Enzyme Science Miscellaneous
Direction: A Nick Translation reaction in a final volume of 100µl was carried out by using 25 µCi of labeled [α-32P]–dCTP for labeling a 1.2 Kb λ-Interferon DNA fragment.
- After completion of Nick translation reaction, 10 µl of reaction was spotted on a glass-fiber filter that upon counting resulted into 4.2 × 104 cpm in reaction. Another 10 µl was processed for TCA precipitation to determine redioisotope incorporation. The TCA precipitated sample gave 2.94 × 104 cpm. What is the percent of [α-32P]-dCTP incorporation into the DNA sample?
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% incorporation is calculated by incorporation into precipitated form. (The polymerized form will precipitate).
= Precipitate from epm total epm = 2.94 × 104 emp × 100 ≅ 70% 4.2 × 104 emp Correct Option: D
% incorporation is calculated by incorporation into precipitated form. (The polymerized form will precipitate).
= Precipitate from epm total epm = 2.94 × 104 emp × 100 ≅ 70% 4.2 × 104 emp
Direction: An enzyme (24000 Da) undergoes first-order deactivation kinetics while catalyzing a reaction according to Michaelis-Menten kinetics (Km= 10–4 M). The enzyme has a turnover number if 104 molecules substrate/min-(molecule enzyme) and a deactivation constant (kd) of 0.1 min–1 at the reaction conditions. The reaction mixture initially contains 0.6 mg/l of active enzyme and 0.02 M of the substrate.
- The maximum possible conversion for the enzymatic reaction will be
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NA
Correct Option: E
NA
- The time required to convert 10% of the substrate will be rapproximately
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NA
Correct Option: E
NA
- In a chemostat, evaluate the dilution rate at the cell wash-out condition by applying Monod’smodel with the given set of data: µmax = 1h–1;Yx/s = 0.5 g g–1; Ks = 0.2 g L–1; S0 = 10 g L–1
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We know, from Monod's equation
μ = μmax [S0/(KS + S0)]
⇒ μ = 1 * [10/(0.2 +10)] = 10/10.2 = 0.98
At wash out, Dilution rate D = μ
Therefore,
the required dilution rate = 0.98 hr–1Correct Option: C
We know, from Monod's equation
μ = μmax [S0/(KS + S0)]
⇒ μ = 1 * [10/(0.2 +10)] = 10/10.2 = 0.98
At wash out, Dilution rate D = μ
Therefore,
the required dilution rate = 0.98 hr–1
- Evaluate the Michaelis constant for the following lipase catalyzed trans-esterification reaction for the production of biodiesel
where, k1 = 3 × 108 M–1 s–1; k = 4 × 10³ s–1 and k2 = 2 × 10³ s–1
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Km = K1 + K2 K1 = 4 × 104S-1 + 2 × 108 S-1 3 × 108M-1S-1 = 4200 3 × 108
= 1.4 × 10-4MCorrect Option: D
Km = K1 + K2 K1 = 4 × 104S-1 + 2 × 108 S-1 3 × 108M-1S-1 = 4200 3 × 108
= 1.4 × 10-4M
