Mother:

Cycle 1: P

Cycle 2: E (note that the peak corresponding to P decreases but do not completely disappear)

Cycle 3: E (the peak E remains very high)

Conclusion:

Mother: 2 HN- __P__ E E -COOH

Same thing for the father:

Father: 2 HN- __P__    __V__ E -COOH

For the child, the second cycle has 2 intense AA-PTH peaks: E and V suggesting that the child has 2 forms of β-hemoglobin, one from the mother (P3 contains E) and one form from the father (P3 contains V).

Therefore:

Child: 2 HN- P _E/V_ E -COOH

b) 4 pts

Peptides are the product of proteolytic digestion using trypsin.  Trypsin cleaves after R (arginine) or K (Lysine).

Therefore, the last residue of P3 is either K or R.

c) 4 pts

The disease is recessive, and only homozygous individual (that is the individual should have two mutant copies of the β-hemoglobin gene) will suffer from sickle cell anemia. Heterozygous individuals (1 copy of the mutant gene and 1 copy of the wild type gene) will be carrier of the mutation but will not suffer from sickle cell anemia.  Heterozygous individuals will synthesize two different forms of the β-hemoglobin chain (WT protein and mutant protein with 1 amino acid substitution) and would be characterized by two types of peptides 3 (WT and mutant).  It appears that the child has 2 types of peptides 3, one WT (...PEE...) inherited from her/his mother and one mutant  (...PVE)  inherited  from  her/his  father.   Therefore,  the  child  will  not  suffer  from  the disease.

d) 4 pts.

The mutant β-hemoglobin is characterized by a single amino acid substitution located in peptide 3. The heterozygous child has 2 types of peptides 3 and 6 spots (1,2,3,3*,4, &5) will be visible.

Question II (17 points)

1) 8 pts

X contains the proteins from the H region

Y contains the protein A

The 2 features that dictate this choice are:

- In size exclusion chromatography, large proteins elute before smaller one.  Therefore, protein A will elute later than the larger proteins from the H region and A should be found in the second peak.

- The height of the peak is proportional to the amount of protein found in the peak.

Based on the SDS-gel, A is more abundant than the protein from the H region and thus the peak containing A should be higher .

2) 9 pts

S1: F

Explanation: In cation exchange chromatography, the stationary phase is negatively charged and retains positively charged proteins. Therefore, protein A must be positive.

S2: T

Explanation: D represents the migration of the electrophoresis front (bromophenol blue) and d represents the migration of protein A. The ratio d/D can be used in conjunction with a standard curve to calculate the size of A. The standard curve would be built by calculating the d/D ratio for all the protein with a known size present in the ladder.

S3: F

Explanation: SDS would denature proteins and give them an elongated form. The diameter of denatured proteins  H and A would increase and as such they apparent size would  be  larger (denatured   and   elongated   proteins   occupy   a   sphere   with   a   larger   diameter   than   the corresponding native forms). If they appear larger, the peaks would be shifted to the left. It is hard to predict whether the distance between the peaks change.

Question III 12 points

1. 6 pts

Mass accuracy = (Measured mass-monoisotopic mass)/monoisotopic mass

The monoisotopic mass of the protein is 14701 Da

The m/z ratio is 14702.5. The comparison between the m/z ratio and the monoisoptic mass suggest that the ion produced in the instrument carries 1 proton.

We also know that m/z = (Measured mass +1)/1

The measured mass is m/z -1 = 14702.5 -1 = 14701.5

The mass accuracy is (14701.5-14701)/14701 = 3.4 x 10-5  = 34 ppm

2. 6 pts

Based on their position, the difference between a and b represents Δm (the peak width at half the peak height)

Therefore, Δm= 0.125 – (-0.125) = 0.25

The resolution is defined by R= m/Δm = 690.2 /0.25 = 2760.8

Question IV (34 points)

1) 4 pts.

The difference between a b and a c-ion is the additional amino group in the c-ion.

Therefore, the mass (and the m/z ratio) of a c-ions is slightly higher than the mass of a b ion, and the c3 ion will be located on the right side of b3.

2)

The difference in mass between bn  (or yn ) and bn+1  (or yn+1 ) is used to identify the residue in position n+1

Ion b1 correspond to A and is not listed in the b-series, therefore the identity of the residue in position 2 cannot be determined from the b-series.  Similarly, b8 is unknown and the last residue of the peptide cannot be determined from the b-series.

In position 6 the difference in mass between b5 and b6 correspond to the mass between Leu or Ile (isomeric residues).

a) From the b-ion series (6 pts)

2 HN-A- ? - G - P - K/G - I/L - ? - ? -COOH

Note the first? (Between A and G) can be identified as V: m/z(b1)- (m/z(A) +1)

y1 is known and correspond to arginine (R) + the expected 19Da (OH, H)

y8 is unknown, but we know from the question that the peptide starts with an A

b) From the y-serie (6 pts):

2 HN-A- ? - G - P - K/G - I/L - N - R -COOH

3) 6 pts.

b7 is an ion corresponding to b6 + the residue in position 7 (N):

Therefore m/z (b7) = 566.37 + 114.10 = 680.47

4) 6 pts.

The position of the peak labeled with a black triangle in the spectrum suggest that the corresponding m/z ratio is higher than m/z(b4) and lower than m/z (y3).

From part c) m/z(b7+) = 680.47 (+ indicates the number of associated protons).

Therefore, the mass of b7, M(b7) can be calculated from the formula:

m/z(b7+) = [M(b7) + 1]/1

that is M (b7) = m/z(b7+) – 1 = 679.47

Therefore,

m/z (b7++) = [M(b7) + 2]/2 = 340.73 (m/z ratio for b7 with 2 charges)

m/z (b7+++) = [M(b7) + 3]/3 = 227.49.

Based on the position of the peak labeled with the black triangle you know that the m/z ratio for this peak must be higher than b4 (325.19), therefore the peak labeled with the black triangle correspond to b7 with 2 associated protons.

5) 6 pts

m/z(black triangle) = 340.73 (see calculation above)

Question V (26 points)

1.

a) 4 pts

At the intercept, the net charge of the protein is 0.   Therefore, the intercept corresponds to the isoelectric point of the protein .

b) 6 pts.

Anion exchange chromatography is based on the purification of negatively charged proteins (the stationary phase is positively charged).  Therefore, to attach to the stationary phase the protein of interest must have net charge that is negative.  The isoelectric plot shows that if the buffer has a pH higher than the isoelectric point (5.66) the protein of interest is negatively charged.

c) 6 pts.

A  higher  pH  will  be  necessary  to  neutralize  the  extra  basic  (positively  charged)  residues . Therefore, the pI will be higher, and the curve will shift toward the right.

2. 10 pts

The steeper the gradient is the closer the two peaks will be.  Therefore, gradient B will reduce the resolution of the two peaks.  With gradient A, once the first protein starts to elute from the column (first peak), the salt concentration is maintained constant preventing the elution of the    second protein.  Then once the first protein is eluted, the salt concentration is increased again to trigger the elution of the 2 proteins. The resolution of the chromatography is improved by delaying the elution of the second protein .

Question VI (28 points)

1)

a) 4pts

The sequence of y5 is C-A-G-A-R-(COOH)

b) 6 pts

The m/z ratio of y5 is 103 + 71 + 57+ 71 +156 + (19) = 477

c) 4 pts

The mass of the peptide before fragmentation is

101 + 71 + 99 + 57 +103 + 71 + 57+ 71 +156 + (1 + 17) = 804 Da

m/z(6) = m/z ratio of the amino acid that occupies position 6

m/z (6) = m/z (b6) – m/z (b5) = m/z (y3)-m/z (y2)

Therefore:

m/z (y3) = m/z (b6) – m/z (b5) + m/z (y2) = 610.39 – 511.32 + 248.16 = 347.23

c) 3 pts

m/z (1) = m/z ratio of the amino acid occupying position 1

m/z (1) = m/z (y8) – m/z (y7)

m/z (1) = 857.54 – 786.50 = 71.04

d) 3 pts

m/z (b1) = m/z (1) + 1 = 71.04 + 1 = 72.04

Why adding 1?

m/z (y8) – m/z (y7) is the m/z ratio of a residue inside a polypeptide chain – [NH-CHR-CO]- see m/z ratio table in the lecture notes .

However, b1 is NH2-CHR-CO, and the extra H adds 1 to the m/z ratio calculated in part c .

e) 4 pts

m/z (2) is the m/z ratio of the amino acid occupying position 2 in the unfragmented peptide

m/z (2) = y7-y6 = 786.5-715.47= 71.03

b2 = b1 + m/z (2) = 143.03

f) 3 pts

m/z (4) is the m/z ratio of the amino acid occupying position 4 in the unfragmented peptide

m/z (4) = y5-y4 545.36 – 446.29 = 99.07

b4 = m/z (4) + b3 = 412.25

g) 4 pts

m/z (Arg) = 156.10

Therefore y1 (Arg) = 156.1 + 19 = 175.1

The 19 da comes from the fact that m/z (arg) read on the table corresponds to the m/z of – [NH- CHR-CO]- , while y1 corresponds to NH2-CHR-COOH + H+

Therefore , you need to consider the added masses of the H (1), the proton (1) and the OH (17): 17 + 1 +1 = 19

m/z (Lys) = 128.09

Therefore y1 (Lys) = 128.09 + 19 = 147.09