Download Bchm 2000 Problem Set 3 Spring 2008 1. You

Bchm 2000
Problem Set 3
Spring 2008
1. You have purified a new enyzme which hydrolyzes ATP. In order to
characterize the catalytic abilities of this enzyme, you have measured the
initial velocity of ATP hydrolysis at 0.5 µM enzyme and increasing
concentrations of ATP with the following results:
[ATP], µM
v0, µM s-1
2
23
4
41
7
63
10
90
15
103
20
115
30
140
50
145
80
150
a. Create a Michaelis-Menten plot for this data set.
b. Estimate from the plot the Michaelis-Menten constant and the catalytic
constant for this enzyme.
c. Calculate the catalytic efficiency of this enzyme.
2. You are investigating an enzyme which is under allosteric control by a
small molecule A. Binding of A to the enzyme lowers both the affinity of
this enzyme for its substrate as well as its catalytic constant.
Draw a Michaelis-Menten plot with curves for the non-inhibited reaction
and the reaction in presence of the inhibitor. Label the curves and the
characteristic features of the graph.
3. The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 10-4 M. At a
substrate concentration of 0.2 M, it has been determined that v0 = 43 µM
min-1 for a certain enzyme concentration. However, with a substrate
concentration of 0.02M, v0 has the same value. How can you explain this
finding?
4. Give brief definitions or unique descriptions of the following terms:
a. Steady-State assumption
b. Michaelis-Menten equation
c. Lineweaver-Burke plot
5. The student working on the project before you, has determined the
following parameters for the enzyme you are now investigating: KM = 50
µM, kcat = 10 s-1. You don’t have enough protein and time to verify these
experiments, therefore you measure the initial velocity at 1µM enyzme at
only three different substrate concentrations: 25 µM, 50 µM and 100 µM.
What are your expectations for the initial velocity at these substrate
concentrations?
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Bchm 2000
Problem Set 3
Spring 2008
6. Nucleic Acid Structure:
a. Draw the stucture of dGMP and label the components, including the
5‘ and 3‘ carbons.
b. Draw the two Watson-Crick base-pairs that can be found in DNA.
c. The third position of the codon-anticodon interaction between tRNA
and mRNA is often a G-U base pair. Draw a plausible structure for
such a base pair.
d. Draw a pCG dinucleotide and provide the full proper name of this
dinucleotide.
e. Deoxyadenosine diphosphate
7. Give brief definitions or unique descriptions of the following terms:
a. Nucleoside
b. DNA denaturation
c. C2’ endo conformation
d. mRNA
8. Provide the missing terms in the following sentences:
a. Adenine and _____________ are derivatives of ______________,
wherease Cytidine and _______________ are derivatives of
____________.
b. According to ___________ rule, DNA has an equal amount of
adenine and ____________ and an equal amount of cytidine and
_________.
c. As detected by increased absorbance at 260 nm, the _________
point of (long) DNA increases with its ____ content.
9. Provide a list of three main differences between DNA and RNA.
10. Provide the short and the long name of the following tetranucleotide:
U
C
G
OH
A
OH
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Bchm 2000
Problem Set 3
Spring 2008
11. Answer the following questions regarding the double helical forms of
DNA:
a. What is the predominant form of double helical DNA found in the
cell?
b. Provide three characteristics which differ between A-DNA and BDNA.
c. Provide two characteristics which are the same in A-DNA and BDNA.
d. What is the name of the third type of double helical DNA next to Aand B-DNA?
12. mRNA processing:
a. List the three main events in eucaryotic mRNA processing and
describe each step in one sentence.
b. Where does eukaryotic mRNA processing take place?
Answers:
1. Michaelis-Menten plot:
a.
vmax = 160 µM s-1
175
v0, µM s
-1
150
125
100
vmax /2
75
50
25
0
0
10
20
KM =
10 µM
30
40
50
60
70
80
90 100
[ATP], µM
b. Michaelis-Menten constant: KM = 10 µM
(estimations from 5 – 15 µM would be OK)
kcat = vmax / [Etotal] = 160 µM s-1 / 0.5 µM = 320 s-1
(estimations from 150 – 180 s-1 would be OK)
c. The catalytic efficiency of this enzyme is
kcat/KM = 320 s-1 / 10 µM = 32 µM-1 s-1 = 3.2 x 107 M-1 s-1
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Bchm 2000
Problem Set 3
Spring 2008
2.
175
vmax
(- inhibitor)
v0, µM s
-1
150
- inhibitor
125
100
vmax
(+ inhibitor)
vmax /2
75
50
+ inhibitor
vmax /2
25
0
0
20
40
60
80
100
120
140
[Substrate], µM
KM (+ inhibitor)
KM (- inhibitor)
3. The investigated enzyme has a KM of 1.0 x 10-4 M = 0.0001 M. At a
substrate concentration of 0.0001 M, the observed initial velocity would be
half of the maximum velocity. The inititial velocity increases hyperbolically
with substrate concentration. Both a substrate concentration of 0.2 M as
well as of 0.02 M are significantly higher than the KM value (2000 fold and
200 fold). Thus the observed initial velocity will be in both cases close to
the maximum veloxity (since we are measuring very far on the right site of
the Michaelis-Menten plot). Probably, 43 µM min-1 is the vmax value for this
reaction.
4. a. The steady-state assumption states the the concentration of the
enzyme-substrate complex is believed to stay constant during the
measuring of an intial velocity of an enzyme-catalyzed reation.
b. The Michaelis-Menten equation is: v0 = vmax [S] / (KM + [S])
c. The Lineweaver-Burke plot is the reciprocal plot of the MichaelisMenten diagram because it plots 1/v0 versus 1/[S].
5. To calculate the expectations at various substrate concentrations, we use
the Michaelis-Menten equation: v0 = kcat [Etotal] [S] / (KM + [S]). Here, kcat =
10 s-1, [Etotal] = 1 µM and KM = 50 µM. Inserting the different values for [S]
yields the following values for v0:
for [S] = 25 µM (i.e. KM/2), v0 = 3.33 µM s-1;
for [S] = 50 µM (i.e. KM), v0 = 5 µM s-1 = vmax/2 = kcat [Etotal] / 2;
for [S] = 100 µM (i.e. 2 KM), v0 = 6.66 µM s-1
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Bchm 2000
Problem Set 3
Spring 2008
6. Nucleic Acid Structure:
a.
b.
c. G-U base pair:
O
O
N
HO
HN
HO
N O
NH O
O N
N
OH
NH2
HO
HO
OH
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Bchm 2000
Problem Set 3
Spring 2008
d. pCG dinucleotide = phosphate-5’-cytidyl-3’,5’-guanosine
NH2
N
O
O P O
O
O
N
O
O
OH
O
O P
O
N
O
NH
N
N
O
HO
NH2
OH
e. Deoxyadenosine diphosphate
H2N
O
O
HO P
O P
O
N
N
O
O
O
N
N
HO
7.
a. A nucleoside consists of a nitrogenous base covalently attached
via a β glycosidic bond to the C1’ of a five carbon sugar.
b. DNA denaturation results in disruption of the double helix and
dissociation of the strands (e.g. by heat).
c. In the C2’ endo conformation of a 5-carbon sugar ring
conformation, the C2’ is out of the plane on the sugar ring and on
the same site as the C5’ atom.
d. mRNA (or messenger RNA) is a short-lived complement of
genomic DNA which directs protein synthesis.
8.
a. Adenine and Guanine are derivatives of purines, wherease Cytidine
and thymine (or uracil) are derivatives of pyrimidines.
b. According to Chargaff’s rule, DNA has an equal amount of adenine
and thymine and an equal amount of cytidine and guanine.
c. As detected by increased absorbance at 260 nm, the melting point
of (long) DNA increases with its GC content.
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Bchm 2000
Problem Set 3
Spring 2008
9.
1. DNA contains deoxyribose, RNA ribose
2. DNA contains thymine, RNA uracil
3. DNA is double-stranded, RNA is single-stranded
10.
UdCGdAp
uridyl-3’,5’-deoxycytidyl-3’,5’-guanyl-3’,5’-deoxyadenyl-3’-phosphate
11.
Double helical structure of DNA:
a. B-DNA is the predominant form of DNA in the cell.
b. In B-DNA the sugar pucker is 2’ endo whereas it is C3’ endo in ADNA. A-DNA (2.6 nm) has a larger diameter than B-DNA (2.0 nm –
exact size not necessary) and thus A-DNA has more base-pairs per
turn than B-DNA (11.6 vs. 10.0, respectively – again numbers not
necessary). In B-DNA the major groove is wide and deep and the
minor groove is narrow and deep, whereas in A-DNA the major
groove is narrow and deep and the minor groove is wide and
shallow.
c. In both B-DNA and A-DNA the bases adopt an anti conformation
and both DNAs are right-handed.
d. The third helical form of DNA is called Z-DNA.
12.
a. Eucaryotic mRNA processing:
• In 5’capping, a 7-methylguanosine is added via a 3 phosphate
linker to the 5’ end of the mRNA.
• During 3’ polyadenylation, poly A phosphorylase adds a variable
number of A nucleotides to the mRNA.
• During splicing (intron removal), the introns of the primary transcript
are removed and the exons are joined to form the mature mRNA.
b. Eucaryotic mRNA processing takes place in the nucleus.
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