The purified OXA-M290 enzyme can now be tested to determine which β-lactamase inhibitor is most effective. This inhibitor could be prescribed in combination with a β-lactam antibiotic to treat the infection caused by the E. coli KGH1 strain. Before testing inhibitors against OXA-M290, the kinetic activity of this enzyme must first be measured. The activity of OXA-M290 is measured using nitrocefin, a chromogenic β-lactam antibiotic. When nitrocefin is hydrolyzed by a β-lactamase, it changes from yellow to red in colour. The nitrocefin hydrolysis product has an extinction coefficient of 20,500 M-1 cm-1 at 486 nm. The hydrolysis of 60 μM nitrocefin by 1 nM OXA-M290 is monitored using a microplate reader. The absorbance of the wells in the plate is measured at 486 nm every 30 seconds. This experiment is carried out with three replicates, generating the following data: Time (min) Absorbance of Replicate 1 Absorbance of Replicate 2 Absorbance of Replicate 3 0.5 0.0984 0.1368 0.1344 1.0 0.2544 0.2256 0.2448 1.5 0.3432 0.3744 0.3648 2.0 0.504 0.4476 0.4944 2.5 0.6732 0.6504 0.4956 3.0 0.732 0.7032 0.7344 3.5 0.8052 0.7968 0.8412 4.0 0.9144 0.9264 0.8928 4.5 0.9636 1.0104 0.9888 5.0 1.08 0.9852 1.0152 5.5 1.0944 1.0632 1.0848 6.0 1.1064 1.1208 1.0944 6.5 1.1352 1.104 1.1472 7.0 1.1472 1.1328 1.1424 7.5 1.1676 1.1436 1.1568 8.0 1.1712 1.1568 1.1616 8.5 1.1772 1.17 1.1892 9.0 1.2024 1.176 1.1904 9.5 1.2072 1.2168 1.1808 10.0 1.2024 1.2048 1.2024 Average absorbance of the three replicates for each of the 20 time points: Time 0.5 min: (0.0984 + 0.1368 + 0.1344) / 3 ≈ 0.1232 Time 1.0 min: (0.2544 + 0.2256 + 0.2448) / 3 ≈ 0.2416 Time 1.5 min: (0.3432 + 0.3744 + 0.3648) / 3 ≈ 0.3608 Time 2.0 min: (0.504 + 0.4476 + 0.4944) / 3 ≈ 0.482 Time 2.5 min: (0.6732 + 0.6504 + 0.4956) / 3 ≈ 0.6064 Time 3.0 min: (0.732 + 0.7032 + 0.7344) / 3 ≈ 0.7239 Time 3.5 min: (0.8052 + 0.7968 + 0.8412) / 3 ≈ 0.8144 Time 4.0 min: (0.9144 + 0.9264 + 0.8928) / 3 ≈ 0.9112 Time 4.5 min: (0.9636 + 1.0104 + 0.9888) / 3 ≈ 0.9876 Time 5.0 min: (1.08 + 0.9852 + 1.0152) / 3 ≈ 1.0261 Time 5.5 min: (1.0944 + 1.0632 + 1.0848) / 3 ≈ 1.0808 Time 6.0 min: (1.1064 + 1.1208 + 1.0944) / 3 ≈ 1.1072 Time 6.5 min: (1.1352 + 1.104 + 1.1472) / 3 ≈ 1.1288 Time 7.0 min: (1.1472 + 1.1328 + 1.1424) / 3 ≈ 1.1408 Time 7.5 min: (1.1676 + 1.1436 + 1.1568) / 3 ≈ 1.156 Time 8.0 min: (1.1712 + 1.1568 + 1.1616) / 3 ≈ 1.1632 Time 8.5 min: (1.1772 + 1.17 + 1.1892) / 3 ≈ 1.1785 Time 9.0 min: (1.2024 + 1.176 + 1.1904) / 3 ≈ 1.1896 Time 9.5 min: (1.2072 + 1.2168 + 1.1808) / 3 ≈ 1.2016 Time 10.0 min: (1.2024 + 1.2048 + 1.2024) / 3 ≈ 1.2032 Using Beer's law and the average absorbance values calculated in the previous question, what is the concentration of hydrolyzed nitrocefin at each of the 20 time points? Assume that the path length is 1 cm, and use the extinction coefficient provided in Question 1. Although you can do each calculation manually, it's far easier to use Microsoft Excel to fill in the formula used for this calculation. Show a sample calculation for the first time point, and make sure to include units in your calculation and your answers. Provide your answers in μM (micromolar) units.
Molecular Techniques
Molecular techniques are methods employed in molecular biology, genetics, biochemistry, and biophysics to manipulate and analyze nucleic acids (deoxyribonucleic acid (DNA) and ribonucleic acid (RNA)), protein, and lipids. Techniques in molecular biology are employed to investigate the molecular basis for biological activity. These techniques are used to analyze cellular properties, structures, and chemical reactions, with a focus on how certain molecules regulate cellular reactions and growth.
DNA Fingerprinting and Gel Electrophoresis
The genetic makeup of living organisms is shown by a technique known as DNA fingerprinting. The difference is the satellite region of DNA is shown by this process. Alex Jeffreys has invented the process of DNA fingerprinting in 1985. Any biological samples such as blood, hair, saliva, semen can be used for DNA fingerprinting. DNA fingerprinting is also known as DNA profiling or molecular fingerprinting.
Molecular Markers
A known DNA sequence or gene sequence is present on a chromosome, and it is associated with a specific trait or character. It is mainly used as a genetic marker of the molecular marker. The first genetic map was done in a fruit fly, using genes as the first marker. In two categories, molecular markers are classified, classical marker and a DNA marker. A molecular marker is also known as a genetic marker.
DNA Sequencing
The most important feature of DNA (deoxyribonucleic acid) molecules are nucleotide sequences and the identification of genes and their activities. This the reason why scientists have been working to determine the sequences of pieces of DNA covered under the genomic field. The primary objective of the Human Genome Project was to determine the nucleotide sequence of the entire human nuclear genome. DNA sequencing selectively eliminates the introns leading to only exome sequencing that allows proteins coding.
The purified OXA-M290 enzyme can now be tested to determine which β-lactamase inhibitor is most effective. This inhibitor could be prescribed in combination with a β-lactam antibiotic to treat the infection caused by the E. coli KGH1 strain.
Before testing inhibitors against OXA-M290, the kinetic activity of this enzyme must first be measured.
The activity of OXA-M290 is measured using nitrocefin, a chromogenic β-lactam antibiotic. When nitrocefin is hydrolyzed by a β-lactamase, it changes from yellow to red in colour. The nitrocefin hydrolysis product has an extinction coefficient of 20,500 M-1 cm-1 at 486 nm.
The hydrolysis of 60 μM nitrocefin by 1 nM OXA-M290 is monitored using a microplate reader. The absorbance of the wells in the plate is measured at 486 nm every 30 seconds. This experiment is carried out with three replicates, generating the following data:
|
Time (min) |
Absorbance of Replicate 1 |
Absorbance of Replicate 2 |
Absorbance of Replicate 3 |
|
0.5 |
0.0984 |
0.1368 |
0.1344 |
|
1.0 |
0.2544 |
0.2256 |
0.2448 |
|
1.5 |
0.3432 |
0.3744 |
0.3648 |
|
2.0 |
0.504 |
0.4476 |
0.4944 |
|
2.5 |
0.6732 |
0.6504 |
0.4956 |
|
3.0 |
0.732 |
0.7032 |
0.7344 |
|
3.5 |
0.8052 |
0.7968 |
0.8412 |
|
4.0 |
0.9144 |
0.9264 |
0.8928 |
|
4.5 |
0.9636 |
1.0104 |
0.9888 |
|
5.0 |
1.08 |
0.9852 |
1.0152 |
|
5.5 |
1.0944 |
1.0632 |
1.0848 |
|
6.0 |
1.1064 |
1.1208 |
1.0944 |
|
6.5 |
1.1352 |
1.104 |
1.1472 |
|
7.0 |
1.1472 |
1.1328 |
1.1424 |
|
7.5 |
1.1676 |
1.1436 |
1.1568 |
|
8.0 |
1.1712 |
1.1568 |
1.1616 |
|
8.5 |
1.1772 |
1.17 |
1.1892 |
|
9.0 |
1.2024 |
1.176 |
1.1904 |
|
9.5 |
1.2072 |
1.2168 |
1.1808 |
|
10.0 |
1.2024 |
1.2048 |
1.2024 |
Average absorbance of the three replicates for each of the 20 time points:
- Time 0.5 min: (0.0984 + 0.1368 + 0.1344) / 3 ≈ 0.1232
- Time 1.0 min: (0.2544 + 0.2256 + 0.2448) / 3 ≈ 0.2416
- Time 1.5 min: (0.3432 + 0.3744 + 0.3648) / 3 ≈ 0.3608
- Time 2.0 min: (0.504 + 0.4476 + 0.4944) / 3 ≈ 0.482
- Time 2.5 min: (0.6732 + 0.6504 + 0.4956) / 3 ≈ 0.6064
- Time 3.0 min: (0.732 + 0.7032 + 0.7344) / 3 ≈ 0.7239
- Time 3.5 min: (0.8052 + 0.7968 + 0.8412) / 3 ≈ 0.8144
- Time 4.0 min: (0.9144 + 0.9264 + 0.8928) / 3 ≈ 0.9112
- Time 4.5 min: (0.9636 + 1.0104 + 0.9888) / 3 ≈ 0.9876
- Time 5.0 min: (1.08 + 0.9852 + 1.0152) / 3 ≈ 1.0261
- Time 5.5 min: (1.0944 + 1.0632 + 1.0848) / 3 ≈ 1.0808
- Time 6.0 min: (1.1064 + 1.1208 + 1.0944) / 3 ≈ 1.1072
- Time 6.5 min: (1.1352 + 1.104 + 1.1472) / 3 ≈ 1.1288
- Time 7.0 min: (1.1472 + 1.1328 + 1.1424) / 3 ≈ 1.1408
- Time 7.5 min: (1.1676 + 1.1436 + 1.1568) / 3 ≈ 1.156
- Time 8.0 min: (1.1712 + 1.1568 + 1.1616) / 3 ≈ 1.1632
- Time 8.5 min: (1.1772 + 1.17 + 1.1892) / 3 ≈ 1.1785
- Time 9.0 min: (1.2024 + 1.176 + 1.1904) / 3 ≈ 1.1896
- Time 9.5 min: (1.2072 + 1.2168 + 1.1808) / 3 ≈ 1.2016
- Time 10.0 min: (1.2024 + 1.2048 + 1.2024) / 3 ≈ 1.2032
Using Beer's law and the average absorbance values calculated in the previous question, what is the concentration of hydrolyzed nitrocefin at each of the 20 time points? Assume that the path length is 1 cm, and use the extinction coefficient provided in Question 1.
Although you can do each calculation manually, it's far easier to use Microsoft Excel to fill in the formula used for this calculation.
Show a sample calculation for the first time point, and make sure to include units in your calculation and your answers. Provide your answers in μM (micromolar) units.
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