PMM/PGM and its substrates
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- PMM/PGM and its substrates
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PMM/PGM and its substrates
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[post_date] => 2024-12-23 07:26:30
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[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Nearly all Gram-negative bacteria possess lipopolysaccharide (LPS), an immunogenic molecule depicted in Figure 1. This structure is highly conserved across various microbes and plays a pivotal role in triggering the innate immune response in host organisms. When purified LPS is injected into experimental animals, it can induce endotoxic shock, often resulting in fatal outcomes.
Figure 1 Structure of LPS
The precursor of LPS is glucose-1-phosphate, which is formed when the enzyme phosphoglucomutase (PGM) rearranges glucose-6-phosphate. Conversely, PGM can also convert ample supplies of glucose-1-phosphate, sourced from glycogen, into glucose-6-phosphate—a fundamental molecule for both glycolysis and the pentose phosphate pathways. PGM has emerged as a potential target for antibacterial therapy due to its pivotal role; inhibiting PGM could disrupt LPS production and impede bacterial utilization of glucose-1-phosphate as an energy source.
In the Gram-negative bacterium P. aeruginosa, the enzyme phosphomannomutase/phosphoglucomutase (PMM/PGM) is essential for the biosynthesis of LPS and alginate, another molecule implicated in bacterial virulence. Phosphomannomutase (PMM) functionality rearranges mannose-6-phosphate into mannose-1-phosphate, a precursor of alginate, believed to encapsulate bacterial cells, shielding them from phagocytosis. PMM/PGM comprises four domains, three of which share a similar three-dimensional fold.
A team of researchers isolated the PMM/PGM enzyme through gel filtration and subjected it to analysis via SDS-PAGE. In Figure 2, lane 1 displays a standard protein ladder, lane 2 depicts the PMM/PGM sample before filtration, and lanes 3 to 5 depict sequential purifications of the enzyme."
Figure 2 Results of SDS PAGE analysis
Researchers also assessed the Km and Vmax values of the PMM/PGM enzyme for its substrates by Lineweaver-Burke analysis. Results are shown in Table 1.
Table 1 Kinetics of the bifunctional PMM/PGM enzyme
Sources: Ye, R. W., Zielinski, N. A., and Chakrabarty, A. M. (1994) Purification and characterization of phosphomannomutase/phosphoglucomutase from Pseudomonas aeruginosa involved in biosynthesis of both alginate and lipopolysaccharide, J Bacteriol 176, 4851-4857.
[post_title] => PMM/PGM and its substrates
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[0] => Array
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[quiz_unique_key] => 602779517
[question] => Where in the cell would LPS likely be found?
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[answer] => 2
[description] => Reason for Correct Answer:
LPS has long hydrocarbon tails that are hydrophobic, so the hydrocarbon tails of LPS will not exist in a polar environment like the cytoplasm.
LPS also has polar groups, so the polar part of the molecule will be able to exist in a polar environment.
LPS is located on the exterior plasma membrane. The structure and shape of LPS is similar to a phospholipid in that there is a polar region and a non-polar region. The fatty acid tails insert themselves into the phospholipid bilayer while the polar glycosylated regions are located outside the cell.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Free-floating in the cytoplasm
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[1] => Array
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[each_answer] => B. Attached to the plasma membrane
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[2] => Array
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[each_answer] => C. Fully integrated into the plasma membrane
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[3] => Array
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[each_answer] => D. Inside vesicles ready for exocytosis
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[1] => Array
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[quiz_unique_key] => 3243476205
[question] => Following the data in Table 1, does this graph accurately depict the enzyme kinetics of PMM/PGM?
[value] => Array
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[answer] => 4
[description] => Reason for Correct Answer:
To answer this question, see if the data shown in the graph matches the relative or absolute Kₘ or Vₘₐₓ values shown for mannose or glucose substrates in Table 1.
Kₘ is a measurement of how quickly a molecule dissociates from the enzyme. The higher the value, the more quickly the dissociation occurs. You can also think of Kₘ in the inverse way, where Kₘ describes an enzyme’s affinity for a molecule. In this reading, a high Kₘ means a low affinity and a low Kₘ means a high affinity.
Table 1 shows that the Kₘ value for mannose-1-P is 15 μM, which is lower than the Kₘ for glucose-1-P (22 μM).
Table 1 also shows the maximum velocity, or Vₘₐₓ value, for each substrate. These values should correspond to the horizontal asymptotes on a kinetics graph: namely, 28 for mannose and 60 for glucose.
As for assessing Kₘ on the graph, note that mathematically, when [substrate] = Kₘ then V₀ = ½ Vₘₐₓ.
Looking at the graph, it appears as though the Vₘₐₓ values for the substrates are inversed. It shows the enzyme reaching a higher velocity with mannose-1-P, instead of with glucose-1-P as the table suggests.
Looking at the substrate concentration values that correspond to the ½ Vₘₐₓ value, this graph has around the same value for both glucose and mannose (around [S] = 50μM). Yet according to Table 1, the Kₘ values for the different substrates should be different.
Accordingly, the graph implies that the enzyme has equal affinity for mannose and glucose (equal Kₘ), and the Vₘₐₓ for each substrate is also not correctly drawn.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Yes. The graph shows the enzyme has higher affinity for mannose than glucose, and the Vₘₐₓ for each substrate is correctly drawn.
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[each_answer] => B. Yes. The graph correctly shows the enzyme has equal affinity for glucose and mannose, and the Vₘₐₓ for each substrate is correctly drawn.
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[each_answer] => C. No. Although the graph does correctly show the enzyme has a greater affinity for mannose than glucose, the Vₘₐₓ for each substrate is not correctly drawn.
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[each_answer] => D. No. The graph implies that the enzyme has equal affinity for mannose and glucose, and the Vₘₐₓ for each substrate is not correctly drawn.
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[2] => Array
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[quiz_unique_key] => 2187790141
[question] => Using the results from the SDS PAGE experiment presented in Figure 2, what is the approximate size in kDa of the enzyme PMM/PGM that is found in Pseudomonas aeruginosa?
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[answer] => 3
[description] => Reason for Correct Answer:
PAGE stands for PolyAcrylamide Gel Electrophoresis. In this process there are small pores that molecules travel through.
It is more challenging for large molecules to traverse the small pores, so larger fragments will generally travel through pores at a slower rate.
Gel electrophoresis results will generally displayed such that fragments that have traveled further appear further down.
You can see the band belonging to the purified protein in lanes 4 and 5. You can compare its position to the reference 45 kDa band in the reference ladder.
50 kDa is the most reasonable answer, and it is actually the real measurement made by the research team. The fragment must be larger than 45 kDa because it traveled less distance. Because the fragment only traveled slightly less distance than the 45-kDa marker, it will not be an order of magnitude larger.
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[each_answer] => A. 15 kDa
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[1] => Array
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[each_answer] => B. 30 kDa
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[each_answer] => C. 50 kDa
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[3] => Array
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[each_answer] => D. 300 kDa
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[3] => Array
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[quiz_unique_key] => 1579828684
[question] => The electrophoresis experiment was repeated under the same conditions, but SDS was excluded. What would be a reasonable location for the enzyme PMM/PGM?
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[answer] => 1
[description] => Reason for Correct Answer:
SDS (sodium dodecyl sulfate) binds to amino acid chains and has a net negative charge in buffer solution.
SDS also interrupts the tertiary structure of the protein, making a straighter chain of the polypeptide.
The net negative charge is what causes a force of attraction between the anode (+) terminal of the electrophoresis machine and the negative charge on the polypeptide.
Without a net negative charge and straight shape, the amino acid chain will not migrate quickly through the gel. The molecule will appear significantly larger than it actually is. Therefore the most reasonable answer near letter A in Figure 2.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Near letter A in Figure 2
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[1] => Array
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[each_answer] => B. Near letter B in Figure 2
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[2] => Array
(
[each_answer] => C. Near letter C in Figure 2
)
[3] => Array
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[each_answer] => D. Near letter D in Figure 2
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[quiz_unique_key] => 1498560436
[question] => PGM converts a glucose-1-phosphate into glucose-6-phosphate. If the glucose-6-phosphate molecule enters the pentose phosphate pathway, which one of the following is produced?
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[answer] => 4
[description] => Reason for Correct Answer:
NAD⁺ and NADH are most prominently used and generated in glycolysis, the citric acid cycle, and the electron transport chain. These molecules do not participate in the pentose phosphate pathway.
The pentose phosphate pathway generates reducing molecules that can be used in reductive biosynthesis of other molecules.
To be a reducing molecule there must be electrons that can be transferred.
NADPH with its extra hydrogen is the reducing molecule produced in the pentose phosphate pathway. Among other things, this pathway also produces the ribose needed to synthesize nucleotides.
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[each_answer] => A. NAD⁺
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[each_answer] => B. NADH
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[each_answer] => C. NADP⁺
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[3] => Array
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[each_answer] => D. NADPH
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[5] => Array
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[quiz_unique_key] => 3413571627
[question] => Researchers verified the molecular weight of PMM/PGM using gel filtration experiments and estimated a weight of 47 kDa regardless of whether the experiments were performed under reducing conditions or native conditions. What does this suggest about the protein?
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
Gel filtration is a technique that separates proteins by size. The question says that “native conditions” are used; this means that the secondary, tertiary, and quaternary structure is preserved (the separation occurs on the basis of the overall size of the native protein).
When gel filtration is run under reducing conditions, disulfide bonds are broken; this means that the secondary, tertiary, and quaternary structure is destroyed (the separation occurs primarily on the basis of the length, or molecular weight, of the polypeptide chain).
Running gel filtration under reducing conditions might involve adding reducing agents like β-mercaptoethanol or dithiothreitol (DTT) to the sample or the running buffer before loading it onto the column. Keep in mind that this modification may disrupt disulfide bonds and native protein structures.
Both native and denaturing experiments indicate a single species with a molecular weight of approximately 50 kDa.
This information tells you that the protein must be a monomer in its native state. If it wasn’t a monomer, the molecular weight would be higher when the protein was run under native conditions than under reducing conditions – because the reducing conditions would separate the protein into multiple smaller subunits.
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[answers] => Array
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[0] => Array
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[each_answer] => A. It exists in several forms with very similar molecular weight
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[1] => Array
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[each_answer] => B. It is a monomer in its native state
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[2] => Array
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[each_answer] => C. It has no disulfide bonds
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[each_answer] => D. It contains multiple domains
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[553720|1] => B
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[553720|4] => A
[553720|5] => D
[553720|6] => B
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