AP Bio help

<p>After an enzyme is mixed with its substrate, the amount of product formed is determined at 10-second
intervals for 1 minute. Data form this experiment is shown from below:</p>

<p>Time (sec) 0 10 20 30 40 50 60
Product formed (mg) 0.00 0.25 0.50 0.70 0.80 0.85 0.85</p>

<p>a. What is the initial rate of this enzymatic reaction?
b. What is the rate after 50 seconds? Why is it different from the initial rate?
c. What would be the effect on product formation if the enzyme were heated to a temperature of 100 oC
for 10 minutes before repeating the experiment? Why?</p>

<p>A and B are easy but I need to write an essay of this whole thing. How can I make my answers more detailed, especially part a and b? I just have numbers for it. Also, what is the answer to part c??? Also, do you guys have an idea on what my thesis can be (and it has to include part A, B, and C)? Thanks so much.</p>

<p>bump (10 characters)</p>

<p>oh and ignore that question for part c it’s supposed to be this: how might altering the substrate concentration affect the rate of the reaction? Why? How might altering the pH affect the reaction? Why?</p>

<p>I am not going to do the essay for you. However:</p>

<p>a. The initial rate of an enzymatic reaction is the linear slope of the reaction when the enzyme is saturated with substrate. In other words, this is when the product formed increases at a constant rate.</p>

<p>b. Let’s put it this way - the rate levels off at around 50 seconds because the enzyme has ravaged through the substrate and made most of it into product. There is little substrate left, so the reaction rate is obviously going to slow down.</p>

<p>c. Substrate concentration wil not affect the reaction rate. It will affect its duration instead. This is because at the beginning, the enzyme soution is saturated. This means that it is converting as much substrate to product as possible. Its rate cannot go any higher under those conditions. As for altering the pH, again, I’m not going to tell you, but it involves the properties of acids/bases and hydrogen bonds breaking/forming. This answer should be in your lab manual.</p>

<p>so for b the rate will be zero and it’s different from the initial because the initial had more substrate to start with and at 50 seconds, the majority of the substrate was absorbed during the reaction. and btw, thanks for the reply.</p>

<p>Correct. And no problem.</p>

<p>Time (sec) 0 10 20 30 40 50 60
Product formed (mg) 0.00 0.25 0.50 0.70 0.80 0.85 0.85</p>

<p>a. 3,128 miles per hour
b. 312.8 miles per hour. Multiply 50 x 2 / 10 = 10. Hence, you divide by 10.
c. There would be no effect, except for Chicken Little’s claims to suddenly be true.</p>