HW # 2, Answers
BCH 4053.0001, Fall 2011
1. (2 pt) Dermatan sulfate was treated with dimethyl sulfate. The product was hydrolyzed under
acidic conditions. Draw the structure of the products.
2. (1 pt) Calculate T, W, and L for a 10000 base pair closed circular plasmid with 5 negative supercoils.
Assume the DNA remains entirely B-form.
T = 10000/10.4 = 961.5; W = – 5; L = T + W = 961.5 – 5 = 955.
3. (4 pts) A primary structure of tRNA(Val) is pGGU UUC GUm1G GUC AG DGG DDA UGG CA CUG
CU IAC ACG CAG AAC m7GDC m5CCC AGT CG m1AUC CUG GGC GAA AUC ACC A. Draw the
secondary structure of tRNA indicate acceptor stem, D loop, T C loop, anticodon loop, variable loop,
and CCA terminus. Draw the full chemical structures of the nucleotides m1G, m7G, m5C, I, , D.
4. (2 pt) Draw the complete chemical structure of the fragment of human serum albumin MKWVTFISLL
FLFSSAYSRG VFRRDAHKSE. Indicate bonds that are hydrolyzed by trypsin, chymotrypsin, and
endopeptidase V8. What is the maximum number of peaks that can be observed in electrospray ionization mass
spectrum of this peptide? Estimate the isolectric point of this peptide.
5. (2 pts) Calculate the percentage of oxygen-carrying capacity of hemoglobin at high altitude, where pO2 = 65
torr. Assume that pO2 in the tissue is 20 torr, p50 (hemoglobin) = 30 torr, n = 2.8. Compare the result with
hemoglobin oxygen-carrying capacity at normal conditions.
6. (3 pts) The following fractional saturation data have been measured for a certain blood sample. On a piece
of graph paper, plot out the data in a Hill plot. What are the Hill constant and the p50 of this blood sample?
pO2, torr
YO2
35
0.131
45
0.275
55
0.428
65
0.557
75
0.658
85
0.753
Calculate log (pO2) and Log (YO2/(1-YO2))
Log (pO2)
Log (YO2/(1-YO2))
1.544068044
-0.821748481
1.653212514
-0.421005313
1.740362689
-0.12595226
1.812913357
0.099451469
1.875061263
0.284199788
1.929418926
0.484098023
Plot the values:
Answer: p50 = 59.4; n = 3.45.
7. (1 pt) Ribosomal protein L32 was cleaved to fragments by two different methods A and B. After
sequencing by Edman degradation, the primary sequence of the fragments was determined (shown
below): What methods were used to fragment the protein? Determine the full sequence of L32.
Method A
1.CAEIAHNVSSKNRKAIVERAAQLAIRVW
Method B
1. AALRPLVKPKIVKKRWKKFIRHQ
2. LVHNVKELEVLLMCNKSY
3. MAALRPLVKPKIVKKRW
4. RKPRGIDNRVRRRF
5. NPNARLRSEENE
6. KGQILMPNIGY
7. KHMLPSGF
8. IRHQSDRY
9. VKIKRNW
10. GSNKKW
11. RKF
12. KKF
2.
3.
4.
5.
SDRYVKIKRNWRKPRGIDNRVRR
RFKGQILM
CNKSYCAEIAHNVSSKNRKAIVE
RAAQLAIRVWNPNARLRSEENE
LPSGFRKFLVHNVKELEVLLM
PNIGYGSNKKWKHM
M
Answer: Method A: Chymotrypsin cleavage, Method B: BrCN cleavage.
Full peptide: MAALRPLVKPKIVKKRWKKFIRHQSDRYVKIKRNWRKPRGIDNRVRRRFK
GQILM PNIGYGSN KKWKHM LPSGFRKFLVHNVKELEVLLM CNKSYCAEIAHNVSSK
NRK AIVERAAQ LAIRVWNPNARLRSEENE.
8. (3 pts) Figure 1 below shows the primary, secondary and the tertiary structures of ribonuclease A
(RNase) a) Draw the complete chemical structure of the fragment boxed in Fig. 1b. Indicate
hydrogen bonds, which are represented by the arrows in Fig. 1b; b) Find and indicate the same
fragment on Fig. 1c.
9. (2 pt) A single myosin head can generate a force of ~ 4 piconewtons (4 pN). Calculate the
efficiency of myosin engine (ratio of energy used to perform useful work to the total energy contained
in the fuel). Assume that G of the reaction ATP ADP +Pi under physiological conditions is - 50
kJ/mol.
10. (3 pts) The kinetic of an enzymatic reaction was measured as a function of substrate concentration in the
absence and in the presence of 0.1 mM inhibitors A and B.
[S], µM
Velocity (µM/minute)
No inhibitor
Inhibitor A
Inhibitor B
5
2.82
2.22
2.22
10
4.02
3.17
2.86
30
5.38
4.38
3.55
90
6.17
5.46
3.80
1/S
0.2
0.1
0.033333
0.011111
No inhibitor
0.35461
0.248756
0.185874
0.162075
Inhibitor A
0.45045
0.315457
0.228311
0.18315
Inhibitor B
0.45045
0.34965
0.28169
0.263158
a)
b) What are the Vmax and Km values in the absence of the inhibitors?
In the absence of inhibitor
Vmax _6.67 µM/min_
Km _6.9 µM_
In the presence of inhibitor A
Vmax 5.7 µM/min ___ Km (app) _8 µM ____
In the presence of inhibitor B
Vmax ____4 µM/min __
Km (app) _3.85 µM ___
b) Determine the type of inhibition: A _non-competitive (mixed)_ B _Uncompetitive_____
c) Calculate the binding constant of the inhibitor.
KI (A) __0.28 mM______ KI ‘ (A) _0.59 mM_______(if applicable)
KI (B) _------_ KI’ (B) __0.127 mM_________(if applicable)
11. (2 pt) Explain why tight binding of a substrate to an enzyme slows down an enzymatic reaction.
Answer: Tight binding corresponds to the low energy of enzyme-substrate complex (ES), which
increases the activation energy of the reaction (ΔG(2) > ΔG(1)).
12. (2 pt) Suggest inhibitors for lysozyme a) reversible b) irreversible
13. (3 pts) Derive dependence of initial velocity of the product (P) accumulation for the standard
Michaelis–Menten reaction scheme assuming that [E] >> S0.