Drafts

When it comes to biodiversity offsetting, there is no accurate way to measure the equivalent gains made after the destruction of one area of biodiversity.  Often times, the destruction of biodiversity is done for capital gains.  With this in mind, the company doing the destruction is going to look for the lowest cost option compensate for the loss.  Depending how the compensation was measured, there could be a loophole that could  be exploited by the company.  As a result, the biodiversity of an area will suffer because of the greed of a company.

    Our world is currently suffering due to the actions of humans.  We are living in a time where the biodiversity might not ever recover.  Although “equivalent gains” sounds good in theory, it cannot fix what was destroyed.  It might make the hurt feel a little bit better but there is no way to completely match the biodiversity lost.  If humans are so interested in maintaining biodiversity, then why are we destroying it in the first place?

    The impact of photobleaching on a specimen can be detrimental and irreversible. Depending on the fluorescent stained utilized, certain fluorophores photobleach at a more rapid rate than others requiring more care. The time lapse images collected from part I reveal the different properties of each fluorophore. Admittedly, the photobleaching rate constants extracted from the trend lines of Figure 4, Figure 5, and Figure 6 suggested unexpected values. Since each of these rate constants were so similar, I felt that no definitive conclusions may be drawn about comparing each of the fluorophores with this data. In future experiments, I will try to increase the number of significant digits in the data collected and increase the overall quantity of numerical data. Therefore, I utilized the mostly qualitative data to reach the following conclusions.

Visually, DAPI appears in Figure 1 to decrease less in intensity than the fluorophores seen in Figure 2 or Figure 3. Furthermore, the overall trend line of the DAPI stained nuclei in Figure 4 exhibits the smallest rate of photobleaching. In comparison, Figure 6 and Figure 7 suggest that the rhodamine stained F-actin and FITC stained tubulin have the same rapid rate of photobleaching. If more trials were conducted then we could have conclusively stated which fluorophore photobleached the fastest. However from these varying trend lines and visual differences, the data would suggest that different fluorophores bleach at different rates. Due to DAPI’s chemical structure it might be less likely to bind to neutral compounds like oxygen that inhibit photon emission. Conversely, Rhodamine and Fluorescein may act as unstable compounds that prefer to bind to other molecules than undergo periodic photon emission and absorption.

At rest, the blood flow to the Gi Tract (kidney) is high, in skeletal muscle it’s low. Once we begin to exercise and contract our muscles, there is an increase in metabolic activity which causes dilation. This leads to TPR (total peripheral resistance) dropping, and thus blood pressure drops. This is sensed by baroreceptors and chemoreceptors and they increase levels of CO2 and H+. The sympathetic nervous system reacts by increasing Heart rate, stroke volume, and TPR. The mean arterial pressure also increases. More blood flows to the skeletal muscle and less goes to the abdominal organs. The skeletal muscle has high flow because local dilators are stronger than central constrictors. The reason for low flow in the kidneys is because central constrictors increase resistance there because we need blood flow to our muscles to exercise.

Identifying the number of plant species

    Our aim is to count the number of different plant species found around a series of local small aquatic ecosystems. 

    We intend to limit the radius within which we count the number of plant species to 2 metres into and 2 metres out of the pond or waterbed edge. Starting at an arbitrary point and marking it, the radius will be subdivided into 2 metre wide plots circling the entirety of the pond. This will be done in order to ensure accuracy of counting and to prevent backtracking. Within each of these plots, the number of plant species visible to the naked eye and the number of individuals of that species will be counted. Then, the counts for each species from each of the plots will be aggregated into a series of totals.

    The data will be analysed by generating a Simpson biodiversity index for each area of study. This is a statistical tool described by the following equation:

D = 1-[(n(n-1))/(N(N-1))]

Where n is the number of individuals of each species and N is the total number of individuals of all species, and D represents the biodiversity index. The higher D is, the higher the biodiversity of a particular area. 

    We expect the results to show the level of biodiversity around local small aquatic ecosystems in order to inform us about one aspect of the ecosystems’ integrities. As there is no data on the biodiversity of these ecosystems, we will have no point of comparison for our data.

Most people believe that coding is difficult, and those people are right. However; the most important thing you need to know about coding is that you can enjoy doing it. Often, others think coding is only fun for people who love computers, but the reality is coding can be applied to so many hobbies, projects, and careers. Not only that, but coding gives you the freedom to build whatever software you would like, so if you ever wanted to create your own website or service, it is possible through coding. Learning to code has never been easier as more resources become available and free to use. Getting started, beginners usually are guided to create puzzles and games centered around key ideas/concepts of the programming world to learn more about it. If I wish to create, there is an endless amount of knowledge on the internet that is free and readily available to help steer anyone to become proficient at what they desire.

    Other limitation of the experiment includes that there was no replication experiment that confirms the result of the experiment, the cross between the unknown mutants were limited, and that the growth of the colonies were left in the incubator for a long time so that determining growth was difficult. The first limitation that there is no replication. Replication of the experiment is important in determining what is happening is not due to chance or due to contamination. However, due to the limited amount of time, a replication was not performed. Another limitation is the cross between the unknown mutants were limited. While unknown alpha 1  and unknown alpha 2 was crossed with unknown a1 nd unknown a2, unknown alpha 1 and unknown a1 did not have the same genotype, as did unknown alpha2 and unknown a 2 because of this, unknown alpha 1 was not able to be crossed with unknown alpha 2 and unknown a2 is not able to be crossed with unknown a2. This limits the amount of analysis that can be done as it may be possible that both as have the mutation at the same gene but it was never shown because those two were never crossed. Another limitation is in determining the growth in MV and MV+adenine plates. Because the plates were left in the incubator for 3 days rather than 1 day like other plates, the growth was hard to determine because there were colonies that had noticeably less growth compared to others. Because those had less growth, it was hard to determine whether those counted as part of the colonies that grew well,especially the lack of growth was so noticeable compared to those that grew well. 

    Future experiments that could be done would be to replicate the experiment, fully cross all combination of mutants and genotype the yeast colonies. The first experiment that could be done is to replicate the experiment. By replicating the experiment, the certainty in the results can be increased, and the possibility that the results were skewed due to contamination or other factors would be limited. Another experiment that could be done is to cross the as and alphas with each other and themselves. By doing this, the genotype of the mutants can be better determined. For example, if unknown a 1 was crossed with unknown a1, and the resulting diploid colony grew well on YED  and MV+adenine plate but poorly on MV plate, it can be determined that even if unknown a 1 does not have a mutation at ade1 or ade2, the mutation is still in the adenine pathway. Yet another experiment that can be done to confirm the site of mutation is to genotype the colonies. By extracting the DNA, running PCR and running the DNA on a gel, a more accurate result can be obtained. In addition, using the purified DNA, the DNA sequence can be determined, and the exact nature of the mutation, whether it is a spot mutation or a missense mutation can be determined.

    Unknown 1a and unknown 2a both does not seem to have a mutation at ade1 or ade2 since both of the crosses yielded white diploid colonies that grew well on MV plate and MV+adenine plate. Because of this, it is most reasonable to say that the unknown a mutants were complemented at every mutation and as a result, the resulting diploid cell behaves exactly as a wildtype mutation would. Because of this it can be reasoned that the red pigment that is present in both unknown a and unknown 2a is due to some other pigment intermediate that happens to look similar to that of AIR rather than there being a mutation in either ade1 or 2 which would result in the over accumulation of AIR. This would make sense since the cross with the known mutants would cause the resulting diploid colonies to complement, and the trait would not be able to be observed in MV and MV+adenine plates.

    In the cross between the four unknown mutations, it is clear that the four mutations does not have mutations in the same gene since the diploid colonies that are the result of the cross behaves in a similar way as a wild type colony in both the YED, MV and the MV+adenine plate. However, there is no guarantee that the two mutant haploid colonies does not have the same mutation and the mutation is in a gene that is not vital in growing on YED, MV and MV+adenine plate and just could not be seen. In addition, the possibility that the two unknown alphas  or two as may have the same mutation was not tested.

    While there are many limitations to the experiment that was done, the fact that both positive control and negative control had result that was similar to previous experiments implies that the properties of YED and MV plate and the identity of the known mutants could be sure if the known mutants were different from what was previously used, the result of the cross between known mutants would have been different from previous experiments. In the YED plate especially, the result indicates that there are no excess adenine present that may skew with the result of the experiment since the cross between ade2 alpha and ade2a indicates that if a colony is white the reason is not at the fault of the plates but with the mutant colonies themselves. In the MV plate, the lack of known haploid cells and the ade2alpha ade2a crossing indicates under the condition of the plate, a haploid colonies and colonies that cannot produce adenine will not grow well, and if there was a mistake in the plate then the control would not turn out similar to previous experiments. However, since MV+adenine plate was not used last time, the cross of known haploid mutant cells were not useful in that plate, which is one of the limitations of the experiment. 

In this experiment, the goal was to figure out the genotype of the unknown haploid mutants through crossing the haploid cells with known and other unknown haploid cells. In doing this, if the resulting diploid colony failed to complement, that would indicate that the two haploid cells had the same mutation. In addition, if the resulting diploid cells had a color that would highly indicate that the place where the mutation is is an enzyme that digest a red intermediate, such as AIR. By using the coloring of the crosses and different properties of the colony that emerges when grown on different plates, a genotype of the haploid mutants can be guessed. 

    Looking at the data, it is highly likely that unknown 1 mutant alpha have a mutation at ade2. This is indicated by the diploid cross between unknown alpha 1 and ade 2a. The resulting diploid colony is very similar in property that of ade 2 alpha and ade2a. Because of this, it can be assumed there is a ade2 mutation in unknown alpha 1. The mutation at ade2 would also explain the unknown alpha 1 an ade2a’s diploid colony’s behavior on the MV and MV+adenine plates. In the MV plate the colony was not able to grow since adenine is crucial to the viability of the colony. However it was able to grow on the MV+adenine plate because the only mutation that the cross had was at that specific gene, and as soon as the product of the pathway was supplied to the colony it was able to grow. 

    In addition to unknown alpha 1, unknown alpha 2 also seems to have a mutation in ade2. This is due to the visible color difference in the cross between unknown alpha 2 and ade2a. However, there is also a difference between this diploid cross and cross between ade2 alpha and ade2a. This is perhaps due to other mutations that unknown alpha 2 have acquired through the UV treatment. If for example another enzyme developed a gain of function mutation that digested AIR, that would explain the pinkish color of the diploid colony, and if the pathway crosses with the adenine pathway again, it would explain why the colony grew well on the MV plate. 

In the cross between the unknown mutants, there was not a specific expectation as to what the result of the cross would be. However, I expected that there would be at least one cross that would result in a mutant diploid colony, assuming that the reason for the red color resulted from the intermediate AIR. However, against expectations, none of the diploid crosses between the unknowns resulted in a visible mutant cell in the YED plate, as shown in figure 1 and table 1. In the replica to MV plate, all of the white diploid colonies were expected to grow well on the MV plate, and this was shown in the MV plate as shown in figure 2 and table 2. In the replica of MV+adenine plate it was expected that all of the diploid colonies would grow well and the result would be expected as shown in figure 3, table 3. 

In the crosses between the known mutants, the expectation of the mutants and their crosses was that ade1alpha, ade2alpha, ade1a, ade1ax ade1alpha, ade1a xade2alpha,and ade2a x ade1alpha would be white or cream colored on the YED plate while ade2a and ade2alpha x ade2a would be red. The result of the YED plate was as expected as shown in figure 1, table 1. In the MV plate, the  expectation from previous experiment was that all of the known haploid mutants and ade1a x ade 1alpha and ade2a and ade2 alpha would not grow well while ade1ax ade 2 alpha ande ade2a and ade1alpha would grow well, and the result was as expected as shown in figure 2 and table 2. 

In general, most of the results from the cross between the different unknown mutants were unexpected, with very few colonies behaving as predicted. One major difference between the expectation and the result is in the MV plate where there was more colonies growing than I expected.  However, all of the control crosses behaved exactly the same as in previous experiments.

In the cross between the known mutants and unknown 1a, the expectation was that since the unknown 1a had a red color that is similar to that of ade2a, in the cross between unknown 1a and ade2alpha the cross would result in a red colony of diploid colony, and the cross with ade1alpha would result in a white diploid colony. While the cross with ade1 alpha was white as expected, the cross with ade2alpha was also white which was unexpected (figure 1, table1). In the replica of the colonies in the MV plate, the expectation was that the cross between unknown 1a and ade2alpha  would not be able to grow well on the MV plate while the cross between unknown 1a and ade1alpha would be able to grow well on the MV plate. While unknown 1a did not grow well on the MV plate, both of the diploid colonies resulting from unknown 1a grew well on the MV plate as shown in figure 2 and table 2. In the MV+adenine plate replica, the expectation was that both of the diploid colonies would be able to grow on the MV+adenine plate. The result was as expected with both of the diploid crosses from unknown 1a and known mutants growing well on the MV+adenine plate as shown in figure 3, table 3. 

In the cross between the known mutants and unknown 2a, the expectation for the cross on the YED plate was that since the unknown a is a pink haploid colony, the cross with ade1 would create a visibly pink diploid colony whereas in the cross with ade2, there would be complementation and result in an white colony. However, neither of the crosses resulted in a visibly mutant colony as shown in figure 1 table 1. When the colonies were transferred to the MV plate, the expectation that the diploid cross with ade1a would not be able to grow well. However, both of the diploid crosses grew well on the MV plate, as shown in figure 2 and table2. In the MV+adenine plate, the expectation is that both of the diploid crosses would grow well. The result was as expected, as both of the colonies grew well on the MV+adenine plates, as seen in figure 3 and table 3.