ANSWERS
 

 

Part   #1.  
0.1 ml of the homogenate show an absorbance of 0.5, therefore it contains 4.0mg of Bovine serum albumin equivalent protein.  Since the total homogenate contain 25 ml, then  the total amount or protein in the homogenate would be [25ml/0.1ml] x 4.0mg 05 = 1,000mg of protein
 #2.   Since a 0.1 ml aliquot of the homogenate converts 48 Ámoles of substrate to product in 8 minutes, then its activity is:   48 Ámoles /  8 minutes = 6 Ámoles per minute or  6 Units ( 1 Unit = 1 Ámoles per minute).
# 3.  The specific activity is the number iof units per mg protein; therefore since 0.1 ml contains 4 mg protein and 6 units of activty, the specific activity is: 6 units / 4 mg protein = 1.5 units per mg protein
 

 

 

PART 2.

# 4 & 5.  Since the maximum velocity of the SDH reaction appears to be approximately 15 units at concentrations of substrate from 15 to 40  Ámoles of substrate minute, then the one-half maximal velocity should be 7.5 units.  The substrate concentration that gives the one-half maximal rate from the table shown is 4 Ámoles of starch substrate, which is likely to be SDH's Km

 

  

 

   
  #1.   The Vmax of B is 10 units of activity

#2.  The Km of A is 0.3 umoles glucose

#3.  The Km for A is .3 umoles of glucose, while the Km for B would be 2.0 umoles; therefore it takes significantly more glucose to reach the same one-half Vmax and enzyme A would have the greater affinity for the substrate glucose
#4.  Non-competitive inhibition of enzyme A by the drug insulineride, would remove a fixed amount of active enzyme from the mix and therefore the Vmax would be less in the presence of insulineride.
   

 

 

Part 3.   

Purification table for "cellbiolase" a new enzyme
 

STEP

Fraction Volume
(ml)

Total
 Protein

(mg)

Activity
(units)

Specific Activity
(units/mg protein)

1.  Homogenate
      supernatant

1,00

1,000

10,000

10,000/1,000= 10

2.  Precipitate
 
from supernatant is
 dissolved in buffer

200

100

impossible to know without measuring

50

3. Ion-exchange chromatography

40

400

80,000

80,000/400 = 200

What level of increase is the purification from step 2 to 3?
the increase in purification is from 50 to 200, or a 4 fold increase.

4. Gel filtration

20

100

60,000

600

5. Affinity    chromatography

6

3

75,000

75,000/3 = 25,000

 

What level of increase is the purification from step 1 to 5?

the level of purification increase from step #1 (10 units per mg) to step #5 (25,000 units per mg) is 2,500 fold purification increase.