Question

DISCUSSION HMW 2 Q1. Ad apted from: Keith J. Chappell, Martin J. Stoermer, David P. Fairlie, and Paul R. Young Insights Substrate Binding and Processing by West Nile Virus N$3 Protease through Combined Modeling. Protease Mutagenesis, and Kinetic St Mosquito borne flaviviruses include West Nile Virus (WNV) and Denge Viruses (Den1-4). One of the most important differences between the sequences of the proteins these two viruses produce is the amino acid identity at position 84 of the NS2B cofactor. WNV has an asparagines at position 84 and Den2 has a Serine at the same position Below are the comparison of the results of enzyme kinetics determined using the wild type and a mutant (NS2B-N84S) form of the WNV and two different substrates (Ac-LKKR-pNA and AC- LKRR-pNA). Using the given information and your general understanding of kinetic parameters state three independent conclusions you can derive from the given data. Make sure to clearly state your reasoning Ac-LKKR PNA Ac LKRR. PNA Mutant Wild type NS2B-N84S 546 ± 5.1 205 ± 16 1644 ± 0041 0682 ± 0.019 30,117 ± 2248 3333 ± 183 195± 13 100 ± 8 2494 ± 0.053 0.158 ± 0.003 12,783 ± 596 1577 ± 104 Q2. The free energy diagram of an enzyme- catalyzed reaction is shown below. Please answer the following questions about this reaction making sure to provide a brief justification for your answer using appropriate biochemical terminology ES E+P progress of reaction a. How many steps are involved in the mechanism of this reaction? b. Write down the enzymatic reaction depicted in the diagram. Make sure to assign rate c. Order the rate constants from smallest to biggest and using this list indicate which step d. If the activation energy of this step was reduced by 30 kJ/mol when the enzyme was constants for each step. in the forward pathway is the rate limiting step added to the system at 25C, calculate how much slower would this step be if the reaction was uncatalyzed. (Remember k (AeEaRT)

This is Biochemistry. I need help with Q1 and Q2. Please be detailed. Thanks

Answer 1

Question 1

K_m is the substrate concentration required to achieve half of the maximum velocity, As we see Km is less for Ac-LKKR-pNa for wild type, meaning it is a better substrate for the enzyme than aC-LKRR=pNa, whereas the opposite is true for Ns2b-N84S.

A larger Kcat means more favored product formation. For wildtype Ac-LKRR=pNa results in more product formation whereas opposite is true for Ns2b-N84S

Kcat/Km is the catalytic efficiency, meaning the ability of the enzyme to convert the bound substrate to product. For wild type Ac=LKKR-pNa has higher ratio meaning it converts a higher proportion of the substrate bound to product, i.e. it has a better enzyme for Ac=LKKR-pNa

for Ns2b-n84S as well Ac=LKKR-pNa has higher ratio meaning it converts a higher proportion of the substrate bound to product, i.e. it has a better enzyme for Ac=LKKR-pNa