Debate and discussion of any biological questions not pertaining to a particular topic.
5 posts • Page 1 of 1
If any of these seem wrong, please do tell me with an explanation. I am still really confused with this all, so I had to guess on a lot of them. The help is greatly appreciated! Thanks!
1. For an amino acid residue in Hb to act as a proton binding site in the Bohr effect,
a) it needs to have a pKa value that is within the blood pH range and the pKa value must change
between deoxy and oxy states
b) it must be negative in charge
c) it must have a pKa value that is above 9
d) all of the above
e) “b” and “c”
2. The peptide sequence ALEPTWFEHDKFVADKFVA :
a) could not form an alpha-helix
b) could form an alpha-helix
c) could partially form an alpha-helix
d) could partially form an alpha-helix with 8 H-bonds
e) could partially form an alpha-helix with 4 H-bonds
3. Some organisms do not use BPG as the central allosteric effector that binds between the two Beta-subunits of Hb. Instead they use:
4. Some species of fish have a modified Hb, where the N-termini of all four subunits are acetylated, i.e. CH3CO-NH-CHR1CONH-CHR2CONH-etc. These fish will have:
a) the same H+ and CO2 portions of the Bohr effect as fish without the acetylation modification
b) a higher H+ portion of the Bohr effect, but no change in the CO2 portion of the Bohr effect, compared to fish without the acetylation modification
c) virtually no CO2 portion of the Bohr effect and a reduced H+ portion of the Bohr effect, compared to fish without the acetylation modification
d) virtually no H+ or CO2 portions of the Bohr effect, compared to fish without the acetylation modification
e) a decreased H+ portion of the Bohr effect, but no change in the CO2 portion of the Bohr effect, compared to fish without the acetylation modification.
5. Imagine that a strange new form of hemoglobin has been discovered. It has the same alpha- and beta-subunits that normal Hb has, but instead of being alpha2beta2 it is alpha3beta3. We would expect this new Hb to have the following number of salt bridges in the deoxy state:
a) 13 salt bridges
b) 15 salt bridges
c) 18 salt bridges
d) 20 salt bridges
e) 30 salt bridges 2 of 3
6. The elution profiles for the first three cycles of Edman chemistry in the protein sequenator for the peptide A-D-K-F-K would be:
a) bis-PTH-K in cycle 1, PTH-F in cycle 2, and bis-PTH-K in cycle 3
b) PTH-A in cycle 1, PTH-D in cycle 2, and PTH-K in cycle 3
c) PTH-A in cycle 1, PTH-D in cycle 2, and bis-PTH-K in cycle 3
d) PTH-K in cycle 1, PTH-F in cycle 2, and PTH-K in cycle 3
e) DNP-A in cycle 1, DNP-D in cycle 2, and bis-DNP-K in cycle 3
7. Amino acid analysis of one mole of a peptide yields two moles of methionine, two moles of cysteine, one mole of arginine, and one mole of alanine. Determine the sequence of this peptide from the following information: a) treatment of the intact peptide with Sanger’s reagent, followed by acid hydrolysis yields DNP-A; b) treatment of the peptide with cyanogen bromide yields a single amino acid (M) and a pentapeptide; c) treatment of the original peptide with trypsin yields a single product; d) treatment with mercaptoethanol also yields a single product; but e) treatment with trypsin followed by treatment with mercaptoethanol yields two products.
The peptide sequence is:
d) either “a” or “c”
e) none of the above
8. Secondary protein structure occurs in the interior of a globular protein because:
a) H-bonds formed between peptide backbone groups lower the energy of the protein
b) H-bonds formed between peptide R-groups lower the energy of the protein
c) ionic bonds formed between peptide R-groups lower the energy of the protein
d) “a” and “b”
e) “b” and “c”
9. Dr. Sanger could not completely remove the contaminant, dinitrophenol, from the
DNP-amino acid that he was trying to isolate. The reason he could not do this was:
a) the pKa of the OH group of dinitrophenol was very similar to the pKa of the N-terminal NH3+ group
b) the pKa of the OH group of dinitrophenol was very similar to the pKa of the COOH group of the DNP-amino acid
c) the pKa of the OH group of dinitrophenol was very similar to the pKa of the side chain R-group of the DNP-amino acid
d) the dinitrophenol contaminant was of similar size to the DNP-amino acid
e) “a” and “d”
1. __false____ The most important weak bond in 2o, 3o, and 4o protein structure is van der Waals interactions.
2. _true_____ Reverse turns are stabilized by a single ionic bond between the R-groups of the first and third amino acid residues that comprise the turn.
__false____ Proline is rarely found in an alpha-helix.
_true_____ The peptide sequence LWYFWHY might form an alpha-helix at pH 5.
_false_____ An alpha-helix in a protein that had 18 residues in it would form 6 turns and have 10 H-bonds.
__true____ Protein Z is a dimer. There is no type of protein 2o structure that could exist between the surfaces of the two subunits.
7. _false_____ The peptide sequence MEGCYALC will always be cut by chymotrypsin into two separate pieces.
8. __false____ Most protein 3o structures do not have 2o structures. This is because the interior of most proteins is polar and the polypeptide chain has no energetic reason to condense into a tightly packed form.
9. __true____ A peptide contains many different amino acid residues. It also happens to have an N-terminal arginine and an internal arginine. Sanger could have differentiated between these two arginines in the following ways: one would have been in the ether layer of a 6 N HCl/ether extraction; both would have been yellow; and detecting each on a TLC plate would be easy using ninhydrin.
__true____ In the protein sequenator, the machine that uses the Edman chemistry, cleavage of the N-terminal residue is done using an aqueous solution of 6 N HCl at room temperature. This allows the free amino acid to be separated and identified using a salt gradient on an ion exchange column.
11. _false_____ Of the negative allosteric effectors of hemoglobin, H+ is the most important and it binds to all four subunits.
12. __true____ Hemoglobin has four different histidine residues that participate, directly or indirectly, in the transfer of O2 from the lungs to the tissues. Of these, the C-terminal histidine on the alpha-subunits is the most important.
_false_____ A mutation in the N-terminal amino acid residue of the alpha-subunits of Hb would seriously affect the H+ portion of the Bohr effect.
14. _true_____ A protein was completely digested with arg-C and an internal peptide (i.e. not derived from the N- or C-terminals) was purified. Its sequence turns out to be C-K-C-K-W-M-C-K-R based on the following information: a) amino acid analysis of one mole of the peptide yields three moles C, one mole M, three moles K, one mole R, and one mole W; b) treatment of the intact peptide with trypsin gives a dipeptide containing R and W, a dipeptide containing K and C; and a tripeptide containing C, M, and K; c) treatment of the intact peptide with Edman’s reagent, followed by acid hydrolysis, yields PTH-C; and d) treatment of the intact peptide with cyanogen bromide gives a pentapeptide containing C, M, and K; and a tetrapeptide containing R, K, W, and C.
Again, explanations would be greatly appreciated!
Thanks a lot!
6 to my knowledge, Edman sequencing is from N-terminus, right? And because proteins should be written from N to C terminus, you should start with A
7 as trypsin leaves only one product, R should be at the C end or followed by P (which is not present)
3 true, P is called structure breaker
Cis or trans? That's what matters.
5 posts • Page 1 of 1
Who is online
Users browsing this forum: No registered users and 6 guests