Question 1 2 3 4 5 6 7 8 9a 9b Total
Marks 2 7 3 8 4 4 5 4 3 4 44
Archaea (or archaebacteria) are single-celled microorganisms. They significantly differ from bacteria and eukaryotes at the molecular level.
Enzymatic reaction of methylamine with water is the major energy source for some archaea. In a particular experiment, an archaea strain was cultivated at pH 7 under anaerobic (oxygen free) conditions with the nutrient medium containing 13СH3NH2 as the only energy source. After a certain incubation period, the gas over the archaea culture was sampled and analyzed. It was found that the gas contains two substances А and B in the molar ratio of 1.00:3.00 correspondingly (the sample density rel. H2 is of 12.0).
1. Calculate the volume fractions (in %) of А and B in the mixture.
Volume ratio of gases A and B is equal to their molar ratio (1:3).
Volume fractions of A and B are 25 and 75%, respectively.
Total:
1 p 1 p 2 p 2. Determine А and B if there is no nitrogen atoms in gas collected.
Your work:
Molecular mass of the A and B mixture equals 12.02.0 = 24.0 g/mol.
The variant of two gases, both with molecular masses of 24.0 g/mol is impossible. Thus, one of the gases is lighter, whereas the other is heavier.
Reaction of 13С-methylamine with water under anaerobic conditions can
theoretically lead to two nitrogen-free gases with the molecular mass lower than 24.0 g/mol: Н2, NH3 or 13CH4. Further considerations are summed up in the table.
1 p
1 p
Light gas Volume fraction of the light gas, %
Molecular mass of the heavy gas, g/mol ammonium ion and does not transfer into gaseous phase. Thus, the only possible variant is: 13С16О2 (A) and 13C1H4 (B).
Total:
5 p 7 p 3. Write down the equation of enzymatic reaction of methylamine with water described in the above experiment using predominant form of each species.
4 13CH3NH3+ + 2H2O → 3 13CH4 + 13CО2 + 4NH4+ Total:
The correct reaction products (even in case of non-protonated)
3 p 1 p
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Methylamine and/or ammonia protonated Right coefficients:
1 p 1 p Enzymes containing the residue of α-amino acid X are found in many archaea capable of methylamine utilization. It is known that X:
is composed of atoms of 4 elements; starting lysine molecules. Different isotope-labeled L-lysines were introduced into a model system to clarify the biosynthetic pathways of X. The results are summarized in the table.
Isotope composition of L-lysine
Molecular mass (rounded to integer) of the X residue [RCH(NH2)CO], bound to
tRNA, g/mol
Normal 238
All carbons 13С, all nitrogens 15N 253
ε-Amino group with 15N 239
4. Determine the molecular formula of X.
Calculations:
Two lysines contain 12 carbons and 4 nitrogens, 16 in total.
From comparison of lines 1 and 2 of the table: 15 of 16 carbons and nitrogens are found in X.
From comparison of lines 1 and 3 of the table: 1 of 2 ε-amino nitrogens is lost during X biosynthesis.
X contains 12 carbons and 3 nitrogens.
The rest of the molecular mass: 255 – 1212 – 314 – 316 = 21 g/mol is due to X is biosynthesized in archaea according to the following scheme (E1–E3 – enzymes):
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At the first step, lysine is transformed into its structural isomer of lysine (α–amino acid, C), whereas D contains a peptide bond, and E a formyl group [
C O
H]. All reaction coefficients in the above scheme equal 1.
5. Give the chemical formula of C, D and E. From the reaction types given hereunder, choose (tick) only one corresponding to the enzyme Е3 catalyzed reaction.
Calculations. plus O. Thus, it is oxidative deamination:
R-CH2-NH2 + [O] → R-CH=O + NH3 (schematically).
Х contains the following fragment:
N with at least one hydrogen atom. Each substituent (H, Me and R) is found only once.
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6. Determine the positions of substituents H, Me, and R.
Your work:
H atom bound to the 4th or 5th C atom would mean the loss of chirality, thus it is unambiguously attributed to the 3rd C atom.
It is needed to decide about the amino group forerunning the heterocyclic nitrogen to attribute the positions of the other two substituents.
Nitrogen is included in the cycle due to the reaction of an amino and formyl groups, the latter appearing as a result of the oxidative deamination.
The size of the cycle suggests it was the α–amino group, thus:
the 3rd position – Н; the 4th position – Me; the 5th position – R. stereocenter of X with either R or S.
Moving backwards (X→D) one gets that C is (3R)-3-methyl-D-ornithine:
R N
Stereochemistry of C is derived from that of the above cyclic fragment with an account that no isomerization occurs on the way from C to X.
Both amino groups of lysine can form the peptide bond with the carboxylic group of C. Still, involvement of only the ε-amino group will provide X as α-amino acid. X is pyrrolysine, the 22nd amino acid of the genetic code:
HOOC(S) H
Penalty for each wrong or no decision about the stereocenter (R or S)
2.5 p
2.5 p
5 p -0.5 p
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Only one codon is responsible for the incorporation of X residues into proteins in archaea. The nitrogen bases forming this codon contain two exocyclic amino groups and three exocyclic oxygen atoms in total.
8. Determine the nucleotide composition of the codon by filling in the hereunder table.
Write down the number of each nitrogen base in the codon encoding X. Tick only one box in each line.
Nitrogen base The number of bases in the codon
1 2 3 0 or 1 1 or 2
A
C
G
U
A has 1 amino group and 0 oxygen atoms C has 1 amino group and 1 oxygen atom G has 1 amino group and 1 oxygen atom U has 0 amino groups and 2 oxygen atoms 2 amino groups per 3 bases suggest one U.
There are 2 amino groups and 1 oxygen atom per two bases left. A is one of these.
Either G or C is the last one.
Total:
In case of more than one box ticked in one line:
1 p residue incorporation into an archaea enzyme:
5’…AAUAGAAUUAGCGGAACAGAGGGUGAC…3’
9a. Using the table of the genetic code, decide how many amino acid residues are incorporated into the enzyme chain due to this fragment translation.
9b. Write down the amino acid sequence translated from this fragment. Note that more than one X residue is found in the fragment.
a. The fragment contains only four U, which can be used as the starting point to determine the reading frame. There should be only one A in the triplet. UGA and UAG are the options, the latter met twice. Both are STOP codons in the table. But the fragment of mRNA represents coding sequence! Within definite nucleotide motives, the STOP codons can be responsible for amino acid incorporation into proteins. Therefore, 8 amino acids encoded in the fragment (if UGA is STOP codone, then 7 amino acids residues:
…AA|UAG|AAU|UAG|CGG|AAC|AGA|GGG|UGA|C…
Number of amino acids = _8_
3 p
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b. Fill in the boxes with the amino acid abbreviations (from N- to C-terminus).
Note that the number of boxes is excessive. If there is more than one possibility, write all separated by “/”. If the translation is stopped in a particular position, write “STOP” and leave all the boxes to the right empty.
Your work:
Since only one codon is responsible for the incorporation of X residues into proteins in archaea, it is UGA or UAG. There are more than one X residue in the polypeptide fragment, thus it is UAG (met twice), while UGA encodes Sec.
(No penalty if STOP instead of Sec).
X Asn X Arg Asn Arg Gly Ses
Total:
4 p 7 p
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