Imagination is more important than knowledge Albert Einstein As first shown in 1993, a type of acido — Physical Chemistry — Kinetics Chemistry Question

Glucose consists of carbon, oxygen and hydrogen. As a result of its fermentation in H₂O the following gaseous (at STP) products could be theoretically formed:
1) Molecular hydrogen,
2) Various hydrocarbons,
3) Formaldehyde,
4) CO and CO₂.

Absence of C^-H bonds in C and D allows excluding variants 2 and 3 from further consideration.

Molar mass of the gas mixture is 10.55 × 2 g mol–1 = 21.1 g mol–1. It is obvious that hydrogen is one of the two gases, whereas either CO or CO₂ is the other one.
CO seems to be an improbable variant; still all the options should be checked by applying the hereunder formula for n:

[10 * M(C) + n * M(D)] / (10 + n) = 21.1

Which can be rewritten as:
n = 10 * (21.1 - M(C)) / (M(D) - 21.1)

Checking the combinations:
- If C = H₂, D = CO₂: n = 12.0
- If C = CO₂, D = H₂: n = 8.3
- If C = H₂, D = CO: n = 9.6
- If C = CO, D = H₂: n = 27.7

Since n is integer in only one case, C and D are attributed to H₂ and CO₂, respectively.

Note that bacterial cultures exist in specific, sometimes solid, nutritious media. Thus, conventional data of gases (in particular, of CO₂) solubility in water may be inapplicable.

With respect to the results in 25.1, the updated reaction (1) is rewritten as:
5 С6H₁₂O₆ + k H₂O → l A + m B + 12 H₂ + 10 CO₂

a) In the case when each of A and B is a saturated monocarboxylic acid, the equation transforms into:
5 С6H₁₂O₆ + k H₂O → l СxH2xO₂ + m СyH2yO₂ + 12 H₂ + 10 CO₂,
where x and y are the numbers of carbon and hydrogen atoms in A and B, respectively.

With account of the balance of the elements numbers, one gets the hereunder system of equations:
- Carbon (C): l·x + m·y = 20
- Hydrogen (H): 18 + k = l·x + m·y
- Oxygen (O): k = 2 l + 2 m – 10

It is seen from the first two equations that k = 2. Thus, the equation for oxygen can be rewritten as l + m = 6.

b) In the case when A is a saturated monocarboxylic and B a saturated dicarboxylic acid (reverse variant is equivalent), the equation transforms into:
5 С6H₁₂O₆ + k H₂O → l СxH2xO₂ + m СyH2y^-2O₄ + 12 H₂ + 10 CO₂

Further analysis provides an analogous system of equations:
- Carbon (C): l·x + m·y = 20
- Hydrogen (H): 18 + k = l·x + m·y – m
- Oxygen (O): k = 2 l + 4 m – 10

There is only one set of integer values corresponding to m = k = 1. Still, then l = 3.5, which is in contradiction with the conditions of the problem.

A and B with higher number of carboxylic groups (for example, two dicarboxylic acids) are impossible, as this results in negative k, l, or m.

l and m are integers, and l + m = 6. This suggests the following possible ratios: 1 : 1 (3 : 3), 1 : 2 (2 : 4) and 1 : 5. Still, l·x + m·y = 20, which makes the ratio of 1 : 1 impossible (both x and y non-integer, 20 / 3 = 6.67). Ratios of 2 : 1 and 5 : 1 are theoretically possible. Thus, the correct variants are b, e and f.

The next step is a search for integer solutions of the equation l·x + m·y = 20 for the ratios established in part 25.3.

For l = 2; m = 4:
x | y
8 | 1
6 | 2
4 | 3
2 | 4

For l = 1; m = 5:
x | y
15 | 1
10 | 2
5 | 3

Since the number of carbon atoms decreases as a result of fermentation (x < 6 and y < 6), only the variants (4, 3), (2, 4) for (l=2, m=4) and (5, 3) for (l=1, m=5) are left for consideration.
These correspond to four unbranched monocarboxylic acids:
- acetic acid (CH₃COOH)
- propanoic acid (CH₃CH₂COOH)
- butyric acid (CH₃CH₂CH₂COOH)
- valeric acid (CH₃CH₂CH₂CH₂COOH)

Further discrimination of the variants based on the available data is impossible.

For your information: A and B are acetic and butyric acids, respectively.

Since Zstart and Zfinish contain the same number of nitrogen atoms, a system of equations can be set up:
a / b = 0.12727 (for Zfinish)
a / (b + n) = 0.12069 (for Zstart)

Solving this yields:
b = 18.34 n

where a is the number of N atoms, whereas b and b+n are the total numbers of atoms in Zfinish and Zstart, respectively.

The given limitation of less than 100 atoms in each of Zstart and Zfinish can be written as n < 6. Variable b is necessarily integer, thus leading to the solely possible combination of b = 55 and n = 3. So, Zstart and Zfinish are composed of 58 and 55 atoms, respectively. This means that Zstart loses 3 atoms in acetyl^-CoA formation.

The difference in the number of hydrogen atoms in Zstart and Zfinish is:
ΔN_H = N_H(Zstart) - N_H(Zfinish) = 58 × 0.43103 - 55 × 0.41818 ≈ 2

Thereby, two of three atoms appearing in acetyl^-CoA from Zstart are hydrogen atoms. Oxygen or carbon can be the third atom lost by Zstart. In the former case, Zstart loses H₂O, and in the latter case a CH₂-group, which is formally equivalent to substituting a CH₃-group with 1 hydrogen atom.

Both variants can be rewritten in a form of equations (4) and (5):
Z^-CH₃ + CoA^-SH + E → Z^-H + CH^3-CO^-SCoA (4)
H^-Z^-OH + CoA^-SH + E → Z + CH^3-CO^-SCoA (5)

Equation (5) is invalid with any E, whereas equation (4) is correct, if E is carbon monoxide CO formed via enzymatic reduction of CO₂.

Since bacteria cultivation proceeds in the presence of isotope-labeled CO₂, the number and isotope distribution of nitrogen atoms in acetyl CoA are not influenced.
Thus, the molecular mass of acetyl^-CoA isotopologues is:
M = 100 * 14.01 * 7 / 12.08 = 811.8 g mol–1

Molecular mass of unlabeled acetyl^-CoA is 809.5. With account of rounding of nitrogen mass fractions, the difference is of 2 g/mol. Two hereunder variants are possible:
1) CO₂ labeled with ¹³C enters the reaction, thus giving acetyl residue with two ¹³C atoms;
2) CO₂ labeled with two ¹⁸O enters the reaction, thus giving acetyl residue with ¹⁸O atom.

It is impossible to distinguish between D₁ and D₂ basing on the available data:
D₁ – 13CO₂ or C₁₈O₂;
D₂ – 13CO₂ or C₁₈O₂.

The above considered acetyl^-CoA biosynthesis is referred to as the Wood^-Ljungdahl pathway.

The initial nucleotide ratio is 1:1, thus the probability of finding G or C at any position equals 1/2. Hence, the probability of any of eight possible codons is of 1/2 * 1/2 * 1/2 = 1/8. Four amino acids are each encoded by two codons composed of only G and/or C. Thus, the ratio between Pro, Arg, Gly, and Ala is 1:1:1:1. However, with account of the limited length of the oligopeptide (about 33 amino acid residues), there could be significant deviations from the above ratio. So, variant 6 is the most correct choice.

Using the table of genetic code, one can write down the nucleotide sequences of the initial and mutant mRNA fragments:
UGG^-CAU/C^-AUG^-GAA/G^-UAU/C (initial);
UGG^-ACN^-UAU/C^-GGN^-GUN (mutant).

Comparison of two sequences suggests that the mutation (insertion of A) occurred right after the first codon. Mutations influencing polypeptide biosynthesis are classified into two groups: the substitution of base pairs and the frameshift. The latter happens upon deletion or insertion of nucleotides in a number not multiple of three. Then, the initial sequence can be rewritten as:

UGGCAUAUGGAGUAU/С

If the mutant protein ends up with the 234rd amino acid residue, the biosynthesis is terminated by a STOP codon present next. Since STOP codons always start with U, the completely deciphered sequence of nucleotides is:

UGGCAUAUGGAGUAU