1. iodoacetamide inactivates enzymes by covalently modifying the r-group of ㅡㅡㅡ? 2.which of the following is true in the presence of competitive inhibitors? vmax decreaseskm decreasesvmax increaseskm increasesboth km and vmax decrease3.if an enzyme is operating at very high substrate levels, and an irreversible site specific inhibitor is added, so that the reaction rate is reduced by 50%, what is true about the resulting condition of the enzyme and/or substrate? half of the substrate does not bind to the enzymethe km of the enzyme has changed by 50%all of the enzyme molecules exhibit a 50% loss of activityhalf of the enzyme molecules are completely active and half of the enzyme molecules are completely inactivethe binding affinity of the substrate is different for all of the enzyme molecules4.which of the following is not true about irreversible enzyme inhibitors? they usually covalently modify the enzymethey often carry reactive groups that combine chemically with protein r-groupsthe protein r-groups modified are often nucleophiles such as sulfhydryl or hydroxyl groups. in many cases they alter the active site of the target enzymein most cases the inhibitor can be removed by dialysis5.which of the following is true for catalysis of a reaction by an enzyme? the equilibrium constant decreasesthe activation energy increasesthe initial velocity decreasesthe standard free energy of the reaction increasesthe initial velocity increases in the presence of an enzyme6.which of the following is true? the vmax is reached when the concentration of substrate is double the kmthe km value decreases in the presence of a competitive inhibitorat high substrate levels the reaction velocity increases in a liner manner with increasing s concentrationat very low substrate levels doubling the s concentration nearly doubles the observed reaction rate so that the rate increases in a linear manner. when the substrate level is equal to the km, the reaction rate does not increase if more substrate is added

1. Iodoacetamide combines with the thiol group of cysteine, thus, the protein cannot produce a disulfide bond. So, iodoacetate is an inhibitor of cysteine peptidase.  

2. In the presence of competitive inhibition, the Km increases. In the presence of a competitive inhibitor -1/Km increases, thus, 1/Km decreases, and hence Km increases. The competitive inhibitors work by combining reversibly with the active site of an enzyme. Thus, more concentration of substrate is required to attain half of the maximum velocity.  

3. In the given case, half of the molecules of the enzyme are completely active and half of the molecules of the enzyme are completely inactive.  

Irreversible inhibition is one of the reasons for the reduction in the rate of an enzyme-catalyzed reaction in the given case. The loss of activity of the enzyme molecules due to irreversible inhibition is time dependent and it is the reason behind the reduction in reaction rate.  

Reduction in the rate of reaction is proportional to the loss of activity. Therefore, a 50 percent reduction in reaction rate will result in 50 percent of the enzyme molecules to be inactive and the other 50 percent of the enzyme molecules to be completely active.  

4. The statement, that is, in the majority of the cases, the inhibitor can be removed by dialysis is not true about irreversible enzyme inhibitors.  

The irreversible inhibitors combine with the enzymes permanently and make covalent bonds with the enzyme. Therefore, it cannot get removed by the process of dialysis in the majority of cases.  

5. The correct statement is that the initial velocity increases in the presence of an enzyme.  

6. The correct statement is that the Vmax is attained when the concentration of the substrate is double the Km.  

The Km is illustrated as the concentration of substrate at which the reaction velocity reaches half of the maximum velocity. Therefore,  

Km = 1/2 Vmax

2 Km = Vmax

Vmax = 2 Km

both male and female gametes are produced by meiosis of the germ cells. therefore, both male and female gametes are haploid. one male and one female gamete unite during the fertilization to form the diploid zygote. in seed bearing plants, pollen grain is the male gametophyte, which produces sperm cells.

i believe the answer is snake.