生化作业(一)
1.写出下列氨基酸的单字母和三字母的缩写符号:精氨酸、天冬酰胺、天冬氨酸、谷氨酰胺、
谷氨酸、赖氨酸、苯丙氨酸、色氨酸和酪氨酸。(《生物化学》王镜岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
Arg, Asn, Asp, Gln, Glu, Lys, Phe, Trp, Tyr.
R, N, D, Q, E, K, F, W, Y
2.计算赖氨酸的ε- NH3+ 20%被解离时的溶液pH。(《生物化学》王镜岩,朱圣庚,徐长法主编第
三版上册, p.155)
答案:
pKa3 = 10.53
pH = pKa + log (ε-NH2 /ε-NH3+)
= 10.53 + log ( 0.2/0.8)
= 9.927
3.计算谷氨酸的γ-COOH三分之二被解离时的溶液pH。(《生物化学》王镜岩,朱圣庚,徐长法主
编第三版上册, p.155)
答案:
pKa = 4.25
pH = 4.25 + log (2/1)
= 4.551
4.计算下列物质0.3 mol/L溶液的pH:(a)亮氨酸盐酸盐,(b)亮氨酸钠盐, 和 (c)等电亮氨酸。
(《生物化学》王镜岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
a/ 当 0.3 mol 的氨基酸盐酸盐溶于1L水, 离子键完全水解, 生成0.3 M 氯负离子与0.3 M NH3+-CHR-COOH.
0.3 M 的NH3+-CHR-COOH 会继续解离:
NH3+-CHR-COOH <--> NH3+-CHR-COO- + H+pKa1 = 2.36
NH3+-CHR-COO- <--> NH2-CHR-COO- + H+pKa2 = 9.60
为了简化, 上述的平衡可写为:
aa+ <--> aa+- + H+pKa1 = 2.36
aa+- <--> aa- + H+pKa2 = 9.60
将 [H+] 记为 x
[aa+-] * x / [aa+] = 10-2.36
[aa-] * x / [aa+-] = 10-9.60
[aa+-] = 10-2.36 *[aa+]/x
[aa-] = 10-9.60 * [aa+-]/x
= 10-2.36 * 1.0-9.60 [aa+]/x2
= 10-11.96 * [aa+]/x2
因为 [aa+] + [aa+-] + [aa-] =总亮氨酸浓度,即0.3 M
[aa+] + 10-2.36 *[aa+]/x + 10-11.96 * [aa+]/x2 = 0.3
[aa+] = 0.3 /(1 + 10-2.36 /x + 10-11.96 /x2 )
= 0.3 x2 / (x2 + 10-2.36 x + 10-11.96 )
[aa+-] = 10-2.36 * 0.3 x / (x2 + 10-2.36 x + 10-11.96 )
[aa-] = 10-11.96 *0.3 / (x2 + 10-2.36 x + 10-11.96 )
带正电的离子有: aa+, H+,
带负电的离子有: aa-, OH-, Cl-
为了保持电荷平衡,
[aa+] + [H+] = [aa-] + 1.0-14/[H+] + [Cl-]
0.3 x2 / (x2 + 10-2.36 x + 10-11.96 ) + x = 10-11.96 *0.3 / (x2 + 10-2.36 x + 10-11.96 ) + 10-14/[H+] + 0.3
解上述方程(你可以用牛顿数值解法, 或用Excel 中的Solver 模块)
x = [H+]
= 10-1.46
pH = 1.46
b/ 当 0.3 mol 的氨基酸钠盐溶于1L水, 离子键会完全水解, 生成0.3 M 钠离子与0.3 M NH2-CHR-COO-. 0.3 M NH2-CHR-COO-会在水溶液中与质子结合, 达成如下平衡:
NH3+-CHR-COOH <--> NH3+-CHR-COO- + H+pKa1 = 2.36
NH3+-CHR-COO- <--> NH2-CHR-COO- + H+pKa2 = 9.60
为了简化, 以上的平衡可写为:
aa+ <--> aa+- + H+pKa1 = 2.36
aa+- <--> aa- + H+pKa2 = 9.60
将 [H+] 记作 x
[aa+-] * x / [aa+] = 10-2.36
[aa-] * x / [aa+-] = 10-9.60
[aa+-] = 10-2.36 *[aa+]/x
[aa-] = 10-9.60 * [aa+-]/x
= 10-2.36 * 10-9.60 [aa+]/x2
= 10-11.96 * [aa+]/x2
因为 [aa+] + [aa+-] + [aa-] = 氨基酸总浓度,
[aa+] + 10-2.36 *[aa+]/x + 10-11.96 * [aa+]/x2 = 0.3
[aa+] = 0.3 /(1 + 10-2.36 /x + 10-11.96 /x2 )
= 0.3 x2 / (x2 + 10-2.36 x + 10-11.96 )
[aa+-] = 10-2.36 *0.3 x / (x2 + 10-2.36 x + 10-11.96 )
[aa-] = 10-11.96 * 0.3 / (x2 + 10-2.36 x + 10-11.96 )
溶液中带正电的离子有: aa+, H+, Na+
溶液中带负电的离子有: aa-, OH-
为保持电荷平衡,
[aa+] + [H+] + [Na+] = [aa-] + 10-14/[H+]
0.3 x2 / (x2 + 10-2.36 x + 10-11.96 ) + x + 0.3 = 10-11.96 * 0.3 / (x2 + 10-2.36 x + 10-11.96 ) + 10-14/[H+]
解上述方程得x:
x = [H+]
= 2.9103 x 10-12
pH = 11.53
c/ 与上述解法大同小异
[aa+] + [H+] = [aa-] + 10-14/[H+]
0.3 x2 / (x2 + 10-2.36 x + 10-11.96 ) + x = 10-11.96 * 0.3 / (x2 + 10-2.36 x + 10-11.96 ) + 10-14/[H+]
解上述方程得x
[H+] = 0.000001039658
pH = 5.983
5.根据表3-3中氨基酸的pKa值,计算下列氨基酸的pl值:丙氨酸、半胱氨酸、谷氨酸和精氨
酸。(《生物化学》王镜岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
pI (Ala) = (2.34 + 9.69)/2 = 6.02
pI(Cys) = (1.71 + 8.33)/2 = 5.02
pI(Glu) = (2.19 + 4.25)/2 = 3.22
pI(Arg) = (9.04 + 12.48)/2 = 10.76
6.向1 L 1 mol/L的处于等电点的甘氨酸溶液加人0.3 mol HCI,问所得溶液的pH是多少?如果加
人0.3 mol NaOH以代替HCI时,pH将是多少?(《生物化学》王镜岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
a/ + NH3-CH2-COO- + HCl <--> + NH3-CH2-COOH + Cl-
t=0 1 mol 0.3 mol
t = ∞ 0.7 mol 0 mol 0.3 mol 0.3 mol
pH = pKa + log (+ NH3-CH2-COO- / + NH3-CH2-COOH )
= 2.34 + log ( 0.7/0.3)
= 2.71
b/ + NH3-CH2-COO- + NaOH <--> NH2-CH2-COO- + Na+ + H2O
t=0 1 mol 0.3 mol
t = ∞ 0.7 mol 0 mol 0.3 mol
pH = pKa + log ( NH2-CH2-COO- / + NH3-CH2-COO- )
= 9.60 + log (0.3 /0.7)
= 9.23
7.将丙氨酸溶液(400 mL)调节到pH 8.0,然后向该溶液中加入过量的甲醛。当所得溶液用碱反
滴定至pH 8.0时,消耗0.2 mol/L NaOH溶液250 ml。问起始溶液中丙氨酸的含量为多少克?(《生物化学》王镜岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
在pH=8 时, Ala 的主要离子形式为+NH3-CH(CH3)-COO-. 加入甲醛 HCOH 后, Ala 的主要化学形式为 (CH2OH)2N-CH(CH3)-COO-.
因为滴定耗去了0.250 L 0.2M NaOH, Ala 的摩尔数mol(Ala) = 0.250L X 0.2 M = 0.05 mol,
Ala 的质量为 mass(Ala) = 0.05 mol * 89.06 g/mol = 4.45 g
8.计算0.25 mol/L的组氨酸溶液在pH 6.4时各种离子形式的浓度(mol/L)。(《生物化学》王镜
岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
组氨酸的质子解离常数为pKa1=1.82, pKa2 = 6.00, 和pKa3 = 9.17
His(2+) <--> His (+) + H+ pKa1 = 1.82
His(+) <--> His (+/-) + H+ pKa2 = 6.00
His(+/-) <--> His (-) + H+ pKa3 = 9.17
将 [H+]记为 x
His(+) = 10^-pKa1 * His(2+) / x
His(+/-) = 10^-pKa2 * His(+) / x
= 10^-pKa2 * 10^-pKa1 * His(2+) / x^2
His(-) = 10^-pKa3 * His (+/-)/x
= 10^-pKa3 * 10^-pKa2 * 10^-pKa1 * His(2+) / x^3
因为 His(2+) + His(+) + His(+/-) + His (-) = 0.25 mol/L
His(2+) + 10^-pKa1 * His(2+) / x + 10^-pKa2 * 10^-pKa1 * His(2+) / x^2 + 10^-pKa3 * 10^-pKa2 * 10^-pKa1 * His(2+) / x^3 = 0.25 mol/L
His(2+)( 1 + 10^-pKa1 / x + 10^-pKa2 * 10^-pKa1 / x^2 + 10^-pKa3 * 10^-pKa2 * 10^-pKa1 / x^3 ) = 0.25 mol/L
His(2+) = 0.25 mol/L / ( 1 + 10^-pKa1 / x + 10^-pKa2 * 10^-pKa1 / x^2 + 10^-pKa3 * 10^-pKa2 * 10^-pKa1 / x^3 )
= 0.25 mol/L / (1 + 10^-1.82/10^-6.4 + 10^-1.82*10^-6.00/ (10^-6.4)^2 + 10^-1.82*10^-6.00*10^-
9.17/(10^-6.4)^3 )
= 0.000001870118 mol/L
His(+) = 10^-1.82 * 0.000001870118 mol/L / 10^-6.4
= 0.0710999 mol/L
His (+/-) = 0.0710999 * 10^-6.00/10^-6.4
= 0.178595 mol/L
His (-) = 0.033266 mol/L * 10^-9.17/10^-6.4
= 0.00030329 mol/L
另一个解题途径为::
His(-)/His(+/-) = 10^(-6.4 + 9.17) = 588.84
His(+/-)/His(+) = 10^(-6.4 + 6.00) = 0.3981
His(+)/His(++) = 10^(-6.4 + 1.82) = 0.00002630
设 His(-) = x, His(+/-) = 588.84x, His(+) = 588.84x*0.3981, His(++) = 588.84x*0.3981*0.00002630
因为 x + 588.84x + 588.84x*0.3981 + 588.84x*0.3981*0.00002630 = 0.25 mol/L
x = 0.0003033
于是有
His (-) = 0.00003033 M
His(+/-) = 0.1786 M
His(+) = 0.07110 M
His(++) = 0.000001870 M
9.说明用含一个结晶水的固体组氨酸盐酸盐(相对分子质量= 209.6;咪唑基pKa=6.0)和1 mol/L
KOH配制1 L pH=6.5的0.2 mol/L组氨酸盐缓冲液的方法。(《生物化学》王镜岩,朱圣庚,徐长法主编第三版上册, p.155)
答案:
pH = pKa + log (-N= / -+NH=)
6.5 = 6.0 + log (-N= / -+NH=)
-N= / -+N= = 100.5
= 3.1622
设侧链带质子的His 摩尔数 n(-+NH=) = x, 那么侧链不带质子的His 摩尔数 n(-N=) = 3.1622 x.
x + 3.1622x = 0.2 moles
n(-+NH=) = x
= 0.048052 moles
n(-N=) = 0.15195 moles
因为起始氨基酸是 Cl - +
NH 3-C(H-R)-COOH, 需要0.2 mole KOH 使羧基解离,还需0.15195 moles KOH 使侧链解离,所以,KOH 的总量为0.2 mole + 0.15195 moles = 0.35195 moles KOH.
0.35195 moles = V * 1 mol/L V = 0.35195 L = 352 mL
His HCl 盐的重量 = 0.2 * 209.6 g/mol = 41.92 g
11. L- 亮氨酸溶液(3.0 g/50 mL , 6 mol/L HCI )在20 cm 旋光管中测得的旋光度为+1.81。
。计算L-亮氨酸在6 mol/L HCI 中的比旋[α]. (《生物化学》 王镜岩,朱圣庚,徐长法 主编 第三版上册, p.155)
答案:
[α]λt
=100·α/(c ·l )
上式中, α 是实验测得的比旋值 (1.81 degree), c 是每100g 溶剂中的溶质克数 (3 g per 50 mL = 6 g per 100 mL), L 旋光测量仪的试管长度,以1 dm 作单位(20 cm = 2 dm).
[α]λt
= 100 x 1.81 /(6 g ·2 dm )
= 15.1 o
12. 标出异亮氨酸的4个光学异构体的(R ,S)构型名称。(生物化学, 王镜岩,朱圣庚,徐长法 主编
第三版上册, p.155)
答案:
COOH
H NH 2H
H 3C
C 2H 5
COOH
H 2N H H
H 3C
C 2H 5
COOH
H NH 2CH 3
H
C 2H 5
COOH
H 2N H CH 3
H
C 2H 5
S S R R R
S
R
S
13. 指出在正丁醇:醋酸:水的系统中进行纸层析时,下列混合物中氨基酸的相对迁移率(假定
水相的pH 为4.5):(1)Ile, Lys; (2)Phe, Ser; (3) Ala, Val, Leu ; (4) Pro, Val; (5) Glu, Asp; (6) Tyr, Ala, Ser, His. (生物化学, 王镜岩,朱圣庚,徐长法 主编 第三版上册, p.156)
答案:
(1). Lys carries positive charge at pH = 4.5, tends to stay in polar layer, therefore, Ile > Lys; (2) Ser is more polar, therefore, tends to stay in polar layer, Phe > Ser;
(3) In terms of the length of the side chain, Leu > Val > Ala, therefore, Leu is more likely to stay in non-polar mobile phase, Leu > Val > Ala;
(4) Compare Val and Pro, Val ’s side chain does not link to the polar amio group, so it is more likely to stay in non-polar phase, therefore, Val > Pro;
(5) Glu has a longer side chain compare to Asp, therefore, Glu is more likely to be in non-polar phase, Glu > Asp;
(6)
Tyr, Ala, Ser, His Tyr has an aromatic ring, so it stays in mobile phase, Ala is more “non-polar ” than Ser and His, therefore the moving rate is Tyr > Ala > Ser ≈ His.
14.(思考题)
以下多肽由九个氨基酸组成。请指出它的N端,C端,氨基酸顺序,以及二硫键的位置。
15. (思考题)
二十种天然氨基酸中,色氨酸(Trp,W)的侧链是最大的。色氨酸侧链环中有一个氮原子,是不是可以据此而认为色氨酸侧链是碱性的?是不是可以据此而认为色氨酸侧链是极性的?侧链氨基上的氢原子能不能参与形成氢键?
16. (思考题)
组氨酸侧链的咪唑基团质子解离常数为pKa = 6.00, 与生理pH=7.4 十分接近,因而组氨酸在与酸碱催化有关的酶中,常处于活心中心。在推测组氨酸与质子结合的化学结构时,有人提出了如下图的两种可能,你认为哪一种是比较可信的?
17. (思考题)
计算由5个赖氨酸组成的多肽, 即 Lys-Lys-Lys-Lys-Lys的等电点.