(整理)上海交大生物化学教材上册习题答案

Chapter 2

1. (a) tropomyosin原肌球蛋白70-kd→35-kd (two-stranded)→318 residues (mean

residue mass 110)→477Ǻ(axial distance of per residue in α-helix is 1.5Ǻ)(b)40 residues→36 (4 for hairpin turn) →18 (two-stranded) →63Ǻ(the rise per residue 3.5Ǻin βsheet) motif基序Ǻ mom基基t

2. contrasting isomer截然不同的异构体dioxane二噁烷in Leu methyl甲基is attached to γ-C atom, while in Ile it to β-C atom(比较靠近主链α-碳原子,造成立体位阻,阻断α-螺旋的形成)

3. Ala＜Val; Ile ＞Gly …Ala……Ile….. with activity (wild type)

…Val……Ile……lose activity (mutant type)

…Val……Gly… regain activity (double-mutant type)

4. shuffle重新洗牌, disulfide-sulfhydryl exchange二硫键-巯基互换; PDI蛋白二硫键异构酶, scrambled ribonuclease杂乱的核糖核酸酶

杂乱的核糖核酸酶在PDI作用下由无序变成有序,而胰岛素则由有序变成无序,说明胰岛素结构并非热力学最稳定的形式;因为胰岛素是由(比它多33个残基的)胰岛素原形成的,后者才是热力学最稳定的形式.

5.stretching a target使目标(蛋白底物)伸展; protease蛋白水解酶

6. Gly has the smallest side chain of AA: tight turns and approaching one another closely

7.guanidinium group胍基of Arg (positive): Glu, Asp, and the terminal carbonyl group (negative)

8.keratin角蛋白烫发: 加入巯基化合物(作为还原剂)和适当的温度使原来(发型中)的二硫键打断,卷曲头发(至所需型态)并加入氧化剂,使之形成新的二硫键并稳定.

Chapter 3

1.(a)phenyl isothiocyanate (PITC) 异硫氰酸苯酯

(b)Dansyl chloride丹黄酰氯(or dabsyl chloride)

(c)Urea尿素; β-mercaptoethanol β-巯基乙醇

(d)chymotrypsin糜蛋白酶(N)……R-‖-X……(C) R: Phe Trp, Tyr

(e)CNBr (N)……Met-‖-X……(C)

(f)Trypsin (N)……R-‖-X……(C) R: Arg or Lys

2. K=[H+][A－]/[HA] →pH=pK +log [A－]/[HA]

3.anhydrous hydrazine无水肼: 处理多肽(蛋白质)发生裂解,羧基末端的氨基酸释放成游离氨基酸,可被鉴定.所有其他氨基酸残基变成酰肼衍生物.

4. ethyleneimine乙烯亚胺+ Cys →S-aminoethyl氨乙基derivatives (S-氨乙烯半胱氨酸)

→与Lys类似Cys*-‖-X

5. incident入射, transmitted透射, εextinction coefficient (molar absorption coefficient)消光系数; I = I0 10－εlc A (log I0/I) =εlc

1mg/ml →5.62 x 10－5 M (MW= 17.8 kd) A = 15000 x 5.62 x 10－5 x 1= 0.84 = log10(I0/I) I0/I=6.96 即14.4%的入射光被透射

6. tropomyosin原肌球蛋白rod shaped

7.s ∝m/f f = 6πηr m ∝r3 →s ∝m2/3

8. y = kx + b log (MW) = k (RM) + b y—log (MW) MW:molecular weight X—RM: relative mobility

log 92000 = 0.41 k + b

log 30000 = 0.80 k + b k= －1.26 b= 5.465

So, the apparent mass of a protein having a mobility of 0.62 is 50 kd

Chapter 4

1.5’ GATCAA 3’

3’ CTAGTT 5’→5’TTGATC3’(书写自左向右5’→3’)

3. 17μm－15μm = 2μm = 20000A÷3.4A = 5800 base pairs

4．A(15N) B(14N)

Conservative replication: AA→AA+BB (F1)→AA+BB+ BB+ BB(F2)

Semiconservative replication: AA→AB+ AB (F1) →AB+ AB +BB+ BB (F2)

6. competence感受态Bacillus subtilis枯草芽胞杆菌

possible reason: 1.not able to take up DNA 2.having deoxiribonuclease脱氧核糖核酸酶3.not able to integrate fragments of DNA into their genome

7. propitious有利的; T2 is used by Hershey and Chase (P78) to separate itself into genetic (DNA) and nongenetic (protein) parts. M13 is not. (它的蛋白外壳将嵌入宿主的内膜,离心后将随细胞沉淀)

8. （a）tritiated thymine (or thymidine)

(b) dNs-P*~P~P

9.

5’_______________3’-OH

3’-OH_______________5’