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AQ 12/5 Perfect Paragraph

Submitted by atquang on Thu, 12/05/2019 - 20:08

Over one week, the germination rate in a variety of seed species was tested between seeds with and without testa. Particular species of seeds provided evidence of fast germination while others did not. Visual analysis sees no significant difference between the germination rates of seeds with or without testa. The germination rate of seeds with testa is equal to the germination rate of seeds without a testa. Uncertainties in our experiment include meeting optimal germination environments, as well as the uneven distribution of water to the seeds.
Furthermore, a small sample size experiment increases the likelihood of falsely accepting the null hypothesis. Future questions that can be asked to follow up on our experiment include whether pH or external factors such as soil conditions, temperature, and fertilizers affect the germination rate. These questions would lead the experiment towards figuring out how optimal germination rates vary by seed species.

AQ 12/5 Draft

Submitted by atquang on Thu, 12/05/2019 - 16:47

Over a one week period, the germination rate in a variety of seed species was tested between seeds with and without testa. Particular species of seeds provided evidence of fast germination while others did not. A t-test concludes that there is no significant difference between the germination rates of seeds with or without testa. The germination rate of seeds with testa is equal to the germination rate of seeds without a testa. Uncertainties in our experiment include meeting the optimal environment for each species of seed, as well as the uneven distribution of water to the seeds. Furthermore, a small sample size experiment increases the likelihood of falsely accepting the null hypothesis. Another question our experiment brings up is whether the difference in germination rate between seeds with and without testa is practically significant. Even if the mean germination rate is truly significantly different from seeds without testa, what would industries have to invest to make this happen, and if so, would it be worth it?

AQ 12/4 Draft

Submitted by atquang on Wed, 12/04/2019 - 12:42

Based on the premise that since all living things have the same DNA, that DNA from one organism can still be read and expressed when transferred to another organism (genetic engineering). Monsanto is a company that produces genetically modified organisms. They have made plants such that round up, a herbicide and carcinogen, would not be effective against vegetation. This is concerning because we are consuming dangerous compounds. An interesting topic to think about is the ruling of GMOs. GMOs are not considered GMOs if they are produced via CRISPR-Cas9, versus a product produced by a "gene gun" which is considered a GMO because we are taking DNA from one organism and inserting it into another. One example in which genetic engineering can be seen as a positive result is the modification of fast food potatoes. McDonald's takes advantage of genetic engineering to get rid of formaldehyde that can synthesize in potatoes when cooked at a high temperature. This lets us consume fast food safely, while letting companies mass-produce safe food for consumption.

AQ 12/4 Draft

Submitted by atquang on Wed, 12/04/2019 - 12:42

Based on the premise that since all living things have the same DNA, that DNA from one organism can still be read and expressed when transferred to another organism (genetic engineering). Monsanto is a company that produces genetically modified organisms. They have made plants such that round up, a herbicide and carcinogen, would not be effective against vegetation. This is concerning because we are consuming dangerous compounds. An interesting topic to think about is the ruling of GMOs. GMOs are not considered GMOs if they are produced via CRISPR-Cas9, versus a product produced by a "gene gun" which is considered a GMO because we are taking DNA from one organism and inserting it into another. One example in which genetic engineering can be seen as a positive result is the modification of fast food potatoes. McDonald's takes advantage of genetic engineering to get rid of formaldehyde that can synthesize in potatoes when cooked at a high temperature. This lets us consume fast food safely, while letting companies mass-produce safe food for consumption.

AQ 12/3 Draft

Submitted by atquang on Tue, 12/03/2019 - 23:01

The first deviation from procedures happens during the MC1R sequencing preparations. After so many PCR reactions, it is not surprising that some groups would lose their test tubes as the heat from the thermocycler erases the labels on tubes. Unfortunately, we were confident that our tubes were gone. Despite remembering the color of our labels as well as where we placed them, none of the tubes from both genetic lab sections matched our description. Loss of the MC1R PCR products resulted in us having to use Dog 5’s MC1R sequencing, which came out as below average, yet useable, quality.

In the comparison between Tasha the Boxer versus our sample dog, EMBOSS Needle showed many gaps between the reference sequence and our sequence. The gaps could be due to the poor quality that came up from the files. MC1R PCR products may not have been amplified to a great extent due to lack of dNTPs or primers. The SNP at nucleotide 790 helps conclude that our dog does not have a melanistic mask because our sequence does not have the same nucleotide (C) at 790 as the reference sequence (G). The SNP at nucleotide 808, 832, and 901 are specific to Siberian Husky dogs, which may suggest our dog is a Siberian Husky, or at least related to a breed of Siberian Husky. The relationship was no surprise because our first BLAST results showed that our MC1R sequence could be related to a Siberian Husky breed with 97.85% identity. MC1R files showed that at the MC1R gene is homozygous at nucleotide 916, which means the dog cannot be yellow because our dog must have two copies of nonfunctional MC1R (the e allele) to not produce eumelanin.

AQ 12/3 Draft

Submitted by atquang on Tue, 12/03/2019 - 00:06

DefB103 shows that we have do not have 3 G’s in a row, (shown in Figure 3). Since the data gets messy after the 2 G’s, we are favorable for the deletion and can conclude that our dog does not have wild-type DefB103 gene. However, at the third G, there is a heterozygous peak (A/G): there is one copy of the DNA strand that has the 5 G’s in a row, while another copy of DNA does not have 5 G’s in a row. Having a copy of both 5 Gs and 2 Gs means that we can conclude that we are heterozygous for β-Defensin, so our dog’s genotype is KB/kY at the K locus.

Schmutz and Dreger (n.d.) provide a table that answers the “phenotypes produced by the genotype interactions of ASIP, DEFB103, and MC1R in dogs.” Understanding that our dog has no e allele for MC1R, no ay allele for ASIP, and is heterozygous for DEFB103, we can look at the table to predict our dog’s phenotype. Figure 6 shows the table that presents our dog’s genotype.



From this table, we can conclude that our dog’s phenotype has a black coat with black lips and a black mouth.

AQ 11/21 Draft

Submitted by atquang on Thu, 11/21/2019 - 17:00

Domestic dogs also have “the K locus, whose genetic characteristics affect the melanocortin pathway.” (Candille et al., 2007) The β-Defensin gene produces a ligand (named β-Defensin) that competitively binds to MC1R. DefB103 or K alleles are discussed in this paper (KB > ky). Although KB is a mutant allele, it is dominant because it has a higher affinity for MC1R than ky. According to a diagram by Candille et al. (2007), in the presence of Agouti and functional MC1R, there is a stronger affinity for the KB allele to bind to the receptor. When bound to MC1R, β-Defensin induces synthesis of eumelanin.

AQ 11/22 Perfect Paragraph

Submitted by atquang on Thu, 11/21/2019 - 16:59

The overall objective of our proposal is to create a phylogenetic tree to determine the reliability of HOXC genes as an indicator of phylogeny. By aligning the sequence, the genes will become easy to compare and allow for the creation of a phylogenetic tree, as proposed. The sequencing data can also be used to determine how conserved the HOXC gene is. By understanding the evolutionary and genetic differences of HOXC genes between different species, the function of HOXC genes, which are currently unknown, can be better understood. The creation of a phylogenetic tree will allow for the determination of reliability to use HOXC genes as an indicator of phylogeny. If the phylogenetic tree proves to be reliable, this would be a phylogenetic tree of vertebrates that can be used in order to trace the evolutionary history of vertebrates. If new species were to be found, its HOXC gene can be sequenced to determine its phylogeny accurately.

AQ 11/20 Draft

Submitted by atquang on Wed, 11/20/2019 - 05:25

DefB103 shows that we have do not have 3 G’s in a row, (shown in Figure 3). Since the data gets messy after the 2 G’s, we are favorable for the deletion and can conclude that our dog does not have wild-type DefB103 gene. However, at the third G, there is a heterozygous peak (A/G): there is one copy of the DNA strand that has the 5 G’s in a row, while another copy of DNA does not have 5 G’s in a row. Having a copy of both 5 Gs and 2 Gs means that we can conclude that we are heterozygous for β-Defensin, so our dog’s genotype is KB/kY at the K locus.

AQ 11/18 Draft

Submitted by atquang on Mon, 11/18/2019 - 23:09

The size of the digested bands must be looked at and referred to the paper by both Berryere et al.’s paper (2005) and Schmutz et al.’s paper (2002), to analyze the gels performed for Agouti and TYRP1. Counting down from the 500 bp band, we cannot see the 100 bp band. A reason for this is that it may have traveled too far down the gel for us to see it. The Agouti gel does not show bands below 200. Berryere et al. (2005) says that "After digestion with BsmA1 the ay-allele yields fragments of 42, 90, 153 bp; other alleles yield fragments of 42 and 243 bp.” Since our digested band yields a length of around 200-300, we can conclude we do not have the ay allele, but still may possess the aw, a+, or a allele. The TYRP1 gel shows bands at 73 bp (visible) and 48 bp (not visible but assumed using paper) for MnIl, and bands of 93 bp (visible) and 28 bp (not visible but assumed using paper) for Acil. Schmutz et al. (2002) says that Acil digestion causes “animals with the premature stop codon cut to bands of 28 and 93 bp” while MnlI digestion causes “animals without the proline deletion cut to band sizes of 48 and 73 bp.” From this, we can conclude that our dog is mutant for the premature stop codon in exon five, and wildtype for the proline deletion in exon five. The study states that “all 43 of the brown group carried two or more of these sequence variants,” while “0 of 34 in the black group carried two or more of these variants” (Schmutz et al., 2002). If our dog has the third variant, we cannot confirm whether the dog will be black or brown from the TYRP1 analysis. However, the variant in exon two is rarely found, suggesting that our dog is more likely to be black in coat color. If we could select one more allele to analyze, we would analyze this third rare variant in exon two because it would conclude whether our dog is black or brown. Analysis of the third variant would be performed by seeing if there is “a serine substitution of a cysteine at amino acid 41 in exon 2” (Schmutz et al., 2002).

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