Draft #1, week 14, causes of RP

Submitted by vvikhrev on Wed, 04/25/2018 - 16:40

Retinitis pigmentosa can be caused by genetic mutations in over 60 genes. More than 20 genes are associated with the autosomal dominant mode of inheritance. Of these 20 genes, the rhodopsin gene, RHO, is the most prevalent to cause autosomal dominant retinitis pigmentosa, and more than 45 mutations in RHO have been identified (Sung et al, 1994). Because there are so many genes that encode various structures in the retina, it is difficult to delegate a single gene to the disease. However, by studying families that show symptoms, researchers are able to locate the mutation, contribute to the scientific community and hopefully provide more explanations that will lead to treatments and cures.
The RHO gene provides instructions for making rhodopsin. Rhodopsin is the photopigment in rods that absorbs photons of light and converts it into an action potential cascade. It is made up of cis-retinal and is important for low-light conditions. When light hits rhodopsin, it begins to decompose, cis-retinal is converted to trans-retinal, and the membrane conductance for Na+ in the outer segment of the rod decreases. The electrical signals are transmitted to the brain. The fovea contains a high density of cones and is located in the macula of the retina. Here, the retinal layers are spread apart to allow light to get to the photoreceptors with minimal interference. This allows for improved resolution and higher acquity for central vision. There is a higher ratio of rods to cones around the fovea that are more responsible for peripheral vision. Since retinitis pigmentosa involves the degeneration of rods and cones, the disease can affect both peripheral, central vision, night blindness and a variety of vision disruptions in low-dimness and bright settings.


Submitted by mglater on Wed, 04/25/2018 - 16:38

We were interested in analyzing the water quality of the stream by the Sylvan residence halls. We found that the levels and diversity of periphyton found in the water could be used as a way to analyze the overall health of the water. Periphyton is a broad term, consisting of algae, cyanobacteria, and other microscopic organisms living in the water. We created a method to collect samples of periphyton by placing three glass slides together in the water. We placed these slides at three different locations in the stream, and collected samples after one week and after two weeks. We found that the levels of periphyton collected increased over the two weeks, and used the Shannon Index to quantify the diversity and abundance. We found that one of the three locations had a significantly higher number of total periphyton.


Intro Paragraph 1

Submitted by tedarling on Wed, 04/25/2018 - 14:42

Over the course of several weeks we carried out multiple related experiments on Saccharomyces cerevisiae, a species of yeast. Yeast is vital to study because it is the simplest of all the eukaryotes and is the basis of our understanding for numerous essential cellular processes. Yeast is a unicellular, eukaryotic fungus that reproduces through budding. To replicate, a bud emerges from the mother cell during S phase of the cell cycle. The bud grows until it is pinched off with newly replicated DNA inside. Yeast can exist in both haploid and diploid states, and both undergo budding.


Human Physiology Notes P3

Submitted by crmckenzie on Wed, 04/25/2018 - 14:07

TRH (Thyrotropin) stimulates TSH release from the pituitary and TRH (Releasing Hormone) “controls the set point” around which negative feedback occurs. Individual variants of thyroid hormone is 10% in comparison to the rest of the population. TSH is a glycoprotein hormone with two subunits, an alpha and a beta. Pure samples of LH FSH and TSH were taken to figure out the subunits. There were subunit switching experiments and alpha is common, but beta requires specificity. Thyroid glands have thryocites that surround ‘juicy centers’ that store the precursors to thyroid hormones. When TSH binds to the receptor and activates the entire mechanism, the production of thyroglobulin is increased. This is then packaged into vesicles where it is released into the colloid.

383 lab 3 methods p 2

Submitted by liamharvey on Wed, 04/25/2018 - 13:58

            Using the online software Primer3, a primer for the gene was found. To do so, the genomic sequence of our unknown gene was put into the first box. The options for left and right primer was left selected. The sequence was named “MHN” in the Sequence ID box. In the box Product Size Range, 500-1000 was selected. Under General Primer Picking Conditions, the parameters were set to: Min = 20, Max = 27, Opt. = 23 for primer size and Min = 65 °C, Max= 70 °C, and Opt. = 67 °C in the “Primer Tm” box. A region of the gene was chosen to be amplified, this region had to include the mutation, was about 200bp from one of the primers and in a restriction site. Primer3 provides five primers, with the top one considered to be the best. With these five options, a primer was chosen based on how well it fit to the previous criteria. The chosen primers were then named “oMHN_L” (GTCCTCCGTCCACCAGTCCTT) and “oMHN_R” (GGTTCCAATGTCCAGCCTCTTGATTT) for the left and right primer’s respectively.

Denitrification Lab Results

Submitted by mkomtangi on Wed, 04/25/2018 - 13:45

For the Denitrification test, the results showed that there was nitrate present in both the rich and poor soil tubes because when nitrate reagent A and B were added to the Nitrate broth tubes, both turned red, meaning it is positive for nitrate. In addition,  a small amount of gas was found in the durham tubes meaning nitrate was broken down into gaseous products.

For both Proteus vulgaris and Pseudomonas aeruginosa nitrate was broken down into gaseous products because there was a large amount of gas in the durham tubes. In addition, when zinc was added to the tubes, there was no color change meaning it was a positive result and nitrate was broken down into a compound other than nitrite such as nitrogen gas.


Bio 288

Submitted by lgiron on Wed, 04/25/2018 - 13:27

In a normal respiratory system with average amounts of blood/blood pressure, the lungs function as oxygen intake centers for the body, as well as excretion of wastes, such as carbon dioxide. Without this constant cycle of intake and expulsion, humans would have a buildup of leftovers that form during the conversion of the nourishment we take in. The lungs do this by readily changing their volume to either force air from the environment in or out through the mouth. The oxygen within the air is then exchanged within the most miniscule parts of the lung, known as alveoli, via the hemoglobin within the red blood cells. Carbon dioxide is then put in its place, allowing blood cells to constantly perform this process and supply the body with the oxygen it need. This monitoring of the carbon dioxide levels within the blood also helps to control blood pH, which has a number of implications within the metabolic activities of other organs and tissues.


Deep Learning Assignment Draft

Submitted by lgorman on Wed, 04/25/2018 - 09:24
  1. After what step are the carbons of a fatty acid completely oxidized? (Lecture 22-Slide 11)

  2. This question is testing the students knowledge on the oxidation of fatty acids. More specifically, it asks the student to recall the steps of the oxidation, and know determine which step is completely oxidized.

  3. New Question: What step produces the most reduced electron carriers, and how many does it make for a (18:0) fatty acid?

    1. Acyl-CoA transport , 4 reduced electron carriers

    2. Beta oxidation, 36 reduced electron carriers

    3. Citric Acid Cycle, 36 reduced electron carriers (CORRECT)

    4. Citric Acid Cycle, 32 reduced electron carriers

  4. Why each answer is wrong/right:

    1. A is incorrect, because Acyl-CoA transport produces no reduced electron carriers, and 4 is the incorrect number.

    2. B is incorrect, because Beta oxidation does not produce the most reduced electron carriers, however 36 is the correct number.

    3. C is correct, because Citric Acid cycle step produces the most electron carriers, and 36 reduced electron carriers is the correct number.

    4. D is incorrect, while Citric acid cycle is the correct answer, 32 reduced electron carriers is the incorrect number.

Tetrahymena Seratonin Lab Introduction

Submitted by benjaminburk on Wed, 04/25/2018 - 04:53

Tetrahymena cells are unicellular predatory ciliates that produce both sexually and asexually. It is an excellent laboratory model because it possesses certain single cell advantages that are beneficial in the laboratory, while also sharing many genes with us multi-cellular animals (Cole 2000-2013). One interesting quality of the Tetrahymena is the way in which they feed. They feed in a process called phagocytosis, which results in the creation of small food vacuoles within the cell membrane (Coyne 2011). For the control group of this experiment the Tetrahymena cells were fed India ink and allowed to feed. Immediately after and then every five minutes for twenty minutes a random sample of Tetrahymena cells were taken and added to 50ul of glutaraldehyde, killing the cells without damaging the tissues. Once mixed, the Tetrahymena were put under a microscope and then ten randomly chosen cells had their black food vacuoles counted. The counts were averaged and the standard deviation calculated. Tetrahymena not randomly chosen were excluded from the experiment. For the experimental group the same exact procedure was followed, except for that initially along with introducing India Ink to the Tetrahymena we also placed 1ul of serotonin into the culture. We still took food vacuole counts of 10 randomly selected cells every five minutes for twenty minutes using the technique described in the control group procedure. In this experiment we decided to test the effects of serotonin on Tetrahymena food consumption. Based on previous experiments we are aware that serotonin stimulates the consumption of nutrients in Tetrahymena cells, therefore we hypothesized that the addition of the serotonin would lead to an increase in the number of food vacuoles observed compared to the amount observed in the control trials. With this being our hypothesis we hoped to find results that lead to higher vacuole count means for the group treated with serotonin, then the control group at all time points.


Submitted by mglater on Wed, 04/25/2018 - 00:53

As expected, the H0 yeast matings served as controls, all resulting in white colonies. The H0 strain is unmutated, so the genes complemented any mutant that they were crossed with. Both of the unknown A type mutants gave the same results, meaning they share the same mutation. When crossed with HB2, they formed a white colony, meaning the genes complement each other. This means that the mutation in the A mutants is in Ade1. This is confirmed by the fact that the cross between the unknown mutants and HB1 results in little to no cell survival. The mutations are in the same gene, and therefore cannot complement each other. The two unknown alpha mutants also gave the same cross results and are therefore the same mutation. The unknown alpha mutants complemented with HA1 to produce a white colony, while they failed to complement with HA2. Following the same logic as with the A type mutants, this means that the mutation in the alpha yeast is in Ade2. Looking at the crosses between the A and alpha unknown mutants further supports this claim. The crosses between the unknown mutants all resulted in surviving white colonies, meaning that the genes complement. The only way for the mutations to complement are if the mutations are on different genes.



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