The results from the gel electrophoresis were inconclusive so students selected their genotype, with the results shown in Table 1. The controls were successful with the (+/+) homozygous, (-/-) homozygous, and heterozygous lanes having bands at 941, 641, and both 941 and 641 respectively. The class allelic frequencies were 56.3% for p and 43.8% for q, indicating that slightly more individuals had the Alu insert on their chromosomes (Table 2). When the Hardy-Weinberg equation was applied, there were some discrepancies as shown by Table 3 between the observed and calculated frequencies. The p frequency was observed at 10% higher than Hardy-Weinberg while the q frequency was observed at 10% lower. Discussion The results partially agreed with the working hypothesis. The student sample genotype was indeterminable, most likely due to PCR errors that …show more content…
Failure could have resulted from improper digestion by the restriction enzymes, forgotten steps in PCR preparation, and within the PCR process itself when replicating the template strands. In order for PCR to work, the cells need to be combined with a Master Mix, which contains DNA template, oligonucleotide primers, individual deoxynucleotide bases, Taq Polymerase, magnesium ions, and a salt buffer. If one of the components is missing, for example the buffer, the environment may not be conducive for successful PCR. Errors could also have occurred within thermal cycling. In the first two cycles the DNA was mostly been replicated and primers had attached to one end, lacking another upstream. In the third cycle the polymerase will extend the primers so that the precise portion between primers is the only part that is amplified (Figure 2). Errors within this process such as failure to attach the primers correctly, could have contributed to the inconclusive results shown in Figure
Digestion of the haemolytic and non-haemolytic cells allowed for easier identification of fragments during electrophoresis analysis. Lane 12 in figure 3 show the size markers of SPP1 digested with EcoR1 while lanes 6 and 7 show samples of pK184hlyA and pBluescript digested with EcoR1 and Pst1. Lane 4 was loaded with plasmid DNA from haemolytic cells digested with EcoR1 and Pst1 while lane 5 was loaded with EcoR1 and Pst1 digested DNA from non-haemolytic cells. There was a lack of technical success in both lanes due to no bands appearing in lane 4 and only a single band appearing in lane 5. Theoretically, two bands should appear in both lanes after successful to allow for fragment identification. A possible explanation for the single, large fragment in lane 5 is that successful digestion did not take place and the plasmid was only cut at one restriction site leaving a large linear fragment of plasmid DNA. The absence of bands in lane 4 could be because there was not enough plasmid loaded into the lane. Another possibility could be that low plasmid yield as obtained when eluting the experimental samples in order to purify it. Lanes 8 and 9 belonged to another group and show technical success as two bands were present in both the haemolytic (lane 8) and non-haemolytic (lane 9) lanes. If the
The product of the reaction of the A primer seems to have failed as no bands were produced apart from the terminal point of the migration which is too small to be considered as either a preintegration site or a retrovirus containing section, not only did my partner seem to have the same problem, most if not all of the submitted gels seem to have no bands for the A set of primers.
When the PCR technique is completed, the tubes are stored at 4°C until analysis of the tubes. To analyze the PCR results with the gel electrophorese, 2.5ul of the 10x loading dye is added to each PCR reaction tube. The gel for the electrophorese consists of 1.5% agarose gel with 0.5x TBE and 200ng/ml ethidium. bromide. The sand is a sand.
In our genes, multiple different alleles determine whether one person will have a certain trait or not. Alleles are what make-up our genotypes and in this lab, we wanted to determine the genotypes of our class in the two loci: TAS2R38 and PV92. The TAS2R38 locus codes for a protein that involves the bitter taste of PTC; the gene determines whether or not a person will taste the PTC paper as very bitter or no taste at all. People with the “T” allele are tasters while those that are homozygous recessive (tt) are non-tasters. The taster locus can be found chromosome 7.3 The two different alleles present in the could be due to the effect of evolution and natural selection because the same can be found in chimps.4 The PV92 locus does not code for any protein but rather involves an Alu element that is 300-bp long. A person with the “+” allele would have the Alu element making that sequence longer while those with the “-“ allele don’t have the element and would have a shorter sequence. This locus can be found on chromosome 16.3 There are multiple Alu sequences found among primate genomes but there are human specific sequences such as the one found on the PV92 locus.1 In the experiment, student DNA was collected from cheek cells and PCR was used to target the loci and amplify the region of DNA. In the taster gene, after amplification, a restriction digest was performed to differentiate between the two alleles. The digest was able to show differentiation because those with the “T” allele would have two bands from gel electrophoresis and those with “t” will have one band because the restriction enzyme doesn’t cut it. For the PV92, we were able to distinguish between the alleles due to the added length of the Alu element. Those...
An individual can be homozygous dominant (two dominant alleles, AA), homozygous recessive (two recessive alleles, aa), or heterozygous (one dominant and one recessive allele, Aa). There were two particular crosses that took place in this experiment. The first cross-performed was Ebony Bodies versus Vestigle Wings, where Long wings are dominant over short wings and normal bodies are dominant over black bodies. The other cross that was performed was White versus Wild where red eyes in fruit flies are dominant over white eyes. The purpose of the first experiment, Ebony vs. Vestigle was to see how many of the offspring had normal bodies and normal wings, normal bodies and vestigle wings, ebony bodies and normal wings, and ebony body and vestigle wings.
Experiment #1: The purpose of this experiment is to investigate the effects of baking soda and light intensity on the rate of photosynthesis of green spinach leave through the observation of floating disk.
The two modes of analysis that will be used to identify an unknown insert piece of DNA would be plating the transformation cells onto LA plates that have either ampicillin or chloramphenicol and PCR. We will use the PCR thermocycler to denature the restriction enzymes that were specifically used to assimilate the vector DNA. It is important to use the PCR thermocycler because denaturation of the restriction enzyme will prevent the restriction enzyme from cutting the vector DNA, after the insert DNA has assimilated to the vector DNA. After the addition of specific primers that complement the base pair to its corresponding target strand, PCR will be used. Subsequently, Taq polymerase will be used to determine whether the insert DNA has been properly assimilated to the vector DNA. Within this specific situation, the target strand will be the insert DNA. After we let the PCR thermocycler run for approximately 2 ½ hours, we will then put our PCR products in the gel and run the gel to completion. After the gel has run to completion, we will then take a photograph of the gel using the UV transilluminator with the assistance of our TA. If the insert DNA was properly assimilated to the vector DNA, then our corresponding gel photo would have one band. After the cells have been transformed, we would g...
If one wanted to know their chance of carrying or having the disease creating a punnet square could help determine that. A normal person without Albinism or the presence of the allele melanin can be represented by capital “A” and another allele that represents the lack of melanin will be represented with lower case “a”. Since Albinism is an autonomic recessive disease, this means a person with a homozygous recessive gene will have the disease. Both parents must be heterozygous dominant and carry the allele; they will have a 25% chance of having a child with albinism and a 70% chance of having a child carrying the disease. If one parent is heterozygous that still carries the flawed gene and the other parent is homozygous dominant there will be a 50% chance their child will carry the disease but wont have a child with Albinism.
Albinism is a genetic condition present at birth, characterized by a small amount of melanin pigment in the skin, hair and eye. Albinism is an occasional inborn sickness related with vision difficult, which affect one in seventeen thousand persons. It is not a contagious disease and cannot be spread over contact. Albinism affects individuals from all races. Most folks with albinism have parents with a normal color of skin. Some may not even recognize that they are Albino until later on in their life. This paper will be based on the study of albinism, causes, types, the genetic transmission and some possible medical problem.
"Polymerase Chain Reaction (PCR) Fact Sheet." National Human Genome Research Institute. 10 Dec. 2007. National Institutes of Health. .
Baharloo, Siamak et al.Absolute Pitch: An Approach for Identification of Genetic and Nongenetic Components. The American Society of Human Genetics. 1998.
The objective of this lab was to make a halftone negative of a small clipart.
National Institute of General Medical Sciences. (2010). "21st-Century Genetics." The New Genetics, p. 74-83. Retrieved from http://publications.nigms.nih.gov/thenewgenetics/chapter5.html
more than half the variation was found to be due to heredity. Among these traits were
An Experiment to Investigate the Effect of Light Intensity on the Rate of Photosynthesis. Introduction Photosynthetics take place in the chloroplasts of green plant cells. It can produce simple sugars using carbon dioxide and water causing the release of sugar and oxygen. The chemical equation of photosynthesis is: [ IMAGE ] 6CO 2 + 6H20 C 6 H12 O 6 + 6O2 It has been proven many times that plants need light to be able to photosynthesize, so you can say that without light the plant would neither photosynthesize nor survive.