112 年 國立臺灣大學生物機電系碩士班丙組《熱力學與反應工程》

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第 1 題10 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.

  1. Describe the meanings and dimensions of r, A, E and T in Arrhenius equation (10%).

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這一題的完整詳解

此題主要考查 Arrhenius 方程式的參數意義與單位。Arrhenius 方程式描述了化學反應速率常數與溫度的關係。

Arrhenius 方程式為：
k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}
在此題目中，將反應速率 rr 直接寫成 r=Ae−ERTr = Ae^{-\frac{E}{RT}}，其中 rr 是反應速率，其單位通常為濃度/時間，例如 mM⋅min−1mM \cdot min^{-1}。

  1. rr (Reaction Rate):

    • 意義: 表示反應進行的速度，即單位時間內生成物濃度增加的量，或反應物濃度減少的量。
    • 單位: 題目中給出的例子是 0.001 mM⋅min−10.001 \, mM \cdot min^{-1} 和 10 mM⋅min−110 \, mM \cdot min^{-1}，所以單位是濃度/時間，例如 M⋅s−1M \cdot s^{-1}、mM⋅min−1mM \cdot min^{-1} 等。

  2. AA (Pre-exponential Factor or Frequency Factor):

    • 意義: 表示當分子具有足夠的能量（活化能）時，碰撞的頻率。它與反應物分子的特性、碰撞的取向等有關。
    • 單位: AA 的單位必須與反應速率 rr 的單位相同，以確保方程式的單位一致性。
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第 2 題5 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
2) The reaction is said to be pseudo-first order, why? (5%)

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這一題的完整詳解

此題主要考查「假一級反應」(Pseudo-first order reaction) 的概念。

一個化學反應的級數是根據反應物濃度的冪次來定義的。例如，對於反應 aA+bB→PaA + bB \rightarrow P，其速率定律可能寫成 r=k[A]m[B]nr = k[A]^m[B]^n，其中 mm 和 nn 分別是 AA 和 BB 的反應級數，總反應級數為 m+nm+n。

當一個反應的級數很高，或者其中一個反應物的濃度遠大於其他反應物時，我們可以將其視為假一級反應。在這種情況下，濃度佔主導地位的反應物的濃度變化對反應速率的影響可以忽略不計，或者其濃度可以被視為常數。

題目中提到，反應速率 rr 由 Arrhenius 方程式預測：
r=Ae−ERTr = Ae^{-\frac{E}{RT}}
這個方程式本身並沒有明確顯示反應物濃度，它描述的是一個「無催化」的未指定反應的速率。

然而，接下來的敘述中提到：
「Under sufficient supply of NAD⁺, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation...」
這裡的關鍵資訊是「Under sufficient supply of NAD⁺」。

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第 3 題15 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
3) Calculate the approximate value of E for the uncatalyzed reaction. (15%)

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這一題的完整詳解

此題要求計算未催化反應的活化能 EE。我們利用題目提供的反應速率與溫度關係來計算。

已知條件：

  1. 反應速率在 27∘C27^\circ C 時，每升高 10∘C10^\circ C 就會加倍。
  2. 在體溫 37∘C37^\circ C 時，反應速率為 0.001 mM⋅min−10.001 \, mM \cdot min^{-1}。
  3. Arrhenius 方程式：r=Ae−ERTr = Ae^{-\frac{E}{RT}}。

首先，將溫度從攝氏轉換為克耳文：
T1=27∘C=27+273.15=300.15 KT_1 = 27^\circ C = 27 + 273.15 = 300.15 \, K
T2=37∘C=37+273.15=310.15 KT_2 = 37^\circ C = 37 + 273.15 = 310.15 \, K
為了簡化計算，我們可以使用 T1=27+273=300 KT_1 = 27 + 273 = 300 \, K 和 T2=37+273=310 KT_2 = 37 + 273 = 310 \, K。

設在 T1=300 KT_1 = 300 \, K 時的反應速率為 r1r_1，在 T2=310 KT_2 = 310 \, K 時的反應速率為 r2r_2。
題目敘述「在 27°C 時，反應速率在每升高 10°C 就會加倍」。這意味著：
如果我們從 27∘C27^\circ C (300 K) 開始，升高 10∘C10^\circ C 到 37∘C37^\circ C (310 K)，反應速率會加倍。
所以，r2=2r1r_2 = 2 r_1。

我們有兩個溫度點：
點 1: T1=300 KT_1 = 300 \, K
點 2: T2=310 KT_2 = 310 \, K

令 r(T)r(T) 為溫度 TT 下的反應速率。
根據題目：「Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature」。
這表示：
r(37∘C)=2×r(27∘C)r(37^\circ C) = 2 \times r(27^\circ C)
r2=2r1r_2 = 2 r_1

我們又知道在體溫 37∘C37^\circ C 時的速率是 0.001 mM⋅min−10.001 \, mM \cdot min^{-1}。
所以，r2=0.001 mM⋅min−1r_2 = 0.001 \, mM \cdot min^{-1}。
因此，r1=r22=0.001 mM⋅min−12=0.0005 mM⋅min−1r_1 = \frac{r_2}{2} = \frac{0.001 \, mM \cdot min^{-1}}{2} = 0.0005 \, mM \cdot min^{-1}。

現在我們有兩個速率和溫度點：
(T1,r1)=(300 K,0.0005 mM⋅min−1)(T_1, r_1) = (300 \, K, 0.0005 \, mM \cdot min^{-1})
(T2,r2)=(310 K,0.001 mM⋅min−1)(T_2, r_2) = (310 \, K, 0.001 \, mM \cdot min^{-1})

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第 4 題10 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
4) Calculate the approximate value of E' for the enzymatic reaction. (10%)

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這一題的完整詳解

核心觀念

本題利用 Arrhenius 方程式比較未催化反應與酵素催化反應的速率：

r=Ae−E/(RT)r=Ae^{-E/(RT)}

在相同溫度且假設前因子 AA 近似不變時，催化劑使活化能由 EE 降低為 E′E'，因此：

renzymeruncatalyzed=Ae−E′/(RT)Ae−E/(RT)=e(E−E′)/(RT)\frac{r_{\text{enzyme}}}{r_{\text{uncatalyzed}}} = \frac{Ae^{-E'/(RT)}}{Ae^{-E/(RT)}} = e^{(E-E')/(RT)}

題目給定飲酒後乙醇濃度約為 20 mM20\ \mathrm{mM}，且酵素反應速率可達：

Vmax⁡=10 mM min−1V_{\max}=10\ \mathrm{mM\,min^{-1}}

未催化反應在體溫 37∘C37^\circ\mathrm{C} 時速率為：

r=0.001 mM min−1r=0.001\ \mathrm{mM\,min^{-1}}

解題方法

1. 先由「升高 10∘C10^\circ\mathrm{C}，速率加倍」求未催化反應的 EE

取：

T1=27∘C=300 KT_1=27^\circ\mathrm{C}=300\ \mathrm{K} T2=37∘C=310 KT_2=37^\circ\mathrm{C}=310\ \mathrm{K}

因為溫度升高 10∘C10^\circ\mathrm{C} 後速率加倍：

r2r1=2\frac{r_2}{r_1}=2

由 Arrhenius 方程式取比值：

ln⁡(r2r1)=ER(1T1−1T2)\ln\left(\frac{r_2}{r_1}\right) = \frac{E}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)

因此：

E=Rln⁡2(1300−1310)E= \frac{R\ln 2} {\left(\frac{1}{300}-\frac{1}{310}\right)}

代入 R=8.3 J mol−1K−1R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}、ln⁡2=0.7\ln 2=0.7：

E=8.3(0.7)(1300−1310)≈5.4×104 J mol−1E= \frac{8.3(0.7)} {\left(\frac{1}{300}-\frac{1}{310}\right)} \approx 5.4\times 10^4\ \mathrm{J\,mol^{-1}}

所以：

E≈54 kJ mol−1E\approx 54\ \mathrm{kJ\,mol^{-1}}

2. 判斷酵素反應速率

Michaelis–Menten 方程式為：

v=Vmax⁡[S][S]+KMv=V_{\max}\frac{[S]}{[S]+K_M}

代入 [S]=20 mM[S]=20\ \mathrm{mM}、KM=0.5 mMK_M=0.5\ \mathrm{mM}：

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第 5 題10 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
5) Calculate the physiological alcohol dehydrogenase level in terms of the international unit (IU)of
enzyme (µmol/min). (10%)

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這一題的完整詳解

核心觀念

本題利用 Michaelis–Menten 動力學與酵素活性的定義。

  • Michaelis–Menten 方程式：

    v=Vmax⁡[S][S]+KMv=V_{\max}\frac{[S]}{[S]+K_M}

  • 題目給定初始血液乙醇濃度：

    [S]0=20 mM[S]_0=20\ \mathrm{mM}

  • 酒精去氫酶的 Michaelis 常數：

    KM=0.5 mMK_M=0.5\ \mathrm{mM}

  • 人體內最大氧化速率：

    v=10 mM min−1v=10\ \mathrm{mM\,min^{-1}}

  • 1 IU1\ \mathrm{IU} 的定義為每分鐘催化生成 1 μmol1\ \mu\mathrm{mol} 產物的酵素活性。

由於題目給定的是血液中「濃度隨時間的變化率」，必須乘上血液體積，才能換算成總酵素活性。

解題方法

剛飲酒時，乙醇濃度為 20 mM20\ \mathrm{mM}。依 Michaelis–Menten 方程式：

v=Vmax⁡2020+0.5v=V_{\max}\frac{20}{20+0.5} v=Vmax⁡2020.5=0.98Vmax⁡v=V_{\max}\frac{20}{20.5} =0.98V_{\max}

因此，初始乙醇濃度已遠大於 KMK_M：

[S]KM=200.5=40\frac{[S]}{K_M}=\frac{20}{0.5}=40

反應速率已達最大速率的約 98%98\%，可視為接近 Vmax⁡V_{\max}。題目直接給出人體內最大反應速率為 10 mM min−110\ \mathrm{mM\,min^{-1}}，故：

Vmax⁡≈10 mM min−1V_{\max}\approx 10\ \mathrm{mM\,min^{-1}}

若採一般成人血液體積約 5.0 L5.0\ \mathrm{L}：

5.0 L=5.0×103 mL5.0\ \mathrm{L}=5.0\times10^3\ \mathrm{mL}

因為：

1 mM=1 μmol mL−11\ \mathrm{mM}=1\ \mu\mathrm{mol\,mL^{-1}}

所以總反應速率為：

10 mM min−1×5.0×103 mL10\ \mathrm{mM\,min^{-1}} \times 5.0\times10^3\ \mathrm{mL}
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第 6 題10 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
6) Under what conditions does the assumption, K=KMK = K_M, hold? (10%)

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這一題的完整詳解

核心觀念

本題要求判斷 Hill–Langmuir 式中的結合常數 KK 何時可以視為 Michaelis–Menten 常數 KMK_M。兩者概念不同：

  • KMK_M 是 Michaelis–Menten 動力學中的速率參數。
  • KK 是蛋白質與配體達到結合平衡時的解離常數。
  • 只有在特定近似條件下，KMK_M 才能代表真正的解離常數。

解題方法與推導

對單一受質、單一活性位點的反應，可寫成：

E+S⇌k−1k1ES⟶kcatE+PE+S \mathop{\rightleftharpoons}^{k_1}_{k_{-1}} ES \mathop{\longrightarrow}^{k_{\mathrm{cat}}} E+P

Michaelis–Menten 常數定義為：

KM=k−1+kcatk1K_M=\frac{k_{-1}+k_{\mathrm{cat}}}{k_1}

其中：

  • k1k_1：酵素與受質結合速率常數
  • k−1k_{-1}：酵素–受質複合物解離速率常數
  • kcatk_{\mathrm{cat}}：複合物進行催化反應的速率常數

若酵素與受質的結合、解離速度遠快於催化反應，即：

kcat≪k−1k_{\mathrm{cat}}\ll k_{-1}

則：

KM=k−1+kcatk1≈k−1k1=KdK_M =\frac{k_{-1}+k_{\mathrm{cat}}}{k_1} \approx \frac{k_{-1}}{k_1} =K_d

而 k−1k1\frac{k_{-1}}{k_1} 正是真正的解離常數。因此，在「快速結合平衡」或「快速平衡近似」下，才可令：

K=Kd≈KMK=K_d\approx K_M

此時 Hill–Langmuir 式中的 KK 才能以 KMK_M 作為近似值。

套用至本題的條件

令 K=KMK=K_M 必須滿足以下條件：

  1. 酵素–受質結合反應接近平衡

    E+S⇌ESE+S\rightleftharpoons ES

    的結合與解離必須比後續催化步驟快很多。

  2. 催化步驟相對緩慢

    kcat≪k−1k_{\mathrm{cat}}\ll k_{-1}

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後續完整解題步驟與【答案】

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第 7 題20 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
7) Describe the way to obtain the Hill plot of the enzymatic reaction and retrieve useful information.
(Hint: 0 = v/v Max) (20%)

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這一題的完整詳解

此題要求描述如何繪製 Hill 曲線圖（Hill plot）以及如何從中提取有用的資訊。題目提示 θ=v/vMax\theta = v/v_{Max}。

Hill 曲線圖是利用 Hill 方程式來分析酶動力學或配體結合數據的一種圖形方法。Hill 方程式描述的是多亞基蛋白質與配體結合的協同效應。

1. 繪製 Hill 曲線圖的步驟：

  • 數據收集: 需要進行實驗，測量不同配體濃度 [L][L] 下，蛋白質的佔有率 θ\theta。在酶動力學的背景下，配體 L 就是底物 S，蛋白質 P 就是酶 E。題目中的 θ\theta 被定義為 v/vMaxv/v_{Max}，這通常用於描述酶與底物的結合程度。
    所以，我們需要測量在不同底物濃度 [S][S] 下的反應速率 vv，並計算 θ=v/vMax\theta = v/v_{Max}。
    已知 Michaelis-Menten 方程式為 v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}。
    則 θ=vvMax=[S][S]+KM\theta = \frac{v}{v_{Max}} = \frac{[S]}{[S]+K_M}。

  • 轉換為 Hill 方程式: 題目給出了 Hill-Langmuir 方程式的對數形式：
    log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
    將 L 替換為 S（底物），將 θ\theta 替換為 v/vMaxv/v_{Max}：
    log⁡(v/vMax1−v/vMax)=nlog⁡([S])−log⁡(K)\log\left(\frac{v/v_{Max}}{1-v/v_{Max}}\right) = n\log([S]) - \log(K)
    log⁡(vvMax−v)=nlog⁡([S])−log⁡(K)\log\left(\frac{v}{v_{Max}-v}\right) = n\log([S]) - \log(K)

  • 繪製圖形:

    • 橫軸：log⁡([S])\log([S]) (底物濃度的對數)
    • 縱軸：log⁡(vvMax−v)\log\left(\frac{v}{v_{Max}-v}\right) (稱為 Hill 座標)

    我們需要從實驗數據中，針對每一個數據點 ([S],v)([S], v)，計算出對應的 log⁡([S])\log([S]) 和 log⁡(vvMax−v)\log\left(\frac{v}{v_{Max}-v}\right) 值。然後將這些點繪製在圖上。

2. 從 Hill 曲線圖提取有用資訊：

  • Hill 係數 (n):
    Hill 曲線圖的斜率就是 Hill 係數 nn。
    Slope=n=Δ(log⁡(vvMax−v))Δ(log⁡([S]))\text{Slope} = n = \frac{\Delta\left(\log\left(\frac{v}{v_{Max}-v}\right)\right)}{\Delta(\log([S]))}

    • 如果 n=1n = 1，表示反應遵循標準的 Michaelis-Menten 動力學，沒有協同效應。
    • 如果 n>1n > 1，表示存在正協同效應 (positive cooperativity)，即一個底物分子與酶結合後，會增加酶與其他底物分子結合的親和力。
    • 如果 0<n<10 < n < 1，表示存在負協同效應 (negative cooperativity)，即一個底物分子與酶結合後，會降低酶與其他底物分子結合的親和力。
      題目中提到，當 n>1n>1 時，蛋白質結合現象是「positive」cooperative。

  • 結合常數 (K):
    當 log⁡(vvMax−v)=0\log\left(\frac{v}{v_{Max}-v}\right) = 0 時，即 vvMax−v=1\frac{v}{v_{Max}-v} = 1，則 v=vMax−vv = v_{Max}-v，所以 2v=vMax2v = v_{Max}，即 v=vMax/2v = v_{Max}/2。
    此時，Hill 方程式變成：
    0=nlog⁡([S])−log⁡(K)0 = n\log([S]) - \log(K)
    log⁡(K)=nlog⁡([S])\log(K) = n\log([S])
    K=[S]nK = [S]^n
    這個 [S][S] 值就是當 v=vMax/2v = v_{Max}/2 時的底物濃度。
    所以，KK 代表了使反應速率達到最大速率一半時的底物濃度 [S][S] 的 nn 次方。
    如果 n=1n=1，則 K=[S]K = [S]，此時 KK 就是 KMK_M (當 v=vMax/2v=v_{Max}/2 時的 [S][S])。
    然而，題目中假設了 K=KMK=K_M。如果 Hill 曲線圖顯示 n≠1n \neq 1，那麼這個假設可能不完全準確，但 KK 仍然可以從圖形上找到。
    更常見的做法是，在 Hill 圖上，當 log⁡(vvMax−v)=−log⁡(K)\log\left(\frac{v}{v_{Max}-v}\right) = -\log(K) 時，log⁡([S])=0\log([S]) = 0，即 [S]=1[S]=1。
    或者，當 log⁡([S])=log⁡(K)/n\log([S]) = \log(K)/n 時，log⁡(vvMax−v)=0\log\left(\frac{v}{v_{Max}-v}\right)=0。

    更直接地尋找 K 的方法：
    在 Hill 圖上，當 yy 軸值為 0 時，即 log⁡(vvMax−v)=0\log\left(\frac{v}{v_{Max}-v}\right) = 0，此時 xx 軸值為 log⁡([S])\log([S])。
    從 0=nlog⁡([S])−log⁡(K)0 = n\log([S]) - \log(K)，我們得到 log⁡([S])=log⁡(K)n\log([S]) = \frac{\log(K)}{n}。
    這個 [S][S] 值是 v=vMax/2v = v_{Max}/2 時的底物濃度。
    因此，我們可以通過找到 y=0y=0 的點，然後計算 [S]=K1/n[S] = K^{1/n} 來得到 KK。
    如果知道 KMK_M 的值（題目中給了 KM=0.5 mMK_M=0.5 \, mM），並且假設 K=KMK=K_M，那麼我們可以將 KK 設定為 0.5 mM0.5 \, mM，然後從圖形上找到對應的 log⁡([S])\log([S]) 值。

  • 協同效應的判斷:

    • n>1n > 1: 正協同效應。
    • n=1n = 1: 無協同效應 (標準 Michaelis-Menten)。
    • n<1n < 1: 負協同效應。

應用於本題的具體步驟：

  1. 獲取數據: 從實驗或題目給出的信息中，獲得不同底物濃度 [S][S] 下的反應速率 vv。題目中提到，ADH 酶遵循 Michaelis-Menten 動力學，並且在飲酒後 VmaxV_{max} 達到。我們也知道 KM=0.5 mMK_M = 0.5 \, mM。
    如果我們有實驗數據，可以計算 θ=v/vMax\theta = v/v_{Max}。
    在沒有具體實驗數據的情況下，我們可以利用 Michaelis-Menten 方程式和 Hill 方程式的關係來描述。
    θ=vvMax=[S][S]+KM\theta = \frac{v}{v_{Max}} = \frac{[S]}{[S]+K_M}。
    Hill 方程式為 log⁡(θ1−θ)=nlog⁡([S])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([S]) - \log(K)。
    將 θ\theta 代入：
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後續完整解題步驟與【答案】

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第 8 題10 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
8) Give possible physiological meaning of the positive cooperative effect of alcohol dehydrogenase.
(10%)

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這一題的完整詳解

此題要求解釋酒精脫氫酶 (ADH) 表現出正協同效應的生理意義。正協同效應意味著一個底物分子與酶（或其亞基）結合後，會增加酶對其他底物分子的親和力。

生理意義：

  1. 靈敏響應酒精濃度的變化:

    • 當血液中的酒精濃度（底物濃度 [S][S]）較低時，由於正協同效應（n>1n>1），ADH 酶對酒精的親和力較低（結合曲線的起始段較為平緩）。這意味著在低濃度下，酶的催化效率相對較低，可以避免對低濃度酒精過度反應。
    • 隨著血液中酒精濃度升高（例如飲酒後），當酒精濃度達到一定閾值時，酶的親和力會顯著增加（結合曲線的斜率變大）。這使得酶能夠更快速、更有效地結合並代謝酒精。這種「開關」效應使得酶系統能夠在酒精濃度升高時迅速啟動並提高代謝速率，有效清除血液中的酒精。

  2. 防止低濃度酒精的過度催化:
    如果酶在低濃度酒精下就表現出很高的親和力，那麼即使是微量的酒精也可能被快速代謝，這在生理上可能不是最優的。正協同效應允許身體在酒精濃度很低時「容忍」較高的濃度，直到濃度升高到一定程度才開始強烈反應。

  3. 調節代謝速率:
    協同效應提供了一種調節機制。在生理範圍內，酶的活性不是簡單地隨底物濃度線性增加，而是呈現 S 形曲線。這使得酶系統能夠在較寬的底物濃度範圍內，對濃度的變化做出更靈敏和更劇烈的響應。當酒精濃度相對較低時，代謝速率緩慢；當濃度升高時，代謝速率迅速增加，直到達到 VmaxV_{max}。

  4. 與信號傳導的類比:
    協同效應在許多生物系統中都有體現，例如血紅蛋白與氧氣的結合。這種效應允許系統在信號（這裡的信號是酒精）強度發生變化時，做出更明顯的響應，從而更有效地傳遞和響應信號。

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後續完整解題步驟與【答案】

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第 9 題10 分

Carefully read the following information and briefly answer the following questions after calculating
with a precision of 2 significant figures. Given that In(2)=0.7, In(10)=2.3.
After drinking alcohol, ethanol (CH3CH2OH) in our blood can soon be built up to 20mM, which will
gradually be oxidized to ethanal (CH3CHO), the chief chemical for the alcohol flush syndrome, with
the following reaction: ethanol + NAD+ → ethanal + NADH + H+. Under sufficient supply of
NAD*, the oxidation rate (r) from ethanol (S) to ethanal (P) can be predicted by Arrhenius equation:
d[S]dt=d[P]dt=r=Ae−ERT\frac{d[S]}{dt} = \frac{d[P]}{dt} = r = Ae^{-\frac{E}{RT}}
where R is the gas constant (~8.3 Jmol⁻¹K⁻¹).
Outside our body, the reaction rate at 27°C doubles for every 10°C rise in temperature (a norm in
high school chemistry); however, the rate at the body temperature (37°C) can only reach
0.001mMmin⁻¹, much slower than the maximum reaction rate in our body, 10 mMmin⁻¹. Learned from
the internet, the reaction in our body was actually catalyzed by a dimeric enzyme, alcohol
dehydrogensase, with one active site on each subunit. The enzyme (KM = 0.5mM) was expected to
follow Michaelis-Menten kinetics with the reaction rate (v) reaching its maximum (VMax) right after
drinking.
v=vMax[S][S]+KMv = v_{Max} \frac{[S]}{[S]+K_M}
However, the in vitro kinetic analysis shows a positive cooperative effect that can be revealed by
Hill-Langmuir equation of Protein(P)-Ligands(L) binding by assuming the dissociation constant
K=KMK = K_M
P+nL⇌PLnP + nL \rightleftharpoons PL_n
θ=Occupied proteinTotal protein=[PLn][P]+[PLn]=[P][L]n/K[P]+[P][L]n/K=[L]n[L]n+K\theta = \frac{\text{Occupied protein}}{\text{Total protein}} = \frac{[PL_n]}{[P]+[PL_n]} = \frac{[P][L]^n / K}{[P]+[P][L]^n / K} = \frac{[L]^n}{[L]^n+K}
log⁡(θ1−θ)=nlog⁡([L])−log⁡(K)\log\left(\frac{\theta}{1-\theta}\right) = n\log([L]) - \log(K)
The logarithmic plot is called Hill plot, and the slope n is called Hill coefficient. The protein binding
phenomenon is "positive" cooperative when n > 1.
9) Give possible physiological meaning of the positive cooperative effect of hemoglobin during
transporting oxygen molecules from lung to tissues. (10%)

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這一題的完整詳解

此題要求解釋血紅蛋白 (Hemoglobin, Hb) 在從肺部運輸氧氣到組織過程中表現出的正協同效應的生理意義。

血紅蛋白是一種蛋白質，它能夠結合並運輸氧氣。它由四個亞基組成（兩個 α\alpha 亞基和兩個 β\beta 亞基），每個亞基都含有一分子血紅素，血紅素中的鐵離子可以結合一分子氧氣。

正協同效應的定義：
當一個分子（例如氧氣）與蛋白質（血紅蛋白）的一個結合位點結合後，會增加蛋白質對其他相同分子（氧氣）的結合親和力。這導致血紅蛋白與氧氣的結合曲線呈現 S 形。

生理意義：

  1. 有效率地從肺部吸收氧氣:

    • 在肺部，氧氣分壓較高。當第一個氧氣分子結合到血紅蛋白的一個亞基上時，會誘發血紅蛋白結構的變化，增加其他三個亞基對氧氣的親和力。
    • 這種親和力的增加使得血紅蛋白能夠在肺部的高氧壓環境下「高效地」飽和結合氧氣，即使氧氣分壓不是非常高，也能夠儘可能多地結合氧氣。這確保了氧氣能夠被有效地轉運到血液中。

  2. 有效地將氧氣釋放到組織:

    • 在組織中，氧氣分壓較低（因為組織細胞消耗氧氣）。當血紅蛋白將一部分氧氣釋放到組織後，血紅蛋白的結構會發生變化，降低其對剩餘氧氣的親和力。
    • 這種親和力的降低使得血紅蛋白能夠在組織的低氧壓環境下「更容易地」釋放氧氣。即使組織的氧氣分壓不是非常低，血紅蛋白也能夠儘可能多地釋放氧氣，以滿足組織細胞的代謝需求。

  3. 精確調節氧氣的供應:

    • S 形的結合曲線提供了一個「觸發點」。在氧分壓稍微降低時，氧氣釋放量會顯著增加；在氧分壓稍微升高時，氧氣結合量也會顯著增加。
    • 這種精確的調節能力使得氧氣的供應能夠根據組織的需求進行動態調整。例如，在劇烈運動時，組織氧氣分壓降低，血紅蛋白會釋放更多氧氣；
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