Question

Can someone please help

The last two questions are the type of calculations introduced in Lab7lp, 61 of the lab manual). There are some worked out problems and practice problems in the LEARN Ste finding the substrate concentration [s. Table Il. Reaction Minxtures for Experiment A 180 ml 10.80 ml 10.60 ml 0.40 ml V2 Buffer Stock ABTS 50150 150 150 1.00 ml 1.00m1.00 mi 1.00 m 3, if the ABTS stock concentration was 20 mM and 150 μ1 was added to the reaction mixture, what is the final concentration of ABTS in the total reaction mixture based on the information shown in the table above? a. Calculate V2 in gl of the total mixture first (same for all trials). b. Calculate the concentration of ABTS (same for all trials). mM 4. The Hy0, stock solution was 3.00 mM. For Run the test tube? 4, what was the concentration of H,02 in mM H2O:l in run 4 Bio 1011 SJU Cells Lab Manual

Answer 1

3. The initial concentration of ABTS = 20mM
Initial Volume $= 150\mu l$
Final concentration of ABTS in total mixture = ?

(a) The final volume is total volume of solutions added in the each trial tube =

Volume of (water+H2O2 stock+ Buffer stock + ABTS + Enzyme)
Final volume = $(1800 + 1000 +150 + 50) \mu l = 3000 \mu l$

(b) from formula of equimolarity

$M_2= \frac{20mM \times 150 \mu l}{3000 \mu l} = 1 mM$

The final concentration of ABTS in each trial tube is 1 mM.

(4) For run 4

The initial concentration of

The initial volume of $H_2O_2 = 1.40 ml = 1400\mu l$

Final concentration in tube = ?

Final volume of solution in tube = $400+ 1400+ 1000+150+50 = 3000\mu l$

Therefore

$M_2=\frac{M_1V_1}{V_2} = \frac{3mM \times 1400 \mu l}{3000\mu l} = 1.4mM$

The concentration of $[H_2O_2]$ in run 4 = = 1.4 mM