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

Question 4. What is the probability that a restriction enzyme will cut DNA if the recognition sequence for the enzyme is 5'-GGATCC-3' ? b) Assuming that the % GC is = 60%. O (2/10)4 (3/10)² O (3/10)4 (2/10)² O (1/6)4 O (1/4)6 O (6/10)4 (4/10)2

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Question:

Resident ER proteins destined to remain in the lumen of the ER are marked by retrieval sequences and if these proteins are exported from the ER to the Golgi, they are recognized by a recycling receptor in the ERGIC or the Golgi and selectively returned to the ER. Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b True False

practice problem genetics

Question in photo

1st answer choices: yes, no 2: yes, no 3:increase, decrease 4: a lot of mRNA or very little mRNA 5: less mRNA, more mRNA, or the same amount of mRNA

Question 1 options: The specificity pocket of the serine protease chymotrypsin, which interacts with Tyr and Phe-containing peptide sequences, contains a Ser residue. A research group is trying to modify chymotrypsin such that it has a low KM with Trp-containing peptides. Enter the name or abbreviation of an amino acid that the Ser could be mutated to that would likely have the desired effect.  (Hint: look at the diagrams of the specificity pockets shown in the course slides, and consider how the Ser would need to change to account for the difference between Tyr/Phe and Trp.)

Is the inhibitor competitive uncompetitive or non competitive? Why?