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Yeast Genetics Discussion PP

Submitted by mrmoy on Wed, 04/18/2018 - 21:09

Haploids HB1, HA1, and HA2 could not grow on their own as they could not synthesize their own adenine. This makes sense because they all had an ade mutation that makes them mutant deficient for adenine biosynthesis. Haploid HA0 could grow on its own, meaning that it can synthesize its own adenine. Again, this makes sense because this haploid strain had no mutations in any of its ade genes. The cross between HB1xHA1 did not proliferate. Because the cross contained one MATa and one MATα, sexual reproduction is possible. However, because it did not proliferate, it means that their alleles are not complementary. Next, the cross of HA1xHA2 did not proliferate. In this cross, both the haploids were MATa, therefore sexual reproduction is not possible and no proliferation is expected. The cross HB1xHA2 did proliferate. Both of these haploids could not grow on their own, however, when they were crossed they were able to proliferate. This cross contains one MATα and one MATa, thus the proliferation is caused by the complementation of their alleles. Finally, the cross of HB1xHA0 was able to proliferate. This cross needs to be furthered examined because although HB1 is MATα and HA0 is MATa, one cannot assume they sexually reproduced. As we know, HA0 is able to proliferate on its own, so either their alleles are complementary and it proliferated or it’s just HA0 proliferating and HB1 dying off.

Yeast Genetics Discussion #3

Submitted by mrmoy on Wed, 04/18/2018 - 21:08

Finally, the cross of HB1xHA0 was able to proliferate. This cross needs to be furthered examined because although HB1 is MATα and HA0 is MATa, one cannot assume they sexually reproduced. As we know, HA0 is able to proliferate on its own, so either their alleles are complementary and it proliferated or it’s just HA0 proliferating and HB1 dying off.

 

Yeast Genetics Discussion #2

Submitted by mrmoy on Wed, 04/18/2018 - 21:08

Next, the cross of HA1xHA2 did not proliferate. In this cross, both the haploids were MATa, therefore sexual reproduction is not possible and no proliferation is expected. The cross HB1xHA2 did proliferate. Both of these haploids could not grow on their own, however, when they were crossed they were able to proliferate. This cross contains one MATα and one MATa, thus the proliferation is caused by the complementation of their alleles.

Yeast Genetics Discussion

Submitted by mrmoy on Wed, 04/18/2018 - 21:07

Haploids HB1, HA1, and HA2 could not grow on their own as they could not synthesize their own adenine. This makes sense because they all had an ade mutation that makes them mutant deficient for adenine biosynthesis. Haploid HA0 could grow on its own, meaning that it can synthesize its own adenine. Again, this makes sense because this haploid strain had no mutations in any of its ade genes. The cross between HB1xHA1 did not proliferate. Because the cross contained one MATa and one MATα, sexual reproduction is possible. However, because it did not proliferate, it means that their alleles are not complementary.

Yeast Genetics Intro #2

Submitted by mrmoy on Wed, 04/18/2018 - 00:11

In this experiment, the yeast mutants requiring adenine was examined. Yeast cells cannot produce their own adenine, but if adenine is in the medium, then cells can convert it to adenosine monophosphate (AMP). The process of converting adenine to AMP requires multiple enzymes. As a result, any single mutation could result in an adenine requiring mutant or a mutant deficient for adenine biosynthesis.

 

Yeast Genetics Intro

Submitted by mrmoy on Wed, 04/18/2018 - 00:11

Throughout the history of science, Saccharomyces cerevisiae (yeast) has served as a model organism. Yeast is a unicellular, eukaryotic microorganism that can exist stably as both a haploid and diploid and can reproduce either sexually or asexually. When yeast are in the haploid state they exist as either the MATa or MATα. Yeast cells can remain in the haploid state, or one MATa and one MATα cell can sexually reproduce to produce a diploid cell.

Esterification and IR Spectroscopy Results PP

Submitted by mrmoy on Thu, 04/12/2018 - 21:57

    N-propyl propanoate (n-propyl propionate) was synthesized using propanoic acid and 1-propanol in the presence of sulfuric acid, resulting in a 62.5 % yield. The initial odor of the mixture of propanoic acid and 1-propanol was very unpleasant and strong. After the reaction was complete, the odor was like that of rubbing alcohol (Table 1). IR spectroscopy was used to determine the purity and identity of the product. The IR spectroscopy revealed the presence of an ester, which correlates to the strong, sharp peak at around 1741 cm-1. In addition, the peak at 2968 cm-1 indicates the presence of the C=O bond commonly found in esters. Lastly,  there was a broad peak at 3550 cm-1, which indicates the presence of -OH groups. Because of the presence of alcohol impurities, this suggests that some of the reagents had not undergone the reaction.

Esterification and IR Spectroscopy Results #2

Submitted by mrmoy on Thu, 04/12/2018 - 21:56

N-propyl propanoate (n-propyl propionate) was synthesized using propanoic acid and 1-propanol in the presence of sulfuric acid, resulting in a 62.5 % yield. The initial odor of the mixture of propanoic acid and 1-propanol was very unpleasant and strong. After the reaction was complete, the odor was like that of rubbing alcohol (Table 1). IR spectroscopy was used to determine the purity and identity of the product.

Esterification and IR Spectroscopy Results

Submitted by mrmoy on Thu, 04/12/2018 - 21:56

There are several ways in which the percent yield could be improved, considering that Le Chatelier's principle and that esterification is a reversible reaction. Esterification is an exothermic reaction and as a result by lowering the temperature during the reflux steps, the reaction will move forward and will produce more ester in the product. Another way to increase the percent yield would be to use a stronger acid than sulfuric acid so that the ester will form faster and water will be removed at a faster rate.

Esterification and IR Spectroscopy #3

Submitted by mrmoy on Thu, 04/12/2018 - 21:54

The contents were transferred into a centrifuge tube that contained water (1 mL) and was mixed thoroughly with a pipet before the lower aqueous layer was removed. Saturated aqueous sodium bicarbonate (1 mL) was added and mixed thoroughly with a pipet before the lower layer was removed. This was repeated once more with sodium bicarbonate (1 mL). The step was then repeated with saturated aqueous sodium chloride (1 mL). To a vial was added the organic layer and 5 spheres of calcium chloride. The vial was then swirled gently. Afterwards, the liquid was transferred into a dry tared capped vial and the odor was observed. The IR spectroscopy was then determined.

 

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