lab 5

Question 1

What reaction does acid phosphatase catalyze?

Answer 1

Hydrolysis of pNPP (p‑nitrophenyl phosphate) into pNP (p‑nitrophenol) and phosphate.

Question 2

Why does pNPP turn yellow during the reaction?

Answer 2

Produced pNP becomes yellow in alkaline conditions, allowing absorbance measurement at 405 nm.

Question 3

What does reaction velocity (v) represent in this lab?

Answer 3

The rate of pNP formation over 15 minutes, expressed as µmoles/min.

Question 4

How is the concentration of pNP determined?

Answer 4

From the absorbance using the standard curve relating A405 to pNP concentration.

Question 5

What is the purpose of the Michaelis–Menten plot?

Answer 5

To show how reaction velocity changes with substrate concentration and estimate Vmax and Km.

Question 6

What shape is the Michaelis–Menten curve?

Answer 6

A rectangular hyperbola that levels off as substrate concentration increases.

Question 7

What does Vmax represent?

Answer 7

The maximal reaction rate when the enzyme is saturated with substrate.

Question 8

What does Km represent?

Answer 8

The substrate concentration at which the reaction reaches half of Vmax.

Question 9

How is Km estimated from the Michaelis–Menten plot?

Answer 9

Find ½Vmax on the y-axis and drop a line to the curve, then down to the x-axis.

Question 10

What happens to Km and Vmax when enzyme concentration is halved?

Answer 10

Vmax is halved; Km stays the same because substrate affinity does not change.

Question 11

What does a Lineweaver–Burk plot graph?

Answer 11

1/v versus 1/[S], producing a straight-line transformation of Michaelis–Menten kinetics.

Question 12

How is Vmax determined from a Lineweaver–Burk plot?

Answer 12

Vmax = 1/(y-intercept).

Question 13

How is Km determined from a Lineweaver–Burk plot?

Answer 13

Km = -1/(x-intercept).

Question 14

What type of inhibitor is phosphate in this lab?

Answer 14

A competitive inhibitor.

Question 15

How does a competitive inhibitor affect Km and Vmax?

Answer 15

It increases Km but does not change Vmax.

Question 16

Why can competitive inhibition be overcome?

Answer 16

High substrate concentration outcompetes inhibitor for the active site.

Question 17

What type of inhibitor is NaF in this lab?

Answer 17

A non-competitive inhibitor.

Question 18

How does a non-competitive inhibitor affect Km and Vmax?

Answer 18

Vmax decreases while Km stays the same.

Question 19

Why can’t non-competitive inhibition be overcome by adding more substrate?

Answer 19

The inhibitor binds allosterically and reduces the amount of active enzyme.

Question 20

What does the Lineweaver–Burk plot look like for competitive inhibition?

Answer 20

Lines intersect on the y-axis; slope increases due to increased Km.

Question 21

What does the Lineweaver–Burk plot look like for non-competitive inhibition?

Answer 21

Lines intersect on the x-axis; Vmax decreases, Km unchanged.

Question 22

Why are Lineweaver–Burk values more accurate than conventional plots?

Answer 22

Linear transformation allows more precise determination of intercepts.

Question 23

What does an increased slope on a Lineweaver–Burk plot indicate?

Answer 23

Either increased Km (competitive inhibition) or decreased Vmax (non-competitive inhibition).

Question 24

Why must reaction rates be calculated over a fixed time?

Answer 24

To ensure comparisons are made while substrate is not depleted and enzyme remains stable.

Question 25

Why are hyperbolic curves used for enzyme kinetics?

Answer 25

They reflect the saturation of enzyme active sites as substrate concentration increases.