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Alcohol Dehydrogenase: From Ethanol To Acetaldehyde

Paper Type: Free Essay Subject: Biology
Wordcount: 2525 words Published: 27th Apr 2017

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(20) 1. Alcohol dehydrogenase (AD) is an enzyme which catalyzes the reaction of its natural substrate ethanol to form acetaldehyde. The Km of AD, from rhinoceros livers, for ethanol is 1 X 10-3M. This enzyme is however somewhat non-specific and will recognize substrates other than ethanol. How would the kinetic plot be affected if AD were to separately catalyze methanol and isopropanol instead of ethanol? Assume that the overall Vmax remains the same in all 3 cases. How would the Km change for methanol compared to ethanol (higher, lower, the same)? How would the Km change for isopropanol compared to ethanol (higher, lower, the same)? How would the Km’s of methanol and isopropanol compare (which higher than the other or about the same). Based upon your knowledge of the mechanisms by which enzymes work, briefly explain how you decided to place your new Km’s. Hint: The alcohols are being added separately. There is not any kind of competition between the alcohols. They are not included in the same reaction. For your reference, the structures of these alcohols are below.

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Because ethanol is the natural substrate of Alcohol dehydrogenase (AD), AD would have a higher affinity and bind more readily to ethanol than other alcohols, including methanol and isopropanol. Because AD has a higher affinity for ethanol than other alcohols, its Km would be lower than methanol and isopropanol. The lower the Michaelis constant (Km) the less substrate required to get to ½ Vmax or ½ the maximum reaction rate and the higher the affinity of the enzyme for the substrate. Higher Km means more substrate concentration to reach ½ Vmax and less affinity of the enzyme for the substrate. Vmax or the maximum reaction rate can be approached, but never actually reached. The Km for methanol would be higher than ethanol, thus requiring more substrate to reach ½ Vmax and demonstrating lower affinity of AD for methanol. The Km for isopropanol would be higher than ethanol, thus requiring more substrate to reach ½ Vmax and demonstrating lower affinity of AD for isopropanol. The Km for methanol would be lower than the Km for isopropanol and show a higher affinity for AD.

The Michaielis-Menten kinetic plot would reflect a Km of 1×10-3M at ½ Vmax for ethanol, a Km greater than 1×10-3M for methanol and a Km greater than the Km of methanol for isopropanol. The overall Vmax is the same for all three, so the ½ Vmax for all three will stay the same. The plotted curve would become less vertical with the initial angle for ethanol becoming more acute and the curve becoming more linear as it changed from ethanol to methanol to isopropanol.

Ethanol is AD’s natural substrate, so based on enzyme mechanisms, it is able to bind more readily to AD due to its size and shape which fits AD’s active site and allows ethanol to get close enough to create hydrogen bonds. The substrate and enzyme change confirmation and become destabilized which stabilizes the transition state, lowers the energy of activation and allows easier formation of the reaction products. Methanol and isopropanol do not bind as well, likely due to their structure or size and shape. Methanol is one carbon shorter which would prevent it from fitting in the AD site as well as ethanol and has fewer numbers of hydrogens, reducing H-bonding potential. Isopropanol is one carbon larger than ethanol which might make it too bulky to effectively bind to AD. Isopropanol is a secondary alcohol, with two carbon atoms attached to the carbon bonded to the OH, creating a bulky Y shape and not a chain alcohol like methanol and ethanol. This conformation and bulky shape prevents isopropanol from binding more readily than methanol, which is similar to ethanol’s linear shape.

(10) 2. Briefly explain the protein cleavage involved in the maturation of an insulin molecule from proinsulin. Briefly explain 3 reasons why it is important that insulin be made as an inactive precursor requiring editing. Hint: Think in terms of things that would be important to the action of insulin (decreasing blood sugar).

Protein cleavage is post-translational processing. Proinsulin is the precursor to insulin. Proinsulin is a polypeptide chain that loops around to form two disulfide bonds between four cysteine amino acids, two near either end. Endopeptidase cuts two molecules by proteolysis to remove the middle portion of the polypeptide. The final disulfide stabilized protein is insulin.

Inactive proinsulin allows for optimal intracellular insulin stores that can be edited or activated quickly if needed to lower blood sugar and quickly prevent hyperglycemia.

Proinsulins can be produced rapidly in response to elevated blood sugar with the post-translational processing switched off quickly; leaving the inactive molecules, once blood sugar is under control.

Proinsulin is important because it is not degraded until it is needed, thus does not cause harmful low blood sugar levels and maintains sustained basal levels of insulin in the body.

(10) 3. Briefly and individually outline the mechanisms of action for covalent, competitive, non-competitive, and uncompetitive enzyme inhibitors indicating how they effect enzyme action. For each type of inhibitor, describe a unique example of how we could learn something valuable, and at least somewhat practical, about an enzyme from each type of inhibitor study.

The mechanism of action for covalent enzyme inhibitors is covalent binding in the enzyme active site and thus preventing substrate binding. This is irreversible and completely deactivates the enzyme requiring more enzymes to be produced to catalyze the reaction. This could tell us what amino acids bind in the enzyme active site by identifying covalent inhibitor modified functional groups and also substrate binding order.

The mechanism of action for competitive enzyme inhibitors is they are shaped like the substrate and can bind in the enzyme active site, blocking the substrate’s binding. Competitive inhibitors can be outcompeted by increasing the substrate concentration and are reversible. Competitive inhibitors could be used to determine enzyme substrate affinities by finding out how much substrate is required and how long it takes to get back to ½ Vmax.

The mechanism of action for non-competitive enzyme inhibitors is they bind in a place other than the enzyme active site, allowing the substrate to bind, but they destabilize the transition state which hinders the enzyme by obstructing its proper performance and reducing Vmax. Non- competitive inhibitors are reversible, but cannot be outcompeted because they do not bind to the active site. Non-competitive inhibitors could be used to determine an enzyme’s induced fit mode of action as the substrate would still be able to bind, but not fully react.

The mechanism of action for uncompetitive enzyme inhibitors is the substrate and inhibitor bind together in multi-substrate enzymes. While substrate binding and Km seem better, velocity is less because the inhibitor acts as part of the substrate. They are reversible. Uncompetitive inhibitors could be used to determine effective drug therapies by inhibiting an enzyme to varying degrees without permanently altering it, counter acting large amounts of the multi-substrate enzyme but not eliminating it from performing other useful functions.

(10) 4. In discussing advances in molecular biotechnology, we mentioned 2 processes whose names sound remarkably similar called RFLP and AFLP. These two processes indeed share some similarities, but have many differences. Briefly explain 2 significant similarities that these share in their processes. Briefly explain 2 significant differences in terms of what these processes are used for.

One similarity in RFLP and AFLP processes is cutting DNA for RFLP and cDNA for AFLP with restriction enzymes to create fragments. Another similarity is that DNA is electrophoresed in RFLP to separate different sized restriction fragments creating unique patterns for organisms or individuals (with the exception of twins) much like fingerprints and used for comparison. PCR products are electrophoresed in AFLP to compare tissues, experiments or expression profiling.

One difference in what these processes are used for is RFLP is used to compare DNA from people or organisms for genetic fingerprinting and forensics, and AFLP is used to profile gene expressions (requiring mRNA to be converted to cDNA) of uncharacterized tissues, organisms or experiments. Another difference is AFLP can be used for Quantitative Trait Loci which help identify multifactorial inheritance of traits and assist in genome mapping, whereas RFLP is not used for QTL, but can be used for identifying a person’s predisposition for a particular disease.

(10) 5. Life on the planet Zornock encodes its genetic info in overlapping nucleotide triplets such that the translation apparatus shifts only one nucleotide at a time. In other words, if we had the nucleotide sequence ABCDEF on Earth this would be two codons (ABC & DEF) whereas on Zornock it would be 4 codons (ABC, BCD, CDE, DEF) and the beginning of two others. Briefly explain and compare the effect of each of the following types of mutations on the amino acid sequence of a protein in 1) an earthling and 2) a Zornocker. A. The addition of one nucleotide. B. The deletion of one nucleotide. C. The deletion of 3 consecutive nucleotides. Assume these all occur in the middle of a gene.

X = added nucleotide, ? = unknown nucleotide

A1. One nucleotide added resulting in ABCXDEF in the earthling would create a frameshift that would produce the original codon ABC, a new codon XDE and one codon beginning F??.

A2. One nucleotide added resulting in ABCXDEF in the Zornocker would create one new codon, making a total of 5 codons, (ABC, BCX, CXD, XDE, DEF) and the beginning of two other codons EF? and F??.

B1. The deletion of one nucleotide resulting in ABCEF in the earthling would create a frameshift that would produce one original codon, ABC and two different beginnings EF? and F??.

B2. The deletion one nucleotide resulting in ABCEF in the Zornocker would result in 3 complete codons, ABC, BCE and CEF and two beginnings EF? and F??.

C1. The deletion of three consecutive nucleotides resulting in ABF in the earthling would create a frameshift that would result in one new codon, ABF.

C2. The deletion of three consecutive nucleotides resulting in ABF in the Zornocker would result in one new codon and two partial codons, ABF and the beginnings BF? and F??.

The insertions and deletions in the earthling would produce a frameshift, creating different codons and a different polypeptide chain from the mutation on. Other effects of the frameshift could be inserting a different AA into the polypeptide or stopping translation altogether. These genotype effects could create non-functioning proteins or fragments, partially functioning proteins or no protein expression.

The insertions and deletions in the Zornocker would add or remove codons at the site of the mutation, but would not alter the polypeptide chain after the mutation due to the overlapping nucleotide triplets.

(10) 6. Imagine that we’ve isolated a new and potentially useful mutation in an existing model plant. Our goal as biotechnologists might be to characterize the mutation, figure out what protein it affects, figure out how it is expressed, figure out how it is controlled, and how to best take advantage of it for crop improvement. Using the techniques that we’ve covered so far, briefly outline a series of experiments and expected results, using at least 5 of the techniques we’ve discussed, to attempt to achieve the above goals. Hint: There is more than one way to do this.

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1 In order to characterize the mutation, we could use Sanger DNA sequencing to determine the amino acid sequence of the mutated gene. We use a primer and DNA polymerase to start DNA synthesis. We then prepare reactions with dideoxynucleotides (ddNTP) for each nitrogenous base, A, T, C and G. We run the reactions with normal nitrogenous bases and one ddNTP nitrogenous base representing either A, T, C or G. The ddNTPs terminate the DNA chains and when all the reactions are electrophoresed on a gel with lanes A, T, C and G, we can read from the bottom up to determine the DNA sequence. We could then compare the DNA sequence to the sequence of the existing model plant to determine the differences in amino acid sequences caused by the mutation.

2. In order to characterize what protein it affects, we could detect gene expression and protein interactions by using qRT-PCR. First we create mRNA by transcribing the mutant DNA genes. Next, we convert the mRNA using reverse transcriptase to cDNA. Then we run a qPCR on the cDNA and add SYBR green to the products. SYBR green intercalates the DNA and we can measure the fluorescence and determine the number of mRNA copies, thus determining which proteins are affected.

3. In order to figure out how it is expressed, we could use DNA microarray and protein microarray analysis. With DNA microarrays we obtain gene chips and hybridize fluorescently labeled cDNA from the tissues containing the mutation. The mutation sample is compared to the model sample in parallel microarrays. A machine then analyzes and overlays the images to measure transcript levels, identify products and determine upregulation and downregulation of many proteins. We could also use protein microarrays which are similar to DNA microarrays, but are used to identify other proteins and compounds a protein interacts with. At times, protein function can be inferred by analyzing the environment in which it is expressed.

4. To figure out how it is controlled, we could use in situ hybridization to locate the mutant gene expression products or RNA molecules produced. First we chemically fix sample tissues to slides. With DNA probes we could localize mRNAs to see which cells and where in these cells the gene is being expressed. We could probe with antibodies to determine which proteins are being translated. We could add or subtract associated enzymes, substrates and cofactors and alter internal and external cell conditions to see how this changes the gene expression and thus determine how the gene is controlled.

5. To determine how best to take advantage of it, we could genetically engineer the model plant with the mutation by inserting the mutant DNA into a Ti plasmid, creating a recombinant Ti plasmid, and have Agrobacterium introduce that into the model plant. The Ti plasmid would recombine with the model plant DNA and create a genetically engineered plant that expresses the new trait. We could then run various experiments on the genetically engineered plant to determine if the trait is expressed as desired and if not, change the variables until we get the advantage we are looking for.


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