108 年 國立臺灣大學生物機電系碩士班丁組《熱力學與反應工程》

Biological reactions (1) __________ by enzymes are generally very (2) __________ to certain chemicals known as
the (3) __________ of the enzymes; the chemicals bind to the (4) __________ site of the enzyme to become an ES
complex with a much lower (5) __________ energy.

$E + S \xrightarrow{k_1, k_{-1}} ES \xrightarrow{k_2} E + P$

The reaction therefore can occur under mild conditions (e.g. temperature and pressure) with sufficient
efficiency. Based on Michaelis-Menten model, the reaction rate is related to the concentration of ES complex as
the following (6) __________ kinetics.

In the (7) __________ state, the concentration of ES complex can be solved as follows.
$\frac{d[ES]}{dt} = k_1[E][S] = (k_{-1} + k_2)[ES]$
$[ES] = \frac{k_1[E][S]}{k_{-1} + k_2} = \frac{[S]}{K_M + [S]} [E]_{total}$
$[ES] = \frac{[S]}{K_M + [S]} [E]_{total}$

An enzymatic reaction rate therefore increases non-linearly with the substrate concentration and reaches a
maximum as the substrate concentration being much higher than (8) __________.
$V = k_2[ES] = k_2[E]_{total} \frac{[S]}{K_M + [S]} = V_{max} \frac{[S]}{K_M + [S]}$

Since the kinetics is not linear, it will be more convenient to analyze the parameters by plotting the (9) __________
values of the reaction rate against those of the substrate concentration. The (10) __________ of the plot will be
proportional to the Michaelis constant.
$\frac{1}{V} = \frac{1}{V_{max}} \left( \frac{K_M + [S]}{[S]} \right) = \frac{1}{V_{max}} \left( \frac{K_M}{[S]} + 1 \right) = \frac{1}{V_{max}} + \frac{K_M}{V_{max}} \frac{1}{[S]}$

The following plots show the existence of (11) __________ inhibitors which will alter the apparent $K_M$ (the slope of
the plot) of the enzyme kinetics without changing the $V_{max}$. The higher the inhibitor concentration, the (12) __________
the apparent $K_M$.

🖼️【此處有附圖，請對照原卷】
(圖示為 Lineweaver-Burk plot，顯示有抑制劑存在時，曲線斜率改變但 y 軸截距 ($1/V_{max}$) 不變。有抑制劑時，曲線較 without inhibitor 的曲線更陡峭。)

According to Arrhenius expression, the rate constant ($k_2$) of an enzyme will be affected by the temperature ($T$)
and the activation energy ($E_a$). The reaction for the substrate possesses an $E_a$ (13) __________ than other chemicals,
which is the origin of the substrate (14) __________ of the enzyme.

$k_2 = Ae^{-E_a/RT}$
$V = k_2[ES] = Ae^{-E_a/RT}[ES]$
$\ln V = \ln A[ES] - \frac{E_a}{RT}$

The best way to calculate the activation energy is to multiply the slope of the semi-logarithmic plot of V
against (1/T) with the negative value of (15) __________.

食品褐變(Food browning)有化學性與酵素性反應。如蘋果接觸空氣中的氧分子,會因為組織中
Polyphenol oxidase的催化,造成Phenol類的氧化而產生黃色的Quinone,濃度一高視覺上就呈現棕色。

🖼️【此處有附圖，請對照原卷】
(圖示顯示 2,4-dihydroxyphenol 在 Polyphenol Oxidase 催化下與氧氣反應生成 ortho-quinone，進一步轉化為 Melanin/Brown Pigmentation。)

Browning catalyzed by phenol oxidation or copper ion

綠茶的澀味主成分的Epigallocatechin gallate (化學構造如下圖)即是茶品褐變的主因。

🖼️【此處有附圖，請對照原卷】
(圖示顯示 Epicatechin (left), gallic acid (middle) and epigallocatechin gallate (right) 的化學結構。)

在茶葉發酵的過程中,上述的Catechin 類化合物被從新鮮茶葉液泡中所釋放出來的Polyphenol oxidase
催化其氧化過程而轉化成 Quinone 類化合物。此時新鮮茶葉組織中的維他命C會將其還原回來。

$\text{Polyphenol} + O_2 \xrightarrow{\text{Oxidase}} \text{Quinone}$
$\text{Ascorbate (Vitamin C)} \xrightarrow{\text{Reduction}} \text{Dehydroascorbate}$

Chemical process during tea withering

日本煎茶的製程,即是在上圖的平衡中,以熱處理讓酵素失活。於是 Quinone一直保持在低含量,沒
有明顯的褐變過程,所以茶湯呈現翠綠色。烏龍茶的萎凋(發酵)過程,則是讓酵素作用,等到內含的維
他命C用完,才開始褐變。

小明分別以放置於25度與85度C保溫箱的熱水泡烏龍茶,因為酵素已於製茶過程中失活,也沒有被
萃取出來,但是由茶色發現褐變還是繼續進行。他將茶繼續放在25度與85度保溫箱中,定時採樣以分光
光度計量測褐變速度,發現85度的褐變速率為25度的3倍。小明心想,100度應該更快吧!於是用煮沸
的水萃取,但發現褐變速度反而比85度C還慢。百思不得其解後,於烏龍茶中擠入些許檸檬,發現褐變
被逆轉了。

小明越想越頭痛吃了顆普拿疼,錠劑在胃內崩解後主成分Paracetamol 於腸道以一定速度($K_{G1}$)被吸收,
於兩小時內血中達到峰值濃度20 µg/ml,頭痛獲得緩解。但是,Paracetamol於生理pH (7.45)中屬於脂溶性,
所以,肝臟進行如下圖三種類的 Biotransformation,將它轉化成水溶性物質由腎臟迅速排除。其中,速率
常數為$K_1$的代謝路徑,是誘發肝毒性的主因。

🖼️【此處有附圖，請對照原卷】
(圖示顯示 Paracetamol 經過 Glucuronidation ($K_3$)、N-Hydroxylation and Sulfation ($K_2$)、NAPQI rearrangement ($K_1$) 等代謝途徑，其中 $K_1$ 路徑產生有毒的代謝物，可能與 Toxic reactions with proteins and nucleic acids 相關。)

Biotransformation of paracetamol in liver

所以,好景不常,超過4小時後,小明頭又痛了,此時血中濃度(C)降為10 µg/ml。

🖼️【此處有附圖，請對照原卷】
(圖示為血中濃度 C(t) 隨時間 t 的曲線圖，在 t=0 時約為 0，在 t=2 時達到峰值 20 µg/ml，然後下降，在 t=4 時降至 10 µg/ml。)

Plasma concentration of paracetamol after oral administration

在請問藥學專家後,專家說普拿疼半衰期只有約2小時,你可以試試普拿疼長效錠。然後給他一個微
分方程式,說剛開始吃藥後腸道還有普拿疼時血中濃度變化如下式:
$\frac{dC}{dt} = K_{G1} - (K_1 + K_2 + K_3)C$
若腸道沒有普拿疼時:
$\frac{dC}{dt} = -(K_1 + K_2 + K_3)C$

小明百思不解,找化工系同學用 Laplace transformation 分別解得:
$C(t) = \frac{K_{G1}}{K_1 + K_2 + K_3} \left(1 - e^{-(K_1 + K_2 + K_3)t}\right) \quad (\text{剛吃藥後})$
$C(t) = C_p e^{-(K_1 + K_2 + K_3)t} \quad (\text{腸道無藥, 血中濃度過峰值 } C_p \text{ 後})$
有一天,小明頭又痛了,看了方程式心理更是霧煞煞。於是,吃了顆普拿疼,配了杯濃茶................

請定性或定量回答以下問題

(16) 如何減低蘋果的褐變過程? (6分)
(17) 如何評估日本煎茶的新鮮度? (6分)
(18) 烏龍茶的萎凋過程中不時翻動茶葉的作用是什麼? (6分)
(19) 烏龍茶有分輕度發酵的如包種、中度發酵的如四季春與高山烏龍、與重度發酵的如鐵觀音。請問如何
以化學方法區分之? (6分)
(20) 如果烏龍茶的化學性褐變速度合乎 Arrhenius expression的推測,試計算其活化能? (6分)
(21) 使用沸騰過的熱水萃取後的烏龍茶,為何褐變速度反而低於用85度C的水來萃取? (6分)
(22) 為何於茶中擠入檸檬會逆轉褐變過程? (5分)
(23) $K_{G1}, K_1, K_2, K_3$ 的單位為何? (6分)
(24) $K_1+K_2+K_3$ 約等於多少? (6分)
(25) 小明發現吃普拿疼配茶,可以讓普拿疼藥效超過5小時,請問是哪個速率常數發生怎樣的改變?試繪
出吃普拿疼配茶與配開水後的血中濃度變化曲線,並比較之。(6分)
(26) 藥劑師說阿斯匹靈傷胃、不宜配茶。普拿疼傷肝,但配茶卻可以減低普拿疼對肝臟的傷害。為什麼?
(5分)
(27) 小明改天吃了普拿疼長效錠,藥效維持12小時,請問是哪個速率常數發生怎樣的改變?試繪出
一般普拿疼與普拿疼長效錠後的血中濃度變化曲線,並比較之。(6分)