Writing in Biology

The order Lampriformes is largely catagorized by morphological features. Fishes in this order have a Mesethmoid bone posterior to their lateral ethmoids, an elongated premaxilla, their first dorsal pterygiphore insterts anterior to their first neural spine and have the absence of a platine prong. Oarfish, Lampris and apahs are all examples of Lampriformes. The giant orfish, of the family Regalecidae is known as the king of the herrings. This fish is a very elongate and slab sided. The giant oarfish has a tube shaped mouth, which it uses to suck in its prey. Furthermore it has a very large eye and red fins. The most notable feature of the Giant Oarfish is its crownlike spines on the top of its head, which are fused with its dorsal fin and follows the entire length of its body. The Giant Oarfish is a very large fish and can reach up to 11 meters in length. Its body stays ridgid as it swims, but sends sin waves down its long dorsal fin to propell itself. There is a notable absence of scales on the oarfish as well.

As a scientist, it is important to have a comprehensive knowledge of the ecological, behavioral, physiological, and evolutionary factors that affect the organisms of an ecosystem and their interactions. For example, if we look at a crab spider on flower in a field, we can study its predator-prey interactions with bees, the evolution and physiology of its color-changing mechanisms, or its defensive behavior towards potential predators like birds. The initial work of this course will build my knowledge on the biology of tropical environments, which will allow me to design and carry out my own project in an actual tropical field site. The ability to develop an experimental design, testing it, and later analyzing the results, is a fundamental skill for graduate school that this course will help me to improve.

It makes sense that G1 cells would be able to replicate while G2 cell would not be able to. Even when a signal for replication is present, G2 will not replicate. I would think that the G2 cells must have a signal or something that indicates that it has passed certain checkpoints, such as the DNA replication checkpoint at the end of S phase. The ORC is a multi-subunit SNA binding complex that binds to the orgins of replication in all eukaryotes. It is made up of 6 subunits and ATP is needed for binding to occur. The subunits are encoded by ORC 1, 2, 3, 4, 5, and 6. The ORC is active at the end of mitosis and early G1, which makes sense for replication. While it is highly unlikely since so many different complexes and proteins are involved, what happens if re-replication occurs after S phase? Is this even possible at all? It makes sense that Cdk activity reduces to zero so chromosomes are ready for a new round of replication. It reminds me of a "reset" button.

The IR test revealed that an alcohol group and an alkene group were present in the product. A large, broad alcohol functional group spike at around 3500 was present proving the presence of cyclohexanol in the product. A small, sharp alkene functional group spike at around 3020 was present proving the presence of cyclohexene in the product. The GC test revealed that there were two components present in the product with a ratio of 1:5 based on their areas. This proves that cyclohexanol and cyclohexene were both present in the product. Cyclohexane spiked first in the GC because it has a lower boiling point and obtained a smaller area. This proves it had a smaller abundance in the product. Cyclohexene spiked after cyclohexane because it has a higher boiling point and obtained a larger area. Cyclohexene has a larger spike in the GC test because it was more abundant in the product than cyclohexanol, obtaining the larger area and higher portion of the ratio.

We conducted an experiment testing the feeding patterns of Tetrahymena thermophila by combining Tetrahymena culture with India Ink (a food of Tetrahymena).  First, we practiced observing the Tetrahymena under a microscope.  We added Methyl Cellulose to slow their movements, but not kill them.  With new Tetrahymena, we added India ink and after set time intervals, we added in Glutaraldehyde, which kills the cells in time so we may observe the number of vacuoles created after a certain time.  We took 5 samples of cells at 0, 10, 20, 30, and 40 minutes and made a wet mount to record the number of vacuoles created. After looking at 10 cells per sample, we took the average of the numbers we collected for each time interval.  The mean of the rows gave us the average number of vacuoles that were observed after the stated time. We noticed that the longer the cells had with the India Ink, the more vacuoles they made on average. I believe that the data shows that the longer the Tetrahymena are in the solution, the more vacuoles they will form.  We did not see any plateau in the data, but I do think that after a certain point the number of vacuoles would not get any larger. We made a table on Excel to organize all of our data and then made calculations to get the mean and standard deviation. To calculate the mean we used the function “=average (cell range)” and highlighted the data.  To calculate the standard deviation we used the function “=stdev (cell range)” and highlighted the data. A line graph was made from all of this data.

Sodium Borohydride Reduction of Benzoin

Purpose:

The main goal of this experiment if to use sodium borohydride to reduce benzoin to 1,2-diphenylethane- 1,2- diol.

Reaction Scheme:

Experimental Procedure:

Benzoin(0.5g, 2.36 mmol) and ethanol (4 mL)were added to a 25 mL erlenmeyer flask. The mixture was swirled at room temperature until the benzoin was completely dissolved.  Then sodium borohydride(0.1 g, 2.64 mmol) was added in several portions over 5 minutes. It was swirled at room temperature for 20 minutes. The mixture was cooled in an ice bath for 15 mins. Then water(5 mL) and HCl(0.3 mL, 6M) were added to the mixture. After 15 minutes,  water(5 mL) was added to the mixture. The mixture was then vacuum filtered to collect the product, using small amounts of ice cold water to wash the flask. The filter was left running with the product to dry for 15 minutes. The crude product was weighed, and the melting point was determined.  Around 1 mg was put aside for TLC later. The rest of the product was recrystallized using minimal acetone, it was washed with hexane during the vacuum filtration process. The crystals were dried overnight. The mass and the melting point of the 1,2 -diphenylethane-1,2-diol were determined during morning hours.  

For the TLC analysis, benzoin(1 mg), crude product(1 mg) and recrystallized product(1 mg) were all dissolved in separate vials with a 9:1 solvent of ethyl acetate.  The TLC plates were then marked and spotted accordingly:

The solvent travelled until around 1 cm from the top of the plate.  The TLC plates were dried and spots were observed and marked under the UV light.  The TLC papers were then set in the iodine chamber to stain the spots. The Rf values were then calculated and recorded.

The TLC analysis provides a means to separate the structures that are present in the sample that was spotted and measure the retention factor to compare amongst the other samples. By spotting the plates as they were, plate one can compare the starting material to the recrystallized product and plate two can compare the starting material to crude product. As seen in Figure 1, the starting material benzoin and the crude product produced one spot. When the two were overlapped, the result was two spots. The eluant traveled 4.8 cm while in the TLC chamber and provided a way to measure the retention factors for how far each spot traveled. Table 1 provides the retention factors and shows that the retention factor generated for benzoin on the first plate was 0.844 and the recrystallized product was 0.533. The mixture of spots located at spot C on the plate produced two marks with Rf factors of 0.844 and 0.644. The similarity seen here is representative of overlap of the spots and is expected to be seen. The spots that share the Rf factor of 0.844 can be identified as benzoin and the second spot can be matched to the recrystallized product. A similar trend was seen in the second plate in that the starting material and crude product produced one spot, where as the mixture produced two. The second plate, however, had Rf factors that were even closer in directly matching the Rf factor of the singular spots. The starting material had an Rf spot of 0.813, which is similar to plate one with the expected difference occurring from the distance travelled by the eluant, and the mixture had an Rf of 0.833. Meanwhile, the crude product had an Rf factor of 0.604 and the mixture had a second Rf value of 0.625. The mixture has Rf values that are slightly higher than the singular spots, but when looking at Figure 1, it appears that the spots from location C have shifted on the paper, which can be explained as that side being exposed to the eluant first. If slight drifting did occur from exposure upon placement in the TLC chamber, then the one side would have been exposed for longer time and have longer time to travel up the paper. Overall, the shifting was not major and the values are close to each other that they are comparable to determine they are the same. The recrystallized product that made spot B on plate one is 0.533 and is lower when compared to the crude product on plate two. The mixture of recrystallized product produces an Rf value of 0.644 that is much more consistent with the crude product results on plate two, 0.604 and 0.625.

In this lab, the synthesis of cyclohexene was performed through the dehydration of cyclohexanol. This was done through distillation and several washes. IR, GC, and chemical tests were performed to reveal the presences of any functional groups and the number of components in the product. Distillation started at around 64 degrees Celsius. This was unexpected because cyclohexene boiling point is 83 degrees Celsius and cyclohexanol boiling point is 161.8 degrees Celsius. The lower boiling point present in the distillation of cyclohexanol may have been due to the weak hydrogen bonds in the alcohol functional group and water, or the weak alkene bonds. Cyclohexanol and cyclohexene have a 1:1 ratio. The initial amount of cyclohexanol used was 2.0 g and the finished amount of cyclohexene obtained was 0.43 g. This resulted in a percent yield of 21.4 %. This may have been due to loss of product during glassware transitions, potential evaporation, or not leaving enough solution in the flask during distillation.

A-site: the ribosomal site most frequently occupied by aminoacyl-tRNA. The aminoacyl-tRNA in the A-site functions as the acceptor for the growing protein during peptide bond formation.P-site: the ribosomal site most frequently occupied by peptidyl-tRNA, i.e. the tRNA carrying the growing peptide chain. The P-site is also referred to as the puromycin-sensitive site. Puromycin is an antibiotic which shows similarities with a part of aminoacyl-tRNA. When puromycin is present in the A-site, the peptide can be linked to puromycin via a peptide bond. Thus, peptidyl-tRNA in the P-site is located in the puromycin-sensitive site. E-site: the ribosomal site harbouring deacylated tRNA on transit out from the ribosome.

Considering the sheer quantity of base pairs that need to be copied during replication, its amazing how accurate the cell can be. In addition, the assembly of the chromosomes via histones is incredible because of how much order there is to it and how much information they contain. 

If cells were still able to replicate DNA, there would be too much DNA in the nucleus. This could lead to a number of issues with the cell, including cell death, or possibly even mutations if the overly copied DNA is able to make it into a fully formed cell. Perhaps this is a way that trisomy can occur. 

This mechanism is crucial for the proper function of all cells in the body. If a cell was allowed to enter the M phase without having all of the nucleotides present, there would be missing genes. Therefore, certain proteins would be mutant or absent and the cell would not be able to function properly. This is the case for many genetic disorders and will almost always result in mutant cells. 

Typically in a positive feedback loop the molecule that is upregulated is the one that results in the deactivation of the pathway. For example, in the hypothalamic pituitary axis, the hypothalamus stimulates the release of cortisol from the adrenal cortex, and the levels of cortisol dictate the amount of CRH released by the hypothalamus. CRH is the precursor in the pathway to the release of cortisol, and its levels are reduced when the levels of cortisol are high.  

I did not realize that M-Cdk performed this function in the process of cellular replication. But, It makes a lot of sense. The phosphorylation of the lamin proteins must lead to a conformational shift that causes the depolymerization of the proteins and the degradation of the nuclear envelope. Its fascinating that the CDKs do so much of the work during M-phase. 

I wonder if the delay is a result of the numerous kinases and pathway members required to activate the Cdc20-APC complex. If they are all in the cytosol or freely floating around, it will take time to transmit the signal. Also if there are a large number of members to the pathway, this could increase the amount of time to activate the complex. Furthermore, there could be a limiting reagent involved as the mechanism of delay. 

This is a great example of a positive feedback loop. The Cdc20-APC complex is activated by the M-Cdks which are activated by the M-cyclins. Since the complex is activated by M-Cdks, the deactivation via ubiquitin-mediated proteolysis is done by the complex which then deactivates its activator. This must occur at a certain level of saturation or something like that because the M-cdks need to be active for a portion of the cell cycle.