{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

BME+321+Homework+4+Solution+-st-dw-st

# BME+321+Homework+4+Solution+-st-dw-st - BME 321 Homework...

This preview shows pages 1–2. Sign up to view the full content.

BME 321 Homework 4 (7 pts) Name: 1. For a standard Michaelis-Menten reaction, k 1 = 5 x 10 7 M -1 s -1 , k -1 = 2 x 10 4 s -1 , and k 2 = 4 x 10 2 s -1 . Calculate Ks (the dissociation constant of the first step in the enzymatic reaction) and Km (the Michaelis constant) for this reaction. (0.5 pts) Does substrate binding achieve rapid equilibrium or steady state or both? (0.5 pts) Note k1 indicates forward reaction while k-1 indicates backward reaction. Ks = = 4 x 10 -4 M Km = = 4 x 10 -4 M Since Km = Ks, the substrate binding will achieve both rapid equilibrium and steady state. 2. The Km of a Michaelis-Menten enzyme for a substrate (same reaction as above) is 1.0 x 10 -4 M. At a substrate concentration of 0.2M, v = 43 µM min-1 for a certain enzyme concentration. However, with a substrate concentration of 0.02M, v has the same value. First, show that this observation is accurate using numerical calculations. (0.5 pts) Then estimate what the best range of [S] is for measuring Km. (0.5 pts) First, calculate Vmax for the reaction: v = v = 43 x 10 -6 M min -1 ; [S] = 0.2M; Km = 1.0 x 10

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

BME+321+Homework+4+Solution+-st-dw-st - BME 321 Homework...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online