17. You work with an enzyme, studiase. Studiase catalyzes the reaction FA. Previous experiments have determined that for this reaction, Km 11.5 μM and Vmax = 625 μM/min. It is known that a molecule, PARTY, inhibits the function of studiase. In order to determine the mechanism of inhibition, you carry out the reaction FA in the presence of studiase and PARTY. Below are the results. [F] (μM) 0.16 Vo (μM/min) 3.95 0.32 7.79 0.50 12.00 0.80 18.75 1.60 35.29 BASED ON THESE RESULTS: a. (14 pts) What type of inhibitor is PARTY (Competitive, Uncompetitive, Noncompetitive, Mixed, or not an inhibitor)? DEFEND YOUR CHOICE WITH BOTH MATH AND A WRITTEN EXPLANATION! If more than = Umg + 151 11.310.16. 625-1.60 = 100 11.66 38.58. one type of inhibitor matches the data, explain your reasoning for each. To determine the tipe of Mhibitor, we need to anditze the relationship b/e km & Vat. Looking at a Umat CK. 625 m/min. Using Vo KM+13/ 9+ [F] = 0.16 cm = 625.0.16 The vois lower than 3.95, which shows that More sustrate is needed se achieve the same velocity. This suggest that the inhibitor is competitive, wow if We take a look at [F] at 1.60 wm, VO = Similarly, this also shows No is lower. this suggest that Km is higher int 13.1 the presence of the inhibitor, which is usually the case with competitive Mnbiter. As a result, base on the higher subtrate concentration reader. a cuiqve a velocity Closer to Umati It's evident that partt is acting as a competitut inhibitor, to 11.571.60 × 1000 = =76.34, b. (6 pts) Predict how the inhibitor will interact with the studiase enzyme. Will it bind to free enzyme only, enzyme bound with F only, or both? DEFEND YOUR CHOICE WITH A WRITTEN EXPLANATION! Given that party is acting as of competitue inhilltor, I'll bind to the Free Enzymy onit, this form of interaction increases the Km because the Ahibitor compete with the subtrate for the Enzyme active site. Since the inhibitor of the subtuler compete for thE SOME binding spot, the presence of the 1941 biler mace It SEAMS as if the enzime has a lower afmity for the subtrate. This can only be overcome by increasing subtracte SITE 1 awing disconcertation displacing the inhibiter at the active

Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
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I recived 16/20 on this portion of my assigment. My professor said "Right idea, use Lineweaver-Burk to calculate the new KM and Vmax and compare to the originals.  Should be noncomp (Vmax changes, but not KM).  For part b, good based on above answer."

Please help me understand this problem ebtter by re-do, re-explain, re-calculate the answer below base on the comment made. And be sure to adjust any other answers that were affected by your corrections.  For example,Part B is base on Part A answer.

17. You work with an enzyme, studiase. Studiase catalyzes the reaction FA. Previous experiments
have determined that for this reaction, Km 11.5 μM and Vmax = 625 μM/min. It is known that a
molecule, PARTY, inhibits the function of studiase. In order to determine the mechanism of inhibition,
you carry out the reaction FA in the presence of studiase and PARTY. Below are the results.
[F] (μM)
0.16
Vo (μM/min)
3.95
0.32
7.79
0.50
12.00
0.80
18.75
1.60
35.29
BASED ON THESE RESULTS:
a. (14 pts) What type of inhibitor is PARTY (Competitive, Uncompetitive, Noncompetitive, Mixed, or not
an inhibitor)? DEFEND YOUR CHOICE WITH BOTH MATH AND A WRITTEN EXPLANATION! If more than
=
Umg + 151
11.310.16.
625-1.60
= 100
11.66
38.58.
one type of inhibitor matches the data, explain your reasoning for each. To determine the tipe of
Mhibitor, we need to anditze the relationship b/e km & Vat. Looking at a Umat CK.
625 m/min. Using Vo
KM+13/ 9+ [F] = 0.16 cm = 625.0.16
The vois lower than 3.95, which shows that More sustrate is needed se
achieve the same velocity. This suggest that the inhibitor is competitive, wow if
We take a look at [F] at 1.60 wm, VO =
Similarly, this also shows No is lower. this suggest that Km is higher int
13.1
the presence of the inhibitor, which is usually the case with competitive
Mnbiter. As a result, base on the higher subtrate concentration reader.
a cuiqve a velocity Closer to Umati It's evident that partt is
acting as a competitut inhibitor,
to
11.571.60
× 1000
=
=76.34,
b. (6 pts) Predict how the inhibitor will interact with the studiase enzyme. Will it bind to free enzyme
only, enzyme bound with F only, or both? DEFEND YOUR CHOICE WITH A WRITTEN EXPLANATION!
Given that party is acting as of competitue inhilltor, I'll bind to the
Free Enzymy onit, this form of interaction increases the Km because the
Ahibitor compete with the subtrate for the Enzyme active site. Since the inhibitor
of the subtuler compete for thE SOME binding spot, the presence of the 1941 biler mace
It SEAMS as if the enzime has a lower afmity for the subtrate. This can only be
overcome by increasing subtracte
SITE 1
awing disconcertation displacing the inhibiter at the active
Transcribed Image Text:17. You work with an enzyme, studiase. Studiase catalyzes the reaction FA. Previous experiments have determined that for this reaction, Km 11.5 μM and Vmax = 625 μM/min. It is known that a molecule, PARTY, inhibits the function of studiase. In order to determine the mechanism of inhibition, you carry out the reaction FA in the presence of studiase and PARTY. Below are the results. [F] (μM) 0.16 Vo (μM/min) 3.95 0.32 7.79 0.50 12.00 0.80 18.75 1.60 35.29 BASED ON THESE RESULTS: a. (14 pts) What type of inhibitor is PARTY (Competitive, Uncompetitive, Noncompetitive, Mixed, or not an inhibitor)? DEFEND YOUR CHOICE WITH BOTH MATH AND A WRITTEN EXPLANATION! If more than = Umg + 151 11.310.16. 625-1.60 = 100 11.66 38.58. one type of inhibitor matches the data, explain your reasoning for each. To determine the tipe of Mhibitor, we need to anditze the relationship b/e km & Vat. Looking at a Umat CK. 625 m/min. Using Vo KM+13/ 9+ [F] = 0.16 cm = 625.0.16 The vois lower than 3.95, which shows that More sustrate is needed se achieve the same velocity. This suggest that the inhibitor is competitive, wow if We take a look at [F] at 1.60 wm, VO = Similarly, this also shows No is lower. this suggest that Km is higher int 13.1 the presence of the inhibitor, which is usually the case with competitive Mnbiter. As a result, base on the higher subtrate concentration reader. a cuiqve a velocity Closer to Umati It's evident that partt is acting as a competitut inhibitor, to 11.571.60 × 1000 = =76.34, b. (6 pts) Predict how the inhibitor will interact with the studiase enzyme. Will it bind to free enzyme only, enzyme bound with F only, or both? DEFEND YOUR CHOICE WITH A WRITTEN EXPLANATION! Given that party is acting as of competitue inhilltor, I'll bind to the Free Enzymy onit, this form of interaction increases the Km because the Ahibitor compete with the subtrate for the Enzyme active site. Since the inhibitor of the subtuler compete for thE SOME binding spot, the presence of the 1941 biler mace It SEAMS as if the enzime has a lower afmity for the subtrate. This can only be overcome by increasing subtracte SITE 1 awing disconcertation displacing the inhibiter at the active
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