# BCH 361 Lecture Notes - Lecture 3: Turnover Number, Cuvette

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BCH 361 Advanced Biochemistry I Fall 2020

1

Group Discussion Questions – Selected Solutions

Week 3 (September 21/22)

5. It’s helpful to recognize that S0 is a blank. You will subtract the background rate from S0 from the

rates in the other tubes.

a.

Table 1. Initial rates of reaction.

Tube

S0

S1

S2

S3

S4

S5

S6

Initial conc. (µmol/mL)

0.00

2.32

3.00

4.55

12.66

38.50

200.00

Initial rate of reaction (µmol mL-1 min-1)

0.2

1.17

1.38

1.65

2.38

2.76

3.18

Corrected vo (µmol mL-1 min-1)

0

0.97

1.18

1.45

2.18

2.56

2.98

b. Using a Lineweaver-Burk plot and keeping all the data points:

vmax = 2.98 (µmol mL-1 min-1)

KM = 4.76 µmol/mL

c. The turnover number, kcat = 59.6 x 103 min-1 or 993 s-1. You needed to know the reaction

volume for this. 1 mL is a standard cuvette size, and therefore the normal reaction volume for

many enzyme assays.

d. The specific activity is 1.99 µmol mL-1 min-1 per µg of enzyme

6. M x 103 is the same as mM. Use the same units for the product in the rate as you use for the

substrate concentration. Use a 1 mL reaction volume. Using a Lineweaver-Burk plot and keeping all the

data points: vmax = 0.181 mM/min

KM = 0.395 mM

7. You should get the same answer regardless of whether you convert from ΔA at the beginning or at

the end of your calculations, however, it is considered better practice to make the conversion at the

beginning. In this data set, one of the points clearly does not fit with the other data. This point is for

0.125 mM NAD+ and 12.5 mM EtOH. You can safely drop this point from your calculations.

a. vmax = 0.0171 mM/min

KMA = 0.191 mM (this is for NAD+)

KMB = 15.4 mM (this is for EtOH)

b. This enzyme follows a sequential mechanism.