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Dictionary » K » Kcat KcatDefinition noun Kcat (1) First-order rate constant, or the catalytic constant is a measure of Vmax / Et expressed in units of inverse time. (2) The turnover number – the number of substrate molecule each enzyme site converts to product per unit time, and in which the enzyme is working at maximum efficiency. (3) The overall catalytic rate of an enzyme, or the maximum number of enzymatic reactions catalyzed per second. ![]()
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Results from our forumEnzyme Kinetics - Turnover... either way in numerous places. However the question i've been given does specifically make a point that it's a homotetramer and asks for the Kcat/turnover number per active site.
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Enzyme Kinetics - Turnover... formed/min". I used this to then plot a lineweaver burke plot to get Vmax and Km values. One of the questions asks me to calculate the Kcat value. This is where i'm stuck. I have the following data at hand: Vmax = 2.442 x 10E-2 μmoles/min Amount of enzyme used each time = 0.05mL Concentration ...
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Re: Gluten-free and Vaccines... - derived peptide PQPQLPYPQPQLPY is .vefold higher than that for its target peptide sequence in .brinogen, its natural substrate (24). 26. kcat/KM440 min 1mM 1 for LQLQPFPQPQLPYPQPQLPYPQPQLPYPQPQPF compared with kcat/KM 82 min 1mM 1 for PQPQLPY (24) and kcat/KM 350 min 1mM 1 for PQPQLPYPQPQLPY. ...
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enzymes in cold temperature... (converts water and carbon dioxide to carbonic acid) the only thing that slows this enzyme down is the rate of diffusion, its turnover number (Kcat) is about 600,000 molecules of product per second. The other perfect enzyme Triose Phosphate Isomerase ( converts DHAP to G3P to avoid a harmful ...
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help!!!!!!!! about enzyme kineticsi've tried but really nothing worthy i lay out the fomula of V(A)andV(B) V(A)= Vmax[A]/{Km(A)+[A]} V(B)= Kcat[Et][B]/{Km(B)+[B]} as you see though [A]=[B] when i put them together it won't get me any constant ,neither to say that [Et] is unknown it seems that the solution to ...
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