2022-1 Enzyme Engineering / Final Exam      75 min (June 20)

1. Amino acid 1, 2, and 3 have R1, R2, and R3 side group, respectively. Draw the

chemical structure of the peptide bond composed of these three amino acids

(10 points).

2. Hexokinase catalyzes the first reaction in glycolysis and phosphorylates D-glucose to      D-glucose 6-phosphate using ATP as the source of the phosphate:

Under conditions of pH 7, 25°C and a hexokinase concentration of 3 nmol/mL, the      KM for hexokinase for the substrate glucose was determined to be 3 x 10-4 M. When     the glucose concentration was set to 160 μΜ, the initial rate of the reaction was        found to be 65 μmol/mL‧s.

(1) What is Vmax for hexokinase under these conditions? 186 μmol/mL‧s

(2) What is the turnover number for hexokinase under these conditions? 62,000/s

(Hint: turnover number=Vmax/total enzyme concentration)

3. Determine the reaction time to convert 80% substrate to product by an enzyme in

100 Liter enzyme reactor, where the enzyme reaction follows competitive inhibition

(20 points). The initial substrate concentration is 20 g/L, the initial enzyme

concentration is 0.01 g/L, and inhibitor concentration is 0.04 g/L. Km=0.5 g/L and

Vmax=1.5 g/L∙min. Ki=0.02 g/L.

Competetive inhibition 식을 적분하여 t=f(S) 식으로 유도한 후 (강의노트 참조),

S=0.2S0, 각각의 kinetic parameter 값들을 대입하면 됨. 정답=12.3 min

4. Determine enzyme volume for the following enzyme reaction (10 points).

- Substrate concentration=50 g/L,   - Reaction volume=500L

- Optimal enzyme dosage=5 units/gram substrate,

- Enzyme activity=400 units/mL

50g/L x 500 L x 5units/g /400 units/mL =312.5 mL

5. Calculate enzyme activity as units/mg for the following reaction (10 points).

Substrate: sucrose (G-F)    -Product: 1-Kestose (G-F-F) (G: glucose, F: fructose, hyphen between each sugar stands for glycosidic bond. Molecular weight of glucose and fructose is 180)

Product formed: 2 g/l for 30 min with 0.2 ml of enzyme, which was prepared by dissolving 0.2 mg of solid enzyme into 10 ml of water.

{2g/L/504 g/L/M x 106 μM/30‧0.2}x10mL/0.2mg=33069 units/mg

6. Fill the blanks (10 points).

When the enzymatic packed-bed reactor (200 liters) is being operated with a

velocity 20 liters/hour, retention time is (200L/20 L/h=10 h① ) and space velocity is     (20L/h/200 L=0.1 h-1 ② ).

7. According the Michaelis-Menten equation, what is the V/Vmax ratio when

[S] = 3 KM?