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yeast lab write up part 5

Submitted by ziweiwang on Thu, 10/24/2019 - 22:52

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. 

 

yeast lab write up part 4

Submitted by ziweiwang on Thu, 10/24/2019 - 22:52

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.

 

yeast lab write up part 3

Submitted by ziweiwang on Thu, 10/24/2019 - 22:50

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.

 

yeast lab write up part 2

Submitted by ziweiwang on Thu, 10/24/2019 - 22:49

In the cross between unknown 1alpha and the known mutant ade1a and ade2a, the expectation was that barring mutation outside of the adenine pathway, the only reason why the unknown alpha1 would be red was because there is either a mutation in ade1, which would cause the cell to be pink, or ade2 which would cause the colony to be a deep red shade. Since unknown alpha 1 had a pink shade, the expectation was that it would have a mutation in ade1, and therefore in the cross between ade 1a resulted in a diploid colony that was pink, while complementation happens with ade2a and resulted in a white diploid colony. This did not happen. Instead  the cross with ade1a resulted in a white colony while a cross with ade2a resulted in a red colony, as seen in figure 1 and indicated in table 1. When the cross with the known mutant was transferred to the MV plate, the expectation was that the cross between unknown alpha 1 and ade2 would grow well and the cross with ade1 would grow poorly since the cross with ade 1 lacks complementation. Again the result did not support the expectation. The cross with ade1 grew well on the MV plate while the cross with ade2 grew poorly as shown in figure 2 and table 2. The expectation for the crosses of unknown alpha 1 was that both of the diploid colony would grow well. However, all three of the colonies grew well on the MV+adenine plates, as shown in figure 3 and table 3. 

In the crosses between the known mutants and unknown alpha 2, the expectation for the crosses on the YED plate was that since unknown alpha 2 was pink, similar to unknown alpha 1, the cross with ade1 would produce a pink diploid colony where cross with ade2 would result in complementation and as a result, have a white diploid colony. The result went against the expectations. The cross between unknown alpha 2 and ade 1a resulted in an white colony whereas the cross with ade2a resulted in a pink diploid colony as shown in figure 1 and table1. When the cross was transferred to an MV plate, the expectation was that the cross with ade1 and unknown alpha 2 would grow poorly while the diploid cross with ade 2 would grow well. The results partially reflected this as both of the diploid cross with both ade1 and ade2 grew well, as shown in figure 2 and table 2. In the MV+adenine plate, the expectation was that both of the resulting diploid crosses would be a small white colony. The resulting two diploid cells were able to grow on the MV+adenine plate which was expected (figure 3, table 3).

 

yeast lab write up part 1

Submitted by ziweiwang on Thu, 10/24/2019 - 22:49

The goal of this experiment is to determine the characteristics of the unknown mutants. 

In this experiment, Yeast cells with unknown mutation that was made using UV light was crossed with other unknown mutant cells and other known mutants in order to characterize the unknown mutant cells. In this experiment, I crossed the unknown mutant colonies with known mutant colonies on an YED plate and then replicate the colonies on a MV plate and a MV+adenine plate. Using the unique characteristics of the different plates and the understanding of complementation, I analyzed the data to characterize the unknown mutants. 

 

RESULTS

    In the experiment that I conducted, in order to know the genotypes of the unknown haploid yeast cells, the unknown yeast cells were crossed with known mutant yeast cells used in previous experiments and with each other on an YED plate. Then the resulting plate was copied onto a MV plate and an MV+adenine plate. In general, the control was expected and there some crosses such as unknown alpha1 x ade 2a that produced a visible mutant, but the majority of the crosses resulted in complementation and as a result did not have a visible phenotype that could be seen in the YED plate. However, there were some mutation that was seen when a seemingly normal colony was transferred into an MV or MV+adenine plate.

 

perfect paragraph - microbio

Submitted by mlabib on Thu, 10/24/2019 - 19:27

There are three types of point mutations. There are missence mutations, nonsense mutations, and silent mutations. Missence mutations happen when there is a change of a single base pair. This tiny change causes te substitution of a different amino acid in the protein. This small letter change may have no effect, if the individual is lucky. However, it may render the protein malfunctional. An example of this would be sickle cell anemia. Second type of point mutation is nonsense mutation. This leads to an early stop codon. Unfortunately, when there is an early stop codon, the protein as a whole cannot be fully developed. This would be very bad if this protein is necessary to the individual. Lastly, silent mutation. This one is very interesting as it actually is not bad for the individual. This kind of mutation changes a base pair, but the amino acid stays the same, as some amino acids have different base pair letterings, but pair for the same protein. This being said, this has no effect on the individual

draft thursday

Submitted by mlabib on Thu, 10/24/2019 - 19:26

Since I want to become an optometrist, I wanted to discover what would happen if you placed a paper over a lense that was shining the image on a screen. The image becomes dimmer. This is becase the paper is covering half the numeber of photons that are coming in through the lens, making the image less bright. Blocking a portion of the lense reduces the number of phtons that are refracted by the lens, which relates to the intensity of the image. Intensity related to brightness which explains the dimmer image. Lenses are very cool to learn about. There are many different kinds of lenses, convergent and divergent. These lenses change the placing of the image, as it could be virtual or real, depending on the type of lens. 

Genetics

Submitted by smomalley on Thu, 10/24/2019 - 18:51

Traits such as hair color, freckles, and certain diseases can be passed from generation to generation. You recieve one copy of a gene from your biological mother, and one copy of a gene from your biological father. alternative forms of a gene are called alleles. The genes inheited can form either dominant or recessive phenotypes. The dominant phenotype is defined as the phenotype that results from the heterozygous gneotype. It is 2/3 likely that you will have the dominant phenotype, and 1/3 likely you will have the recessive phenotype. The dominant phenotype is not stronger or better than the recessive, it is more likely because it is haplosufficient: one copy of the dominant allele is sufficient for the dominant phenotype. 

Perfect Paragraph: Food

Submitted by asalamon on Thu, 10/24/2019 - 18:49

 Throughout the evolution of humans, we moved out of our environment of evolutioary adaptiveness and into a new, novel environment.  One major shift humans made was in their diet conposition and  as a result, humans have become vulnerable to different diseases and conditions we were not vulnerable to before.  For humans, the major physiological change change was an increased brain size occurred during the Paleolithic Era which lasted for 2 million years.  Because a larger energy input was needed to sustain the growth of a large brain, our diet played a key role in giving us the energy and nutrients that enabled these changes.  The human diet is something that has changed considerably from the environment of evolutionary adaptiveness and we are now facing a slew of health consequences as a result.  One of major heath consequences is obesity, which often leads to the other conditions are at risk of developing.  There are two aspects of the the human diet which varies from the environment of evolutionary adaptiveness: the eating patterns of humans and the composition of the diet.  With both these factors different in the novel environment of humans, humans are now vulnerable to a variety of detrimental health conditions.

Cyclohexanol

Submitted by asalamon on Thu, 10/24/2019 - 18:44

In the experiment run, cyclohexanol was dehydrated using phosphoric acid and heat as catalysts to form the produce cyclohexene and produced a 40% yield.  The dehydration reaction utilizes 1 mole of the alcohol to produce 1 mole of the alkene.  To determine the purity of the product, infrared spectrometry and gas chromatography was used.  If the product was pure, there would be peaks at around 3000-27000 cm-1 which indicate a carbon bonded to a hydrogen.  In addition, there will be a peak at about 1700 cm-1 indicating the carbon double bonded to another carbon.  In the IR, both these trends were seen.  Importantly, there was no peaks in the stretch before the 3000-27000 cm-1 band which would have indicated the presence of an alcohol group.  The alcohol is found in the starting material and a sign of a pure product and complete reaction would be its removal in the product.  In gas chromatography, the product was injected into the column where it is heated and is pushed through the column with helium.  The lower boiling product will move quicker through the column and have a longer retention time.  In the results of the gas chromatography of the sample, only one peak was found indicated a single product as 100% of the area of the results can be found under the curve indicating a pure product.  Based on the results testing the purity of the cyclohexene product, the sample is pure.   

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