Download BIOS 1700 Dr. Tanda Week 6, Session 1 1. What two substrates can

BIOS 1700
Dr. Tanda
Week 6, Session 1
1. What two substrates can bind to B-galactosidase? Which one is “real” and which one is
“fake”?
2. When lactose binds to B-galactosidase, B-galactosidase breaks lactose into:
3. IPTG is a “____________” substrate for B-galactosidase.
4. T/F IPTG is stuck with B-galactosidase forever.
5. T/F IPTG will never be free or separate from B-galactosidase
6. T/F B-galactosidase will never be free or separate from glucose and galatose
7. Is B-galactosidase and IPTG a reversible or irreversible inhibition?
8. In non-competitive inhibition, are the binding sites of the substrate and inhibitor the
same or different?
9. In competitive inhibition, the ratio of substrate to inhibitor _________.
matters
does not matter
10. In non-competitive inhibition, the ration of substrate to inhibitor _______________.
matters
does not matter
11. The number of ________ matters regardless of the number of substrates.
12. Enzymes are remarkable biological molecules that catalyze chemical reactions
without burning ourselves. The site for chemical reactions in an enzyme is called (A) site
and consists of several amino acids. In many cases, these amino acids are not next to each
other in its (B) structure, but assemble in one functional unit in its (C) structure.
Therefore, folding enzymes into the right configuration is essential.
A:
B:
C:
13.
Competitive Inhibition
Non-Competitive Inhibition
Definition
Binding Site
(Hint: same or allosteric)
Structure
______________________ to
that of substrate.
___________________ to
that of substrate.
Effect on Binding
(Hint: one blocks, one alters)
14. When Molecule 1 has more energy than Molecule 2 in a chemical reaction G1 (
G2. Therefore, this reaction can be written as follows.
)
G1 + ( ) = G2 (put either DG or -DG in parenthesis)
Or
G1 = G2 - (
) (put either DG or -DG in parenthesis)
Thus, this reaction (
) energy. (put either produces or loses in parenthesis)
15. Threonine dehydratase is the first enzyme in the process of making isoleucine from
threonine. The activity of this enzyme is regulated positively and negatively. This is a
typical example to understand how the cell (or the body) controls biological chemical
reactions. The enzyme has two (A) sites, one for threonine, and the other for isoleucine.
To make a balance of threonine and isoleucine, the enzyme becomes more active when
the cell has excess of threonine because binding of threonine to one of (A) sites
stimulates the catalytic activity of threonine dehydratase. On the other hands, when
isoleucine accumulates in the cell, the enzyme becomes less active because binding of
isoleucine to the other (A) sites inhibits the catalytic activity of this enzyme. Threonine
and isoleucine are called (B) regulators. The effect of isoleucine on threonine dehydratase
is also called (C).
A:
B:
C:
16. The table below explains how non-competitive inhibition works. Fill in the blank
cells with 0, 25, 50, 75, or 100. Assume the number of enzymes is the same in all
conditions.
Number of substrate
Number of inhibitor
Enzyme activity
molecules
molecules
100
0
100
50
50
50
50
100
0
100
100
200
100
15
100
200
50
17. The table below explains how competitive inhibition works. Fill in the blank cells
with 0, 25, 50, 75, or 100.
Number of substrate
Number of inhibitor
Enzyme activity
molecules
100
0
100
50
50
50
25
75
25
100
100
100
300
150
50