Exercise 10 - DNA Fingerprinting Introduction Deoxyribonucleic acid (DNA) is a double stranded genetic molecule consisting of many monomers called nucleotides, hence DNA is a polynucleotide. The two strands of DNA are connected to one another by hydrogen bonds between the nitrogenous bases of each strand. The DNA base pair sequence and DNA quantity (base pair total) vary from species to species. There would be less, but still measurable differences among conspecifics. Indeed, no two organisms of the same species, unless they are clones or identical twins, share exactly the same base pair sequence. It is the differences in these base pair sequences that allows for the identification of genetic similarities between DNA from two sources using a process known as DNA fingerprinting. This laboratory exercise will investigate the basic concepts in DNA fingerprinting involving techniques such as polymerase chain reaction (PCR) and the analysis of short tandem repeats (STR’s). Materials Equipment scissors marking pen or pencils Part A: Polymerase Chain Reaction (PCR) The most current form of DNA fingerprinting begins with a technique known as polymerase chain reaction (PCR). The advantage of PCR is that only a tiny amount of DNA is needed and the sample can be old, stored under less than ideal conditions or even partially degraded. PCR involves the following steps: 1. The DNA sample is placed in a small test tube with a solution of deoxyribonucleotides, small pieces of DNA to act as primers, and the enzyme DNA polymerase. The mixture is then placed in a thermal cycling device, which will raise and lower the temperature of the tube at precisely timed intervals. 2. Denaturing – occurs to the DNA when the mixture is raised to 94qC. The hydrogen bonds between the nitrogenous bases break down from the heat and the two complementary strands of DNA separate. 3. Annealing – the primers attach themselves to the long pieces of DNA through complimentary base pairing (A-T; G-C) when the temperature is lowered to 65qC. 4. Extending – DNA polymerase extends the new strands of DNA from the primers as the temperature is raised to 72qC. 5. The process (denaturing, annealing, extending) is repeated. The PCR process amplifies the original amount of DNA in a very short period of time (~ 2 minutes per cycle) so investigators will have much more DNA to use for subsequent analyses with little delay.
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Exercise 10 – DNA Fingerprinting
Procedure 1. On the following page is a “sample” of DNA and some primers 2. Use scissors to cut out all the pieces and lay them on the table 3. Starting with the two strands next to one another and lined up so the complementary base pairs are aligned 4. Denaturing - slide the two strands of DNA away from one another 5. Annealing - move two of the primers in between the DNA strands, lining them up to form complementary base pairs 6. Extending - use a marking pen or pencil to write in the complementary bases to the new DNA strand until both strands are complete After one round of PCR, one molecule of DNA consisting of two complementary strands yields _____ molecules of DNA for a total of _____ strands. How many DNA molecules would exist after 2 PCR cycles? _____ 5 cycles? _________________ 10 cycles? _________________ 20 cycles? _________________ 30 cycles? _________________ Write a formula to calculate the number of DNA molecules which will be created for a given number of PCR cycles.
Are all the DNA molecules created identical? _____
Lake-Sumter State College, Leesburg Laboratory Manual for BSC 1010C
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Exercise 10 – DNA Fingerprinting
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Exercise 10 – DNA Fingerprinting
Part B: Short Tandem Repeat (STR) Analysis Once the amplification of the original DNA through PCR has occurred, analysis of the DNA fingerprint can begin. Although estimates of the differences in DNA between individuals are very small (~ 1/10 of one percent), the sheer volume of DNA an individual possesses results in about 3 million bases pairs of unique sequence (i.e., each person differs by about 3 million DNA base pairs). The analysis of short tandem repeats examines some of this individual variation. Humans, like other eukaryotes, contain interruptions in the protein-coding genes, called introns. Because introns do not contain protein synthesis information, the base sequences in these regions tend to be repetitive. The same sequence of four, five, or six bases repeats itself over and over again. For example, intron 3 of the human α fibrinogen (a blood clotting protein) gene contains a sequence of bases “TTTC”, which repeats: TTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTC Because these repetitive sequences are short (4-6 bases) and occur side-by-side (in tandem) they are termed short tandem repeats (STR’s). The objective of DNA fingerprinting is to determine how many times a sequence of an STR is repeated in a DNA sample. How many times does the STR “TTTC” repeat itself in the above DNA sequence? _____ For intron 3 in the human α fibrinogen gene, an individual has between a 5 and 20% chance of sharing the same number of repeats with another individual. Although that seems like a relatively low chance, it must be remembered that the human population is quite large, so there would still be quite a few people in this group. Therefore, knowing the number of STR’s for just one particular intron is not sufficient enough to develop a unique DNA fingerprint. To achieve that, more than one STR must be examined. For example, intron 1 of the human tyrosine hydroxylase gene repeats the sequence “AATG”. Like the human α fibrinogen gene, only about 5-20% of people will repeat the “AATG” sequence the same number of times. Combining both these STR percentages narrows the field considerably. As an illustration, assume that 1 out of every 20 people (1/20 or 5%) repeats the intron 3 of the human α fibrinogen gene STR (“TTTC”) 14 times and that 1 out of every 5 people (1/5 or 20%) repeats the intron 1 of the human tyrosine hydroxylase gene STR (“AATG”) 10 times. The chances that two individuals will share the same number of repeats for both STR’s would be calculated as such: Chance of sharing the same number of repeats in both STR’s = chance of sharing the same number of repeats in the first STR x chance of sharing the same number of repeats in the second STR For this example: Chance of sharing the same number of repeats in both STR’s = 1/20 x 1/20 = 1/400 So, an individual has a 1 in 400 chance (0.25%) of sharing the same number of repeats in both STR’s. The more STR’s examined, the more that chance decreases until, within statistical certainty, a unique DNA fingerprint is assembled. Lake-Sumter State College, Leesburg Laboratory Manual for BSC 1010C
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Exercise 10 – DNA Fingerprinting
Since 1997 the Federal Bureau of Investigation (FBI) has set standards for DNA fingerprinting analysis for forensic and law enforcement purposes. To meet those standards, 13 specific genes areas (loci; singular locus) are evaluated. These loci are found on autosomes (non-sex chromosomes). A 14th locus is used to determine the sex of the individual and measures STR’s on the X and Y chromosomes. Requiring a match of at least 14 loci virtually guarantees a conclusive identification as the odds of two, unrelated person matching all 14 loci is approximately 1 in 1,000,000,000,000,000 (1 quadrillion). After PCR, the scientist or technician must find out how many repeats occur at each loci examined. There are number of ways this is accomplished, but most methods compare the DNA molecules produced by PCR to DNA fragments of known lengths. DNA fragments are created by cutting the DNA sample with special molecules known as restriction enzymes. Part C: Restriction Enzymes The preparation of a sample of DNA for fingerprinting involves treating the DNA with specific restriction enzymes (endonucleases). Restriction enzymes cut both strands of DNA at specific locations known as recognition sites (Table 10.1). The original intent of these enzymes is to provide protection for bacteria against some invading bacteriophages. Bacteriophages are viruses that attack bacteria. Cutting DNA with restriction enzymes results in smaller DNA fragments (restriction fragments) of various base pair lengths. Table 10.1 Recognition Sites for Various Restriction Enzymes Restriction Enzyme BstEII
EcoRI
HindIII
Recognition Site
---GpGTNACC-----GGTNACC-----CCANTGG-----CCANTGnG---
---GpAATTC-----GAATTC-----CTTAAG-----CTTAAnG---
---ApAGCTT-----AAGCTT-----TTCGAA-----TTCGAnA---
Digested DNA Fragments
---GXXXXGTNACC-----CCANTGXXXXG--(N = any nucleotide)
---GXXXXAATTC-----CTTAAXXXXG---
---AXXXXAGCTT-----TTCGAXXXXA---
A DNA Restriction Map provides the exact locations of the recognition sites for a particular restriction enzyme on a DNA sample. The map indicates distances from the origin where the enzyme cut the DNA. Size of each DNA fragment produced by restriction enzyme treatment can be measured in base pair (bp) units. Figure 10.2 illustrates a map of a DNA molecule of 50,000 bp in length. This DNA molecule has been treated with three different restriction enzymes. Procedure 1. Examine Fig. 10.2 and notice that each enzyme (A, B, and C) cuts the DNA sample at different recognition sites (for example, enzyme A cuts the DNA at 15,010 bp, again at 24,650 bp, and a third time at 30,003 bp) 2. Calculate the base pair (bp) size for each DNA fragment and write that number above the fragment
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Exercise 10 – DNA Fingerprinting
Fig. 10.2 Restriction Map of 50,000 Base Pair DNA Sample
uncut DNA
enzyme A
enzyme y B
enzyme y C
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Exercise 10 – DNA Fingerprinting
Part D: Gel Electrophoresis and Fingerprint Analysis Electrophoresis means to “carry with an electric current.” The digested DNA fragments created by the restriction enzymes are loaded into an agarose gel in which an electric current will flow through. Because DNA is a negatively charged molecule (due to the presence of large numbers of phosphate groups in its backbone), it will migrate through the gel towards the positive pole of the electrophoresis chamber. This procedure will separate the digested DNA fragments based upon molecular size. Smaller fragments will migrate further through the gel than larger ones. The different fragments can then be stained to make them visible. The resultant pattern of stained DNA “bands” or fragments constitutes a DNA fingerprint (see Fig. 10.3 as example). Fig. 10.3 Sample DNA Fingerprint
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Exercise 10 – DNA Fingerprinting
Part D1: Restriction Maps Procedure 1. Compare the Restriction Map in Fig. 10.2 to the DNA fingerprint in Fig. 10.4. Does the number of bands or fragment created by each restriction enzyme in the fingerprint coincide with the number of fragments illustrated in the Restriction Map? __________ 2. Label each of the DNA bands in Fig. 10.4 with its actual size in base pairs. Remember, in electrophoresis fragments are separated upon size, with smaller fragments moving farther from the wells at the negative end of the chamber than larger ones
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Exercise 10 – DNA Fingerprinting
Fig. 10.4 DNA Fingerprint of 50,000 Base Pair DNA Sample
uncut DNA
enzyme A
-
enzyme B
enzyme C
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Exercise 10 – DNA Fingerprinting
Part D2: Using a DNA Fingerprint to Determine the Number of STR’s The examination of banding patterns in gel electrophoresis can be used to determine the number of STR’s in a given DNA sample. Intron 3 of human α fibrinogen and intron 1 of the human tyrosine hydroxylase genes may result in a banding pattern such as Fig. 10.5. Fig 10.5 STR Banding Patterns for Two Human Genes α fibrinogen locus
-
tyrosine hydroxylase locus
size markers 72 bp
68 bp
64 bp
60 bp
56 bp
52 bp
48 bp
44 bp
40 bp
+ Why is there only one band for the human α fibrinogen loci? (recall an organism can be homozygous or heterozygous for a particular gene)
To determine the number of STR’s for each band, divide the size of the DNA fragment by the number of bases that repeat (usually 4-6) in that particular STR. For example, a DNA fragment of an STR which repeats the 5 bases “AGGTA” that is 55 base pairs long would repeats that STR 55/5 or 11 times.
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Exercise 10 – DNA Fingerprinting
For each allele in Fig. 10.5, how many times does the STR in intron 3 of the human α fibrinogen gene repeat? _____ _____ For each allele in Fig. 10.5, how many times does the STR in intron 1 of the human tyrosine hydroxylase gene repeat? _____ _____ An example of a DNA fingerprint from 14 different loci is presented in Table 10.2. Table 10.2 Sample DNA Fingerprint from 14 Loci Locus D3S1358 vWA FGA Genotype 15, 18 16, 16 19, 24 Locus Genotype
D13S17 10, 11
D7S820 10, 10
D16S539 11, 11
D8S1179 12, 12
D21S11 29, 31
D18S51 12, 13
D5S818 11, 11
THO1 9, 10
TPOX 8, 8
CSF1PO 11, 12
AMEL X, Y
Is the person with this DNA fingerprint male or female? __________ {10.9} Which loci are heterozygous?
Which loci are homozygous?
Procedure 1. Assume the DNA fingerprint in Table 10.2 above comes from a blood sample collected at a crime scene. Law enforcement authorities currently hold three suspects. A sample of DNA has been collected from each suspect and their DNA fingerprints for the loci in Table 10.2 have been determined (Table 10.3)
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Exercise 10 – DNA Fingerprinting
Table 10.3 DNA Fingerprints for Three Suspects Suspect “A” Locus D3S1358 vWA FGA Genotype 15, 18 16, 17 20, 24
D8S1179 12, 14
D21S11 29, 31
D18S51 10, 12
D5S818 11, 13
Locus Genotype
D13S17 12, 12
D7S820 13, 13
D16S539 11, 12
THO1 9, 9
TPOX 7, 8
CSF1PO 12, 12
AMEL X, Y
Suspect “B” Locus Genotype
D3S1358 15, 15
vWA 16, 16
FGA 19, 25
D8S1179 12, 12
D21S11 30, 31
D18S51 10, 11
D5S818 11, 13
Locus Genotype
D13S17 12, 13
D7S820 9, 10
D16S539 11, 11
THO1 9, 10
TPOX 9, 9
CSF1PO 11, 14
AMEL X, Y
Suspect “C” Locus Genotype
D3S1358 15, 18
vWA 16, 16
FGA 19, 24
D8S1179 12, 12
D21S11 29, 31
D18S51 12, 13
D5S818 11, 11
Locus Genotype
D13S17 10, 11
D7S820 10, 10
D16S539 11, 11
THO1 9, 10
TPOX 8, 8
CSF1PO 11, 12
AMEL X, Y
Based on these data, which of these suspects is the most likely culprit? __________ How would the answer change if it was found out Suspect “C” had recently been the recipient of a bone marrow transplant? (hint: what are the functions of bone marrow?)
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Exercise 10 – DNA Fingerprinting
Practice Problems and Review Questions 1. Define or describe the following: DNA
STR
PCR
locus (loci)
annealing
2. In DNA, __________ binds to thymine, guanine binds to __________ through __________ bonds.
3. How many loci are required by the FBI to positively identify an individual? Is it reasonably possible for two, non-related persons to share the same genotypes at all these loci? Why or why not?
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Exercise 10 – DNA Fingerprinting
Lake-Sumter State College, Leesburg Laboratory Manual for BSC 1010C
blood stain on victim’s clothes
victim
size markers
suspect 2
suspect 1
4. The following set of DNA fingerprints is from an actual crime scene. It contains a set of size markers and banding patterns from blood samples collected from the victim, two suspects, and blood found on the victim’s clothing. From what is shown here, determine which suspect’s blood more likely matches the blood found on the victim’s clothes. Justify your decision.
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Exercise 10 – DNA Fingerprinting
5. DNA fingerprinting can also be used in issues of family relationships. Since each person receives half their chromosomes from each parent, half of a person’s DNA fingerprint should match the mother and half should match the father. Consider the following DNA fingerprints as evidence in a paternity case. Is it reasonably possible the male in question is the baby’s biological father? __________ How would the answer change if the male in question had an identical twin?
Mother Locus Genotype
D3S1358 13, 18
vWA 15, 15
FGA 18, 24
D8S1179 12, 12
D21S11 28, 31
D18S51 13, 13
D5S818 11, 11
Locus Genotype
D13S17 10, 10
D7S820 11, 11
D16S539 10, 11
THO1 9, 9
TPOX 10, 11
CSF1PO 11, 11
AMEL X, X
Baby Locus Genotype
D3S1358 15, 18
vWA 15, 16
FGA 19, 24
D8S1179 12, 12
D21S11 29, 31
D18S51 12, 13
D5S818 11, 11
Locus Genotype
D13S17 10, 11
D7S820 10, 11
D16S539 11, 11
THO1 9, 10
TPOX 8, 10
CSF1PO 11, 12
AMEL X, X
Alleged Father Locus D3S1358 Genotype 13, 15
vWA 16, 16
FGA 18, 19
D8S1179 12, 14
D21S11 29, 30
D18S51 12, 12
D5S818 10, 11
D7S820 10, 10
D16S539 11, 11
THO1 10, 10
TPOX 8, 8
CSF1PO 11, 12
AMEL X, Y
Locus Genotype
D13S17 11, 12
6. Calculate the number of repeats in the DNA fragments for the STR’s given. a.
STR: “AAGCTA”
DNA fragment length: 72 bp
# of repeats: _____
b.
STR: “TTAT”
DNA fragment length: 52 bp
# of repeats: _____
c.
STR: “TAGGG”
DNA fragment length: 105 bp
# of repeats: _____
d.
STR: “AAGCT”
DNA fragment length: 60 bp
# of repeats: _____
e.
STR: “GAGGCT”
DNA fragment length: 144 bp
# of repeats: _____
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