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Set up:

About 1,000 frog embryos at the early cleavage stage

GFP injection

Label one half of the embryos with GFP, the other half will be unlabeled

Procedure: 

After one half of the embryos have been labeled with GFP, let all of the embryos grow for 6 hrs until blastula stage is reached (multicellular).  The non-injected embryos’ animal pole would be removed and cultured alone to serve as a control.   The recombination will be of the GFP vegetal mass (endoderm) and the non-labeled animal cap (ectoderm).  The mesodermal tissue will be discarded.  The embryos will then develop for two days until the neurulation stage is reached.

Results:

At the neurulation stage, the non-recombinant embryos’ animal pole would give rise to the surface ectoderm, which is expected.  The recombinant embryos would have a GFP labeled epidermis AND GFP labeled notochord.

Conclusions:

Because both the ectoderm and the mesoderm are labeled with GFP, it suggests that the animal pole cells that would usually become ectoderm derivatives were induced to also become mesoderm derivatives.  The GFP labeled ectoderm and notochord indicates that the vegetal endoderm is what induces the ectoderm.

Experiment one tested the effects of varying substrates on survival. In each of the three
tanks, 20 random guppies, 5 Rivulus and 5 Blue Acara were added. Each tank represents a
different substrate, the three substrates are sandy, muddy and vegetative. Spot brightness was
measured after 1000 days.

Experiment two tested the effects of varying levels of predation on survival. Only tank 1
was used, representing a constant sandy substrate. Five various levels of predation were tested
separately in this tank. For each trial 20 random guppies were added. Then 0, 4, 8, 12, or 16
predators were added. An equal number of Blue Acara and Rivulus were added, so with eight
predators, there were four Blue Acara and four Rivulus. Spot brightness was measured after 500
days.

This method of enhancement can allow users advanced levels of focus and help them stay alert longer to prolong studying. Clearly this presents a great attraction to college students who are desperate to meet deadlines and receive good grades. But the problem arises of whether it is truly fair for some students to abuse this medication to do better in their classes. The problem reflects a watered-down version of the anabolic steroids in professional sports debate. How can it be fair for “natural” athletes or students to compete with these enhanced individuals? If 30% of students are able to pull off all-nighters thanks to these medications, students who are not “enhanced” are faced with quite the handicap. This becomes especially problematic when considering that if these students who are taking ADHD medication are then able to do better than they would without the drugs, this would likely raise the class average performances. If teachers see that their students on average are doing much better on exams, they may feel the need to make the material more difficult.

An extremely focused laser has a very strong electric field at its narrowest point, known as the “beam waist.” The particles are attracted to the strongest field, the center of the beam, which allows for the use of the laser to move particles around. Away from the center of the beam is higher potential, so the particles move to the lower potential in the center of the beam.

Photosynthesis is an essential reaction in order to sustain life. The products of the essential reaction include oxygen and glucose. Besides the obvious benefits in the production of glucose and oxygen these products prove to have many important functions within the cell. The glucose produced from the reaction is converted into storable starch. This starch can then enter the plasma membrane where it interacts with the cellulose synthesis complex. The product of this interaction is cellulose, the most prominent and strongest component of the cell wall. Considering that cellulose is the most prominent component of the cell wall and is present in the majority of plants in the world many scientists consider it one of the most abundant organic compound on earth. Cellulose is also a large source of fiber in the average humans diet.

2 grams of cyclohexanol was added to a 250ml round-bottomed flask. .5ml of phosphoric was then added to the flask as well. A distillation of the mixture was then performed with the first drop occurring at 60 degrees Celsius and the plateau occurring at approximately 68 degrees Celsius. Lower layer of the remaining liquid was removed for waste. A washing procedure was followed with 1 ml of water, then 1 ml of NaOH and finally 1 ml of brine. After each washing sequence the lower layer of the liquid was removed and added to the waste beaker. The remaining liquid was then transferred to a new vial and CaCl2 was then added and contents were left to dry for approximately 7 minutes. The liquid in the beaker was then removed and placed in a new tarred test tube. The remaining liquid was then weighed and a final weight and percent recovery was recorded. Finally a potassium permanganate, bromine dichloromethane, IR and gas chromatography test were performed on the liquid to test for purity and contents.

Purpose of this lab is to dehydrate cyclohexanol and synthesizing cyclohexene by taking all that we’ve done so far. These processes include taking the melting points, boiling points, molecular weight densities, and figuring out what the hazardous properties are. By separating and purifying the substance using combination of extraction, distillation, crystallization, chromatography, and more the lab is complete. There are many processes that we learned so far in the lab shown right above, and honestly it was all pretty simple. There are step by step display on what we have to do on the organic chemistry 269 website. Lab reports are fairly simple as well, it just takes some time to complete them. My favorite one is crystallization because different crystals are formed whether it be different color, different weight, and more factors.

Purpose: Dehydrating cyclohexanol and synthesizing cyclohexene by taking all that we’ve done so far including MPs, BPs, MWs densities, hazardous properties. Separate and purify using combination of extraction, distillation, crystallization, chromatography, and more. There are many processes that we learned so far in the lab, and honestly it was all pretty simple. There are step by step display on what we have to do. Lab reports are fairly simple as well, it just takes a really long time to do them. My favorite one is crystallization because different crystals are formed whether it be different color, different weight, and more factors. 

            The 3rd premiere was called The Garden Maze by Eleanor Devlin. This piece was very interesting because it was a quartet. There were two violins, one viola, and one cello. The piece started with a cello solo in a way, it was harmonic when all 4 instruments joined in but it created resonance in sound. There were lots of same note playing and trills, it was a song that wasn’t too comfortable because it created tension to the listeners but it was very interesting in a weird way. The dynamic increase and decrease were very well drawn out as I could tell when there was a crescendo and diminuendo. Violin parts were usually high as cello parts were usually low. Viola played both high and low. It wasn’t too fast but it wasn’t slow either, I would say the piece had a medium tempo. You can relate this concert with physics on how different pitches are created due to the length and the thickness of the string. concert bass / cello being the bigger instrument creates lower sound and to violin / viola.

Post Lab Questions:

1. Trimyristin is a saturated fat because saturated fat contains no carbon at all. Monounsaturated contain only one carbon double bond as polyunsaturated fats contain two or more.
2. In a crystallization process, if it is cooled too fast then smaller crystals are formed compared to slow cooling that has bigger crystals. Quick cooling causes impure and defiled crystals as compared to slow cooling that has more pure crystals.
3. To calculate stearic acid upon hydrolysis of 1.8 g of tristearin, take the moles of tristearin. 1.8 / 891.48 = 0.002 moles. Then multiply by 3 to get 0.006 moles. Then multiply by the molecular mass of staeric acid which is 284.5 g/mol to get 1.7 g of stearic acid.
4. Heating up a mixture for a long period can boil the mixture speeding up the reaction time. Also, the mixture will mix well as it boils.
5. Heating something at not so high temperature for 45 minutes will just separate the compounds in the mixture itself.