David Daige
Biol 1030H
Due Date: April 1, 2016
Professor Laura Pacey
Lab Report 5: Molecular Biology Techniques
Abstarct
The aim of this study is to accurately measure small volumes using the micropipettes and to use virtual STR and PCR programs to test the paternity of various cases. During the lab, 18uL of different volumes were successfully measured into a tubule using the micropipette. During the lab, fruit juice was accurately extracted from a solution using centrifugation and using the micropipette to remove the solute without removing the fruit juice. Using the virtual STR and PCR software, the identity of the father of the children was discovered. Childs 1 and 2 belonged to father 1 and child 3 belonged …show more content…
Steitz, 1996). Primers are short sequences of DNA with varying nucleotides. They are made in labs and can be custom to specific needs. The primers are attached to the ends of the segment of the desired DNA, they must be specific to the nucleotides of the DNA desired and is necessary because DNA polymerase cannot attach to any place and begin copying. As DNA primase crosses the primers, they attach themselves to the strand and begin adding nucleotides.
The marker band sizes for the PCR are as follows: the smallest band size is 100, the largest is 1000 and the steps/increments of the markers increase by 100. 0.1pM of each sample from patients were used. The annealing temperature was set to 55oC and the denaturing temperature was 96oC. It is at 96oC because the DNA polymerase breaks down at 95oC therefore it is the temperature needed to separate two complementary strand in a test tube ( Monk, M., & Kinross, J, 1972). The annealing temperature 0.5uM of each primer was used. The annealing temperature is important because it affects the efficiency of the primers. If the temperature is too high,If the temperature is too high the primers will not anneal efficiently, and if the annealing temperature is too low the primers may anneal nonspecifically. Generally the temperature is chosen to be about 5oC below the Tm of the primer …show more content…
3 shows that father_1 shares the same bands as all children and father_2 only shares a similar band as child_3. This immediately disqualifies father_2 as the father for child_1 and child_2 but still leaves the question whether he is the father of child_3 or if father_1 is the father of them all. Unfortunately, the band sizes of the children are inferred and therefore may not be accurate.
Observing fig.4, we see that father_1 shares similar bands with child_1 and child_2 and father_1 now shares similar bands with all three of the children. Fig.4 shows that all the cases have a single band far above the rest it is approximately 20 in size. The mother has two bands that is approximately 50 and one abo in size. Father_2 has two other bands that are 100 and 50(from top to bottom). Child_1 and child_2 has two other bands about 50 in size each. Child_3 has two other bands about 100 and 50 (from top to bottom) . Father_1 has a stacked band and using child_1 and child_2 we can determine that both bands are approximately 50 each. Now, father_1 is not the father of child_3 and father_2 still shares the bands with child_3. Another test was conducted to increase
Biol 1030H
Due Date: April 1, 2016
Professor Laura Pacey
Lab Report 5: Molecular Biology Techniques
Abstarct
The aim of this study is to accurately measure small volumes using the micropipettes and to use virtual STR and PCR programs to test the paternity of various cases. During the lab, 18uL of different volumes were successfully measured into a tubule using the micropipette. During the lab, fruit juice was accurately extracted from a solution using centrifugation and using the micropipette to remove the solute without removing the fruit juice. Using the virtual STR and PCR software, the identity of the father of the children was discovered. Childs 1 and 2 belonged to father 1 and child 3 belonged …show more content…
Steitz, 1996). Primers are short sequences of DNA with varying nucleotides. They are made in labs and can be custom to specific needs. The primers are attached to the ends of the segment of the desired DNA, they must be specific to the nucleotides of the DNA desired and is necessary because DNA polymerase cannot attach to any place and begin copying. As DNA primase crosses the primers, they attach themselves to the strand and begin adding nucleotides.
The marker band sizes for the PCR are as follows: the smallest band size is 100, the largest is 1000 and the steps/increments of the markers increase by 100. 0.1pM of each sample from patients were used. The annealing temperature was set to 55oC and the denaturing temperature was 96oC. It is at 96oC because the DNA polymerase breaks down at 95oC therefore it is the temperature needed to separate two complementary strand in a test tube ( Monk, M., & Kinross, J, 1972). The annealing temperature 0.5uM of each primer was used. The annealing temperature is important because it affects the efficiency of the primers. If the temperature is too high,If the temperature is too high the primers will not anneal efficiently, and if the annealing temperature is too low the primers may anneal nonspecifically. Generally the temperature is chosen to be about 5oC below the Tm of the primer …show more content…
3 shows that father_1 shares the same bands as all children and father_2 only shares a similar band as child_3. This immediately disqualifies father_2 as the father for child_1 and child_2 but still leaves the question whether he is the father of child_3 or if father_1 is the father of them all. Unfortunately, the band sizes of the children are inferred and therefore may not be accurate.
Observing fig.4, we see that father_1 shares similar bands with child_1 and child_2 and father_1 now shares similar bands with all three of the children. Fig.4 shows that all the cases have a single band far above the rest it is approximately 20 in size. The mother has two bands that is approximately 50 and one abo in size. Father_2 has two other bands that are 100 and 50(from top to bottom). Child_1 and child_2 has two other bands about 50 in size each. Child_3 has two other bands about 100 and 50 (from top to bottom) . Father_1 has a stacked band and using child_1 and child_2 we can determine that both bands are approximately 50 each. Now, father_1 is not the father of child_3 and father_2 still shares the bands with child_3. Another test was conducted to increase