染色7
最大度为5的图用7种颜色染色
Also available at http://amc.imfm.siISSN1855-3966(printed edn.),ISSN1855-3974(electronic edn.)ARS MATHEMA TICA CONTEMPORANEA4(2011)153–164 Graphs with maximum degree5are acyclically7-colorable∗Alexandr V.Kostochka†Department of Mathematics,University of Illinois,Urbana,IL61801,USA Sobolev Institute of Mathematics,Novosibirsk,630090,RussiaChristopher StockerDepartment of Mathematics,University of Illinois,Urbana,IL61801,USAReceived12May2010,accepted21January2011,published online24March2011AbstractAn acyclic coloring is a proper coloring with the additional property that the union of any two color classes induces a forest.We show that every graph with maximum degree at most5has an acyclic7-coloring.We also show that every graph with maximum degree at)-coloring.most r has an acyclic(1+ (r+1)24Keywords:Acyclic coloring,maximum degree.Math.Subj.Class.:05C15,05C351IntroductionA proper coloring of the vertices of a graph G=(V,E)is an assignment of colors to the vertices of the graph such that no two adjacent vertices receive the same color.A proper coloring of a graph G is acyclic if the union of any two color classes induces a forest. The acyclic chromatic number,a(G),is the smallest integer k such that G is acyclically k-colorable.The notion of acyclic coloring was introduced in1973by Gr¨u nbaum[8]and turned out to be interesting and closely connected to a number of other notions in graph coloring.Several researchers felt the beauty of the subject and started working on problems and conjectures posed by Gr¨u nbaum.Michael Albertson was among the enthusiasts and wrote in total four papers on the topic[1,2,3,4].∗Dedicated to the memory of Michael Albertson†Supported in part by NSF grants DMS-0650784and DMS-0965587and grant09-01-00244-a of the Russian Foundation for Basic Research.E-mail addresses:kostochk@(Alexandr V.Kostochka),stocker2@(Christopher Stocker)Copyright c 2011DMFA Slovenije154Ars Math.Contemp.4(2011)153–164In particular,Gr¨u nbaum studied a(r)–the maximum value of the acyclic chromatic number over all graphs G with maximum degree at most r.He conjectured that for every r,a(r)=r+1and proved that his conjecture holds for r≤3.In1979,Burstein[6] proved the conjecture for r=4.This result was proved independently by Kostochka[10]. It was also proved in[10]that for every k≥3,the problem of deciding whether a graph is acyclically k-colorable is NP-complete.It turned out that for large r,Gr¨u nbaum’s con-jecture is incorrect in a strong sense.Albertson and Berman mentioned in[1]that Erd˝o s proved that a(r)=Ω(r4/3− )and conjectured that a(r)=o(r2).Alon,McDiarmid and Reed[5]sharpened Erd˝o s’lower bound to a(r)≥c r4/3/(log r)1/3and proved thata(r)≤50r4/3.(1.1) This established almost the order of the magnitude of a(r)for large r.Recently,the prob-lem of estimating a(r)for small r was considered again.Fertin and Raspaud[7]showed among other results that a(5)≤9and gave a linear-time algorithm for acyclic9-coloring of any graph with maximum degree5.Furthermore,for everyfixed r≥3,they gave a fast algorithm that uses at most r(r−1)/2colors for acyclic coloring of any graph with maximum degree r.Of course,for large r this is much worse than the upper bound(1.1),but for r<1000,it is better.Hocquard and Montassier[9] showed that every5-connected graph G with∆(G)=5has an acyclic8-coloring.Kotha-palli,Varagani,Venkaiah,and Yadav[12]showed that a(5)≤8.Kothapalli,Satish,and Venkaiah[11]proved that every graph with maximum degree r is acyclically colorable with at most1+r(3r+4)/8colors.This is better than the bound r(r−1)/2in[7]for r≥8.The main result of this paper isTheorem1.1.Every graph with maximum degree5has an acyclic7-coloring,i.e.,a(5)≤7.We do not know whether a(5)is7or6,and do not have a strong opinion about it.Our proof is different from that in[7,9,12]and heavily uses the ideas of Burstein[6]: he started from an uncolored graph G with maximum degree4and colored step by step more and more vertices(with some recolorings)so that each of partial acyclic5-colorings of G had additional good properties that enabled him to extend the coloring further.The proof yields a linear-time algorithm for acyclic coloring with at most7colors of any graph with maximum ing this approach we also show that for everyfixed r≥6, there exists a linear-time algorithm giving an acyclic coloring of any graph with maximumdegree r with at most1+ (r+1)24 colors.This is better than the bounds in[7]and[11]cited above for every r≥6.In the next section we introduce notation,prove two small lemmas and state the main lemma.In Section3we prove Theorem1.1modulo the main lemma.In Section4we derive linear-time algorithms for acyclic coloring of graphs with bounded maximum degree.In the last section we give the proof of the main lemma.2PreliminariesLet G be a graph.A partial coloring of G is a coloring of some subset of the vertices of G.A partial acyclic coloring is then a proper partial coloring of G containing no bicolored cycles.A.V.Kostochka and C.Stocker:Graphs with maximum degree5are acyclically7-colorable155Given a partial coloring f of G,a vertex v is(a)rainbow if all colored neighbors of v have distinct colors;(b)almost rainbow if there is a color c such that exactly two neighbors of v are colored with c and all other colored neighbors of v have distinct colors;(c)admissible if it is either rainbow or almost rainbow;(d)defective if v is an uncolored almost rainbow vertex such that at least one of the two of its neighbors receiving the same color is admissible.A partial acyclic coloring f of a graph G is rainbow if f is a partial acyclic coloring of G such that every uncolored vertex is rainbow.A partial acyclic coloring f of a graph G is admissible if either f is rainbow or one vertex is defective and all other uncolored vertices are rainbow.In these terms,a color-ing is rainbow if it is admissible and has no defective vertices.Note that both,rainbow and admissible colorings are partial acyclic colorings where additional restrictions are put only on uncolored vertices.The advantage of using admissible colorings is that they pro-vide a stronger induction condition that places additional restrictions only on coloring of neighbors of uncolored vertices.So,the fewer uncolored vertices remain,the weaker these additional restrictions are.All colorings in this section will be from the set{1,2,...,7}.Lemma2.1.Let v be a vertex of degree4in a graph G with∆(G)≤5.Let f be an admissible(respectively,rainbow)coloring in which v is colored with color c1,each of the neighbors of v is colored,and exactly3colors appear on the neighbors of v.If at least one of the two neighbors of v receiving the same color and one of the other two neighbors of v each have a second(i.e.,distinct from v)neighbor with color c1,then we can recolor v and at most one of its neighbors so that the coloring remains admissible(respectively, rainbow).In particular,the new partial acyclic coloring has no new defective vertices. Moreover,if we need to recolor a vertex other than v,then we may choose a vertex with5 colored neighbors and recolor it with a color incident to v in f.Proof.Let N(v)={z1,z2,z3,z4},f(z1)=f(z2)=c2,f(z3)=c3,f(z4)=c4.Let z2 and z3be the neighbors of v with colors c2,and c3that are also adjacent to another vertex of color c1.We may assume that z2is adjacent to a vertex of color c5,since otherwise when we recolor v with c5,no bicolored cycles appear and the coloring remains admissible (respectively,rainbow).Similarly,we may assume that z2is adjacent to vertices of colors c6and c7.Then we may recolor z2with c3and repeat the above argument to get that z3also is adjacent to vertices with colors c5,c6,and c7.In this case,we may change the original coloring by recoloring z3with c2and v with c3.So,in this case only v and z3change colors.Note that either only v changes its color,or z2receives color c3,or z3receives color c.For partial colorings f and f of a graph G,we say that f is larger than f if it colors more vertices.Lemma2.2.Let v be a vertex of degree4in a graph G with∆(G)≤5.Let f be a rainbow coloring in which v is colored with color c1,the neighbors z1,z2,and z3of v receive the distinct colors c2,c3,and c4,the neighbor z4of v is an uncolored rainbow vertex.Then either G has a rainbow coloring f1that colors the same vertices and differs from f only at v,or G has a rainbow coloring f larger than f.Moreover,if the former does not hold,156Ars Math.Contemp.4(2011)153–164then z4has degree5and exactly one uncolored neighbor,say z4,4,and we can choose the larger coloring f so that all the following are true:1.Every vertex colored in f is still colored.2.Vertex z4is colored.3.The only uncolored vertex apart from z4that may get colored is z4,4,and it does onlyif it has neighbors of colors c1,c2,c3,and c4.4.Apart from v,only one vertex w may change its color,and if it does,then(a)w is aneighbor of z4,(b)w has four colored neighbors,(c)it changes a color in{c5,c6,c7} to another color in{c5,c6,c7},and(d)z4gets the former color of w.In particular, v is admissible in f .Proof.Let v,z1,z2,z3,and v4be as in the hypothesis.We may assume that z4is adjacent to a vertex z4,1of color c5:otherwise,since v4is rainbow,when we recolor v with c5, the new coloring will be rainbow.Similarly,we may assume that z4is adjacent to vertices z4,2,and z4,3of colors c6and c7.If z4has no other neighbors,then we can recolor v with c5and color z4with c1.So,assume that z4has thefifth neighbor,z4,4.If z4,4is colored,then f(z4,4)∈{c2,c3,c4},since z4is rainbow.In this case,we let f (z4)=c1 and f (v)=c5.So,we may assume that z4,4is not colored.If z4,4has no neighbor of color c2,then coloring z4with c2leaves the coloring rainbow and makes it larger than f. Thus,we may assume that z4,4has a neighbor of color c2and similarly neighbors of colors c3and c4.If z4,4has no neighbor of color c1,then we let f (z4)=c1and f (v)=c5.So, let z4,4have such a neighbor.If z4,1has no neighbor of color c2,then by coloring z4with c2and z4,4with c5,we get a rainbow coloring larger than f.So,we may assume(by symmetry)that z4,1has neighbors of colors c2,c3,c4.If z4,1has no neighbor of color c1,then we let f (z4)=c1, f (z4,4)=c5,and f (v)=c6.Finally,if z4,1also has a neighbor of color c1,then we let)=c6and f (z4)=c5.f (zThe next lemma is our main lemma.We will use it in the next section and prove in Section5.Lemma2.3.Let f be an admissible partial coloring of a5-regular graph G.Then G has a rainbow coloring f that colors at least as many vertices as f.3Proof the the TheoremFor convenience,we restate Theorem1.1.Theorem.Every graph with maximum degree5has an acyclic7-coloring.Proof.Let G be such a graph.If G is not5-regular,form G from two disjoint copies of G by adding for each v∈V(G)of degree less than5an edge between the copies of v.Repeating this process at mostfive times gives a5-regular graph G∗containing G as a subgraph.Since an acyclic7-coloring of G∗yields an acyclic7-coloring of its subgraph G,we may assume that G is5-regular.Let f be an admissible coloring of G from the set{1,2,...,7}with the most colored vertices.By Lemma2.3,we may assume that f is rainbow.A.V.Kostochka and C.Stocker:Graphs with maximum degree5are acyclically7-colorable157Let H be the subgraph of G induced by the vertices left uncolored by f.Let x be a vertex of minimum degree in H.We consider several cases according to the degree d H(x).Case1:d H(x)=0.Since f is rainbow,any color in{1,2,...,7}−f(N G(x))can be used to color x contradicting the maximality of f.Case2:d H(x)=1.Since f is rainbow,we may assume that x is adjacent to vertices of colors1,2,3,and4.Let y be the uncolored neighbor of x.Since y is rainbow,coloring x with5gives either a rainbow coloring or an admissible coloring with the defective vertex y having the admissible neighbor x,a contradiction to the maximality of f.Case3:d H(x)=2.We may assume that x is adjacent to vertices with colors1,2,3, and two uncolored vertices y1and y2.Since in our case y1is adjacent to at most3colored vertices,some color c∈{4,5,6,7}does not appear on the neighbors of y1.Coloring x with c then yields either a rainbow coloring,or an admissible coloring with defective vertex y2and its admissible neighbor x,a contradiction to the maximality of f.Case4:d H(x)=3.We may assume that x is adjacent to vertices of colors1and2. By the choice of x,each uncolored vertex of G has at most2colored neighbors.Since the three uncolored neighbors of x have at most6colored neighbors in total,some color c∈{3,4,5,6,7}is present at most once among these6neighbors.Then coloring x with c again yields an admissible coloring,a contradiction to the maximality of f.Case5:d H(x)≥4.Since each vertex of G has at most one colored neighbor,at most5colors are used in the second neighborhood of x.Hence x may be colored to give a rainbow coloring with more colored vertices.We conclude that H is empty and that f is an acyclic5-coloring of G.4AlgorithmsTheorem4.1.There exists a linear time algorithm forfinding an acyclic7-coloring of a graph with maximum degree5.Proof.The proof of the Theorem1.1,along with Lemmas2.1–2.3gives an algorithm. In order to control the efficiency of the algorithm we make the following modification: whenever the proof checks whether a vertex v is in a two-colored cycle,we check only for such a cycle of length at most12,and if we do notfind such a short cycle,then check whether two bicolored paths of length6leave v.This is enough,since the existence of such paths already makes the proofs of Theorem1.1and all the lemmas work.So,we need only to consider a bounded(at most56)number of vertices around our vertex.It then suffices to compute the running time of this algorithm.Let n be the number of vertices in G.The process of creating a5-regular graph takes O(n)time since we apply this process at most 5times,each time on at most25n vertices,each of degree at most5.We may now assume that G is a5-regular graph.We then create and maintain6databases D j,j=0,1,...,5 (say doubly linked lists),each for the set of vertices with degree j in the current H.At the beginning,all vertices are in D5,and it is possible to update the databases in a constant amount of time each time a vertex gains or loses a colored neighbor.Since there are at most 25n possible searches for a vertex with the minimum number of uncolored neighbors,all the searches and updates will take O(n)time.Note that the processes of Lemma2.1and Lemma2.2also take a constant amount of time to complete.Observe that each of the cases in Lemma2.3eitherfinds a rainbow coloring,orfinds an admissible coloring with more colored vertices,or reduces to a previous case in an amount of time bounded by a constant. Also when Lemma2.3processes a defective vertex,it yields either a rainbow coloring,or158Ars Math.Contemp.4(2011)153–164a larger admissible coloring and the next defective vertex in a constant time.Finally,since we start from an uncolored graph and color each additional vertex in a constant time,the implied algorithm colors all vertices in O(n)time.For a partial coloring f of a graph G and a vertex v∈V(G),we say that u∈V(G)is f-visible from v,if either vu∈E(G)or v and u have a common uncolored neighbor. Theorem4.2.For everyfixed r,there exists a linear(in n)algorithmfinding an acycliccoloring for any n-vertex graph G with maximum degree r using at most1+ (1+r)24colors.Proof.We start from the partial coloring f0that has no colored vertices,and for i= 1,...,n at Step i obtain a rainbow partial acyclic coloring f i from f i−1by coloring one more vertex(without recoloring).The algorithm proceeds as follows:at Step i,choose an uncolored vertex v i with the most colored neighbors.Greedily color v i with a colorαi inC:={1,...,1+ (1+r)24 }that is distinct from the colors of all vertices f i−1-visible fromv i.We claim that we always canfind suchαi in C.Suppose that at Step i,v i has exactly k colored neighbors.Then it has at most r−k un-colored neighbors,and each of these uncolored neighbors has at most k colored neighbors. So,the total number of vertices f i−1-visible from v i is at mostk+(r−k)k=k(r+1−k)≤(r+1)2=|C|−1,and we canfind a suitable colorαi for v i.It now suffices to show that for each i,the coloring f i is rainbow and acyclic.For f0, this is obvious.Assume now that f i−1is rainbow and acyclic.Since v i is rainbow in f i−1, coloring it withαi does not create bicolored cycles.Thus,f i is acyclic.Also sinceαi is distinct from the colors of all vertices f i−1-visible from v i,f i is rainbow.For the runtime,note that at Step i the algorithm considers only v i and vertices at distance at most2from v i.As in the proof of Theorem4.1,it is sufficient to maintain r+1 databases each containing all vertices with a given number of colored neighbors.This allows a constant time search for a vertex with the greatest number of colored neighbors. Moving a vertex as its number of colored neighbors changes takes a constant amount of time.Choosing and coloring v i together with updating the databases then takes O(r2) time.Hence the running time of the algorithm is at most c r n,where c r depends on r.5Proof of Lemma2.3We will prove that under the conditions of the lemma,either its conclusion holds or there is an admissible coloring f larger than f.Since G isfinite,repeating the argument eventually yields either an acyclic coloring of the whole G or a rainbow coloring.In both cases we do not have defective vertices.Let H be the subgraph of G induced by the uncolored vertices.Let x be the sole defective vertex under f and let y1,y2,...,y5be its neighbors.By the definition of a defective vertex,x has two neighbors of the same color.We will assume that f(y1)= f(y2)=1and that y1is admissible.When more then two neighbors of x are colored,we assume for i=3,4,5that if y i is colored,then f(y i)=i−1.Also for i=1,...,5,the four neighbors of y i distinct from x will be denoted by y i,1,...,y i,4(some vertices willA.V.Kostochka and C.Stocker:Graphs with maximum degree5are acyclically7-colorable159 have more than one name,since they may be adjacent to more than one y i).We consider several cases depending on d H(x).Case1:d H(x)=0.First we try to color x with colors5,6,and7.If this is not allowed,then for j=5,6,7,G has a1,j-colored y1,y2-path.This forces that both of y1 and y2have neighbors with colors5,6,and7,each of which is adjacent to another vertex of color1.In particular,both y1and y2are admissible.For i=1,2and j=1,2,3,we suppose that f(y i,j)=j+4and y i,j is adjacent to another vertex of color1.Case1.1:For some i∈{1,2},y i,4is colored and f(y i,4)/∈{5,6,7}.By symmetry, we may assume that i=1and f(y1,4)=2.Recolor y1with3and call the new admissible coloring f .If we can now recolor y2so that the resulting coloring f is rainbow on G−xy2−xy1or the only defective vertex in f on G−xy2−xy1is y2,4,then we do this recoloring and color x with1.Since y1and y2have no neighbors of color1apart from x,we obtained an admissible coloring of G larger than f.If we cannot recolor y2 to get such a coloring,then y2,4is colored with a color c∈{5,6,7}.Moreover,in this case by Lemma2.1applied to y2in coloring f of G−xy2−xy1,we can change the colors of only y2and some y∈{y2,1,y2,2,y2,3,y2,4}to get an admissible coloring f1of G−xy2−xy1.Moreover,by Lemma2.1,f1(y)∈{5,6,7}.Then by coloring x with1 we obtain a rainbow coloring of G,as above.Case1.2:y1,4is not colored.By Lemma2.2for vertex y1in G−xy1,either G−xy1 has a rainbow coloring f that differs from f only at y1(in which case by symmetry,we may assume that f (y1)=3and proceed further exactly as in Case1.1),or G−xy1has a larger rainbow coloring f satisfying statements1)–4)of Lemma2.2.In particular,by4), none of y2,y3,y4,y5changes its color and y1remains admissible.Thisfinishes Case1.2.By the symmetry between y1and y2,the remaining subcase is the following.Case1.3:f(y1,4)=5and f(y2,4)=c∈{5,6,7}.By Lemma2.1applied to y1 in G−xy1,we can recolor y1and at most one other vertex(a neighbor of y1)to obtain another admissible coloring f .If f (y1)∈{5,6,7},then f is a rainbow coloring,as claimed.So,we may assume that f (y1)=c1∈{2,3,4}.If all the colors5,6,7are present on neighbors of y2,then again by Lemma2.1(applied now to y2in coloring f of G−xy2),G has an admissible coloring f that differs from f only at y2and maybe at one neighbor of y2.Then coloring x with1we get a rainbow coloring.So,some color in{5,6,7}is not present in f (N(y2)).By Lemma2.1,this may happen only if y1,1is a common neighbor of y1and y2,and c=f(y2,4)=5.In particular,in this case,y1,1has neighbors of colors1(they are y1and y2),2,3,and4.Since c=5,we may assume that c=6.By the symmetry between y1and y2,we conclude that,in f,vertex y2,2also is a common neighbor of y1and y2and has neighbors of colors1(they are y1and y2),2,3,and 4.Returning to coloring f ,we see that y2has no neighbors of color5,and its neighbors y1,1(formerly of color5)and y2,2(by the previous sentence)also have no neighbors of color5.So,recoloring y2with5yields an admissible coloring of G.Now coloring x with 1creates a larger rainbow coloring.Case2:d H(x)=1.Wefirst try to color x with4.If no bicolored cycle is formed, then either we have a rainbow coloring or an admissible coloring with defective vertex y5and an admissible neighbor x.Hence we may assume that coloring x with4creates a bicolored cycle.This then gives each of y1and y2a neighbor of color4.A similar argument gives each of y1and y2a neighbor of color5,6,and7,i.e.,both y1and y2 are rainbow.Recoloring y1with color2allows us to repeat the argument at y3.Then y3also has neighbors of each of the colors4,5,6,and7.If y5has no neighbor of color160Ars Math.Contemp.4(2011)153–1642,then recoloring(in the original coloring f)y3with1,and coloring x with2yields a rainbow coloring.So,by the symmetry between colors1,2,and3,we may assume that for i∈{1,2,3},f(y5,i)=i.Since y5is rainbow,by the symmetry between colors4,5,6, and7,we may assume that either f(y5,4)=4,or y5,4is not colored.In both cases,recolor (in the original coloring f)y3with1,color x with2and y5with5.We get an admissible coloring larger than f,where only y5,4may be defective.Case3:d H(x)=3.If one of the uncolored neighbors y3,y4,y5(say,y3)of x has4 colored neighbors,then we may color y3with some c/∈f(N(y3))∪{1}and thus create an admissible coloring larger than f.Hence we may assume that each of y3,y4,and y5has at most3colored neighbors.Case3.1:One of y1and y2has three neighbors of different colors such that each of these neighbors has another neighbor of color1.Suppose for example that for j=1,2,3, f(y1,j)=1+j and y1,j has another neighbor of color1.If y1has a fourth color,say c, in its neighborhood,then we recolor y1with a color c /∈{1,c,5,6,7}and get a rainbow coloring of G.Suppose now that color c∈{5,6,7}appears twice on N(y1).Then by Lemma2.1applied to y1in G−xy1,we can change the color of y1and at most one other vertex that is a neighbor of y1not adjacent to uncolored vertices to get another rainbow coloring of G−xy1.Then this coloring will also be a rainbow coloring of G.Finally, suppose that y1has an uncolored neighbor y1,4.Applying Lemma2.2to y1in G−xy1 we either recolor only y1and get a rainbow coloring of G(finishing the case),or obtain a rainbow coloring f of G−xy1larger than f satisfying the conclusions of the lemma. Since each of y3,y4and y5has at least two neighbors left uncolored by f,none of them may play role of z4or z4,4in Lemma2.2when they get colored.Then f is an admissible coloring of G where only x could be a defective vertex with admissible neighbor v.This proves Case3.1.Let T be the set of colors c such that more than one of the vertices y3,y4and y5has a neighbor of color c.Since y3,y4and y5have in total at most9colored neighbors,|T|≤4.Case3.2:|T|≤3.By symmetry,we may assume that T⊆{2,3,4}.If coloring x with c∈{5,6,7}does not create a bicolored cycle,then it will yield an admissible coloring larger than f.So,we may assume that each of y1and y2has in its neighborhood vertices of colors5,6,and7,each of which is adjacent to another vertex of color1.So,we have Case3.1.Case3.3:|T|=4.Let T={2,3,4,5}.As in Case3.1,we may assume that each of y1and y2is adjacent to vertices of colors6and7,each of which have another neighbor of color1.Let y3have exactly3colored neighbors labeled y3,1,y3,2,y3,3with colors2,3,4.Let y3,4be the uncolored neighbor of y3.Then if y3,4has no neighbor of color5,we may color y3with5to get a new admissible coloring.Hence y3,4is adjacent to a vertex of color5. Similarly,y3,4has neighbors of color6and7.By symmetry,we may assume that a vertex of color2is adjacent to at most one of y4and y5.Case3.3.1:y3,4has no neighbor of color1.We try to color y3with1and x with2. If this does not produce a new admissible coloring,then one of y1or y2,say y1,has a neighbor of color2that is adjacent to another vertex of color1.So,we again get Case3.1.Case3.3.2:y3,4has a neighbor of color1.If y3,1has no neighbor of color1,then we again try to color y3with1and x with2,but also color y3,4with2.Then we simply repeat the argument of Case3.3.1.So,suppose that y3,1has a neighbor of color1.If y3,1has no neighbor of some colorα∈{5,6,7},then we color y3,4with2and y3withα.Thus y3,1A.V.Kostochka and C.Stocker:Graphs with maximum degree5are acyclically7-colorable1614,14,24,34,45,15,25,35,4Case4.3.Case4.3.4.3.Figure1:Cases4.3and4.3.4.3from the proof of Lemma2.3.has neighbors of colors1,5,6,7.Then we recolor y3,1with3and color y3with2.Case4:d H(x)=2.As at the beginning of Case3,we conclude that each of the uncolored vertices y4and y5has at least one uncolored neighbor besides x.Let B be the set of colors appearing in the neighborhoods of both,y4and y5.By the previous paragraph,|B|≤3.Case4.1:|B|≤1.We may assume that{4,5,6,7}∩B=∅.Try to color x with 4.By the definition of B,either a two-colored cycle appears,or we get a new admissible coloring larger than f.Hence we may assume that coloring x with4creates a bicolored cycle.Since this cycle necessarily goes through y1,y1is adjacent to a vertex with color 4.Similarly,y1is adjacent to vertices with colors5,6,and7.Then recoloring y1with3 yields a rainbow coloring of G.Case4.2:|B|=2.If1∈B or2∈B,then the argument of Case4.1holds.Assume that B={3,4}.Similarly to Case4.1,we may assume that for i=1,2and j=1,2,3, y i is adjacent to a vertex y i,j of color j+4that is adjacent to another vertex of color1(in particular,y1and y2may have a common neighbor of color j+4).If y1is rainbow,then uncoloring y1and coloring x with7gives Case1or Case2.Thus we may assume that y1and(by symmetry)y2are not rainbow.So,we may assume that for i=1,2,the fourth neighbor y i,4of y i distinct from x has color c i∈{5,6,7}.By symmetry,we may assume that c1=5.Similarly to Case1.3,by Lemma2.1applied to y1in G−xy1,we can recolor y1and at most one other vertex(a neighbor of y1)to obtain another rainbow coloring f of G−xy1.If f (y1)∈{3,4,5,6,7},then f is a rainbow coloring of G,as claimed.So,we may assume that f (y1)=2.Now practically repeating the argument of Case1.3,wefind a promised coloring.Case4.3:|B|=3(see Figure1).If2∈B,then we can repeat the argument of Case 4.2for B =B−{2}.Hence we may assume that B⊆{1,3,4,5,6,7}.Case4.3.0:1∈B.Let B={1,3,4}.Then some color in{5,6,7},say7,is not present on N(y4)∪N(y5).Again,we may assume that for i=1,2and j=1,2,3,y i is162Ars Math.Contemp.4(2011)153–164adjacent to a vertex y i,j of color j+4that is adjacent to another vertex of color1.If y1is rainbow,then we may uncolor y1and color x with7to get Case1or Case2.Suppose now that y1and y2are not rainbow.By Lemma2.1applied to y1in G−xy1,we can recolor y1 and at most one other vertex(a neighbor of y1)to obtain another admissible coloring f . If f (y1)∈{3,4,5,6,7},then f is a rainbow coloring,as claimed.So,we may assume that f (y1)=2.But then we can use the argument of Case4.2with the roles of y3and y2 switched.This proves Case4.3.0.So,from now on,B={3,4,5}.For i=4,5and j=1,2,3,let y i,j be the neighbor of y i of color j+2.We write the neighbor,since y4and y5are rainbow.As observed at the beginning of Case4,y4and y5each have another uncolored neighbor,call them y4,4and y5,4.In particular,y4and y5have no neighbors colored with6or7.If x can be colored with either of6or7without creating a two-colored cycle,then we obtain a rainbow coloring. Hence we assume that for i=1,2and j=1,2,f(y i,j)=j+5and y i,j has a neighbor of color1distinct from y i.Case4.3.1:One of y1or y2,say y1,is rainbow.If y4,4has no neighbor of color c∈{6,7},then we can color y4with c,a contradiction to the maximality of f.If y4,4has no neighbor of color c ∈{1,2},then by uncoloring y1and coloring y4with c and x with 6,we obtain an admissible coloring larger than f.So,f(N(y4,4))={1,2,6,7}.Then we may color y4,4with3and uncolor y1to get a new admissible coloring as large as f with one defective vertex y4,for which Case2holds.Thisfinishes Case4.3.1.So,below y1and y2are not rainbow and hence each of them is adjacent to at least three colored vertices.Case4.3.2:One of y1or y2,say y1,is adjacent to an uncolored vertex y1,4=x.We may assume that f(y1,1)=f(y1,2)=6and f(y1,3)=7.First,we try to color x with7 and y1with3.Since the new coloring has at most one defective vertex,we may assume that a two-colored cycle is created.Hence each of y1,1and y1,2is adjacent to a vertex of color3.The same argument gives these vertices neighbors of colors4and5.Recall that one of y1,1and y1,2,say y1,1,has another neighbor of color1.Then recoloring y1,1with2 gives an admissible coloring in which y1is rainbow.Hence Case4.3.1applies to this new coloring.So,from now on each of y1and y2has4colored neighbors.Since y1is admissible we may assume w.l.o.g.that y1is adjacent either to the colors5,6,6,7or the colors5,5,6,7.Case4.3.3:y1has one neighbor of color5and three neighbors with colors6or7.We may assume that f(y1,1)=5,f(y1,2)=f(y1,3)=6,and f(y1,4)=7.If coloring y1 with3or4yields an admissible coloring,then we are done;so we may assume that a two-colored cycle is formed in each case.It follows that each of y1,2and y1,3has neighbors colored with3and4.By the symmetry between y1,2and y1,3,we may assume that y1,3 has a neighbor of color1other than y1.If y1,3is almost rainbow,then we can uncolor it, recolor y1with3,and color x with7:this will give an admissible coloring with the same number of colored vertices as in f,and the only defective vertex y1,3.Then either Case1 or Case2applies to this new coloring.Hence we may assume that y1,3has two neighbors other than y1that receive the same color.Then since y1,3has no neighbor of color2,y1 may now be recolored with color2without creating a bicolored cycle.Repeating the above argument we derive that y1,2has neighbors of colors2,3,and4,and one of these colors appears twice on N(y1,2)−y1.By Lemma2.1applied to y1,3in the graph G−y1,3y1for the original coloring,we can change its color and the color of at most one other vertex(that is a neighbor of y1,3,all of whose neighbors are colored)to get an admissible coloring of。
酸性橙7染料的合成及应用
酸性橙7染料的合成及应用酸性橙7染料是一种常见的有机染料，也称为染料橙7或亚甲基橙7。
它的化学名为亚甲基橙，是一种偶氮染料。
这种染料具有酸性性质，因此可溶于水。
本文将详细介绍酸性橙7染料的合成方法以及其在不同领域的应用。
酸性橙7染料的合成方法有多种。
其中一种常见的方法是通过对亚甲基橙和酸性染料的还原反应得到。
首先，将亚甲基橙和酸性染料混合，加入还原剂，如亚硫酸钠。
在适当的温度和pH条件下，触发还原反应。
还原剂将染料分子中的氮氧键还原成氨基。
这样，亚甲基橙的分子结构就被改变为酸性橙7染料的结构。
经过适当的提取和净化，即可得到纯净的酸性橙7染料。
酸性橙7染料具有鲜艳的橙色，因此在染料行业中有广泛的应用。
首先，它常被用作纺织品染料。
酸性橙7染料的鲜艳色彩可以使纺织品具有艳丽的颜色，因此被广泛应用于制造彩色衣物、窗帘和家居用品等。
其次，酸性橙7染料也可用于制作颜料。
将染料溶解在合适的溶剂中，可以制成水性或油性的颜料，用于绘画、印刷等领域。
此外，酸性橙7染料还可用于彩色墨水的制备，广泛应用于打印和绘图。
最后，酸性橙7染料还可用于染料敏化太阳能电池。
通过将染料敏化剂与二氧化钛电极结合，可以将可见光转化为电能。
除了上述应用外，酸性橙7染料还有一些其他的特殊应用。
例如，它可以作为生物染料用于活细胞的染色。
酸性橙7染料可与细胞中的核酸结合，从而使细胞呈现出橙色。
这种染料的特性使得细胞在显微镜下更易观察和分析。
另外，酸性橙7染料还可用于污水处理。
由于其酸性特性，它可以用作pH指示剂，监测和调节污水处理过程中的酸碱度。
总的来说，酸性橙7染料是一种常见的酸性有机染料，具有鲜艳的橙色。
其合成方法可以通过亚甲基橙和酸性染料的还原反应得到。
酸性橙7染料在纺织、绘画、印刷等行业中有广泛应用，可以用于染色、颜料制备、墨水以及染料敏化太阳能电池等领域。
此外，它还可以作为生物染料和污水处理中的pH指示剂。
酸性橙7染料的广泛应用使得它在现代化学及相关领域中具有重要的地位。
7细菌的革兰氏染色 (1)
微生物实验报告细菌的革兰氏染色及形态观察一、目的要求：1、熟悉光学显微镜的使用方法。
2、掌握革兰氏染色法。
3、掌握大肠杆菌、枯草芽孢杆菌和金黄色葡萄球菌的染色结果和形态特征二、器材和器皿：1、大肠杆菌、枯草芽孢杆菌、金黄色葡萄球菌2、革兰氏染色液、香柏油、二甲苯3、显微镜、擦镜纸、酒精灯、载玻片、盖玻片、接种环等三、实验原理：革兰氏染色法是细菌学中广泛使用的一种鉴别染色法，1884年由丹麦医师Gram 创立。
未经染色的细菌，由于其与周围环境折光率差别很小，在显微镜下极难观察。
染色后细菌与环境形成鲜明对比，可以清楚地观察到细菌的形态、排列及某些结构特征，而用以分类鉴定。
革兰氏染色属复染法。
该染色法所以能将细菌分为G+菌和G-菌，是由这两类菌的细胞壁结构和成分的不同所决定的。
G-菌的细胞壁中含有较多易被乙醇溶解的类脂质，而且肽聚糖层较薄、交联度低，故用乙醇或丙酮脱色时溶解了类脂质，增加了细胞壁的通透性，使初染的结晶紫和碘的复合物易于渗出，结果细菌就被脱色，再经蕃红复染后就成红色。
G+菌细胞壁中肽聚糖层厚且交联度高，类脂质含量少，经脱色剂处理后反而使肽聚糖层的孔径缩小，通透性降低，因此细菌仍保留初染时的颜色，呈现蓝紫色。
四、实验步骤：1、取载玻片用纱布擦干，涂菌的部位在火焰上烤一下，除去油脂。
2、涂片：液体培养基：左手持菌液试管，在酒精灯火焰附近5cm左右打开管盖；右手持接种环在火焰中烧灼灭菌，等冷却后从试管中沾取菌液一环，在洁净无脂的载玻片上涂直径2mm左右的涂膜，最后将接种环在火焰上烧灼灭菌。
固体培养基：先在载玻片上滴一滴无菌水，再用接种环取少量菌体，涂在载玻片上，使其薄而均匀。
3、晾干：让涂片在空气中自然干燥。
4、固定：让菌膜朝上，通过火焰2-3次固定（以不烫手为宜）。
5、染色：在菌膜上滴加草酸铵结晶紫液，染1min。
6、水洗：用蒸馏水轻轻冲洗涂片上的染色液，用吸水纸吸干。
简单染色结束可观察细胞形态。
mdc染色原理
mdc染色原理
MDC染色原理是指使用一种特殊染料——MDC（7-羟基-9,10-二甲氧基-7,8,9,10-四氢-6-苯并[cd]异噻吩-3,4-二酮甲酯）对细胞或组织进行染色。
MDC属于荧光染料，可以与细胞内的各种器官和分子发生特异性染色。
MDC的染色原理是利用其分子结构中的双键和杂环结构，在细胞内的膜、溶酶体、内质网、高尔基体等结构中发生荧光染色。
MDC能够形成具有荧光性的复合物，并且只与一些特定的细胞器结构发生结合，因此可用于研究细胞器的形态和功能。
此外，MDC染色还可以用于检测自噬的程度，因为自噬过程中产生的自噬体可以与MDC形成特异性的染色。
总之，MDC染色原理是基于其分子结构和特异性染色作用，用于研究细胞器形态和功能以及自噬过程的染色方法。
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想做多重免疫荧光收好指南教你一张片子染出七种颜色
想做多重免疫荧光？收好指南，教你一张片子染出七种颜色想做多重免疫荧光？需要先思考这几个问题：问题1：你认得全动物园里的动物们吗，比如大鼠，小鼠，兔，山羊，绵羊，鸡，荷兰猪（豚鼠）？不同种属的抗体需闹清楚。
问题2：你确定能分得清字母表并做得了十以内得加减法吗？有不同亚型鉴定的过的一抗，比如：IgG1，IgG2a，IgG2b，IgG2c，IgM，IgE 不知道怎能行？问题3：你确定能将动物和数字字母们对号入座，并找对门牌吗？有对应种属特异反应性的Fab 二抗，比如：F(ab')2 Fragment （避免Fc 反应性出现的非特异染色），Cross adsorbed（抗体纯化过程交叉吸附，避免物种交叉反应带来的背景）。
问题4：你确定能将重峦叠嶂移出两山排闼吗？选择合适的荧光素，明辨激发光和发射光，且不会导致荧光背景增强，比如：FITC、PE、Cy、Alexa Fluor、DyLight。
就算上面的问题你都能搞定，订购来了各种试剂，准备来了片子，你就一定能染出来多重颜色？你只是迈开了万里长征的第一步而已。
因为要得到理想的荧光实验结果，每一个步骤都需深(w ā) 度(kēng) 优(mái) 化(tǔ)，固定通透方法、浓度配比、抗体共孵育、抗体稀释液、封闭条件等等。
此时，有人在你发际线还没有升高，头发还乌黑浓密的时候，在你的耳边悄悄跟你说：做多重荧光简单，有种快捷省时的方法。
先让你们随意感受一下这种方法染出来的图片：接下来我们将仔细介绍该方法实验原理和实验步骤。
多重荧光免疫组化（mIHC）实验原理上述方法采用TSA 技术(Tyramide Signal Amplification?，酪胺信号放大技术)：简单来说，用该方法做多重免疫荧光是利用二抗上带有的HRP（而不是直接偶联荧光素），来催化后续添加入体系的非活性荧光素。
荧光素在HRP 和过氧化氢的作用下被活化，跟临近蛋白的酪氨酸残基共价偶联，使得蛋白样品与荧光素稳定结合。
实验七 还原染料染色
实验七 还原染料染色一、实验目的掌握还原染料还原和染色（浸染和轧染）的一般方法。
二、实验原理还原染料是一类非水溶性染料，主要用于纤维素纤维的染色，其分子中不含有水溶性基团，但含有两个或两个以上的羰基，能在碱性条件下被强还原剂还原成水溶性的隐色体钠盐而上染纤维。
上染纤维后，再经氧化处理，使染料恢复为非水溶性的色淀固着在纤维上，染色反应过程如下：C=O [H],[OH -]C O-[O]C=O染料 还原为隐色体后上染纤维 氧化后恢复为染料还原染料还原为隐色体后，与直接染料上染纤维素纤维相似，对纤维有很高的直接性，其直接性的来源，目前还没有充分的理论解释，但一般认为是范德华力和氢键所致。
因还原染料分子结构大，芳环共平面性好，对纤维素纤维的亲和力较大，因此初染率较高，移染性较差，易产生染色不匀的现象。
还原染料还原为隐色体，常用的还原剂是保险粉（连二亚硫酸钠，Na 2S 2O 4）。
在碱性条件下保险粉具有很强的还原能力，还原反应如下：C=O + Na 2S 2O 4 + 4 NaOH 2 Na + 2Na 2S 2O 3 + 2H 2O2 C O 还原染料的还原操作有全浴法和干缸法两种方法。
全浴还原主要用于还原速度较快、隐色体溶解度较小或容易过度还原的染料品种。
干缸还原采用高浓度还原剂对染料进行还原，适用于还原速度较慢的染料。
还原染料不同品种之间的还原性能差异很大，具体染料使用何种方法进行还原应查阅还原染料的应用资料。
还原染料可以采用浸染或轧染方法进行染色。
浸染方法采用隐色体染色工艺，先将染料还原，配制成隐色体溶液，然后上染纤维。
为了获得较高的匀染性和上染率，根据各染料适宜的染色温度，浸染方法可分为甲、乙、丙和特别法四类。
常用的是前三类方法。
其中，甲法染色温度较高，60～65℃，适用于隐色体对纤维亲和力高，容易聚集的染料；丙法染色温度较低，30～35℃，适用于亲和力较低，聚集倾向较小的染料；乙法染色温度为50～55℃，介于上述两者之间。
高中生物必修一第一章到第四章测试题
高二生物第一章---第四章综合测试一、单项选择题（本题包括30小题，1-20每小题1分，21-20每小题2分，共40分。
每小题给出的四个选项中，只有一个选项最符合题目要求。
多选、错选均不得分。
）1.细胞内含量最多的有机化合物是（）A糖类B.核酸2.下列四组生物中，都属于真核生物的一组是A.病毒和柳树B.细菌和草履虫C.脂类D.蛋白质（）C .蓝藻和酵母菌D .青蛙和酵母菌3.生物体的基本组成物质中，作为生命活动主要承担者的是A .无机盐B .水4.糖类、脂肪、蛋白质、核酸共有的元素是A . C、H、0、N、P C .蛋白质()B.D .核酸C、H、0、NC. C、H、O5.染色体和染色质的关系是（）A .不同时期，不同物质的不同形态C.同一时期，同一物质的不同形态6 .玉米叶肉细胞中，具有色素的一组细胞器是D. 0、HB .不同时期，同一物质的不同形态D .同一时期，不同物质的不同形态（）C、中心体和核糖体A、线粒体和高尔基体B、叶绿体和液泡7 .食醋中的醋酸成分是活细胞不需要的小分子物质，蔗糖不是活细胞需要的大分子物质。
用食醋和蔗糖可将新鲜的大蒜头很快地腌成糖醋蒜，其原因是（）A.醋酸和蔗糖分子均能存在于活细胞的间隙中D、内质网和液泡B.醋酸和蔗糖分子均能被吸附在活细胞的表面C.醋酸能固定和杀死活细胞，细胞膜失去了选择性D.因腌的时间过久，两种物质均慢慢地进入活细胞8.下列核苷酸中，在DNA结构中可能具有的是（）0^0A.C D . O19.蛋白质、DNA和RNA的基本组成单位依次是（）A .氨基酸、核苷酸、核糖核苷酸C .氨基酸、核糖核苷酸、脱氧核苷酸B .核苷酸、脱氧核苷酸、核糖核苷酸D .氨基酸、脱氧核苷酸、核糖核苷酸（）10.生物体进行生命活动的主要能源物质和细胞内良好的储能物质分别是A .蛋白质、脂肪B.糖类、脂肪C.脂肪、糖类 D .糖类、蛋白质11 .下列哪一项不属于细胞膜的功能A.控制物质进出细胞B.（）将产生的抗体分泌到细胞外C .维持细胞内的小环境的稳定 D. 使细胞内的各种生物化学反应加快12 .观察DNA和RNA在细胞中的分布，所运用的原理是A 、单独利用甲基绿对细胞染色，可显示DNA在细胞中的分布，从而推知B、单独利用吡罗红对细胞染色，可显示RNA在细胞中的分布，从而推知（）RNA勺分布DNA勺分布C 、利用甲基绿和吡罗红混合染色剂对细胞染色，同时显示DNA和RNA在细胞中的分布D 在细胞核和细胞质内可以分别提取到 DNA 和RNA 由此说明DNA 和RNA 的分布 13.当植物由代谢旺盛的生长期转入休眠期时，体内结合水与自由水的比值通常会(A .升高B .下降C .无变化D .产生波动14. “观察DNA 和RNA 在细胞中的分布”实验中，正确的实验步骤是(A. 取口腔上皮细胞制片7水解7冲洗7染色7观察B. 取口腔上皮细胞制片7染色7冲洗7水解7观察C.取口腔上皮细胞制片7水解7染色7冲洗7观察D.取口腔上皮细胞制片7冲洗7水解7染色7观察15. 人体某些白细胞可以吞噬病菌，这一生理过程的完成依赖于细胞膜的 A .选择透过性B .主动运输16. 有一种物质能顺浓度梯度进出细胞膜， 出人细胞膜的方式是17 .血红蛋白分子中含有 4条多肽链，共由574个氨基酸构成，则血红蛋白分子中含有的肽键和至少含 有的游离氨基和羧基数分别是( )C .保护性D .流动性但却不能顺浓度梯度进出无蛋白质的磷脂双层膜。
7aad和annexin v 染色原理
7aad和annexin v 染色原理
7-AAD/Annexin V染色技术是一种细胞凋亡检测技术，它可以检测凋亡细胞、早期凋亡细胞和活细胞之间的分布情况，是研究细胞凋亡过程的常用方法。
它简单、快速，有临床价值及作用。
7-AAD/Annexin V染色技术原理：
7-AAD/ Annexin V染色技术是一种利用特定抗原二聚体光学染色技术，可同时识别凋亡细胞和活细胞，从而了解细胞凋亡过程的分布情况。
7-AAD/ Annexin V染色技术的基本原理是：试剂中的7-AAD(7-Aminoactinomycin D)与凋亡细胞的外源性DNA结合，而另一种特异性抗原Annexin V与凋亡细胞的内建核蛋白结合，从而使凋亡细胞获得一种特定的染色光谱。
7-AAD/ Annexin V染色技术的具体技术步骤是：前处理：将细胞按照一定浓度悬液处理，脱模；制备细胞：将上述悬液浓缩，放入比较宽的银杯中， or 5mL 毕氏管中；加入染色液：加入适量的7-AAD指示染料，1μg/ml Annexin V-Fluorescein，10 mmol/L HEPES, pH7.4;拌匀：拌匀染色液，放置黑暗室或室外低温器中，静置15-30分钟；加入末端备份液：以1×PBS/BSA分数极性液为末端液，可以使染色效果更稳定；旋转滚筒技术检索：将混合液用自动滚筒（Rotator）旋转，滚筒速度175转/min，旋转技术以检索凋亡细胞；流式分析：将滚筒旋转出的混合液放入流式分析仪，进行计数和定量分析，如有需要可以还可进行细胞图像的统计分析。
7-AAD/ Annexin V染色技术可以用于生物体内针对不同环境及药物因素改变细胞凋亡机制的检测，它有助于研究细胞凋亡活性调节及靶向药物设计，为肿瘤等疾病的研究奠定基础。
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酸性含媒染料教学目标：知识目标：1、酸性含媒染料的定义、结构特点以及分类2、酸性含媒染料的染色能力目标：培养学生自主学习的能力。
情感目标：培养学生持之以恒的学习态度。
教学重点：酸性含媒染料基本特点及分类教学难点：酸性含媒染料的染色教学方法：讲授法教学过程：一、组织教学二、复习导入上节课已经讲过酸性媒介染料染色的相关知识。
这次课，我们开始讲另一种染料——酸性含媒染料。
三、新授（一）定义：染料分子中已含有金属鳌合结构的酸性染料，即合成时已将金属离子引入染料，也称金属络合染料（二）特点：色泽鲜艳度介于酸性媒介染料和酸性染料之间对蛋白质纤维有直接性，不需铬媒处理，染色工艺过程比酸性媒介染料简便染色牢度优良，与酸性媒介染料相似，具有较高的耐洗、耐日晒和耐缩绒等牢度，是羊毛和蚕丝染色常用染料，在锦纶染色也有较多应用（三）分类根据染料分子和金属离子的比例关系，分为1：1型和1：2型⏹1：1型酸性含媒染料在强酸性条件下染色，称为酸性络合染料，主要用于羊毛染色⏹1：2型酸性含媒染料在弱酸性或近中性条件下染色，称为中性络合染料，简称中性染料，可用于羊毛、蚕丝、锦纶、维纶的染色1：1型酸性含媒染料⏹1：1型主要用于羊毛染色，染色方法与强酸性相似，不过用酸量较大，沸染时间长，元明粉的匀染作用很小⏹染色原理与一般强酸性染料不同⏹染料磺酸基与离子化的氨基形成离子键结合⏹染料上的铬原子与纤维羧基或氨基形成配价键⏹如果上述结合方式同时迅速进行，必然造成上染过快，染色不匀⏹必须加入较多硫酸，降低结合形成速率，达到缓染和匀染的目的⏹酸促使氨基离子化，抑制羧基离子化，减少配价键结合，延缓和减弱染料和纤维之间的结合，起缓染作用，促进匀染。
⏹水洗去酸后，再发生络合作用1：2型酸性含媒染料染色⏹染料分子中金属原子已与染料完全络合，不能与纤维上的基团发生配价键结合⏹染色原理与弱酸性染料相似⏹染料分子复杂，相对分子质量大，对纤维亲和力高，初染速度快，在纤维内扩散速度慢，染料移染性差，对纤维微结构差异引起的染色疵病的遮盖性差。
⏹染浴pH值控制在微酸性和中性，控制上染速度。
小结：主要讲了酸性含媒染料的定义、结构特点、染色原理以及酸性染料的染色等基本知识。
作业：简述酸性含媒染料的分类。
板书设计：酸性含媒染料一、定义二、特点三、分类及染色四、小结五、作业课后记：这节课主要讲了酸性含媒染料的定义、结构特点、染色原理以及酸性染料的染色等基本知识。
接受情况较好，无迟到、旷课现象。
参考资料：染整概论张洵栓中国纺织出版社分散染料染色教学目标：知识目标：分散染料的定义、结构特点。
能力目标：培养学生自主学习的能力。
情感目标：培养学生持之以恒的学习态度。
教学重点：分散染料的结构特点教学难点：三个结构特点教学方法：讲授法教学过程：一、组织教学二、复习导入上节课已经讲过酸性含媒染料染色的相关知识。
这次课，我们开始讲另一种很重要的染料——分散染料。
三、新授（一）概念●分散染料的分子结构中不含离子化水溶性基团，仅含少量极性基团，相对分子质量较小，结构简单、在水中溶解度很低的非离子染料●染色时，分散染料以细小颗粒状态均匀分散在染液中，是涤纶染色和印花用的主要染料。
●分散染料的染色工艺主要有高温高压染色法、热熔染色法和载体染色法●除染涤纶外，还可以染锦纶、腈纶、氨纶、丙纶、芳纶等合成纤维，但色泽较淡●分散染料一般可按化学结构和按应用性能分类，主要依据耐升华牢度（即耐热稳定性）分类●分散染料含极性基团少，相对分子质量小，结构简单，分子间作用力弱，耐升华牢度较差。
●染料生产厂家一般都按染料耐升华性能来命名，不同厂家命名规则不同（二）分散染料的结构特点●分散染料是非离子染料，分散染料不含强离子化基团，如-SO3Na和-COONa，而含有-OH，-NH2，-NO2，-CN等极性基团在水中溶解度较小。
●商品分散染料中已加入大量分散剂，染浴中还需加入适量分散剂。
●分散剂使分散染料以细小晶体均匀地分散在染液中，形成稳定的悬浮液，部分染料溶解在分散剂所形成的胶束里。
1. 染料的溶解性●染料应具有一定的溶解度，保证染色均匀和提高染色速率●溶解度过低——染料上染速率会很慢●溶解度过大——染料与水作用力增大，与纤维亲和力降低，上染速率和上染百分率下降●溶解度影响因素：●（1）分子量小、极性基团多的，溶解度大●（2）晶体小、不稳定晶型的，溶解度大；反之，则小●（3）分散剂具有增溶作用●（4）染液温度提高，溶解度提高●（5）若染浴中的化学试剂能引起染料结构变化，将影响溶解度●（6）染浴温度太高或研磨颜料温度太高，染料晶型将从不稳定晶型向稳定晶型转变●（7）染料晶体颗粒小，溶解度大●（8）染液温度不匀或突然降温，发生晶体增长现象，染料难于扩散，上染速率降低，易产生染色不匀（色斑）等疵点2. 染料的分散稳定性●（1）分散剂影响稳定染浴，抑制染料发生晶型转变和晶体增长起增溶作用，染色时能及时释放单分子染料●（2）颗粒大小染料颗粒直径在0.5～2μm之间3.化学稳定性●（1）酸碱稳定性●在高温碱性条件，含酯基、酰胺基、氰基等基团的分散染料，会发生水解——染料色光变化，上染百分率变化●碱浴中，含羟基的染料会发生羟基离子化D-OH+OH- →D-O-+H2O影响染色性能及染色产品色光●pH值过低，氨基发生质子化，D-NH2+H+ D-NH3+影响染色性能及染色产品色光●分散染料染色时，染浴pH值控制在5～6，染色织物的上染百分率高，色泽鲜艳度好●（2）耐还原剂稳定性分散染料基团中约有50％为偶氮类结构的染料偶氮基团不耐还原剂，会导致染料结构变化，发生消色作用，分解产物对纤维亲和力降低●利用● A..不耐还原剂的分散染料可用作拔染印花的地色染料● B. 还原清洗利用分散染料在稀碱性还原液中易被还原分解而去除的性质，来去除浮色（并去除染色时涤纶织物析出的低聚物等杂质）以提高染色牢度小结：主要讲了分散染料的定义、结构特点。
作业：简述酸性含媒染料的分类。
板书设计：分散染料（一）一、定义二、特点1、溶解性2、分散稳定性3、化学稳定性三、小结四、作业课后记：这节课主要讲了分散染料的定义及结构特点。
接受情况较好，无迟到、旷课现象。
参考资料：染整概论张洵栓中国纺织出版社分散染料染色（二）教学目标：知识目标：分散染料的染色员里。
能力目标：培养学生自己分析问题、解决问题的能力。
情感目标：培养学生持之以恒的学习态度。
教学重点：分散染料的染色原理教学难点：原理教学方法：讲授法教学过程：一、组织教学二、复习导入上节课已经讲过分散染料染色的相关知识。
这次课，我们继续讲解其染色原理。
三、新授（一）染色基本原理1、n1.聚酯纤维的染色性能●涤纶属于疏水性纤维，纤维缺乏能与染料发生结合的基团，不能使用水溶性染料染色，只能使用分子量小、不含强离子性水溶性基团、溶解度较低的非离子分散染料染色●涤纶结构紧密，常压沸染，染料难以扩散进入纤维内部，把纤维染透●在玻璃化温度以上时，纤维大分子链段发生剧烈运动，聚合物分子间空隙增大，自由容积增加，提高染色速率●染料按自由体积模型扩散●可采用升高染色温度（高温高压染色法、热熔染色法）或使用纤维膨化剂及染色促进剂（载体染色法）两种途径实现。
2. 染料与纤维间的作用力●氢键——涤纶中酯基含量约46％，分散染料含有-OH、-NH2，此外涤纶苯环也可形成氢键●范德华力●疏水键力——纤维与染料间通过疏水部分相互作用根据相似相溶原理，染料可看作“溶解”在固体纤维上的无定形区中。
3.上染速率（1）温度：升高，A. 纤维大分子运动加剧，大分子链发生剧烈转动，纤维自由容积增大，扩散空间阻力减少，扩散速率增大；B. 染料扩散动能增加，有利于更多染料向纤维内部扩散；C. 染料溶解度提高，提高染色速率染色温度从85 ℃提高到100℃，扩散速率增加近48倍。
（2）纤维膨化剂或染色促进剂：●膨化剂有利于纤维膨化，降低纤维的玻璃化温度，使纤维自由容积增大，从而提高染料向纤维内部扩散●促进剂促进染料纤维表面吸附，增加纤维表面染料浓度，提高内外浓度差，提高染色速率●部分载体同时具有纤维膨化剂和染色促进剂的双重作用●适量渗透剂可缩短染色时间，得到良好的染色效果（3）染料溶解度的影响●提高溶解度，有利于单分子染料的吸附和扩散，提高染色速率●但染料溶解度不能太高，否则染料亲水能力大于亲纤维的能力，平衡上染百分率降低。
小结：主要讲了分散染料染色原理。
作业：简述分散染料的染色原理。
板书设计：分散染料（二）一、染色原理1、n1.聚酯纤维的染色性能2. 染料与纤维间的作用力3.上染速率（1）温度：（2）纤维膨化剂或染色促进剂：（3）染料溶解度的影响二、小结三、作业课后记：这节课主要讲了分散染料的染色原理，内容较深，希望同学们课后好好理解掌握。
无迟到、旷课现象。
参考资料：染整概论张洵栓中国纺织出版社分散染料染色（三）教学目标：知识目标：分散染料的染色员里。
能力目标：培养学生自己分析问题、解决问题的能力。
情感目标：培养学生持之以恒的学习态度。
教学重点：分散染料的染色原理教学难点：原理教学方法：讲授法教学过程：一、组织教学二、复习导入上节课已经讲过分散染料染色的相关知识。
这次课，我们继续讲解其染色原理。
三、新授染色方法1.高温高压染色●染色织物色泽鲜艳、手感好●织物品种适应性好，适用的染料范围广，染料利用率高●对设备要求高，需密闭●间歇式，染色时间长，生产效率较低●染液中含有分散染料、分散剂、高温匀染剂、pH调节剂等●高温高压染色，染料易出现晶型转变●温度控制不匀，出现晶体增长现象●高温染色，表面易析出低聚物●部分染料发生水解或还原●染料要求：良好的分散稳定性、化学稳定性和较高的耐升华牢度，以保证均匀、鲜艳的染色效果●染色过程●（1）40 ℃温水化料，过滤●（2）慢慢升温至120～130 ℃，在升至染色转变温度以上时，上染速率迅速提高，应严格控制升温速率，保证染料均匀上染●（3）继续保温染色45～60min●（4）然后降温，进行还原清洗，水洗等后处理，彻底去除浮色，以提高染色产品的染色牢度和鲜艳度●染色温度选择在120～130℃为宜，温度不能超过145℃，否则纤维会造成损伤，特别是毛涤混纺织物●染色温度控制分三个阶段：●（1）染色升温阶段：在70～110℃温度区间，染料上染速率很快，要严格控制1～2℃/min，保证染料均匀吸附●（2）染色保温阶段●染料向纤维内部扩散，并增进染料移染，染匀染透●（3）降温阶段●在玻璃化温度以上，降温速率应适当慢些，其后快速降温，避免引起织物产生折皱和手感粗糙●pH值●控制在5～6之间，色光纯正、色泽鲜艳，上染百分率高；●高温碱性，染料水解，●碱性太强，织物损伤大，手感差●但酸性染色，齐聚物容易沉积在织物表面●浴比：●小，节能节水，但易使织物产生折皱、擦伤、染色不匀●大，耗能耗水●匹染：10：1～30：1●分散剂影响●匀染剂（载体、非离子型活性剂）●（1）缓染剂通过助剂对染料或纤维亲和力，延缓染料上染，使染料能均匀地吸附在纤维的各个部分，从而染色均匀，但会导致上染百分率下降●（2）移染●匀染剂一般为载体、非离子型聚氧乙烯类表面活性剂●载体有毒●非离子表面活性剂在染色温度高于浊点，会产生沉淀，一般需与阴离子表面活性剂复配● 2.热熔染色●在185～215℃干热条件下使染料在纤维上发生固着的连续轧染的染色方法●连续化生产，生产效率高，用水量少，污水少，使用的染料要求耐升华牢度较高●染色时织物所受张力较大，织物手感及色泽鲜艳度不及高温高压染色法●一般多用于涤棉混纺织物中涤纶染色●浸轧液含有分散染料、分散剂、渗透剂、防泳移剂等●染料颗粒要特别匀细，渗透性及分散稳定性要好●二浸二轧，室温，涤棉织物轧余率一般为50～60％●焙烘固色（热熔）温度根据染料升华牢度来选择，一般为180～220℃，时间2～1min染色工艺条件●（1）润湿剂或渗透剂●润湿剂可以排除织物中的空气，有利于染料渗透到织物内部●织物前处理充分，可不加●（2）泳移现象及防止●因染料对涤纶缺乏亲和力，容易发生烘干现象●合理烘干方式●提高浸轧效果，减少带液率，涤棉带液率降低到20％～30％后，不会发生明显泳移●加入适量防泳移剂●热熔染色时间和时间选择●织物烘干后，染料停留在纤维表面，没有与纤维发生固着，通过高温热焙烘，染料向纤维内部扩散，把纤维染透，完成固色——热熔固色●温度高——染料升华较多●拼色时，染料耐升华性应相近● 3.载体染色●利用载体助剂对涤纶的增塑膨化性能，在常压100℃条件下染色●设备简单，适合毛涤织物染色●另外，载体对染料具有亲和性，对染料溶解能力高，提高染料浓度，提高吸附和扩散；过高，也不好●但载体有毒性，存在环境污染，残留载体不易洗净，影响染色牢度小结：主要讲了分散染料几种染色方法。