Chem 453 Quiz 1 !!! Do in Pairs!
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NAMES
KEY
1) We will soon be purifying and analyzing a protein called Alkaline Phosphatase (AP)from E.
coli. Go to http://ca.expasy.org/ and find and copy the amino acid sequence of this protein (or
its precursor with signal).
sp|P00634|PPB_ECOLI Alkaline phosphatase precursor (EC 3.1.3.1) (APase) - Escherichia coli.
MKQSTIALALLPLLFTPVTKARTPEMPVLENRAAQGDITAPGGARRLTGDQTAALRDSLS
DKPAKNIILLIGDGMGDSEITAARNYAEGAGGFFKGIDALPLTGQYTHYALNKKTGKPDY
VTDSAASATAWSTGVKTYNGALGVDIHEKDHPTILEMAKAAGLATGNVSTAELQDATPAA
LVAHVTSRKCYGPSATSEKCPGNALEKGGKGSITEQLLNARADVTLGGGAKTFAETATAG
EWQGKTLREQAQARGYQLVSDAASLNSVTEANQQKPLLGLFADGNMPVRWLGPKATYHGN
IDKPAVTCTPNPQRNDSVPTLAQMTDKAIELLSKNEKGFFLQVEGASIDKQDHAANPCGQ
IGETVDLDEAVQRALEFAKKEGNTLVIVTADHAHASQIVAPDTKAPGLTQALNTKDGAVM
VMSYGNSEEDSQEHTGSQLRIAAYGPHAANVVGLTDQTDLFYTMKAALGLK
2) Go to Prot-Param database (http://ca.expasy.org/tools/protparam.html) and find the pI
(isoelectric point) of AP(without precursor signal sequence).
5.54
What charge would this protein have at pH 7.0?
Negative (above pI)
3) ) What is the the extinction coefficient estimate of AP (without signal) at 280 nm ( at the
same location) and calculate the following. Assume no half-cysteines:
 =31150 M-1 cm-1
You have purified AP and have 76.0 mL of a solution giving an absorbance of 0.87 at 280
nm. How many MOLES of AP do you have?
0.87=31150*1*c
c= 2.8 X 10-5 mol/L
2.8 X 10-5 mol/L* .076 L= 2.1 X 10-6 moles
How many GRAMS of AP do you have?
2.1 X 10-6 moles X 52745.3 g/mol = 0.110 g or 110 mg