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Samenvatting Verbetering van alle oefeningen van centrifugatie van het vak scheiding en zuivering van biomoleculen (BMW: behaald resultaat 18/20) $7.04   Add to cart
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Samenvatting Verbetering van alle oefeningen van centrifugatie van het vak scheiding en zuivering van biomoleculen (BMW: behaald resultaat 18/20)
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  • Course
  • Scheiding En Zuivering Van Biomoleculen
  • Institution
  • Universiteit Antwerpen (UA)

Verbetering van alle oefeningen van centrifugatie van het vak scheiding en zuivering van biomoleculen (BMW: behaald resultaat 18/20): Dit is een samenvatting van het vak Scheiding en Zuivering van Biomoleculen gegeven door Xaveer Van Ostade. Het bevat een volledige verbetering van alle oefeningen v...

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Document information

  • November 3, 2021
  • 2
  • 2021/2022
  • Summary

Subjects

  • xaveer van ostade
  • biomedische wetenschappen
  • scheiding en zuivering
  • zuivering
  • scheiding
  • centrifugatie
  • oefeningen
  • oplossingen
  • met oplossingen
  • scheiding en zuivering van biomoleculen

Written for

  • Universiteit Antwerpen (UA)
  • Biomedische Wetenschappen
  • Scheiding En Zuivering Van Biomoleculen
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Oefening 1

SW27: max RPM = 27000

16,1cm/7,5cm= 2,14
ln 2,14 = 0,76

Hoe lang centrifugeren om een partikel van 2S te pelleteren?



Uitrekenen via rps:

t1 = k/s waarbij k = ln (rmax/rmin)/ω2


27000 RPM of 27000 x 2π/60 = 2826 rad/sec

ω2 = 28262 = 7.986.276 rad2/sec2

k (sec2) = 0.76/7.986.276 = 951x10-10 sec2/rad2 (k is constante)

Voor een partikel van 2S:

2S = 2x10-13 sec

t = k/s
t = 951x10-10 /2x10-13 sec
t = 951x10-10 x 2x1013 sec
t = 475x103 sec = 475.000 sec = 131 hr



Uitrekenen via RPM:

t1 = K/s (in S) waarbij
2,53 x 1011 ln (rmax/rmin)
K=
(RPM)2



K = 2,53 x1011 x 0,
K = 1,922 x ,29 x 108
K = 0,26 x 103 = 260 (zie tabel)

t (in hr) = K/s (in S) = 260/2 = 130 hr

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