drop of the saliva saline mixture from tube 8B and drop it into 16B . Expel the remainder of the saliva BACK INTO THE SOURCE TUBE AND DISCARD THE PIPETTE. You will now dilute the saliva saline mixture in tube 16 by vortexing or by drawing the contents into the pipette and expelling it back into the tube several times. 16
10. Using a clean pipette, take one drop of the saliva saline mixture from tube 16A and drop it into tube 32A, then take one drop of the saliva saline mixture from tube 16B and drop it into 32B Expel the remainder of the saliva BACK INTO THE SOURCE TUBE AND DISCARD THE PIPETTE. You will now dilute the saliva saline mixture in tube 32 by vortexing or by drawing the contents into the pipette and expelling it back into the tube several times. 11. In the last step using a clean pipette, withdraw the entire contents of tube 32. Expel ONE DROP of the solution back into tube 32. Discard the remaining saliva mixture with the pipette. TESTING THE SAMPLES 12. Add one drop of anti‐A serum to each test tube labeled with an A, then add one drop of anti‐B to each of the test tubes labeled with a B. Shake each tube and allow it to stand for 15 minutes. Then add one drop of the group A red blood cells to each A tube, and one drop of group B cells to each of the B tubes. Let each tube stand for 5 minutes and then centrifuge all of your tubes at high speed for 20 seconds. 13. Inspect each of the tubes for the presence or absence of red blood cell clumping. This is done by picking up each tube near its top and flicking the bottom with your finger. DO NOT SHAKE THE TUBES. 14. Hold each test tube in turn up to a light and determine if there is an even suspension of particles or if there are large clumps of coarse particles. This can be confirmed using a microscope to determine the degree of aggregation. 15. To prepare the samples for microscopic examination, pour the contents of each tube into a separate compartmentalized plastic tray. Next to each compartment, mark the tube from which each of the samples originated. 16. Under low power, search the entire field of fluid from each tube. Are all the red blood cells discrete, (individual), and evenly distributed in the surrounding fluid. If there is no agglutination mark the sample “O”. If there are clumps present mark a + by the tube if there are no free cells remaining mark ++++. 17. Make sure to record your results for each tube. Typically in a secretor tube, the tubes with the most concentrated saliva will have only slight aggregation of any. Whereas the tubes with more dilute saliva 8, 16, and 32 will have an appreciable amount of aggregation. Why? Because the number of antigens and antibodies more closely match each other and the overabundance of the antigens does not overwhelm the number of antibodies in the anti‐ serum. 17
Checklist for Analysis of Saliva to Determine Blood Type: 1. Locate and identify tube number 12 and 13 2. Stand each tube upright in a test tube rack in a hot water bath for 10 minutes 3. Allow tubes to cool 4. Centrifuge the tubes at high speed for three minutes 5. Place two rows of six micro‐centrifuge tubes into a test tube rack and label them CA, 2A,4A,6A,8A,16A,32A.and in a second row CB,2B,4B,6B,16B,32B 6. Place one drop of saline into each of the labeled tubes 7. Use a clean pipette and place one drop from tube 12 into tube 2A. Expel remainder of solution in the pipette back into the source tube. 8. Use a clean pipette and place one drop from tube 13 into tube 2B. Expel remainder of solution in the pipette back into the source tube. 9. Use a clean pipette and place one drop from tube 2A into 4A. Expel remainder of solution in the pipette back into the source tube. 10. Use a clean pipette and place one drop from tube 2B into 4B. Expel remainder of solution in the pipette back into the source tube. 11. Use a clean pipette and place one drop from tube 4A into 8A. Expel remainder of solution in the pipette back into the source tube. 12. Use a clean pipette and place one drop from tube 4B into 8B. Expel remainder of solution in the pipette back into the source tube. 13. Use a clean pipette and place one drop from tube 8A into 16A. Expel remainder of solution in the pipette back into the source tube. 14. Use a clean pipette and place one drop from tube 8B into 16B. Expel remainder of solution in the pipette back into the source tube. 15. Use a clean pipette and place one drop from tube 16A into 32A. Expel remainder of solution in the pipette back into the source tube. 18
16. Use a clean pipette and place one drop from tube 16B into 32B Expel remainder of solution in the pipette back into the source tube. 17. Add one drop of anti‐A serum to each of the tubes marked with a letter A. Shake each tube and allow them to stand for 15 minutes. 18. Observe each tube for the presence or absence of coagulation (clumping) 19. Add one drop of anti‐B serum to each of the tubes marked with a letter B. Shake each tube and allow them to stand for 15 minutes. 20. Observe each tube for the presence or absence of coagulation (clumping) Tube Number CA 2A 4A 6A 8A 16A 32A CB 2B 4B 6B 8B 16B 32B Coagulation Results + for present –for absent 21. Add one drop of group A red blood cells to all of the tubes marked with an A 22. Add one drop of group B red blood cells to all of the tubes marked with a B 23. Let all tubes stand for 5 minutes 24. Centrifuge all tubes on high speed for 20 seconds. FROM THIS POINT FORWARD DO NOT SHAKE THE TUBES 19
21. Inspect each of the tubes for the presence or absence of clumping or coagulation 22. Examine the contents of each tube microscopically and record your results. If the saliva of a secretor is mixed with the antiserum of lectin specific for its blood group substance then most of the antibody in the antiserum will bind to the blood group substance in the saliva. So when you add the red blood cells for that type no clumping or very little clumping should be observed. This is the opposite of what you would see during a traditional blood test. Data Table Two: Tube Number Coagulation Results: Secretor Yes or No Record “+” for some cells clumped Record “++ for half the cells clumped Record “+++” for most of the cells clumped Record “++++” for all cells clumped CA 2A 4A 6A 8A 16A 32A CB 2B 4B 6B 8B 16B 32B Use the following table to determine the secretor status of each of the tubes and record in the table above. Antiserum A B Red Group Cell A B Type 1 No clumps Clumps Type 2 Clumps No clumps Type 3 No clumps No Clumps Type 4 Clumps Clumps Type 5 Type 6 Lectin H O No Yes The tube is: Secretor group A Secretor group B Secretor group AB Indeterminate Blood group O NON SECRETOR 20
Unknown Tissue Gender or Blood Type
Sample Number Male/Female 8 9 Male/Female 10 Male/Female Genetic Anomaly 11 Male/Female Genetic Anomaly 12 Blood Antigen Present/Absent A/B/O/H 13 Blood Antigen Present/Absent A/B/O/H 14 DNA Present/DNA Absent 15 DNA Present/DNA Absent Procedure used for determination Based on the information that you have available can you identify the two sets of remains that the authorities have found? What conclusions can you make? 21
DNA Fingerprinting: The following activity is designed to demonstrate the techniques used by molecular geneticists and forensic pathologists. Today you will perform an activity known as DNA fingerprinting or electrophoresis. You have been provided with samples of mitochondrial DNA sequences. These samples are designed to represent one‐half of a strand of replicated DNA. The base sequences found within these strands are unique to you and your maternal relatives and can be used as a means to identify a person from a sample of bodily fluid. With nuclear DNA, a child receives half of its DNA from each parent in any give strand, half of the sequence will match the mother’s DNA and the other half will match the father’s DNA. In the case of mitochondrial DNA, an exact copy is inherited from your mother as it has been passed from all of the female relatives on your mother’s side of the family. During the actual process of electrophoresis, DNA is removed from a tissue sample, (such as blood or hair). This DNA contains hundreds of copies of each DNA strand. The DNA molecules, hundreds of thousands of bases long, are cut into smaller pieces using several different restriction enzymes. These enzymes only cut the DNA where specific sequences of bases occur. The result of the enzyme activity is a pool of DNA fragments are sorted by size. This is accomplished by taking advantage of the electrical charge on each one of the fragments. As an electric current is passed through a semi‐permeable agarose gel the various fragments are pulled through the gel at different rates. The rate and distance migrated, depends on the charge and the length of each of the pieces. The DNA is then treated so that the two strands come apart, exposing single strands of DNA bases. The gel is then transferred to a thick sturdy piece of paper. The paper is then soaked in a solution containing tiny single stranded DNA fragments. These fragments are radioactive markers, called probes, which will attach to complementary 22
bases in the cut up DNA strands. The treated paper will allow the probes to stick to the appropriate bases, and the remainder of the probes is washed away. The piece of paper is then placed on X‐ray film, and the film is developed. A dark spot appears wherever a radioactive probe stuck to the DNA. The result is a unique pattern of spots, called the DNA fingerprint. This fingerprint is unique to the individual, and can be repeated with tissue samples from any bodily fluid, with the exception of red blood cells, and achieve the same result. Why not red blood cells? Mature red blood cells do not contain a nucleus, and as a result do not contain enough reliable DNA to produce a true test. If red blood cells are the only source of DNA, the DNA must be cloned using a process known as PCR (polymerase chain reaction) so that enough DNA is present in order to get an accurate test. You will now use the materials provided to simulate such a test. Remember the concept of heteroplasmy, where a very small portion of the mitochondrial DNA might be a contribution from the male parent. This occurs only if the tail of the sperm enters the egg during the process of fertilization. The head of the sperm is an empty vessel that contains the DNA of the male parent, while the tail of the sperm must contain mitochondria to supply energy for the process of locomotion. MATERIALS 1.
DNA sequence from Rebekah 2.
DNA sequence from maternal relative Aaron 3.
DNA sequence from Aaron’s mother 4.
DNA sequence from Rebekah’s mother 5.
DNA sequence standard to ensure the tests accuracy 23
Procedure: 1. Review your data table and you will see the labels that should be attached to create 5 separate columns, as seen in the example below. Rebekah Aaron Aaron’s Mother Rebekah’s STANDARD Mother MAKE SURE THAT YOU KEEP EACH PERSON’S DNA SEPARATE THROUGHOUT THIS ACTIVITY. 2. To simulate the action of radioactive probes use a highlighter and cover the letters CAT in each of your segments. 3. With your PENCIL you are going to simulate the action of a restriction enzyme. Scan you DNA strips until you find the letters “GG CC”. MARK across the strip between the letter G and C, you will be forming a fragment that ends with a GG and another that begins with a CC. (Each sample has 5 of these sites so after cutting you will have 6 pieces. The standard contains 7 of these sites so you will have 8 fragments) 4. Count the number of bases in each fragment, (the number of letters). Match this number to the numbers in the chart provided. Then write the letters that occur in each piece in the box that corresponds to the correct persons name and the number of letters in each piece. Use your highlighter to mark the position of each of the CAT sequences in each piece. THIS IS AN EXAMPLE ONLY 24
Rebekah 20 ATCCATCAT 18 6. Compare the DNA of Rebekah to all of the possible relatives. To whom is she related? How do you know? Are you sure? (Use this space to write your essay response) 25
Rebekah CCACATCAGTTAGACCGAGGCCAAGGCCAACCGACGGCAAGGCCCGACAGGCCAAAGACGGCCATATAGGGGG Aaron CCTAGACGGCCAGGCACAAGCCAGGCCATGGCCACATCAGTTAGACCGAGGCCGAATCAGGCCTTATTGCAGG Aaron’s Mother CCTAGACGGCCAGGCACAAGCCAGGCCATGGCCACATCAGTTAGACCGAGGCCGAATCAGGCCTTATTGCAGG Rebekah’s mother CCACATCAGTTAGACCGAGGCCAAGGCCAACCGACGGCAAGGCCCGACAGGCCAAAGACGGCCATATAGGGGG 26
STANDARD CCAAGACATTATGCAGATGGCCAATAGACATTACGGCCATACCAGAGGCCAACATGGCCAAACACACC CATCAGGCCATGGCAGACAGGCCATACGG 27
Rebekah Aaron
Aaron’s Mother
Rebekah’s Mother
STANDARD
Number of Bases From Longest to Shortest Number of Bases From Longest to Shortest Number of Bases From Longest to Shortest Number of Bases From Longest to Shortest Number of Bases From Longest to Shortest 20 20 20
20 20
18 18 18
18 18
16 16 16
16 16
14 14 14
14 14
12 12 12
12 12
These numbers indicate the possible number of Base Pairs in each of the pieces of DNA that can occur after the action of restriction enzymes
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Rebekah Aaron
Aaron’s Mother Rebekah’s Mother STANDARD
10 10 10
10 10
9 9 9
9 9
8 8 8
8 8
6 6 6
6 6
These numbers indicate the possible number of Base Pairs in each of the pieces of DNA that can occur after the action of restriction enzymes
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