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Samenvatting Excel verslagen vele practica's biochemie en biotechnologie $17.96
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Samenvatting Excel verslagen vele practica's biochemie en biotechnologie

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Verslagen (excel als word) enkele practica's van het vak practicum biochemie en biotechnologie. Alle practica's waren > 7/10. De practica's die erin zitten zijn MM, endocellulase, PCR-voorbereiding, pDNA en Lowry.

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  • January 4, 2024
  • 23
  • 2023/2024
  • Summary
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1. Bepaling van de enzymatische activiteiten in functie van de substraatconcentratie
A) Opstellen van de ijklijn
Proefbuis
volume p-nitrofenol = V1 (mL)
volume buffer (pH = 10,5) (mL)
NaOH 0,2 N (mL)

totaal volume = V2 (mL)
x-as = concentratie p-nitrofenol (µM)
concentratie p-nitrofenol stockoplossing = C 1 (µM)


Proefbuis
y-as = extinsie 420 nm



B) Experimentale bepaling van de enzymatische activiteit in functie van de substraatconcentratie
Proefbuis
Substraat (mL)
Buffer (pH=10,5) (mL)
Alkalisch fosfatase (mL)
NaOH 0,2 N (mL)
Totaalvolume (mL)
concentratie p-nitrofenol (µM) = C1

substraatconcentratie (µM) = C2 = xH
concentratie p-nitrofenol (µM)
reactiesnelheid (µM/min) = v
1/[S] (1/µM) = x(LB)
1/v (min/µM) = y(LB)
[S]/v (min) =yH

Proefbuis
y-as = extinsie 420 nm


C) Berekening van de enzymatische activiteit en van de kinetische parameters
Lineweaver-burk
intercept y-as = 1/vmax
vmax (µM/min)
rico = Km/vmax
Km (mM)

, Hanes
intercept y-as = Km/vmax
vmax (µM/min)
rico = 1/vmax
Km (mM)




2. Experimentele bepaling van de enzymatische activiteit in aanwezigheid van verschillende hoeveelheden inh
Proefbuis
volume substraat = V1 (mL)
volume buffer (pH = 10,5) (mL)
volume inhibitor (mL)
volume alakalisch fosfatase (mL)
volume NaOH 0,2 N (mL)
Totaalvolume (mL)
concentratie p-nitrofenol (µM) = C1

substraatconcentratie (µM) = C2 = xH

concentratie p-nitrofenol (µM)

reactiesnelheid (µM/min) = v
1/[S] (1/µM) = x(LB)
1/v (min/µM) = y(LB)

[S]/v (min)

Proefbuis
y-as = extinsie 420 nm



Lineweaver-burk met inhibitor
intercept y-as = 1/vmax
vmax (µM/min)
rico = Km/vmax
Km (mM)

bepaling van Ki
volume inhibitor (mL) = V1
concentratie inhibitor (µM) = C2
Ki = [I]/(Km,inh/Km)-1

, concentratieinhibitor (µM) = C1
totaal volume (mL) = V2


Opmerkingen
Km is een beetje verhoogd
Vmax klopt volgens mij niet, maar ik denk dat deze onveranderd zou moeten zijn…
In dit geval spreken we bij mijn onbekend enzym van een competitieve enzyminhibitie

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