People have discovered the chemical process of metabolizing fatty acids in organisms, which is called β-oxidation. After one cycle, the number of carbons in the R group decreases, and the fatty acid is partially degraded. $PPi$ is pyrophosphate, which is quickly hydrolyzed into two molecules of phosphate. This process can be represented by the following catalytic cycle diagram:

<MultiModal>![First, [R]CCCC(O)=O (where R refers to a longer chain alkyl group) reacts with coenzyme A ($HSCoA$) under the action of $ATP$ to generate [R]CCCC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O (fatty acyl $CoA$), while $ATP$ is broken down into $AMP$ and pyrophosphate; fatty acid $CoA$ produces $\bf{1}$ and $FADH_2$ under the action of $FAD$, then $\bf{1}$ reacts with water to generate [R]CC(O)CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O, which then reacts with $NAD^+$ to generate $\bf{2}$ and $NADH$ and hydrogen ions, and $\bf{2}$ then reacts with one molecule of coenzyme A to generate acetyl $CoA$ and partially degraded fatty acid $CoA$, entering the next cycle.](/media/uploads/99e3cd03-1339-45a7-ad18-2cb1560aff29.png)</MultiModal>

Select the correct option from the following:
人们发现了生物体内代谢脂肪酸的化学过程，这一过程被称为 β-氧化反应。经过一次循环后，R基的碳数减少，脂肪酸被部分降解。$PPi$是焦磷酸，很快水解为两分子磷酸。该过程可由下面的催化循环图表示：

<MultiModal>![首先[R]CCCC(O)=O（其中R指代更长链的烷基）在$ATP$作用下与辅酶A（$HSCoA$）反应生成[R]CCCC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O（脂肪酰$CoA$），同时$ATP$被分解为$AMP$和焦磷酸；脂肪酸$CoA$在$FAD$作用下产生$\bf{1}$和$FADH_2$，随后$\bf{1}$与水反应生成[R]CC(O)CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O，进而与$NAD^+$反应生成$\bf{2}$和$NADH$与氢离子，$\bf{2}$再与一分子辅酶A反应生成乙酰$CoA$和部分降解的脂肪酸$CoA$，进入下一步循环。](/media/uploads/99e3cd03-1339-45a7-ad18-2cb1560aff29.png)</MultiModal>

选出以下选项中正确的一项：
All other options are incorrect
Undecanoic acid and stearic acid produce the same product species when directly degraded in the cycle.
CCCCCCCC(C)(C)CC(O)=O can be directly metabolized through the above pathway.
For $\bf{1}$ produced by saturated fatty acids, there are no carbon-carbon double bonds conjugated with the carbonyl group.
If $\bf{2}$ is produced by stearic acid in the first cycle, then its chemical formula under physiological conditions is $\mathrm{C_{39}H_{64}N_7O_{18}P_3S^{4-}}$
$1$ $\mathrm{mol}$ of $\mathrm{FADH_2}$ and $\mathrm{NADH}$ are re-oxidized during biological oxidation to release energy, generating $1.5$ $\mathrm{mol}$ and $2.5$ $\mathrm{mol}$ of $ATP$, respectively. $\mathrm{CH_{3}COSCoA}$ is oxidized to $\mathrm{CO_2}$ in the tricarboxylic acid cycle, simultaneously generating $10$ $\mathrm{mol}$ of $\mathrm{ATP}$. Given that the molar heat of combustion of palmitic acid is $9800$ $\mathrm{kJ·mol^{-1}}$ and the energy storage of each high-energy phosphate bond is $30.5$ $\mathrm{kJ·mol^{-1}}$, the energy storage efficiency of the organism during the complete oxidation of palmitic acid is $33.6\%$
Fatty acyl-CoA reacts with protein lysine residues to form fatty acylation modifications, which always require enzyme participation.
其他选项均不正确
十一烷酸和硬脂酸在循环中降解直接产生的产物物种相同
CCCCCCCCC(C)(C)CC(O)=O可直接通过上面的途径被代谢
对于饱和脂肪酸产生的$\bf{1}$，其中不存在与羰基共轭的碳碳双键
假如$\bf{2}$是由硬脂酸在第一次循环中产生的，则其在生理条件下的化学式为$\mathrm{C_{39}H_{64}N_7O_{18}P_3S^{4-}}$
$1$ $\mathrm{mol}$的$\mathrm{FADH_2}$ 和$\mathrm{NADH}$在生物氧化过程中重新被氧化放出能量，分别生成 $1.5$ $\mathrm{mol}$和$2.5$ $\mathrm{mol}$的$ATP$。 $\mathrm{CH_{3}COSCoA}$ 在三羧酸循环中被氧化为 $\mathrm{CO_2}$ ，同时生成$10$ $\mathrm{mol}$的$\mathrm{ATP}$，已知软脂酸的摩尔燃烧热为 $9800$ $\mathrm{kJ·mol^{-1}}$ ，每根高能磷酸键的储能为$30.5$ $\mathrm{kJ·mol^{-1}}$，则软脂酸完全氧化过程中生命体的储能效率为$33.6\%$
脂肪酰$\mathrm{CoA}$与蛋白赖氨酸残基反应形成脂肪酰化修饰一定需要酶的参与
The oxidation cycle is analyzed. First, fatty acids react with coenzyme A to generate fatty acyl $CoA$. Under the action of $FAD$, $\bf{1}$ is produced, losing a molecule of hydrogen. $\bf{1}$ can react with water to generate [R]CC(O)CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O. It is not difficult to deduce that an α, β-unsaturated carbonyl compound is produced. Then the structure of $\bf{1}$ is [R]C/C=C/C(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O, which has a carbon-carbon double bond conjugated with the carbonyl group.

<Checkpoint>[The structure of $\bf{1}$ is [R]C/C=C/C(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O, which has a carbon-carbon double bond conjugated with the carbonyl group. Option D is incorrect.](1)</Checkpoint>

[R]CC(O)CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O reacts with $\mathrm{NAD^+}$ to produce $\bf{2}$ and $\mathrm{NADH}$ and hydrogen ions. $\bf{2}$ then reacts with a molecule of coenzyme A to generate acetyl $\mathrm{CoA}$ and partially degraded fatty acid $\mathrm{CoA}$. $\bf{2}$ is produced by a one-step oxidation, and it is easy to deduce that its structure is [R]CC(CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O)=O.

<Checkpoint>[The structure of $\bf{2}$ is [R]CC(CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O)=O](1)</Checkpoint>

The chemical formula of stearic acid is $\mathrm{{C_{18}H_{36}O_2}}$. Under physiological conditions, all phosphates of coenzyme A are ionized, and the chemical formula is $\mathrm{C_{21}H_{32}N_7O_{16}P_3S^{4-}}$. Then the chemical formula of $\bf{2}$ produced by stearic acid in the first cycle is $\mathrm{C_{39}H_{64}N_7O_{18}P_3S^{4-}}$.

<Checkpoint>[The chemical formula of $\bf{2}$ produced by stearic acid in the first cycle is $\mathrm{C_{39}H_{64}N_7O_{18}P_3S^{4-}}$. Option E is correct.](1)</Checkpoint>

Stearic acid has 18 carbons. Each β-degradation reduces two carbons, producing $\mathrm{FADH_2}$, $\mathrm{NAHD}$, and acetyl $\mathrm{CoA}$. In the last step of degradation, stearic acid will be completely degraded to acetyl $\mathrm{CoA}$. Undecanoic acid has 11 carbons. In the penultimate cycle, three carbons will remain, and it cannot be completely degraded to acetyl $\mathrm{CoA}$ (propionyl $\mathrm{CoA}$ will actually be produced, which is converted to succinyl $\mathrm{CoA}$ and then enters the citric acid cycle for degradation), and the types of degradation products produced are different.

<Checkpoint>[Stearic acid is degraded to acetyl $\mathrm{CoA}$. Undecanoic acid is degraded to propionyl $\mathrm{CoA}$, and the degradation products are different. Option B is incorrect.](1)</Checkpoint>

β-oxidation requires hydrogen at the β-position of the fatty acid. CCCCCCCCC(C)(C)CC(O)=O cannot undergo β-oxidation.

<Checkpoint>[CCCCCCCCC(C)(C)CC(O)=O cannot undergo β-oxidation. Option C is incorrect.](1)</Checkpoint>

Fatty acyl $\mathrm{CoA}$ has a highly active thioester structure, which can modify lysine residues without enzymatic catalysis in some scenarios.

<Checkpoint>[Fatty acyl $\mathrm{CoA}$ has a highly active thioester structure, which can modify lysine residues without enzymatic catalysis in some scenarios. Option G is incorrect.](1)</Checkpoint>

Palmitic acid is a saturated fatty acid with 16 carbons. Complete degradation will produce 8 acetyl $\mathrm{CoA}$, 7 $\mathrm{FADH_2}$, and 7 $\mathrm{NADH}$, which can generate a total of $7 × (10 + 1.5 + 2.5) + 1.5 + 2.5 = 108$ $\mathrm{ATP}$. The reaction of palmitic acid with coenzyme A requires a molecule of $\mathrm{ATP}$ to be degraded into $\mathrm{AMP}$. $\mathrm{AMP}$ will react with another molecule of $\mathrm{ATP}$ to generate $\mathrm{ADP}$, and pyrophosphate will quickly hydrolyze, consuming two high-energy phosphate bonds. The actual yield is 106 $\mathrm{ATP}$, so the energy storage efficiency is: $ 106 × 30.5 / 9800 = 33.0 \%$

<Checkpoint>[Complete degradation of palmitic acid produces 108 $\mathrm{ATP}$, and the reaction of palmitic acid with coenzyme A consumes two high-energy phosphate bonds, and the actual yield is 106 $\mathrm{ATP}$, so the energy storage efficiency is $ 33.0 \% $. Option F is incorrect.](1)</Checkpoint>

Therefore, choose option E.
对氧化循环进行分析，首先脂肪酸与辅酶A反应生成脂肪酰$CoA$，在$FAD$作用下产生$\bf{1}$，失去一分子氢气，$\bf{1}$可与水反应生成[R]CC(O)CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O，不难推断出产生了α, β-不饱和羰基化合物，则$\bf{1}$的结构为[R]C/C=C/C(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O，其中存在与羰基共轭的碳碳双键。

<Checkpoint>[$\bf{1}$的结构为[R]C/C=C/C(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O，其中存在与羰基共轭的碳碳双键，选项D错误](1)</Checkpoint>

[R]CC(O)CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O与$\mathrm{NAD^+}$反应生成$\bf{2}$和$\mathrm{NADH}$与氢离子，$\bf{2}$再与一分子辅酶A反应生成乙酰$\mathrm{CoA}$和部分降解的脂肪酸$\mathrm{CoA}$，$\bf{2}$经一步氧化产生，很容易推测得其结构为[R]CC(CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O)=O。

<Checkpoint>[$\bf{2}$的结构为[R]CC(CC(SCCNC(CCNC(C(O)C(COP(OP(OCC(C(OP([O-])([O-])=O)C1O)OC1N(C=N2)C3=C2C(N)=NC=N3)([O-])=O)([O-])=O)(C)C)=O)=O)=O)=O](1)</Checkpoint>

硬脂酸化学式为$\mathrm{{C_{18}H_{36}O_2}}$，辅酶A在生理条件下所有磷酸均电离，化学式为$\mathrm{C_{21}H_{32}N_7O_{16}P_3S^{4-}}$，则由硬脂酸在第一次循环中产生的$\bf{2}$化学式为$\mathrm{C_{39}H_{64}N_7O_{18}P_3S^{4-}}$

<Checkpoint>[硬脂酸在第一次循环中产生的$\bf{2}$化学式为$\mathrm{C_{39}H_{64}N_7O_{18}P_3S^{4-}}$，选项E正确](1)</Checkpoint>

硬脂酸有18个碳，每次β降解会减少两个碳，产生$\mathrm{FADH_2}$、$\mathrm{NAHD}$和乙酰$\mathrm{CoA}$，在最后一步降解中硬脂酸会被完全降解为乙酰$\mathrm{CoA}$。十一烷酸有11个碳，在倒数第二个循环会剩余三个碳，最终无法被完全降解为乙酰$\mathrm{CoA}$（实际会产生丙酰$\mathrm{CoA}$，转化为琥珀酰$\mathrm{CoA}$后进入柠檬酸循环被降解），产生的降解产物种类不同。

<Checkpoint>[硬脂酸降解为乙酰$\mathrm{CoA}$。十一烷酸降解为丙酰$\mathrm{CoA}$，降解产物不同，选项B错误](1)</Checkpoint>

β氧化需要脂肪酸β位有氢，CCCCCCCCC(C)(C)CC(O)=O无法进行β氧化。

<Checkpoint>[CCCCCCCCC(C)(C)CC(O)=O无法进行β氧化，选项C错误](1)</Checkpoint>

脂肪酰$\mathrm{CoA}$具有高活性的硫酯结构，在一些情景下无需酶促即可对赖氨酸残基进行修饰。

<Checkpoint>[脂肪酰$\mathrm{CoA}$具有高活性的硫酯结构，在一些情景下无需酶促即可对赖氨酸残基进行修饰，选项G错误](1)</Checkpoint>

软脂酸是具有16个碳的饱和脂肪酸，完全降解会产生8个乙酰$\mathrm{CoA}$、7个$\mathrm{FADH_2}$和7个$\mathrm{NADH}$，共可产生$7 × (10 + 1.5 + 2.5) + 1.5 + 2.5 = 108$个$\mathrm{ATP}$，而软脂酸与辅酶A反应需要一分子$\mathrm{ATP}$降解为$\mathrm{AMP}$，$\mathrm{AMP}$会与另一分子$\mathrm{ATP}$反应生成$\mathrm{ADP}$，焦磷酸很快水解，消耗两根高能磷酸键，实际生成106个$\mathrm{ATP}$，则储能效率为：$ 106 × 30.5 / 9800 = 33.0 \%$

<Checkpoint>[软脂酸完全降解产生108个$\mathrm{ATP}$，而软脂酸与辅酶A反应消耗两根高能磷酸键，实际生成106个$\mathrm{ATP}$，则储能效率为$ 33.0 \% $，选项F错误](1)</Checkpoint>

故选选项E