JACS-2007-2607-碱基催化

Three-Dimensional Porous Coordination Polymer Functionalized with Amide Groups Based on Tridentate

Ligand:Selective Sorption and Catalysis

Shinpei Hasegawa,Satoshi Horike,Ryotaro Matsuda,Shuhei Furukawa,

Katsunori Mochizuki,Yoshinori Kinoshita,and Susumu Kitagawa* Contribution from the Department of Synthetic Chemistry and Biological Chemistry,Graduate School of Engineering,Kyoto Uni V ersity,Katsura,Nishikyo-ku,Kyoto,615-8510,Japan

Received October14,2006;E-mail:kitagawa@sbchem.kyoto-u.ac.jp

Abstract:To create a functionalized porous compound,amide group is used in porous framework to produce attractive interactions with guest molecules.To avoid hydrogen-bond formation between these amide groups our strategy was to build a three-dimensional(3D)coordination network using a tridentate amide ligand as the three-connector part.From Cd(NO3)2?4H2O and a three-connector ligand with amide groups a3D porous coordination polymer(PCP)based on octahedral Cd(II)centers,{[Cd(4-btapa)2(NO3)2]?6H2O?2DMF}n(1a), was obtained(4-btapa)1,3,5-benzene tricarboxylic acid tris[N-(4-pyridyl)amide]).The amide groups,which act as guest interaction sites,occur on the surfaces of channels with dimensions of4.7×7.3?2.X-ray powder diffraction measurements showed that the desolvated compound(1b)selectively includes guests with a concurrent flexible structural(amorphous-to-crystalline)transformation.The highly ordered amide groups in the channels play an important role in the interaction with the guest molecules,which was confirmed by thermogravimetric analysis,adsorption/desorption measurements,and X-ray crystallography. We also performed a Knoevenagel condensation reaction catalyzed by1a to demonstrate its selective heterogeneous base catalytic properties,which depend on the sizes of the reactants.The solid catalyst1a maintains its crystalline framework after the reaction and is easily recycled.

Introduction

In recent years,numerous studies of porous coordination polymers(PCPs),also called metal-organic frameworks(MOFs), have been reported1because of their applications in gas adsorption,2molecular storage,3and heterogeneous catalysis.4 PCPs have characteristic features that include(1)well-ordered porous structures,(2)flexible and dynamic behaviors in response to guest molecules,and(3)designable channel surface func-tionalities.Although channel surface modification is essential for creation of functionalized porous structures,application of this synthetic approach to the PCP system has received little attention.Two types of strategies are used to functionalize channel surfaces:immobilization of coordinatively unsaturated (open)metal sites(OMS)5and introduction of organic groups to provide guest-accessible functional organic sites(FOS).4b,6 There is growing interest in the use of OMS for Lewis acid catalysis and specific gas adsorption,but less attention has been devoted to the study of FOS,despite their importance.The paucity of information on FOS is because of the difficulty of producing guest-accessible FOS on the pore surface:these

(1)(a)Yaghi,O.M.;Li,H.;Davis,C.;Richardson,D.;Groy,T.

L.Acc Chem.

Res.1998,31,474-484.(b)Batten,S.R.Curr.Opin.Solid State Mater.

Sci.2001,5,107-114.(c)Moulton,B.;Zaworotko,M.J.Chem.Re V.

2001,101,1629-1658.(d)Janiak,C.Dalton Trans.2003,2781-2804.

(e)Kitagawa,S.;Kitaura,R.;Noro,S.Angew.Chem.,Int.Ed.2004,43,

2334-2375.(f)Rosseinsky,M.J.Microporous Mesoporous Mater.2004, 73,15-30.

(2)(a)Eddaoudi,M.;Kim,J.;Rosi,N.;Vodak,D.;Wachter,J.;O’Keeffe,

M.;Yaghi,O.M.Science2002,295,469-472.(b)Kitaura,R.;Kitagawa, S.;Kubota,Y.;Kobayashi,T.C.;Kindo,K.;Mita,Y.;Matsuo,A.;

Kobayashi,M.;Chang,H.C.;Ozawa,T.C.;Suzuki,M.;Sakata,M.;Takata, M.Science2002,298,2358-2361.(c)Fe′rey,G.;Latroche,M.;Serre,C.;

Millange,F.;Loiseau,T.;Gue′gan,https://www.wendangku.net/doc/0e19149638.html,mun.2003,2976-2977.(d)Chu,H.;Dybtsev,D.N.;Kim,H.;Kim,K.Chem.Eur.J.2005, 11,3521-3529.(e)Rowsell,J.L.C.;Yaghi,O.M.Angew.Chem.,Int.

Ed.2005,44,4670-4679.(f)Matsuda,R.;Kitaura,R.;Kitagawa,S.;

Kubota,Y.;Belosludov,R.V.;Kobayashi,T.C.;Sakamoto,H.;Chiba, T.;Takata,M.;Kawazoe,Y.;Mita,Y.Nature2005,436,238-241.(g) Dincaˇ,M.;Long,J.R.J.Am.Chem.Soc.2005,127,9376-9377. (3)(a)Yaghi,O.M.;Li,H.J.Am.Chem.Soc.1996,118,295-296.(b)

Dalrymple,S.A.;Shimizu,G.K.H.Chem.Eur.J.2002,8,3010-3015.(4)(a)Fujita,M.;Kwon,Y.J.;Washizu,S.;Ogura,K.J.Am.Chem.Soc.

1994,116,1151-1152.(b)Seo,J.S.;Whang,D.;Lee,H.;Jun,S.I.;Oh, J.;Jeon,Y.J.;Kim,K.Nature2000,404,982-986.(c)Wu,C.D.;Hu,

A.;Zhang,L.;Lin,W.J.Am.Chem.Soc.2005,127,8940-8941.(d)Lor,

B.G.;Puebla,E.G.;Iglesias,M.;Monge,M.A.;Valero,

C.R.;Snejko,

N.Chem.Mater.2005,17,2568-2573.(e)Sato,T.;Mori,W.;Kato,

C.N.;Yanaoka,E.;Kuribayashi,T.;Ohtera,R.;Shiraishi,Y.J.Catal.

2005,232,186-198.(f)Dybtsev,D.N.;Nuzhdin,A.L.;Chun,H.;

Bryliakov,K.P.;Talsi,E.P.;Fedin,V.P.;Kim,K.Angew.Chem.,Int.

Ed.2006,45,916-920.(g)Uemura,T.;Kitaura,R.;Ohta,Y.;Nagaoka, M.;Kitagawa,S.Angew.Chem.,Int.Ed.2006,45,4112-4116.

(5)(a)Noro,S.;Kitagawa,S.;Yamashita,M.;Wada,https://www.wendangku.net/doc/0e19149638.html,mun.2002,

222-223.(b)Kitaura,R.;Onoyama,G.;Sakamoto,H.;Matsuda,R.;Noro, S.;Kitagawa,S.Angew.Chem.,Int.Ed.2004,43,2684-2687.(c)Chen,

B.;Fronczek,F.R.;Maverick,A.W.Inorg.Chem.2004,43,8209-8211.

(d)Maggard,P.A.;Yan,B.;Luo,J.Angew.Chem.,Int.Ed.2005,44,

2-5.(e)Chen,B.;Ockwig,N.W.;Millward,A.R.;Contreras,D.S.;

Yaghi,O.M.Angew.Chem.,Int.Ed.2005,44,4745-4749.(f)Cho, S.H.;Ma,B.;Nguyen,S.T.;Hupp,J.T.;Albrecht-Schmitt,T.E.Chem.

Commun.2006,2563-2565.

(6)(a)Kitaura,R.;Fujimoto,K.;Noro,S.;Kondo,M.;Kitagawa,S.;Angew.

Published on Web02/09/2007

organic groups tend to coordinate metal ions via a self-assembly process,resulting in frameworks in which FOS are completely blocked.

The advantage of FOS is that the base-type catalyst is easy to create.There are a variety of organic functional groups that can serve as active base sites.In this report,we describe an approach for the preparation of a PCP using base-type FOS and demonstrate its base catalytic properties in a heterogeneous reaction.These observations will contribute to the development of new types of catalysts constructed from metal -organic frameworks.

PCPs with amide groups are candidate compounds that have FOS as guest interaction sites.7The amide group is a fascinating functional group because it possesses two types of hydrogen-bonding sites:the -NH moiety acts as an electron acceptor and the -C d O group acts as an electron donor.8These multifunctional moieties of amide groups tend to form hydrogen bonds among themselves and interact negligibly with guest molecules after construction of PCPs.If a PCP could be prepared without forming amide -amide interactions,these groups would constitute attractive interaction sites for selective sorption and/or catalysis inside the channel.To retain amide groups as guest-interaction FOS inside the network,we employed a three-connector ligand containing amide groups.Three-connector ligands are more useful for construction of 3D PCPs with coordination bonds than bidentate or monodentate ligands.9We employed a three-connector ligand containing three pyridyl groups as coordination sites and three amide groups as guest interaction sites (1,3,5-benzene tricarboxylic acid tris[N -(4-pyridyl)amide]10(4-btapa,Scheme 1)and succeeded in produc-ing a new base-type catalytic pore framework.

Experimental Section

General Information.4-Aminopyridine and 1,3,5-benzene tricar-boxylic acid trichloride were obtained from Tokyo Kasei Industrial.Cd(NO 3)2?4H 2O was obtained from Wako without purification.El-emental analysis (EA)was carried out with a Thermo Finnigan EA1112.1

H NMR spectra were measured in DMSO-d 6with a (500MHz)JEOL

JNM-A500FT NMR system,and chemical shifts are reported in δ(ppm)values.Solid-state magic angle spinning/cross polarization (CPMAS)113Cd NMR experiment was carried out on a (300MHz)JEOL JNM-LA300WB spectrometer.IR spectra were recorded on a Perkin-Elmer 2000FT -IR spectrophotometer with samples prepared with KBr pellets.Thermogravimetric analyses (TGA)were performed using a Rigaku Thermo plus TG 8120apparatus in the temperature range between 298and 723K in a N 2atmosphere and at a heating rate of 10K min -1.X-ray powder diffraction (XRPD)data were collected on a Rigaku RINT-2200HF (Ultima)diffractometer with Cu K R radiation.The adsorption isotherm of nitrogen at 77K and methanol at 298K were measured with BELSORP18volumetric adsorption equipment from Bel Japan,Inc.

Ligand (4-btapa).A solution of 4-aminopyridine (17.0g,180mmol)and distilled triethylamine (27mL,194mmol)in distilled THF (200mL)was added dropwise to a solution of 1,3,5-benzene tricarboxy-trichloride (16.1g,60.0mmol)in THF (60mL)at 0°C.Triethylamine (9.0mL,65mmol)was added,and the reaction mixture was stirred for 7h.The temperature was allowed to step up to room temperature.The brown crude product of 4-btapa was collected by filtration and washed with THF.The product was recrystallized from DMSO (700mL)and H 2O (1500mL)by stirring for 1h.The product was washed with acetone (1000mL)by stirring for 1day.The yellowish white powder was collected by filtration,washed with acetone,and dried under vacuum for 35h at room temperature (21.4g,81%).1H NMR (500MHz,DMSO-d 6):δ10.95(s,3H,NH),8.75(s,3H,-H 2,4,6-ph),8.52(d,6H,J H -H )5.0Hz,H 3,5-py),7.82(d,6H,J H -H )5.0Hz,H 2,6-py).FAB-MS (m /z ):calcd for C 24H 18N 6O 3,438.14;found,438.Anal.Calcd for C 24H 18N 6O 3(438.5):C,65.75;H,4.14;N,19.17.Found:C,64.84;H,4.05;N,18.81.

{[Cd(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a).A solution of Cd-(NO 3)2?4H 2O (9.3mg,0.030mmol)in MeCN (1.5mL)was carefully layered on a solution of in 4-btapa (20mg,0.045mmol)in DMF (1.5mL)with MeCN/DMF (0.75mL/0.75mL)placed between the two layers.Colorless single crystals began to form in a few days.One of these crystals was used for X-ray crystallographic analysis.The bulk product was obtained by the following procedure.A solution of Cd-(NO 3)2?4H 2O (0.93g,3.0mmol)in DMF (50mL)was added to a solution of 4-btapa (2.0g,4.5mmol)in DMF (50mL).The reaction mixture was stirred for 3.5h.The crystalline yellowish white powder 1a was obtained by filtration,washed with DMF,and dried under vacuum for 1day at room temperature (2.86g,93%yield based on 4-btapa).Anal.Calcd for {[Cd(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a )C 54H 62CdN 16O 20(1367.58):C,47.42;H,4.57;N,16.39.Found for single crystal:C,48.23;H,4.66;N,16.75.Found for bulk product:C,48.29;H,4.30;N,16.52.For X-ray crystallographic analysis some guest molecules that show positional disorder could not be fixed in the structure model like the early compounds.5b,11The formula {[Cd-(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a )was assigned by EA,TGA,1H NMR,IR,as well as single-crystal X-ray diffraction studies.12Further details of experimental studies are given in the Supporting Information.{[Cd(4-btapa)2]?2NO 3}n (1b).The amorphous yellowish white powder 1b was obtained by drying 1a under vacuum for 7.5h at 140°C.Removal of all solvent molecules was checked by TGA.The amorphous state of 1b was checked by XRPD.Anal.Calcd for [Cd-(7)(a)Uemura,K.;Kitagawa,S.;Kondo,M.;Fukui,K.;Kitaura,R.;Chang,

H.C.;Mizutani,T.

Chem.Eur.J.2002,8,3586-3600.(b)Uemura,K.;Kitagawa,S.;Fukui,K.;Saito,K.J.Am.Chem.Soc.2004,126,3817-3828.(c)Tzeng,B.C.;Chen,B.S.;Yeh,H.T.;Lee,G.H.;Peng,S.M.New J.Chem.2006,30,1087-1092.

(8)(a)Lee,C.M.;Kumler,W.D.J.Am.Chem.Soc.1962,84,571-578.(b)

Bent,H.A.Chem.Re V .1968,68,587-648.(c)Huyskens,P.L.J.Am.Chem.Soc.1977,99,2578-2582.

(9)(a)Batten,S.R.;Murray,K.S.Coord.Chem.Re V .2003,246,103-130.

(b)Chae,H.K.;Siberio-pe ′rez,D.Y.;Kim,J.;Go,Y.;Eddaoudi,M.;Matzger,A.J.;O’Keeffe,M.;Yaghi,O.M.Nature 2004,427,523-527.(c)Ohmori,O.;Kawano,M.;Fujita,M.J.Am.Chem.Soc.2004,126,(11)(a)Hoskins,B.F.;Robson,R.J.Am.Chem.Soc.1989,111,5962-5964.

(b)Ohmori,O.;Fujita,https://www.wendangku.net/doc/0e19149638.html,mun.2004,1586-1587.(c)Luo,T.T.;Tsai,H.L.;Yang,S.L.;Liu,Y.H.;Yadav,R.D.;Su,C.C.;Ueng,C.H.;Lin,L.G.;Lu,K.L.Angew.Chem.,Int.Ed.2005,44,6063-6067.(12)Some guest molecules that show positional disorder could not be fixed in

the structure model.During the final stages of refinement,several Q peaks were found,which probably correspond to highly disordered molecules.However,each guest molecule was detected by EA,TGA,IR,and 1H NMR.Crystal data for 1a :C 48H 36CdN 12O 12,M r )1085.30,cubic,Ia -3(#206),a )24.756(4)?,V )15171.6(43)?3,Z )8,F (000))4423.68,F calcd )0.952gcm -3,μ)0.342mm -1,GOF )0.904.A total of 84868reflections Scheme 1.1,3,5-Benzene Tricarboxylic Acid Tris[N -(4-pyridyl)amide](4-btapa)

A R T I C L E S

Hasegawa et al.

(4-btapa)2(NO 3)2](1b )(1113.30):C,51.78;H,3.26;N,17.61.Found for bulk product:C,49.52;H,3.57;N,16.66.

{[Cd(4-btapa)2(NO 3)2]?6MeOH ?yH 2O }n (1c).The methanol-exchanged colorless compound 1c was obtained by exposing 1b on the methanol vapor for 30h at room temperature.The amount of adsorbed methanol molecules was checked by TGA and 1H NMR.X-ray Structure Determination.The colorless single crystal of 1a was mounted on glass fibers with epoxy resin.X-ray data collection for the single crystal was carried out on a Rigaku Mercury diffracto-meter with graphite-monochromated Mo K R radiation (λ)0.71070?)and a CCD two-dimensional detector at 238K in a cold nitrogen stream.The X-ray condition for 1a was 50kV ×40mA.The structure was solved by direct method using the SHELXS-97program and refined by the full-matrix least-squares method on F 2with appropriate software implemented in the crystals program package.The final cycles of the full-matrix least-squares refinements were based on the observed reflections [I >3σ(I )].All calculations were performed using the crystal structure program package.All non-hydrogen atoms except for those of disordered solvent molecules in 1a were refined anisotropically.Hydrogen atoms were added at their geometrically ideal positions and refined isotropically.Further refinement was unsuccessful because counteranions and some solvent molecules were highly disordered,and the high symmetric position in 1a would make occupancies of these atoms decrease.

Selective Accommodation and Activation of Reactants.For each reagent (malononitrile,ethyl cyanoacetate,and cyano-acetic acid tert -butyl ester),the accommodation procedure was carried out.The as-synthesized 1a (0.10g)and a reagent (2mmol)in dehydrated benzene (10mL)were stirred for 6h.The precipitates were filtrated and dried in the open air for 1.5h and then identified by IR (KBr pellet)and 1H NMR (DMSO-d 6)spectroscopies.

Knoevengael Condensation Reaction Catalyzed by 1a.A solution of benzaldehyde (0.21mL,2.1mmol)and malononitrile (or ethyl cyanoacetate or cyano-acetic acid tert -butyl ester)(0.132g,2.0mmol)in benzene (10mL)was stirred for 5min.Then 1a (0.10g,0.08mmol,4mol %)was added.The suspension was stirred at room temperature for 12h.The progress of the reaction was monitored by 1H NMR Results and Discussion

Structural Description.The coordination environment around Cd(II)of {[Cd(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a )is shown in Figure 1.The Cd(II)center is octahedrally coordinated to six pyridyl nitrogen atoms from different bridging (4-btapa)connectors.To the best of our knowledge,1a is the first compound in which Cd(II)is coordinated to six pyridyl groups as the ideal octahedral coordination environment.All the Cd -N bond lengths and angles are similar to each other (Table 1).The solid-state cross-polarization magic-angle spinning (CP-MAS)113Cd nuclear magnetic resonance (NMR)spectrum for 1a also supports the ideal octahedral coordination environment of Cd(II).The spectrum has one resonance centered at 183.6ppm (Figure 2).It is well known that each aromatic nitrogen donor ligand in the CdN 4plane contributes approximately +50ppm.13Therefore,in this compound the chemical shift should have a value of ca.+200ppm directed normal to the 4N plane.The observed value of 183.6ppm is in agreement with this value.Furthermore,one can safely state that the one single resonance peak indicates a highly symmetrical coordination environment around the Cd(II)center.

3-D Porous Framework Functionalized with Amide Groups.The framework of 1a was constructed from the Cd(II)center as a six-connected node using 4-btapa as a three-connected linker.This arrangement produces a large six-membered ring composed of three octahedral Cd(II)moieties linked together by three 4-btapa units.The six-membered ring is distorted because of the rotation of 4-btapa between the benzene ring and the amide group (Figure 3a).The framework expands in

Figure 1.ORTEP drawing of 1a with ellipsoids probability 30%(asterisk indicates atoms that are symmetrically related).Cd(II)centers are octahe-drally coordinated to N atoms of different six 4-btapa.H 2O molecules and hydrogen atoms are omitted for clarity.

Figure 2.Solid-state 113Cd CPMAS NMR spectrum of 1a at a MAS spinning rate of 4382Hz.

Table 1.Selected Bond Distances (?)and Angles (deg)for {[Cd(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a )a

Cd (1)-N1 2.372(7)Cd1-N11 2.372(8)Cd1-N12 2.372(8)Cd1-N13 2.372(8)Cd1-N14

2.372(8)Cd1-N15

2.372(7)N11-Cd1-N190.8(3)N12-Cd1-N190.8(3)N13-Cd1-N1N14-Cd1-N189.2(3)N15-Cd1-N1N12-Cd1-N1190.8(3)N13-Cd1-N1189.2(3)N44-Cd1-N4189.2(3)N15-Cd1-N11180.0(4)N13-Cd1-N1289.2(3)N14-Cd1-N12180.0(4)N15-Cd1-N1289.2(3)N14-Cd1-N1390.8(3)N15-Cd1-N1390.8(3)

N15-Cd1-N14

90.8(3)

a

Symmetry transformation used to generate equivalent atoms:1)(y ,z ,x );2)(z ,x ,y );3)(-y ,-z ,-x );4)(-z ,-x ,-y );5)(-x ,-y ,-z );6)(-z +1/2,-x ,y +1/2).

Selective Sorption and Catalysis A R T I C L E S

three directions,with a Cd(II)???Cd(II)separation of 17.5?,to form a 3D network (Figure 3b).The two frameworks mutually interpenetrate with a nearest-neighbor Cd(II)???Cd(II)distance of 12.4?between Cd(II)atoms in the adjacent frameworks,producing three-dimensionally running channels (so-called 3D pores)with dimensions of 4.7×7.3?,2which are available for guest accommodation and exchange (Figure 4).The channel is not straight but zigzags and occupied by nitrate anions and H 2O and DMF molecules.Two nitrate anions,six H 2O molecules,and two DMF molecules per formula unit have been confirmed by elemental analysis (EA),thermogravimetric analysis (TGA),infrared spectroscopy (IR),and 1H NMR,14in addition to the crystallographic results.1a also has another type of zigzag channel,with dimensions of 3.3×3.6?,2running in three directions (Figure 5).In 1a ,43.8%of the void space is accessible to the solvent molecules.15,16The nearest N(amide)-to-N(amide)distances are about 24.8?in the network and 4.0?between the two adjacent networks.The distance of N(amide)-to-O(amide)is about 9.7?.Consequently,the amide groups,which are highly ordered on the surfaces of the channels,could interact with guest molecules.The X-ray crystallographic results clearly show that H 2O molecules interact weakly via hydro-gen bonds with the NH moieties of all the amide groups (O[water]???N[amide])ca.3.1?)(Figure 6).The amide groups inside the pores would act as important FOS in the interaction between the host and guest molecules.

Thermogravimetric Analyses of 1a and 1c.As seen in the crystal structure analysis and 1H NMR results,1a includes H 2O and DMF molecules as guests.The desorption process was monitored by TGA.The observed weight loss of the guest molecules is in agreement with that calculated for the corre-sponding crystal structure.The TGA data for 1a are indicative of two stages of weight loss for the guest molecules:six H 2O molecules and two DMF molecules per formula unit (Figure 7a).The resultant species,[Cd(4-btapa)2(NO 3)2]n (1b ),is stable up to 250°C and then gradually decomposes with the loss of 4-btapa.The TGA curve showed that the weight loss below 150°C of the methanol-exchanged compound 1c corresponds to three methanol molecules per 4-btapa ligands (Figure 7b).The amount of methanol adsorbed is in good agreement with the number of amide groups in the host framework.

Recoverable Collapse on Desolvation and Resolvation.1a ,1b ,and 1c are insoluble in water or common organic solvents and only soluble in dimethyl sulfoxide.Removal of the guest molecules causes a significant structural change in the frame-work of 1a .As-synthesized crystalline 1a changes to the amorphous state 1b after heating under reduced pressure.The

(14)More detailed measurement results are available in the Supporting

Information.

(15)The void space was calculated by PLATON.16The value was taking the

volume of NO 3-(77.0?3)into account.The size is calculated by Figure 3.Crystal structure of {[Cd(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a ).(a)The distorted 6-membered ring composed of three octahedral Cd(II)linked together by three 4-btapa units.(b)The framework expanding three directions to form a 3-D framework.Orange and sky blue represent Cd(II)and the framework.H atoms and guests are omitted for clarity.

Figure 4.(a -c)Two-fold interpenetrating 3-D crystal structure of 1a to form three-dimensionally running channels of dimensions 4.7×7.3?.2The perspective view is down from the b axis.Guests and H atoms are omitted for clarity.

Figure 5.Crystal structure of 1a to form another type of zigzag channels with dimensions of 3.3×3.6?.2Guests and H atoms are omitted for clarity.

Figure 6.View of the ordered amide groups on the channel surface.H 2O molecules are interacted by hydrogen bonds with the NH group of the amide groups.Light brown,red,blue,gray,and purple represent channel surface,Cd(II),O,N,C,and H,respectively.

Figure 7.Thermogravimetric analyses of (a)1a and (b)1c over the temperature range from 25to 450°C at a heating rate of 10°C min -1under N 2atmosphere.Amount of guests in figures is based on one 4-btapa ligand of each compound.

A R T I C L E S Hasegawa et al.

X-ray powder diffraction (XRPD)patterns for 1a and 1b are shown in Figure 8.This crystalline-to-amorphous transformation is accompanied by loss of the H 2O and DMF guest molecules in the channels,and the structural behavior has also been observed in other compounds.17However,when amorphous 1b is immersed in methanol or exposed to methanol vapor,it changes immediately to the original crystalline phase (1c ),which was detected by XRPD (Figure 8).We can rule out the possibility of recrystallization because of the low solubility and the results of experiments on vapor exposure.The amorphous-to-crystalline transformation is also induced by H 2O or DMF.18These types of flexible frameworks provide pore structures that are suitable for given guest molecules and much more useful for molecular recognition or selective guest inclusion than are robust porous structures.19

Selective Binding of Alcohols Inspired by Hydrogen Bonding.The desolvated host 1b can easily include certain alcohols via hydrogen bonding with the amide groups.1b was immersed in a few drops of various alcohols,and the XRPD was measured;several of these experiments are shown in Figure 9.Short-chain alcohols (methanol,ethanol,n -propanol,and n -butanol)exhibited guest inclusion with structural transforma-tion.Conversely,long-chain alcohols (n -pentanol and n -hexanol)caused no change in the XRPD patterns,indicating that 1b did not undergo a structural change with guest adsorption.This selective inclusion system reflects the size of the guest molecule.1b also exhibits unique selectivity responsive to functional groups of guests.Figure 10shows the XRPD patterns of 1b that was immersed in n -butanol,n -pentane,or n -pentene.Unlike its response to n -pentane and n -pentene,which have no OH groups,1b immersed in n -butanol exhibited guest inclusion with structural transformation.Considering the fact that n -butanol,n -pentane,and n -pentene are similar in size and shape,this selective inclusion system reflects the presence/absence,in the guest,of OH groups that would be available for hydrogen bonding.On the other hand,in IR spectra for 1a ,1b ,and 1a

with other guests,the NO symmetric stretching vibration band of NO 3-can be seen at 1385cm -1(Figure S1),supporting no coordination or strong interaction of nitrate anions with the amide moieties in the framework.20In the case of the chloride derivative of 1b ,the Cl -ions clearly interact with the amide moieties inside the channels,resulting in no porous properties such as sorption or catalysis.7c Therefore,weaker participation of anions such as NO 3-in this study might be important for successful formation of functional PCP system with the amide groups.

Gas Sorption Properties with Dynamic Structural Trans-formation.The mechanism of guest binding,with structural transformation,should be investigated in detail.The adsorption and desorption behaviors of some guests were examined.1b showed no N 2adsorption at 77K,as shown in Figure 11.This result indicates that 1b does not retain channels available for N 2(kinetic diameter )3.64?).21Because 1b undergoes an amorphous-to-crystalline transformation when exposed to metha-nol vapor,methanol adsorption and desorption experiments were carried out on 1b .The adsorption and desorption isotherms for

(17)(a)Maspoch,D.;Ruiz-Molina,D.;Wurst,K.;Domingo,N.;Cavallini,M.;

Biscarini,F.;Tejada,J.;Rovira,C.;Veciana,

J.Nat.Mater.2003,2,190-195.(b)Carlucci,L.;Ciani,G.;Moret,M.;Proserpio,D.M.;Rizzato,S.Angew.Chem.,Int.Ed.,2000,39,1506-1510.

(18)The XRPD patterns are shown in Figures S3and S4.

(19)(a)Lee,E.Y.;Jang,S.Y.;Suh,M.P.J.Am.Chem.Soc.2005,127,6374-6381.(b)Maji,T.K.;Mostafa,G.;Matsuda,R.;Kitagawa,S.J.Am.Chem.(20)(a)Newman,S.P.;Jones,W.J.Solid State Chem.1999,148,26-40.(b)

Figure 8.XRPD patterns of (a)1a simulation based on the single-crystal structure,(b)the as-synthesized 1a ,(c)the desolvated compound 1b ,which is obtained from 1a under vacuum for 7.5h at 140°C,and (d)the methanol-exchanged compound 1c obtained by exposing 1b to methanol vapor for 30h.

Figure 9.XRPD patterns for desolvated compound 1b immersed in a few drops of (a)methanol,(b)ethanol,(c)n -propanol,(d)n -butanol,(e)n -pentanol,and (f)n -hexanol solution.

Figure 10.XRPD patterns for desolvated compound 1b immersed in a few drops of (a)n -butanol,(b)n -pentane,and (c)n -pentene solution.

Selective Sorption and Catalysis A R T I C L E S

methanol at 298K are shown in Figure 11.The adsorption isotherm showed gradual uptake to P /P 0)0.9.Conversely,the desorption isotherm did not trace the adsorption profile and decreased slightly until a sudden drop at P /P 0)0.1,with a large-range hysteresis loop.This characteristic adsorption profile shows the conversion of amorphous 1b to the crystal 1c .The interaction between the host framework and methanol is strong enough to transform and maintain the channel structure,so that the large hysteresis profile appears.The amount of adsorbed methanol is 3.2molecules per 4-btapa ligand at P /P 0)0.9,which is close to that calculated from the TGA measurement for 1c .This result also supports the proposition that the adsorbed methanol is bound to the amide groups of 1c in a 1:1ratio.From these results,we conclude that the amide groups on the channel surfaces function effectively and give rise to an attractive interaction with the methanol guest molecules via hydrogen bonding.The structural transformation is inspired by the interaction between the guest molecules and the host framework.

Selective Inclusion and Activation of Reactants in Selective Catalytic Reactions.Amide groups have two sites that are involved in their function in different ways.In 1a ,the -NH moiety acts as a hydrogen donor for short-chain alcohols in the channels,whereas the -C d O moiety acts as an electron donor.Recently,a study reported a PCP with base carbonyl oxygen atoms as the catalytic interaction sites on the channel wall.4g We also expected the -C d O moiety in the channels of 1a to provide guest interaction sites as base sites.To confirm the selective accommodation and activation of reactants by 1a ,selective adsorption experiments were performed to identify the substrates of 1a .The substrates chosen for the reaction were malononitrile (molecular size,6.9×4.5?2),ethyl cyanoacetate (10.3×4.5?2),and cyano-acetic acid tert -butyl ester (10.3×5.8?2).22As-synthesized 1a was immersed in benzene contain-ing each reagent and stirred.The powder was filtered off and dried in air,and then its 1H NMR and IR spectra were acquired.The 1H NMR spectra showed that 1a adsorbs 2.9,0.7,and 0.6molecular stoichiometric amounts of malononitrile,ethyl cy-anoacetate,and cyano-acetic acid tert -butyl ester per 4-btapa ligand of 1a ,respectively (Figure 12).The inclusion amount of malononitrile for 1a was 4-5times larger than those of the other reagents,implying that the smallest molecule,malononi-trile,was most easily introduced in the channels of 1a .From

the IR spectra,only malononitrile interacted with the host 1a .In general,when an active methylene compound interacts with part of the R -H atom,the IR band attributed to the C t N stretching vibration shifts to a lower wavenumber and υs CN -υas CN splitting appears.23The υs CN -υas CN splitting at 2180and 2200cm -1was observed for malononitrile in 1a ,whereas the other reagents showed only the original peak (Figure 13).Malono-nitrile in the channels of 1a may interact with the -C d O moiety of the amide groups via hydrogen bonding.Unfortunately,the shift in the peak for the -C d O stretching vibration was not large because the amide groups of the original 1a had already interacted with the guest molecules,which maintain the channels of 1a .Therefore,it was difficult to observe the shift in the peak of the -C d O stretching band.It is worth noting that the IR spectrum of 1a containing malononitrile shows two types of guests:one interacting with the host 1a and the other not interacting.These 1H NMR and IR results indicate that the smallest molecule,malononitrile,is selectively adsorbed and activated in the channels of 1a .

Figure 11.Gas adsorption/desorption isotherms for the uptake of N 2,MeOH.P 0is the saturated vapor pressure,102.3kPa,of N 2(77K),and 16.94kPa,of methanol (298K).

Figure 12.1H NMR (DMSO-d 6)spectra of 1a that adsorbed each guest molecule:(a)malononitrile,(b)ethyl cyanoacetate,and (c)cyano-acetic acid tert -butyl ester.The peaks marked with an asterisk indicate the 4-btapa.

Figure 13.IR spectra in the region of C t N stretching vibration bands at room temperature of (a)malononitrile (reactant),(b)1a containing malononitrile,(c)1a containing ethyl cyanoacetate,and (d)1a containing cyano-acetic acid tert -butyl ester.The bands marked with an asterisk indicate the C t N stretching vibration bands due to each substrate that is not activated.

A R T I C L E S Hasegawa et al.

Base Catalytic Properties of 1a.Although several studies about PCPs with acid properties have been report-ed,4a,c -e,5e -f,24studies of PCPs with base properties have been few.4b,4g,25To characterize the base-type properties of 1a ,a Knoevenagel condensation reaction catalyzed by 1a was per-formed.The Knoevenagel reaction is well known,not only as a weak base-catalyzed model reaction but also as a reaction that generates a C -C bond.26However,to our knowledge,the reaction has never been reported in the context of PCP.Knoevenagel condensation reactions of benzaldehyde with each of the active methylene compounds (malononitrile,ethyl cy-anoacetate,and cyano-acetic acid tert -butyl ester)were catalyzed by 1a .As a result,the malononitrile was a good substrate,producing 98%conversion of the adduct,whereas the other substrates reacted negligibly (Table 2).This guest-selective reaction suggests that the reaction occurs in the channels and not on the surface of 1a .

Only as-synthesized 1a promoted the reaction with a good yield compared with desolvated 1b or the 4-btapa ligand (Figure 14a).Desolvated 1b has no channels,and the 4-btapa ligand has no free amide groups because it forms an intermolecular hydrogen bond among themselves.10We also checked that no reaction occurred with Cd(NO 3)2?4H 2O.These results also support that the reaction occurs within the channels function-alized with the amide group of 1a .The heterogeneity and recyclability of 1a in the Knoevenagel condensation of benzal-dehyde and malononitrile were examined.After the reaction mixture was stirred for 3h in the presence of 1a ,1a was removed by filtration.After removal of 1a from the reaction system,the reaction did not proceed further,demonstrating the

catalytic activity of 1a .The catalyst 1a shows good recyclability.1a is easily isolated from the reaction suspension by filtration alone and can be reused without loss of activity.The XRPD patterns of the 1a before and after the reactions were the same,indicating the high stability of 1a .Even when 1a changes its form to the amorphous state 1b ,it is easily recovered in the crystalline state by immersing it in DMF or short-chain alcohols,which cause the amorphous-to-crystalline structural transforma-tion of the host.Therefore,this compound containing effective FOS,amide groups,can be considered to exhibit a sufficient catalytic activity with base property.Under solvent-free condi-tions,remarkable improvements were observed in the conver-sion,via Knovenagel condensation of benzaldehyde with malononitrile,catalyzed by 1a (Figure 14b).The reaction of malononitrile (2mmol)with liquid benzaldehyde (10mL)was performed without benzene,and 1a achieved a conversion to the adduct of 100%after a 2.5h reaction time.

Conclusion

This work describes the construction of a 3D PCP containing guest-accessible amide groups and characterizes the selective inclusion of guest molecules with the structural transformation of the host.Moreover,we demonstrated a PCP that acts as a heterogeneous base catalyst based on a FOS located on its channel surfaces.

First,we successfully synthesized a 3D PCP functionalized with amide groups,{[Cd(4-btapa)2(NO 3)2]?6H 2O ?2DMF }n (1a ),from the reaction between Cd(NO 3)2?4H 2O and a three-connector-type amide ligand (4-btapa).The amide groups of 1a are ordered uniformly on the channel surfaces.1a undergoes reversible dynamic structural transformation via an amorphous state (1b )in response to the removal and rebinding of guest molecules.This is inspired by hydrogen bonding between the guest molecules and amide groups.We observed selective guest inclusion via the hydrogen bond,which is based on not only the size and shape of the incoming guest but also its affinity for the amide group,as demonstrated by X-ray crystallographic analysis and adsorption measurements.

Second,1a selectively accommodated and activated guests in its channels because of the active amide groups.As a result,the Knoevenagel condensation reaction,which is a well-known base-catalyzed model reaction,was selectively promoted in good yield.This selectivity depends on the relationship between the (24)Vimont,A.;Goupil,J.M.;Lavalley,J.C.;Daturi,M.;Surble ′,S.;Serre,

C.;Millange,F.;Fe ′rey,G.;Audebrand,N.

J.Am.Chem.Soc.2006,128,3218-3227.

(25)Shin,D.M.;Lee,I.S.;Chung,Y.K.Cryst.Growth Des.2006,6,1059-1061.

(26)(a)The number of published articles containing the keyword “Knoevenagel

condensation”is shown in Figure S5.(b)Rao,P.S.;Venkataratnam,R.V.Tetrahedron Lett.1991,132,5821-5822.(c)Reddy,T.I.;Varma,R.S.Tetrahedron Lett.1997,38,1721-1724.(d)Rodriguez,I.;Iborra,S.;Corma,A.;Rey,F.;Jorda ′,https://www.wendangku.net/doc/0e19149638.html,mun.1999,593-594.(e)Harjani,J.R.;Nara,S.J.;Salunkhe,M.M.Tetrahedron Lett.2002,43,Table 2.Knoevenagel Condensation Reaction of Benzlaldehyde with Substrates,Catalyzed by 1a

a

All reaction were performed with 2mmol substrates in 10mL of benzene using 0.1g (0.08mmol,4mol %)of the as-synthesized 1a for 12h.b Determined by 1H NMR,based on starting materials.

Figure 14.(a)Conversion (%)vs time (h)for Knoevenagel condensation reaction of benzaldehyde with malnonitrile in benzene catalyzed as-synthesized,1a (blue square),desolvated,1b (green circle),and 4-btapa ligand (black triangle).(b)Reaction of solvent-free condition (red circle)is much faster than that in benzene solution (blue square).Each conversion was detected by 1H NMR based on starting material.

Selective Sorption and Catalysis

A R T I C L E S

after complete guest removal,the porous crystalline structure can be easily recovered by guest adsorption and demonstrated the same catalytic activity as the initial one.

This research is particularly relevant in the context of porous solid-state chemistry in the generation of new materials with unusual properties.This result suggested that the combination of metal ions and designed organic ligand can provide PCP with FOS showing base catalytic performance.PCPs functionalized with FOS herald the next advance in porous materials that can be used as a fascinating class of adsorbents and heterogeneous catalysts.

Acknowledgment.This work was supported by a Grant-In-Aid for Science Research in a Priority Area“Chemistry of Coordination Space”(#434)and a CREST program from the Ministry of Education,Culture,Sports,and Science and Technology,Government of Japan.

Supporting Information Available:General experimental details and characterization data.This material is available free of charge via the Internet at https://www.wendangku.net/doc/0e19149638.html,.

JA067374Y

A R T I C L E S Hasegawa et al.

碱基互补配对规律试题

碱基互补配对规律的有关计算 设在双链DNA分子中一条为1链，另一条为2链，根据碱基互补配对原则，有：A1= ，A2= ，C1= ，C2= 。 由此可以推知： 公式一： = ， = ； 即在双链DNA分子中，配对的碱基数相等。 公式二： + = + 或 + = + =50% 即在双链DNA分子中，不互补的两碱基含量之和是相等的（嘌呤之和与嘧啶之和相等），占整个分子碱基总量的50%。 结论： 例1.某DNA的碱基中,鸟嘌呤的分子数占22％,那么胸腺嘧啶的分子数占多少 练习1：某DNA双链中,若腺嘌呤有P个,占全部碱基数的比例为N/M（M>2N）则该DNA分子中含胞嘧啶数是多少 设在双链DNA分子1链中：（A1+G1）/（T1+C1）=m 则：（A1+G1）/（T1+C1）=（T +C ）/A +G ）=m 因此2链中：（A2+G2）/（T2+C2）= 结论： 例的一条单链中A+G/T+C =,在互补单链和整个DNA分子中分别是多少 练习2：在双链DNA分子中,当T+C/A+G 在一条多核苷酸链上的比例是时,则在另一条互补链和整个DNA分子中,这种比例应分别是多少 设1链上的（A1+T1）/（C1+G1）=n，则其互补链2链的（A2+T2）/（C2+G2）= 结论：

例3. DNA的一条单链中A+T/G+C =,上述比例在其互补单链和整个DNA分子中分别是多少 （四）在一个双链DNA分子中,某碱基占碱基总量的百分数等于它在每条链中的平均百分数,若在其中一条链多占A％,则在另一条链应少占A％ 例4.在一个双链DNA分子中,G占碱基总量的18％,其中一条链中G占20％,那么此链中C 应为多少 练习4：在一个双链DNA分子中,A和T之和占碱基总量的56％,其中甲链中G占24％,那么乙链中G占多少 （五）A= ，T= ，C= ，G= （注意：此公式特指各碱基所占百分比，即A、T、C、G占整个双链DNA分子的百分比，A1 、A2 、T1 、T2、 C1、 C2、 G1 、G2占各单链的百分比。）例5：某分子中A与T之和占整个DNA分子碱基总数的54%，其中一条链中的G占该链碱基总数的22%，则另一条链中G占该链碱基总数的（） A、22% B、24% C、27% D、32% （六）整个DNA分子互补的两种碱基之和（A+T或C+G）所占比例=其每一单链中这两种互补碱基之和占单链中碱基数之比。 （七）1链中互补碱基之和等于2链中该互补碱基之和，又等于双链中该互补碱基之和的一半。 例7：已知双链DNA分子，其中一条单链中，A+T=n,其互补链中A+T= 在整个DNA分子中A+T= 知识拓展： （八）DNA分子复制前后某种碱基数量的计算 若某DNA分子含某种碱基x个，则该DNA分子进行n次复制，需含该碱基的脱氧核苷酸分子数=互补碱基的脱氧核苷酸分子数=(2n-1)x个。 例8：某DNA分子中碱基T的数量为a,若此DNA分子连续复制n次，所需游离的胸腺嘧啶脱氧核苷酸数为多少个 巩固练习 1.（1992年全国）下列哪项对双链DNA分子的叙述是不正确的？ A.若一条链A和T的数目相等,则另一条链A和T的数目也相等 B.若一条链G的数目为C的2倍,则另一条链G的数目为C的倍

浅议核酸中碱基含量的计算

浅议核酸中碱基含量的计算 在高中生物的试题中，常见有关核酸中碱基的计算问题，但中学生物教材对这部分内容未见叙述。以下是笔者近年来对这类题型解法所作的一些归纳。 遗传的物质基础告诉我们，在双链DNA分子中，碱基之间的连接是遵循碱基互补配对原则的，在DNA分子的复制及信息RNA的转录过程中也是遵循这一原则。由此可见，要解决碱基含量计算的问题，关键就是在于对碱基互补配对原则的理解及运用。 碱基互补配对原则对双链DNA分子中四种碱基的关系作了明确的阐述：嘌呤与嘧啶配对，且A=T，G=C。由此，我们可引伸出如下关系： 公式一：A+G=T+C或A+C=T+G 即在双链DNA分子中，不互补的两碱基含量之和是相等的，占整个分子碱基总量的50%。例1：已知一段信息RNA有30个碱基，其中A和G有12个，那么，转录成信息RNA 的一段DNA分子中，应有C和T是 A、12个 B、18个 C、24个 D、30个 分析：信息RNA有30个碱基，则其模板DNA分子中应有60个碱基，因T+C=A+G，故而T+C=60×50%=30，应选答案D。 例2：一个具有1000个碱基对的DNA分子，如果腺嘌呤的含量为20%，则含胞嘧啶的碱基对为 A、200个 B、300个 C、400个 D、600个 分析：根据碱基互补配对原则，A=T=20%，则G=C=30%，G+C=30%+30%=60%，然则含胞嘧啶的碱基对为60%，即1000×60%=600个，应选答案D。 公式二：即在双链DNA分子中，一条链中的嘌呤之和与嘧啶之和的比值与其互补链中相应的比值互为倒数。 例3：DNA的一条单链中，A和G的数量与T和C的数量之比为0.4，上述比例在其互补链中和整个DNA分子中分别是 A、0.4和0.6 B、2.5和1.0 C、0.4和0.4 D、0.6和1.0 分析：已知A左+G右/T左+C左=0.4，根据公式二，可得A右+G右/T右+C右=1/0.4=2.5，即其互补链中同样的比例值为2.5；而根据公式一，可得整个DNA分子中的比值应为1，故应选答案B。 公式三：即在双链DNA分子中，一条链中的两种碱基对的比值（A+T/G+C）与其在互补链中的比值和在整个分子中的比值都是一样的。 推理过程：根据碱基互补酸对原则，A左=T右，T左=A右，G左=C右，C左=G右，则同理可得：, 两式合并成为公式三。 例4：已知某DNA分子G与C占全部碱基数的48%，它的一条模板链中C与T分别占该链碱基总数的26%和24%，问由它转录的RNA链中尿嘧啶与胞嘧啶分别占碱基总数的 A、24%和22% B、28%和26% C、28%和22% D、48%和52%

碱基互补配对原则的有关计算

碱基互补配对原则的有关计算 一、规律总结 设在双链DNA分子中一条为α链，另一条为β链，根据碱基互补配对原则，有： ① ② ③ 由此可以推知： 1. 在双链DNA分子中： 结论1：在一个双链DNA分子中，嘌呤碱基总数＝嘧啶碱基总数（），各占全部碱基总数的50%。 2. 设在双链DNA分子α链中： 则： 因此β链中： 结论2：在一个双链DNA分子中，两条互补单链中的的值互为倒数。 3. 设在双链DNA分子α链中： 则： 整个DNA分子中： 结论3：在一个双链DNA分子中，互补配对的碱基之和在两条单链中所占比例与整个DNA分子中所占比例相同。 4. 不同生物的DNA分子中，其互补配对的碱基之和的比值不同，代表了每种生物DNA分子的特异性。（也可以作为结论4） 二、例题剖析

例1.若DNA分子的一条链中，则其互补链中该比值为（） A. a B. 1/a C. 1 D. 解析：已知链中，则未知链的A＋T和已知链的A＋T相等，未知链的G＋C与已知链的G＋C相等，故未知链的。 参考答案：A 例2. 一个DNA分子中有100个碱基对，其中有40个腺嘌呤，如果该DNA分子连续复制两次，则参与到DNA分子的复制中的游离的胞嘧啶脱氧核苷酸有（） A. 40个 B. 80个 C. 120个 D. 180个 解析：已知，由于A＝T，C＝G，且，∴C＝60。一个DNA分子复制两次得到4个DNA分子，新增加3个DNA分子，需要游离的胞嘧啶脱氧核苷酸为个。 参考答案：D 例3. 在DNA的一条单链中，，上述比例在其互补链和DNA分子中分别是（） A. 0.4和0.6 B. 2.5和1.0 C. 0.4和0.4 D. 0.6和1.0 解析：因为双链DNA分子一条链中两个不互补配对的碱基之和的比值等于另一互补链中这一比值的倒数，所以，如果DNA的一条单链中，则与其互补的另一条链中的这一比值应为。又因为在双链DNA分子中，两个不互补配对的碱基之和相等（即），所以在整个DNA分子中。 参考答案：B 例4. 下列哪项对双链DNA分子的叙述是错误的（） A. 若一条链A和T的数目相等，则另一条链A和T的数目也相等 B. 若一条链G的数目为C的两倍，则另一条链G的数目为C的0.5倍 C. 若一条链A：T：G：C＝1：2：3：4，则另一条链的相应碱基比为2：1：4：3 D. 若一条链A：T：G：C＝1：2：3：4，则另一条链的相应碱基比为1：2：3：4 解析：双链DNA分子中的碱基配对遵循碱基互补配对原则，A与T配对，G与C配对，即A＝T，G＝C。若一条链中A＝T，那么其互补链中也必定是A＝T；若一条链中G＝2C，由于其互补链中的C与G分别等于已知链中的G与C，因此互补链中的G（1/2）C；若已知链中的A：T：G：C＝1：2：3：4，根据碱基互补配对原则，其互补链中的相应碱基比就等于已知链中的T：A：C：G＝2：1：4：3。 参考答案：D

高考生物碱基计算公式及技巧

碱基计算（咼考生物重难点） 对于碱基计算的这类题目，关键是抓住几个基础的关系。 1、如果把DNA的两条链分别定为I链和n链的话，那么根据碱基互补配对原则，1链上的 腺嘌呤（A）一定等于n链上的胸腺嘧啶（T）, I链上的鸟嘌呤（G）—定等于n链上的胞嘧啶（C）,反之亦然。可简写为：A i = T n, T i = A n , Ci = Gn , Gi = C n o 2、根据A I= T n, T I= A n, C I= Gn, G = O,对于整个DNA分子来说，A的总量等于T的总量，C的 总量等于G的的总量。可简写为A总=丁总，G总=C总。 3、若DNA的I链中A+T的量占I链碱基总量的a %,由A】=T n、T】= A n及I链的碱基总量等于 H链的总量，可得在DNA的H链中A+T的量也占H链碱基总量的 a %。同理，可得m RNA中的A+U的量占m RNA碱基总量的a%。对于整个DNA分子来说，A总+「总=Ai+ A n + T n+ T I = 2 （Ai + T I），整个DNA分子的碱基总量等于2倍I链碱基总量，所以A总+ T总的量占整个DNA分子碱基总量的a %。可简写为：若（A I +T I） / I总=a% ,则（A n + T n）/ □总=a%,（ A 总+ T 总）/DNA 总=a% （A+U）/ m RNA= a% C+G同样有此关系：若（C I + G） / I 总=b% ,则（Cn+ Gn）/「总=b%,（6+ G总）/DNA总=b% （C+G / m RNA=b% 4、公式一：A+G=T+C或A+C=T+G 即在双链DNA分子中，不互补的两碱基含量之和是相等的，占整个分子碱基总量的50% 5、公式二： 丸星+ G左_ T右+ C右_ 1 耳+ Q A/G石A疳q 即在双链DNA分子中，一条链中的嘌呤之和与嘧啶之和的比值与其互补链中相应的比值互为倒数。6、公式三： 电左+耳_ A右+ T右_ A + T G/C左G右+ C右G + C 即在双链DNA分子中，一条链中的两种碱基对的比值（A+T/G+C）与其在互补链中的比值和在整个分子中的比值都是一样的。 推理过程：根据碱基互补酸对原则，A左=T右，T左=A右，G左=C右，C左=G右，则

习题课：DNA分子中的计算类型归纳

课时课题：第3章 DNA分子中的计算类型归纳课型：习题课

能力点运用碱基互补配对原则分析问题、解决问题自主探究点DNA分子中的碱基计算规律 易错易混点1、碱基比例的计算 2、DNA复制与细胞分裂结合的计算 训练点运用碱基互补配对原则分析问题 拓展点DNA分子中的碱基计算规律 教学流程 教学环节教师活动学生活 动 设计意图 目标展示：出示习题导学案，展示本节课的知识目标、能力目标、情感态度和价值观目标。 温故知新：复习提问： 1．DNA分子的基本单位是什么？2．DNA分子的复制方式和过程？3．DNA分子精确复制的原因？ 题型一：碱基互补配对原则的基础应用——基础知识引导学生合作探究： 应用碱基互补配对原则，总结导学案上的碱基规律。在黑板上画出DNA的两条链，标出1链、2链。 如下图

规律一：DNA双链中的两种互补的碱基相等。 即A=T，G=C。(最基本的规律) 规律二：DNA双链中，嘌呤碱基之和与嘧啶碱基之和相等， 任意两个不互补的碱基之和相等，占碱基总数的50%。 即 A+G=T+C = A+C=T+G=50% 规律三：按照碱基互补原则，DNA分子一条链中腺嘌呤A1在 该链中的比例等于互补链中的互补碱基胸腺嘧啶T2在互补链 中的比例。 即A1 = T2 T1 = A2 G1 =C2 C1 = G2 规律四：在DNA双链中，每条单链上任意两个互补的碱基和 之比相等，与DNA双链中的此比值相等。 即（A1+ T1）／单链=（A2+ T2）／单链=（A+T）／双链 （C1+G1）／单链=（C2+G2）／单链=（C+G）／双链例题讲解例1：已知某双链DNA分子中，G与C之和占全部碱基总数的34%，其一条链中的T与C分别占该链碱基总数的32%和 18%，则在它的互补链中，T和C分别占该链碱基总数的（ ） A．34%和16% B．34%和18% C．16%和34% D．32%和18% 解析：解此题目时，应先绘出两条链碱基符号，并注明含量，这样非常直观，便于推导和分析来理清解题思路，寻 求解决方法。明确DNA的两条链中含碱基数目相同，且A＝ T，G＝C，一条链中的A1、T1、G1、C1数量等于另一条链 中的T2、A2、C2、G2数量。

DNA分子中有关碱基的计算

5.方茴说："那时候我们不说爱，爱是多么遥远、多么沉重的字眼啊。我们只说喜欢，就算喜欢也是偷 偷摸摸的。" 6.方茴说："我觉得之所以说相见不如怀念，是因为相见只能让人在现实面前无奈地哀悼伤痛，而怀念 却可以把已经注定的谎言变成童话。" 7.在村头有一截巨大的雷击木，直径十几米，此时主干上唯一的柳条已经在朝霞中掩去了莹光，变得普普 通通了。 1."噢，居然有土龙肉，给我一块！" 2.老人们都笑了，自巨石上起身。而那些身材健壮如虎的成年人则是一阵笑骂，数落着自己的孩子，拎着骨棒与阔剑也快步向自家中走去。 DNA 分子中有关碱基的计算 周李邻（四川省邻水县第二中学） 一、 RNA 与DNA 的区别 RNA 有碱基U ,DNA 有碱基T 以此来区别 二、双链DNA 与单链DNA 的区别 有少数病毒是单链DNA 。双链DNA 一定是A=T 、C=G ， 如果A ≠T 、C ≠G 则为单链DNA 。 三、双链DNA 中的碱基关系 1 双链DNA 中A=T 、C=G 或A+G=T+C 解释：双链DNA 中 ①互补碱基相等 ②两个不互补碱基之和衡等 2 A+G=T+C=A+C=T+G=双链DNA 中的碱基总数的50% 解释：双链DNA 中任意两个不互补碱基之和占总碱基的50%。 例1 某基因中，有腺嘌呤300个，占全部碱基的30%，求基因中胞嘧啶有多少个？ 3在DNA 分子的两条互补链之间 ①A 1+G 1 T 1+C 1 A 2+G 2 T 2+C 2

5.方茴说："那时候我们不说爱，爱是多么遥远、多么沉重的字眼啊。我们只说喜欢，就算喜欢也是偷 偷摸摸的。" 6.方茴说："我觉得之所以说相见不如怀念，是因为相见只能让人在现实面前无奈地哀悼伤痛，而怀念 却可以把已经注定的谎言变成童话。" 7.在村头有一截巨大的雷击木，直径十几米，此时主干上唯一的柳条已经在朝霞中掩去了莹光，变得普普 通通了。 1."噢，居然有土龙肉，给我一块！" 2.老人们都笑了，自巨石上起身。而那些身材健壮如虎的成年人则是一阵笑骂，数落着自己的孩子，拎着骨棒与阔剑也快步向自家中走去。 解释DNA 分子的两条互补链之间A+G T+C 互为倒数关系 ② A 1+T 1G 1+C 1 A 2+T 2G 2+C 2 A+T G+C 解释 DNA 分子的两条互补链之间A+T G+C 的比值相等并且与整 个DNA 比值相等。 4整个DNA 分子与其子链之间 ①A 1+T 1 A 1+T 1+C 1+G 1 A 2+T 2+C 2+G 2 A+T A+T+C+G A 2+T 2 解释：整个DNA 分子中互补碱基之和占总碱基的比例=每一条单链中互补碱基之和占该单链总碱基的比例。 例2 某双链DNA 片段中，A 占23%，其中一条链中C 占该单链的24%，问另一条链中C 占多少？ 例3 一个DNA 分子中G 和C 之和占全部碱基的46%，又知在该DNA 分子的H 链中A 和C 分别占碱基数的28%和22%，则该DNA 分子H 链的互补链中A 和C 分别占该链碱基的比例为 。

碱基互补配对的解题方法总结

碱基计算题目的解题技巧 对于碱基计算的这类题目，关键是抓住几个基础的关系。只要能掌握三个关系式，灵活运用这三个关系式，就可解决一般的碱基计算的题目。这三个关系式分别是： 1、如果把DNA的两条链分别定为Ⅰ链和Ⅱ链的话，那么根据碱基互补配对原则，Ⅰ链上的腺嘌呤（A）一定等于Ⅱ链上的胸腺嘧啶（T），Ⅰ链上的鸟嘌呤（G）一定等于Ⅱ链上的胞嘧啶（C），反之亦然。 可简写为：AⅠ＝TⅡ，TⅠ＝AⅡ，CⅠ＝GⅡ，GⅠ＝CⅡ。 2、根据AⅠ＝TⅡ，TⅠ＝AⅡ，CⅠ＝GⅡ，GⅠ＝CⅡ，对于整个DNA分子来说，A的总量等于T的总量，C的总量等于G的的总量。 可简写为A总=T总，G总=C总。 3、若DNA的Ⅰ链中A+T的量占Ⅰ链碱基总量的a％，由AⅠ＝TⅡ、TⅠ＝AⅡ及Ⅰ链的碱基总量等于Ⅱ链的总量，可得在DNA的Ⅱ链中A+T的量也占Ⅱ链碱基总量的a％。同理，可得ｍRNA中的A+U的量占ｍRNA碱基总量的a％。对于整个DNA分子来说，A总＋T总＝A Ⅰ＋AⅡ＋TⅡ＋TⅠ＝2（AⅠ＋TⅠ），整个DNA分子的碱基总量等于２倍Ⅰ链碱基总量，所以A总＋T总的量占整个DNA分子碱基总量的a％。 可简写为：若（AⅠ＋TⅠ）/Ⅰ总＝a％,则（AⅡ＋TⅡ）/Ⅱ总＝a％，（A总＋T总）/DNA总＝a%，（A+U）/ｍRNA＝a%。C+G同样有此关系：若（CⅠ＋GⅠ）/Ⅰ总＝b％,则（CⅡ＋G Ⅱ）/Ⅱ总＝b％，（C总＋G总）/DNA总＝b%，（C+G）/ｍRNA＝b%。 下面利用这三个关系式解几个有代表性的题目： 1、（1991年全国高考试题） DNA的一个单链中（A+G）/（T+C）＝0.4，上述比例在其互补链和整个DNA子中分别是（）A、0.4和0.6 B、2.5和1.0 C、0.4和0.4 D、0.6和1.0 解析在DNA分子中AⅠ＝TⅡ、TⅠ＝AⅡ、CⅠ＝GⅡ、GⅠ＝CⅡ，据题意的得（AⅠ＋GⅠ）/（TⅠ＋CⅠ）＝0.4，那么（TⅡ＋CⅡ）/（AⅡ＋GⅡ）＝0.4,即（AⅡ＋GⅡ）/（TⅡ＋CⅡ）＝2.5。因为A总=T总、G总=C总，对于整个DNA分子来说（A总＋G总）/（T总＋C总）＝1.0。 ［答案］B。 2、（1999年上海高考题） 若DNA分子一条链中（A+T）／（C+G）＝c，则其互补链中该比例为（） A、c B、1/c C、1 D、1－1/c 解析方法1：由AⅠ＝TⅡ，TⅠ＝AⅡ，CⅠ＝GⅡ，GⅠ＝CⅡ得（AⅠ＋TⅠ）/(GⅠ＋CⅠ)＝（TⅡ＋AⅡ）/（CⅡ＋GⅡ）＝c。 方法2：（AⅠ＋TⅠ）/Ⅰ总＝a％, （CⅠ＋GⅠ）/Ⅰ总＝b％,已知（AⅠ＋TⅠ）/（CⅠ＋G Ⅰ）＝a％/b％=c,则由（AⅡ＋TⅡ）/Ⅱ总＝a％，（CⅡ＋GⅡ）/Ⅱ总＝b％得（AⅡ＋TⅡ）/（CⅡ＋GⅡ）＝a％/b％=c。同时还可得（A总＋T总）/（C总＋G总）＝c。 ［答案］A。 3、（2002年山东省会考模拟题）

有关碱基计算的练习题!! 超级棒!!(⊙o⊙)哦!!

i求配对的碱基名称 1．在DNA分子中，能与腺嘌呤互补配对含氮碱基是（） A．腺嘌呤或胞嘧啶 B．胸腺嘧啶 C．胞嘧啶 D．鸟嘌呤 2．DNA分子的双链在复制时解旋，这时下述哪一对碱基从氢键连接处分开（）A．鸟嘌呤与胸腺嘧啶 B．鸟嘌呤与尿嘧啶 C．鸟嘌呤与胞嘧啶 D．腺嘌呤与尿嘧啶 3．DNA分子复制时，解旋酶作用的部位应该是（） A．腺嘌呤与鸟嘌呤之间的氢键 B．腺嘌呤与胞嘧啶之间的氢键 C．鸟嘌呤与胞嘧啶之间的氢键 D．脱氧核糖与含氮碱基之间的化学键 4．以DNA的一条链“—A—T—C—”为模板，经复制后的子链是（） A．—T—A—G— B．—U—A—G— C．—T—A—C— D．—T—U—G— 5．DNA分子的一条母链上的部分碱基排列顺序已知为ACGT，那么以另一条母链为模板，经复制后子链的碱基排列顺序应是（） A．TGCA B．ACGT C．UGCA D．ACGU ii求DNA分子中碱基和其它的数量 1．在一个DNA分子中共有碱基200个，其中一条链含胞嘧啶为20个，其互补链共有胞嘧啶26个，问这个DNA分子中含T多少（） A．54个 B．92个 C．108个 D．46个 2．在某个DNA分子的片段中，测得A占碱基总数的21%，那么C占碱基总数的百分比为（）A．21% B．29% C．79% D．无法计算 3．某DNA分子片段中胞嘧啶有240个，占全部碱基的30％，则该片段中腺嘌呤有（）A．240个 B．48个 C．800个 D．160个 4．DNA分子中的某一区段上有300个脱氧核糖和60个胞嘧啶，那么该区段胸腺嘧啶的数量是（） A．90 B．120 C．180 D．240 5．已知某DNA分子中共有个x个脱氧核苷酸，A+T/G+C=n，则该DNA中有腺嘌呤（）A．nx/(2n+2)个 B．nx/(n+1)个 C．(x+1)/n个 D．(x﹣1)/n个 6．从分析某DNA分子的成份得知，含有胸腺嘧啶脱氧核苷酸20％，其数目为400个。那么该DNA分子中有C—G碱基对是（） A．600 B．1200 C．1000 D．2000 7．一个DNA分子中有腺嘌呤1500个，腺嘌呤与鸟嘌呤之比是3﹕1，则这个DNA分子中含有脱氧核糖的数目为（） A．2000个 B．3000个 C．4000个 D．8000个 8．在一DNA分子片段中有200个碱基对，其中腺嘌呤有90个。因此在这个DNA片段中含有游离的磷酸基的数目和氢键的数目依次为（） A．200个和400个 B．2个和510个 C．2个和400个 D．400个和510个9．DNA分子中所含的嘌呤数必定等于（） A．T和C的总数 B．A和U的总数 C．C和G的总数 D．A和T总数

生物遗传选择题计算规律

生物遗传选择题计算规律（及例题） 生物试题中的计算题主要是通过计算考查学生对生物知识的理解程度。高中生物教材中多处涉及计算问题却没有详细的讲解，试题变化多样，所以是学习的难点。突破难点的方法是总结规律。 一、遗传物质基础的有关计算 1．有关碱基互补配对原则的计算 双链DNA分子中A=T，G=C，A+G=T+C，(A+G/T+C＝1)。DNA分子中互补碱基之和的比值【（A＋T）/（G＋C）】和每一个单链中的这一比值相等；DNA分子中一条链中的两个不互补碱基之和的比值【（A＋G）/（C＋T）】是另一个互补链的这一比值的倒数。 例题.某DNA分子的一条链（A+G）/（T+C）=2，这种比例在其互补链和整个DNA分子中分别是（） A.都是2 B.0.5和2 C.0.5和1 D.2和1 解析：根据碱基互补配对原则A=T C=G，整个DNA分子中（A+G）/ （T+C）=1；已知DNA分子的一条链（A+G）/（T+C）=2，推出互补链中（T+C）/（A+G）=2，（A+G）/（T+C）=1/2。 答案：C 2．DNA复制的有关计算 公式：X=A(2n-1) X代表DNA复制过程中需要游离的某脱氧核苷酸数；A代表亲代DNA中该种脱氧核苷酸数，n表示复制次数。

例题．某DNA分子共有a个碱基，其中含胞嘧啶m个，则该DNA分子复制3次，需要游离的胸腺嘧啶脱氧核苷酸数为 （） A. 7(a－m) B. 8(a－m) C. 7(a /2－m) D. 8(2a－m) 解析：根据碱基互补配对原则A=T C=G，该DNA分子中T的数量是(a- 2m)/2, 该DNA分子复制3次，形成8个DNA分子，共有T的数量是4(a-2m)，复制过程中需游离的胸腺嘧啶脱氧核苷酸数是：4 (a-2m)－[(a-2m)/2]= 7(a /2－m)。 答案: C 3．基因控制蛋白质合成的有关计算 信使RNA上决定一个氨基酸的三个相邻碱基称为一个密码子，决定一个氨基酸，信使RNA是以DNA（基因）一条链为模板转录生成的，所以，DNA分子碱基数：RNA分子碱基数：氨基酸数=6：3：1 例题：一段原核生物的mRNA通过翻译可合成一条含有11个肽键的多肽，则此mRNA分子至少含有的碱基个数及合成这段多肽需要的tRNA个数，依次为 A．33 11 B．36 12 C．12 36 D．11 36 解析：一条含有11个肽键的多肽是由12个氨基酸缩合形成的。mRNA上三个碱基决定一个氨基酸，则此mRNA分子至少含有的碱基36个，一个氨基酸需要一个tRNA转运，共需要12个tRNA。 答案:B 二、有关遗传基本规律的计算

碱基互补配对原则的公式及应用

碱基互补配对原则的公式及应用 1．最基本规律 在数量上，两个互补的碱基相等，任意两个不互补的碱基之和恒等。 即A甲链= T乙链，T甲链= A乙链C甲链= G乙链G甲链= C乙链 A+G=C+T A+C=G+T 2．在双链DNA分子中，任意两个不互补的碱基之和占总碱基数的50％。 即 【例1】在一个双链DNA分子中，含有35％的腺嘌呤，它所含的胞嘧啶应该是： A、15％ B、30％ C、35％ D、70％ 解： 因此答案为A。 3、在双链DNA分子中，配对的两种碱基的含量之和，在整个DNA分子及每条链中都相等A% + T% = A甲链％+ T甲链％= A乙链％+ T乙链％ C% + G% = C甲链％+ G甲链％= C乙链％+ G乙链％ 【例2】某细菌的一个DNA分子中含有20％的A+T，那么由它转录合成的信使RNA中的G+C

的含量应是 A、80％ B、60％ C、40％ D、20％ 解：∵A% + T% = A甲链％+ T甲链％= A乙链％+ T乙链％＝20％ ∴G% + C% = G甲链％+ C甲链％= G乙链％+ C乙链％＝80％ ∴信使RNA中G+C的含量应是80％。 因此答案为A。 4、在双链DNA分子中，某种碱基在两条链中的含量之和等于该碱基在整个DNA分子中含量的2倍。 2·A％＝A甲链％+ A乙链％；2·T％＝T甲链％+ T乙链％； 2·G％＝G甲链％+ G乙链％；2·C％＝C甲链％+ C乙链％ 【例3】某DNA 分子中，A＝20％，其中一条链中A＝10％，以 这条链为模板合成RNA分子中A占： A、10％ B、20％ C、30％ D、40％ 解：∵2·A％＝A甲链％+ A乙链％ ∴A乙链％＝2·A％－A甲链％＝2×20％－10％＝30％

DNA分子中有关碱基的计算

DNA分子中有关碱基的计算 周李邻（四川省邻水县第二中学） 一、RNA与DNA的区别 RNA有碱基U ,DNA有碱基T以此来区别 二、双链DNA与单链DNA的区别 有少数病毒是单链DNA。双链DNA一定是A=T、C=G，如果A≠T、C≠G则为单链DNA。 三、双链DNA中的碱基关系 1 双链DNA中A=T、C=G或A+G=T+C 解释：双链DNA中①互补碱基相等②两个不互补碱基之和衡等 2A+G=T+C=A+C=T+G=双链DNA中的碱基总数的50% 解释：双链DNA中任意两个不互补碱基之和占总碱基的50%。 例1某基因中，有腺嘌呤300个，占全部碱基的30%，求基因中胞嘧啶有多少个？ 3在DNA分子的两条互补链之间 ①A1+G1 T1+C1A 2+G2 T2+C2 解释DNA分子的两条互补链之间 A+G T+C互为倒数关系

② A 1+T 1G 1+C 1A 2+T 2G 2+C 2A+T G+C 解释 DNA 分子的两条互补链之间A+T G+C 的比值相等并且与 整个DNA 比值相等。 4整个DNA 分子与其子链之间 ①A 1+T 1 A 1+T 1+C 1+G 1 A 2+T 2+C 2+G 2 A+T A+T+C+G A 2+T 2 解释：整个DNA 分子中互补碱基之和占总碱基的比例=每一条单链中互补碱基之和占该单链总碱基的比例。 例2 某双链DNA 片段中，A 占23%，其中一条链中C 占该单链的24%，问另一条链中C 占多少？ 例3 一个DNA 分子中G 和C 之和占全部碱基的46%，又知在该DNA 分子的H 链中A 和C 分别占碱基数的28%和22%，则该DNA 分子H 链的互补链中A 和C 分别占该链碱基的比例为 。 ②A 1 A 1+T 1+C 1+G 1 A 2+T 2+C 2+G 2 A A+T+C+G A 2 1 2 变通 A 1 A 1+T 1+C 1+G 1 A+T+C+G 12 A 1+T 1+C 1+G 1 G 1 A 1G 1 解释：整个DNA 分子中某一碱基所占的比例=该碱基在每一单链中所占比例之和的一半。 例 4一个DNA 分子的一条单链上腺嘌呤比鸟嘌呤多40%，

碱基互补配对原则的应用

碱基互补配对原则的应用 设DNA 一条链为1链，互补链为2链。根据碱基互补配对原则 可知：A1=T2 ， A2=T1， G1 = C2 ， G2 =C1。 则在DNA 双链中： A = T ， G = C 与DNA 结构有关的碱基计算 （A+G ）/（T+C ）= （A+C ）/（T+G ）= （A+G ）/（A+T+G+C ）= （A1+T1）/（A2+T2）= （G1+C1）/（G2+C2）= （A1+G1）/（T1+C1）=a ，则（A2+G2）/（T2+C2）= （A+T ）/（A+T+G+C ）=a ，则（A1+T1）/（A1+T1+G1+C1）= 1.DNA 分子的一条单链中（A+G ）／（T+C ）=0.5，则另一条链和整个分子中上述比例分别等于（ ） A ．2和1 B 0.5和0.5 C ．0.5和1 D ．1和1 2.实验室里，让一个DNA 分子（称第一代）连续复制三次，问：在第四代DNA 分子中，有第一代DNA 分子链的DNA 分子占（ ） A.100% B. 75% C. 50% D. 25% 3、某双链DNA 分子中，G 占23%，求A 占多少？________________________ 4、在DNA 的一个单链中，A+G/T+C=0.4，上述比例在其互补链和整个DNA 分子中分别是多少？ _______________________若DNA 的一个单链中，A+T/G+C=0.4，上述比例在其互补链和整个DNA 分子中分别是多少？__________________________ 5、某双链DNA 分子中，A 与T 之和占整个DNA 碱基总数的54%，其中一条链上G 占该链碱基总数的22%。求另一条链上G 占其所在链碱基总数的百分含量?_____________ 6、某DNA 分子中A+T 占整个DNA 分子碱基总数的34%，其中一条链上的C 占该链碱基总数的28%，那么，对应的另一条互补链上的C 占该链碱基总数的比例是多少？__________ 7.分析一个DNA 分子时，发现30%的脱氧核苷酸含有A ，由此可知，该分子中一条链上G 含量的最大值可占此链碱基总数的多少？_________________ 8．用15N 标记某噬菌体DNA ，然后再浸染细菌，设细菌破裂后共释放出16个噬菌体，问这些噬菌体中有几个不含15N ？____________________ 9．已知双链DNA 分子中的一条单链中m C T G A =++，求： （1）在另一互补链中上述比例为_________，据此可得推论：___________。 （2）在整个DNA 分子中上述比例为__________，据此可得推论：___________。 若在一单链中，n C G T A =++时，求： （3）在另一互补链中上述比例为_________； （4）在整个DNA 分子中上述比例为_____________。 据（3）和（4）可得推论_____________。 10．在试管内合成DNA 的实验过程是：先把高能磷酸基团接到四种脱氧核苷酸上，然后加入DNA 解旋酶和DNA 聚合酶，最后放入一个带有15 N 标记的DNA 分子，根据下述实验结果，请回答：

碱基互补配对原则演绎的计算规律专题训练(理科)

碱基互补配对原则演绎的计算规律专题训练(理科) 碱基互补配对原则 规律一： DNA双链中的两个互补的碱基相等，即双链中A=T，G=C； 双链中任意两个不互补的碱基之和恒等，占碱基总数的50%，即A+G=T+C=A+C=T+G=50%， （A+G）/（T+C）=（A+C）/（G+T）=（T+C）/（A+G）=1。 一般情况下，A+T≠G+C。 规律二： 在DNA双链中，一条单链的（A1+T1）/（G1+C1）的值，与另一条互补链的（A2+T2）/（G2+C2）的值是相等的，也与整个DNA分子的（A+T）/（G+C）的值是相等的。 规律三： 在DNA双链中，一条单链的（A1+G1）/（T1+C1）的值，与另一条互补单链的（A2+G2）/（T2+C2）是互为倒数的。 规律四： 在双链DNA及其转录的RNA之间，有下列关系： 在碱基数量上，在DNA单链和RNA的单链中，互补碱基的和相等，且等 于双链DNA的一半，即A1+T1= A2+T2=RNA分子中（A+U）=1/2DNA双链中（A+T），G1+C1= G2+C2=RNA分子中（G+C）=1/2DNA双链中（G+C）。 典型例题分析 例1已知一段双链DNA分子中碱基的对数和腺嘌呤的个数，能否知道这段DNA中4种碱基有比例和（A+C）：（T+G）的值（） A.能 B.否 C.只能知道（A+C）：（T+C） D.只能知道4种碱基的比例 〖解析〗：本题着重考查对双链DNA结构碱基互补配对原则的全面掌握，强调分析推理能力。根据题意，已知碱基总数和A的个数，由于A=T，G=C，A+G=50%，因此，4种碱基的比例均可以求出，且任何DNA中（A+C）：（T+G）=1。〖答案〗：A 例2某DNA分子含腺嘌呤520个，占碱基总数的20%，该DNA分子中含胞嘧啶（） A.350 B.420 C.520 D.780 〖解析〗：考查DNA分子中碱基之间的等量关系.在DNA分子中A=T,G=C,A=T=520.则A+T=40%, G+C=1-40%=60%,所以G=C=30%.在DNA分子中碱基总数为520÷20%=2600个,则DNA分子中含胞嘧啶为:2600×30%=780个. 〖答案〗D 例3一个DNA分子的一条链上,腺嘌呤比鸟嘌呤多40%,两者之和占DNA分子碱基总数的24%,则这个DNA分子的另一条链上,胸腺嘧啶占该碱基数目的( ) A.44% B.24% C.14% D.28% 〖答案〗：D 例4 .下列哪项对双链DNA分子的叙述是不正确的（） A.若一条链G的数目为C的2倍，则另一条链G的数目为C的0.5倍 B.若一条链A和T的数目相等，则另一条链A和T的数目也相等 C.若一条链的A∶T∶G∶C=1∶2∶3∶4，则另一条链相应碱基比为2∶1∶4∶3 D.若一条链的G∶T = 1∶2，则另一条链的C∶A= 2∶1 解析：考查DNA分子中4种碱基的比例关系。根据双链DNA分子碱基互补配对原则，DNA分子一条链上的碱基A、T、G、C分别与互补链上的T、A、C、G一一对应形成碱基对。则有一条链上G／C值与互补链上的C／G值相等；一条链上A／T值与互补链上T／A值相等；一条链上A∶T∶G∶C等于互补链上T∶A∶C∶G；一条链上G／T值等于互补链上C／A值。 答案：D

高二生物物碱基计算思路解析

高二生物物碱基计算思路解析 高二生物物碱基计算思路 根据碱基的互补配对原则，在双链DNA分子1和2表示DNA分子的两条互补链中有以下碱基数量关系： A=T C=G A1=T2，A2=T1 C1=G2，C2=G1 不难看出：只要知道任意两种不互补的碱基在双链中的各自总量和在任一单链中的各自数量条件，就能得出每种碱基在每条链中的数量结果;如果只知道任意两种不互补的碱基在双链中的各自总量和某一碱基在任一单链中的各自数量条件，那也能得出每种碱基在双链中的总量，以及该“某一碱基”和其互补碱基在每条链中的数量结果，而与其不互补的碱基在每条链中的数量无法得出。如果将DNA中全部碱基相对总量看成100%，那么同理可得不同碱基在DNA双链或者单链中的百分比了。 当然,命题条件“任意两种不互补的碱基在双链中的各自总量和在任一单链中的各自数量”可以间接给出。常见的是给出DNA分子碱基总量和某一碱基总量或所占比例。 解答有关碱基计算的问题，通常通过画图表征题目条件，使问题形象明了。 例108年上海高考题某个DNA片段由500对碱基组成，A+T占碱基总数的34%，若该DNA片段复制2次，共需游离的胞嘧啶脱氧核苷酸分子个数为 A.330 B.660 C.990 D.1320 答案：C 解析：A+T占碱基总数的34%，A就占17%，C就占33%，该DNA片段中胞嘧啶脱氧核苷酸分子个数就为500×2×33%=330，若该DNA片段复制2次，共得4个相同DNA片段，因为是半保留复制，故共需游离的胞嘧啶脱氧核苷酸分子个数为330×4—1=990。 例2.在一个DNA分子中，胞嘧啶脱氧核苷酸和鸟嘌呤脱氧核苷酸之和占碱基总量的46%。其中一条链上的腺嘌呤脱氧核苷酸占所在链碱基总量的28%，另一条链上腺嘌呤脱氧核苷酸占所在链碱基总量的__________。 答案：26%

碱基互补配对规律

碱基互补配对原则的一般规律及相关计算 1）一个双链DNA分子中，A=T，C=G，A+G=C+T 故(A+G)% = (T+C)%=(A＋C）％=(T+G)％= 50% （A＋T）/（C＋G）≠1 反应了生物的特异性 例：某DNA分子含腺嘌呤520个，占碱基总数的20％，则该DNA分子中含胞嘧啶（ 780 ） 2）双链DNA分子一条链中(A+G)/(C+T)比值等于其互补链中该种碱基的比值的倒数。 例：已知在DNA分子中的一条单链（A+G）/(T+C)=m,求： （1）在另一条互补链中这一比例是 （2）这个比例在整个DNA分子中是 3）双链DNA分子中(A1+T1)％＝（A2+T2)%=(A+T)%＝（A＋U）％ 同理：(C1+G1)％＝（C2＋G2）％＝（C＋G）％ 如：若在一双链DNA分子中鸟嘌呤和胞嘧啶之和占碱基总和的44％，在其中的一条链中A和C分别占该链碱基数的22％和30％，那么在另一条链中腺嘌呤和胞嘧啶分别占该链碱基数的比值为： 解：C＋G＝44％， A1＝22％，T1 ＝30％ 求： A2＝？ C2 ＝？ 某双链DNA分子中，鸟嘌呤与胞嘧啶之和占全部碱基的54%，其中a链的碱基中，22%是腺嘌呤，28%是胞嘧啶，则b链中腺嘌呤占该链碱基的比例和胞嘧啶占整个DNA分子碱基的比例分别为 A．24%、13% B．23%、27% C．48%、26% D．22%、28% 有关DNA中的碱基计算 ①一个DNA连续复制n次后，共有多少个DNA？多少条脱氧核苷酸链？母链多少条？ 子链多少条？

练习 1、假定大肠杆菌含14N 的DNA 的相对分子质量为a ，若将其长期培养在含15N 的培养基中，便得到含15N 的DNA ，相对分子质量为b ，现将含15N 的DNA 大肠杆菌再培养在含14N 的培养基中，那么，子二代DNA 的相对分子质量平均为：（3a ＋b ）/4 2、假设将含有一对同源染色体的精原细胞的DNA 分子用15N 标记，并供给14N 的原料，该细胞进行减数分裂产生的4个精子中，含15N 标记的DNA 的精子所占比例为：100％ 根据半保留复制和碱基互补配对原则 ①一个DNA 连续复制n 次后，共有多少个DNA ？多少条脱氧核苷酸链？母链多少条？子链多少条？ DNA 分子数= 2 n 脱氧核苷酸链数= 2n +1 母链数= 子链数= 2 2 n +1 ﹣2 “ 半保留复制”和“碱基互补配对原则”，并图示分析。 A T G C C G A T A T G C C G A T A T G C C G A T A T G C C G A T A T G C C G A T A T G C C G A T 连续第一次复制 连续第二次复制 连续第n 次复制 解： 答：一个DNA 连续复制n 次后，共有2n 个DNA ，2n +1 条脱氧核苷酸链，母链2条，子链2n +1 ﹣2条 有关DNA 中的碱基计算

DNA复制和碱基计算问题专题

DNA复制和碱基计算问题专题 1．（）一个DNA片段（如图所示），下列相关叙 述正确的是 A．把该DNA放在含14N的培养液中复制两代，子代 DNA中不含有15N的DNA占3/4 B．②处碱基对的缺失将导致染色体变异 C．作用于①和③的酶的种类肯定是不一样的 D．如该DNA片段含120个磷酸基团，则其指导合成的蛋白质最多含40个氨基酸 2．（）一个双链均被32P标记的噬菌体DNA上有x个碱基对，其中腺嘌呤有m个。用这个噬菌体侵染只含31P的大肠杆菌，共释放出128个子代噬菌体。下列叙述正确的是 A．该过程至少需要64(x－m)个胞嘧啶脱氧核苷酸 B．噬菌体增殖需要细菌提供原料、模板和酶等 C．只含31P与含32P的子代噬菌体的比例为63∶1 D．若该DNA上的基因发生突变，则其长度一定改变 3．（）将玉米的一个根尖细胞放在含3H标记的胸腺嘧啶脱氧核苷酸的培养基中完成一个细胞周期，然后将子代细胞转入不含放射性标记的培养基中继续培养。下列关于细胞内染色体的放射性标记分布情况的描述，正确的是 A．第二次分裂结束只有一半的细胞具有放射性 B．第二次分裂结束具有放射性的细胞可能有4个 C．在第二次分裂的中期每条染色体的两条单体都被标记 D．在第二次分裂的中期只有半数的染色体中一条单体被标记 4．（）用32P标记了玉米体细胞(含20条染色体)的DNA分子双链，再将这些细胞转入不含32P 的培养基中培养，在第二次细胞分裂完成后每个细胞中被32P标记的染色体条数是 A．0条 B．20条 C．大于0小于20条 D．以上都有可能 5．（）假设一个双链均被32P标记的噬菌体DNA由1000个碱基对组成，其中腺嘌呤占全部碱基的20%。用这个噬菌体侵染只含31P的大肠杆菌，共释放出100个子代噬菌体。下列叙述正确的是 A．该过程至少需要29 700个鸟嘌呤脱氧核苷酸 B．噬菌体增殖需要细菌提供模板、原料和DNA聚合酶等 C．含32P与只含31P的子代噬菌体的比例为1∶49 D．若该DNA发生突变，则其控制的性状即发生改变

“碱基互补配对原则”规律归纳与应用

“碱基互补配对原则”规律归纳与应用 关于高中生物第二册第六章《遗传和变异》中的“碱基互补配对原则”，笔者发现许多同学不容易理解、不会灵活应用。 “碱基互补配对原则”规律是：A（腺嘌呤）一定与T（胸腺嘧啶）配对；G（鸟嘌呤）一定与C（胞嘧啶）配对，（嘌呤内部和嘧啶内部都不能进行配对）。并且在以下关系中都有此规律：DNA——DNA、DNA——RNA、RNA——RNA等。由此，笔者总结出如下公式并举例说明： 公式一：A=T，G=C；A 1=T 2， A 2 =T 1 ，C 1 =G 2 ，C 2 =G 1 即在双链DNA分子中，配对的碱基数相等。 公式二：A+G=T+C或A+C=T+G=50% 即在双链DNA分子中，不互补的两碱基含量之和是相等的（嘌呤之和与嘧啶之和相等），占整个分子碱基总量的50%。 例1：某信使RNA的碱基中，U占20%，A占10%，则作为它的模板基因DNA分子中胞嘧啶占全部碱基的（） A、70% B、60% C、35% D、17.5% 分析：信使RNA中，A+U=10%+20%=30%，则G+C=70%，根据公式五，得知模板基因DNA分子中也有同样比例，即G+C=70%，由公式一，则C=1/2×70%=35%，由此可得答案为C。 公式三： 即在双链DNA分子中，不互补的两碱基数之和的比值等于1。 例2：已知一信使RNA有碱基30个，则转录该信使RNA分子中C与T有多少个？分析：单链RNA是以DNA的一条链为模板，按照碱基互补配对原则合成的，又RNA中有碱基30个，所以该DNA分子中有碱基60个，由公式三得出A+G=T+C= 60/2=30个。 公式四： 即在双链DNA分子中，一条链中的嘌呤之和与嘧啶之和的比值与其互补链中相应的比值互为倒数。 例3：若某DNA分子的一条链中（A+G）/（T+C）=2.5，则1：其互补链中，（A+G）/（T+C）为多少？2：该DNA分子中（A+G）/（T+C）为多少？ 分析：设已知链为1链，未知链为2链，1：由公式四，得出=2.5，所以（A 2+G 2 ） /（T 2+C 2 ）=1/2.5=0.4； 2：由公式三，得出（A+G）/（T+C）=1 公式五： 即在双链DNA分子中，一条链中的两种碱基对的比值与其在互补链中的比值和在整个分子中的比值都是一样的。 例4：已知某DNA分子G与C占全部碱基数的48%，它的一条模板链中C与T分别占该链碱基总数的26%和24%，问由它转录的RNA链中尿嘧啶与胞嘧啶分别占碱基总数的 A、24%和22% B、28%和26% C、28%和22% D、48%和52%