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Practicum 3 ADH

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Volledige uitwerking van het practicum van ADH-zuivering in deel 1 van de cursus Moleculen. De ADH-activiteit in wel en niet gezuiverde preparaten is bepaald aan de hand van een activiteitsmeting. In het verslag is de ADH-activiteit daarna berekend en verwerkt.

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  • April 2, 2018
  • 6
  • 2017/2018
  • Essay
  • Unknown
  • 8

2  reviews

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By: indymarlie • 2 year ago

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By: michelleweesie • 2 year ago

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Hi Indy! Are you running into something regarding the Practicum?

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By: indymarlie • 2 year ago

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Review went by accident! Will try to adapt it, because I'm happy!

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By: giadakruidenier • 4 year ago

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Practicum 3: ADH-zuivering
Door Rosa Wassenaar (groep 9, 6093345) en Michelle Weesie (groep 10, 6157904)
Assistent: Lisa en Dorien

Experiment A: Zuivering van paardenlever ADH
Bepaal de ADH-activiteit per ml volume. Bepaal de eiwitconcentratie per ml volume.

Net als in practicum 2:
1,0 ml Glycine-NaOH buffer
0,1 ml NAD+
0,5 ml ethanol
0,5 ml leverextract (zuiver en onzuiver, in duplo)

Onzuiver ADH Zuiver ADH
A340 nm 1 2 (duplo) 1 2 (duplo)
15 s 0,600 0,586 0,184 0,186
30 s 0,659 0,645 0,203 0,206
45 s 0,712 0,694 0,220 0,226
60 s 0,766 0,736 0,241 0,243
75 s 0,807 0,796 0,259 0,265
90 s 0,864 0,837 0,275 0,282
105 s 0,915 0,884 0,295 0,302
120 s 0,951 0,920 0,312 0,318
135 s 0,999 0,976 0,332 0,340
150 s 1,028 1,002 0,351 0,359
Tabel 1: Absorptie ADH per 15 seconden bij 340 nm.


Grafiek van de tabel: zie volgende bladzijde.

, ADH-activiteit: absorptie bij 340 nm uitgezet tegen tijd
y = 0,0032x + 0,5657
y = 0,0031x + 0,5517
1,000


0,800
Absorptie bij 340 nm




0,600


0,400 y = 0,0013x + 0,1677

y = 0,0012x + 0,1657
0,200


0,000
0 15 30 45 60 75 90 105 120 135 150
Tijd (seconden)

Grafiek 1: ADH-activiteit. Absorptie is uitgezet tegen de tijd. Hieruit is de DA/min te bepalen.

Legenda:
Oranje = ongezuiverd ADH lijn 1
Blauw = ongezuiverd ADH lijn 2 (duplo)
Geel = gezuiverd ADH lijn 2 (duplo)
Groen = gezuiverd ADH lijn 1




Uit de grafiek kan de ADH-activiteit berekend worden:

Ongezuiverd
DA/min gemiddeld = 0,2085
Dc/min per ml enzym:
DA/min
𝑐=
𝑎* ∗ 𝑙

Am is bepaald in het vorige practicum = 5,81 x 103 L/mol cm
L = 1,00 cm

0,2085 is per 0,5 ml enzym. 0,2085 x 2 = 0,4170

0,4170
𝑐= = 71,8 µ𝑚𝑜𝑙/ min∗ 𝑚𝑙
5,81 x 105 ∗ 1,00

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