Answer: **Part 1.**

Answer: **Bisect 1.** **Instructions restraint frameting LB Frame:**

(The order here in restraint MS Word 2016) –

1. Penetadmonish [S] and V appraises in abound equivocation in separeprimand rows.

2. Genereprimand 1/ [S] and 1/V appraises in abound equivocation.

3. Select 1/[S] and 1/V columns and click on “Insert” tab proper to ‘Home’ tab.

4. Select ‘scatter frame’ -displayed as rare dots in the graph

5. Select trendline liberty

6. Add straight trendline and bridle the liberty restraint ‘trendline equation’.

**#** **Determination of V _{max} and K_{m} using LB Frame**”

Lineweaver-Burk frame gives an equation in from of **Y = m X + c**

where, **y** = 1/ V_{0}, **x** = 1/ [S],

Intercept, **c** = 1/ V_{max} ,

Slope, m = K_{m}/ V_{max}

Trendline (straight retreat) equation restraint “No- inhibitor” from LB frame **y = 0.0758x + 0.0047.**

**#b. Swell, m = 0.0758 s**

**y-intercept, c = 0.0047 L s pmol ^{-1}**

**#c.** Now, from Intercept, c = 1/ V_{max}

Or, **0.0047**= 1/ V_{max}

Or, V_{max} = 1/ **0.0047** = **212.77**

Hence, **V _{max} = 212.77 pmol L^{-1} s^{-1}**

Now,

Km = m x Vmax = 0.0758 x 212.77 = **16.13**

Thus, **Km = 16.13 umol L ^{-1}**

Note: There is no insufficiency of ace transmutation betwixt umol and pmol. Just representation the appertaining aces of Vo and [S] restraint fitted appraises of Vmax and Km.