Reconstructing tumor genome architectures Martin Strauch-Seminar”Selected Topics in Bioinformatics”
T¨u bingen,May,18th,2004
Abstract
It has been observed that cancer genomes typically show not only an increased rate of single nucleotide mutations but also an increase in the number of genome rearrange-ments.As sequencing whole genomes is still a time-consuming task a method has been proposed for reconstructing the cancer genome from the known sequence of the human genome by?nding the breakpoints alongside which the rearrangements occurred.The method involves end sequence pro?ling(ESP)of the cancer genome which is avail-able in BAC library vectors.Mapping the end sequences to the human genome,the candidates for the breakpoints can be identi?ed.For the reconstruction of the cancer genome G’the events have to be found that rearrange the human genome G into G’. This task is speci?ed in the de?nition of the ESP Genome Reconstruction Problem [ESPGRP]and,additionally,a heuristic approach providing a solution will be given. Finally,the method is applied the MCF7breast cancer genome resulting in a map of the rearranged cancer genome.
This work is based in large parts on an article by Raphael et al.(2003),accompanied by supporting material and background information from other texts.
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Contents
1Biological and Bioinformatical Background3
1.1Cancer (3)
1.2Genome Rearrangements (4)
1.2.1Representation of a genome (4)
1.2.2Events that transform a genome (5)
1.3End Sequence Pro?ling(ESP) (6)
2The Genome Reconstruction Problem8
2.1The ESP Sorting Problem (8)
2.1.1Concepts (8)
2.1.2A formal de?nition (9)
2.2The actual Genome Reconstruction Problem (11)
2.2.1Assigning orientations to the BES (11)
2.2.2Multiple chromosomes (11)
2.2.3Events other than reversals (12)
2.2.4Coping with chimeric reads (12)
2.2.5The ESP Genome Reconstruction Problem (13)
3A heuristic approach to the Genome Reconstruction Problem14
3.1Generating the input data (14)
3.2Finding a path of rearrangements (14)
4Results for the MCF7breast cancer genome16 5References20
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1Biological and Bioinformatical Background
1.1Cancer
At the beginning of each cancer stands a neoplasm,i.e.a clone of a new type of cells which usually arises from one or more somatic mutations.What causes us to distinguish a benign neoplasm,which is the most common case,from a cancer,is its characteristic phenotype. While the majority of somatic mutations merely results in a dysfunctional cell,the acquisition of a cancer phenotype does confront us with potentially immortal cells showing the following properties(Burgoyne,2003):
?Cell division and multiplication occur independently of negative or positive signals,i.e.
signals to stop the cell cycle or to trigger cell division.
?They don’t respond to apoptose signals or the telomeric counter.
?Again ignoring regulation mechanisms,in this case tissue localisation,they acquire the ability to metastasise,which is the step that makes the cancer malignant.
?Additionally,in the?nal stages of a cancer,the cells may loose their di?erentiated functions and develop a chaotic cytology.
The above de?nition generalizes and thereby describes all cancer phenotypes.The set of cancer phenotypes is still a complex one,but basically they all behave according to the de?nition given.Cancer genotypes,however,are much more diverse,being individual for each cancer,although predispositions and common candidates for oncogenes exist.Oncogenes are the genes involved in a cancer genotype and there may be di?erent sets of oncogenes which are responsible for the same phenotype.
An initial mutation leading to an oncogene is usually due to the e?ects of biological,chemical or physical mutagens.However,there are circumstances associated with a special genotype which increase the probability of a mutation actually to occur.The oncogenes involved are named cancer susceptibility genes and typically located in biological integrity systems,such as DNA repair mechanisms or free-radicle scavengers.
Oncogenes which are a direct cause of a cancer are termed”true oncogenes”although the use of the de?nition is not always consistent.The role of dominant true oncogenes,such as the growth-stimulating autocrines and their receptors stands undoubted,but the recessive true oncogenes such as p53,which is involved in cell cycle control,are sometimes referred to as cancer susceptibility genes as well.The reason behind distinguishing the recessive true oncogenes from the susceptibility genes is the”stand-alone potential”of cell cycle control genes to cause genomic disorder,while the susceptibility genes will not lead to cancer in the absence of mutagens.
Recessive true oncogenes like p53are especially important for our purposes,as they perform regulating tasks in cell division and proliferation.Chromosomal anomalies and rearrange-ments will most likely take place during the phases of chromosome duplication and mitosis
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and greatly increase the risk of further mutations.One commonly speaks of the self-inducing and ongoing process of an”error cascade”,which in the end will result in a chaotic cytology and a rearranged genome.
A similar e?ect may be visible due to defects in DNA repair mechanisms.Proneness to mutation or even chromosome breakages combined with faulty repair systems is likely to result in rearranged genomes as well.
For breast cancer phenotypes two genes,BRCA1and BRCA2,respectively,have been iden-ti?ed to play a major role forming the phenotype(Scully,2000).They are suspected of promoting chromosomal translocations and known to be able to interact with p53.At least BRCA2is also involved in DNA repair and recombination defects in BRCA1/2mutant mice have been observed(Patel et al.,1998).
Thus,cancer seems to be accompanied not only by increased single nucleotide mutation rates as it has already been known,but also by increased genomic rearrangement rates(Loeb et al.,2003).Consequently it must be of interest to gain information about the overall structure of cancer genomes in addition to what is known about single mutations.
To achieve the goal of having a high resolution map for a speci?c cancer genome,the ESP technique(Volik,Zhao,et al.2003)has been employed.Being an e?cient mapping technique, it will,together with the algorithms involved,be presented in the following chapters.
1.2Genome Rearrangements
1.2.1Representation of a genome
According to(Pevzner,2000)1a genomeΠis de?ned as a set of N chromosomesπ(i),each of which contains a sequence of n i genes:
Π={π(1)···π(N)},withπ(i)=π(i)1···π(i)n
(1)
i
The direction of whole chromosomes is of no relevance in this representation since it is equally possible to view a chromosome from left to right as(π(1),π(2),π(3))or from right to left as(?π(1),?π(2),?π(3)).Thus,?ipping a chromosome is always possible for representation purposes.
This task is performed by a so called?ip vector s=(s(1),···,s(N)|s(i)∈{?1,+1})
We can use the?ip vector for creating a concatenate,which combines the chromosomes into a permutation
Π(s)=s(1)π(1)+···+s(N)π(N)(2) 1All the de?nitions in section2.2are taken from this book.
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Figure1:The illustration(similar to Pevzner,2000,p.216),gives an example for the mim-icking of a translocation by a reversal once we regard a concatenated genome,where entire chromosomes are allowed to be?ipped.Above(1→1 )chromosomes A and B are trans-formed by a translocation event,which is shown to be the same as reversing the concatenate of A and B.This works because on the concatenate blocks that actually are located on di?erent chromosomes,e.g.A2and B1(in arrangement1),suddenly become neighbours in the single virtual chromosome represented by the concatenate(arrangement2).Thus,the reversal is applicable to[A2,-B2]in arrangement2,which then is transformed into arrangement2’,that itself is the concatenated version of1’.
1.2.2Events that transform a genome
When looking at genome rearrangements,events that transform the structure of a genome play an important role.Mutation events do not usually contribute to a change in the overall structure of the genome.The interesting events for our purposes therefore are fusions,?ssions,translocations and reversals.Blocks of genes which remain untouched by these events,i.e.”atomic”intervals[i,j]that stay together as a single unit throughout those events,we call synteny blocks.
The reversal will become the most important event because all the other ones mentioned can be modelled by reversals.A reversal of an interval[i,j]with i≤j of a chromosomeπ, which is referred to asρ(π,i,j)performs a rearrangement onπsuch that the interval[i,j]is inverted in order and orientation:
π1···πi?1πi···πjπj+1···πnρ(π,i,j)
→π1···πi?1?πj···?πiπj+1···πn(3) As we will see in section3,Raphael et al.(2003)employ a special de?nition of the reversal that describes its e?ects on a sequence x we know and use as a marker to follow the course of the synteny blocks.The properties of the reversal will,however,remain una?ected by this,
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so that everything that is shown for the general de?nition of the reversal will be applicable to the special version.
A further event,the translocationρ(π,σ,i,j),swaps genes between chromosomesπandσ:
→(π1···πi?1σj···σm),(σ1···σj?1πi···πn)(4)π,σρ(π,σ,i,j)
In a concatenated genome,where entire chromosomes can be?ipped”for free”,translocations can be modelled by reversals.An example is given in(?gure1).
The two remaining events,fusions and?ssions,are actually special cases of translocations and therefore they,too,can be modelled by reversals.
A fusion is de?ned as a translocationρ(π,σ,n+1,1),that combines chromosomesπandσinto a single chromosome and leaves an empty chromosomeφbehind:
→(π1···πnσ1···σn),φ(5)
π,σρ(π,σ,n+1,1)
Similarly,we de?ne the?ssionρ(π,φ,i,1)for1 →(π1···πi?1),(πi···πn)(6) πρ(π,φ,i,1) Given these de?nitions and a concatenated genome,we can now,for reasons of simplicity, assume in later models that there is only a single kind of event relevant for large scale genome rearrangements:the reversal.Fusions and?ssions are special cases of translocations and translocations themselves can be modelled as reversals.Further information about this topic can be found in Hannenhalli,Pevzner(1995). 1.3End Sequence Pro?ling(ESP) In order to obtain a detailed map of a cancer genome,sequencing is unavoidable.However, more e?cient and less time-consuming ways to perform this task exist than the sequencing of a complete genome.The basic idea behind the End Sequence Pro?ling approach(ESP)is to detect only the breakpoints and thus the loci where rearrangements occurred.The technique is not aimed at?nding single mutations but at yielding an overall picture of the rearranged genome we?nd in cancer cells.Given the sequence of the human genome,which is already known,and the indicators for rearrangements obtained from ESP data,the sequence of the cancer genome can be constructed. The End Sequence Pro?ling technique makes use of a BAC library(Bacterial Arti?cial Chromosome)of the rearranged genome to be analysed and takes the normal sequence of the genome as a reference.A BAC library consists of numerous vectors each containing an insert of about100-200kb taken from a genome.Details about the construction of BAC libraries 6 Figure2:Mapping BAC end sequences(BES)to the human genome as a reference.L is the insert length,x’and y’are the end sequences,while x and y are the corresponding regions in the human genome.A perfect match is depicted in a),whereas b)shows the mapping of a BES containing a breakpoint.For the latter case there is no continuous matching sequence in the reference genome,indicated by the distance between x’and y’being greater than L. can be found in(Asakawa et al.,1997)or in the support material section of(Volik,Zhao,et al.,2003).Detailed protocols for BAC sequencing are available at(Caltech protocol). Genomic rearrangements can be detected with the ESP approach by sequencing BAC end sequences(BES),i.e.the sequences at both ends of the inserts,and subsequently mapping them to the normal version of the respective genome.Due to the nature of the BAC vectors the end sequences will appear convergently in opposite orientations. As shown in(?gure2),rearrangements can be spotted by mapping the BES from the rear-ranged genome against the reference genome.For the case of no rearrangements at all it will be possible to map each pair of BES(x,y)(both ends of a speci?c BAC insert)to the ends of a sequence in the reference genome.Should a rearrangement have occurred,it will result in a breakpoint in one of the BACs,such that the two BES of one pair BES(x’,y’)who lie close together in the cancer genome belong to di?erent parts of the reference genome.Con-sequently,it will not be possible to match this pair BES(x’,y’)to a sequence in the reference genome delimited by x and y,which we thus take as indicative of a rearrangement. A formal de?nition for this approach and further details concerning the genomic rearrange-ment problem will be given in the following chapters. 7 2The Genome Reconstruction Problem 2.1The ESP Sorting Problem The ESP Sorting Problem(ESPSP)that has been introduced by(Raphael et al.,2003)is a ?rst formal approach to the reconstruction of a rearranged cancer genome with the help of ESP data and a normal version of the genome that does not contain rearrangements.Based on the the ESPSP we will later de?ne the actual Genome Reconstruction Problem,which is a re?ned and more realistic variant of the same problem de?nition. 2.1.1Concepts Now,let G be the reference genome and G’the cancer genome which has evolved from G by a series of events.For this restricted model we assume that there is only a single kind of event, the reversal.A reversal of a region in a genome reverses both the order of subsequences and their orientation that is denoted by a sign(+or-).Thus,a sequence”A B C”is transformed into”-C-B-A”by a reversal. As seen in(?gure2),rearrangements in the cancer genome are disclosed by mapping the BES to the reference genome.A breakpoint in the insert causes the sequences x and y in G which correspond to the pair of BES(x’,y’)from G’to be further apart than the insert length L. This motivates the de?nition of validity.We call a pair of BES(x,y)in G valid,if|y?x|≤L, and invalid else,where L is the maximum insert length. The selection of L as a constant value is reasonable since the large scale architecture of the genome is targeted and therefore the risk of missing a few invalid BES pairs because |y i?x i|>L i As rearrangements have occurred in the process of transforming G into G’,there will be some pairs of BES(x,y)in G which are invalid according to the de?nition above.However, the corresponding pairs in the rearranged genome G’are valid,because for G’the rearranged structure is the correct one.Hence,rendering the pairs that are invalid in G valid is a way of reconstructing the rearrangements on the way from G to G’so that in the end,after all invalid pairs have been eliminated by the speci?ed events(reversals in this case),the result is the architecture of G’. A problem which can appear during this process is the false classi?cation of pairs as in-valid because chimeric DNA,a phenomenon that frequently occurs in sequencing,is hard to distinguish from the composite DNA which arises from breakpoints and thus chromoso-mal rearrangements.In chimeric DNA non-contiguous regions of the genome are randomly joined,which,if it happens to be incorporated into one of the BACs,will equally result in invalid BES pairs. 8 Figure3:Example for the reversalρs,t(x).a)shows the case x 2.1.2A formal de?nition The ESPSP assumes for reasons of simplicity that there are no chimeric BACs.Nonetheless the issue will of course be dealt with during the heuristic reconstruction approach in chapter 4. Under this assumption let G be unichromosomal with a range from0to1:G[0,1].Now de?ne the BES(x,y)as points in G,with x For an interval[s,t]the reversal[0,1]→[0,1]is de?ned as ρs,t(x)= x if x t?(x?s)else (7) The reversalρs,t(x)reverses the interval delimited by s and t and changes the position of an end sequence x only if it is located inside the interval(see?gure3).Thus,by reversing intervals containing one or more of the end sequences,their respective position can be altered in a way that,given the appropriate set of reversals,sequences x and y can be brought into positions where the distance between them is smaller than L.Finding the most parsimonious way to transform invalid BES pairs into valid ones therefore is a method to reconstruct G’. Of course we cannot be sure that the most parsimonious path to G’,i.e.the one with the smallest numbers of reversals,is the one actually taken by evolution.Yet,as we are only interested in the result,the structure of G’,the actual path taken is irrelevant.What we need to know to reconstruct G’is simply how G has to be rearranged such that the invalid BES(x,y)from G,that correspond to the valid BES(x’,y’)from G’,become valid in G. The above is,of course,only true,if the result of the path of reversals taken is unambiguous. If there are more than one possible structures for G’it is impossible to name the correct one,as they all present arrangements where the BES are valid.Let us therefore assume that the result is unambiguous,but also keep in mind that the possibility of numerous correct 9 Figure4:The example taken from(Raphael et al.,2003)illustrates the application of the ESPSP approach on a genome and the di?culties that can be encountered.In both a)and b)two reversals are needed to bring the x i and y i into distances closer than L.A problem arises as both possibilities are equally parsimonious but present di?erent results.While the arrangement of the x i,y i,which we use as indicators for the rearrangements of the blocks A,B,C,D and E,is relatively similar,the structure for G’given by the reversals in a)is di?erent from the one given by the reversals in b). architectures for G’has to be taken into consideration for the re?ned de?nition of the Genome Reconstruction Problem(T2). Finally,the time has come to de?ne the ESP Sorting Problem,based on the reversal(formula 8)as the only possible event and the assumptions we have made about the absence of chimeric DNA and the genome being unichromosomal. (T1)ESP Sorting Problem[ESPSP](Raphael et al.,2003):”Given BES pairs (x1,y1),...,(x n,y n),?nd the minimum number of reversalsρs 1,t1,...,ρs k,t k such that if ρ=ρs 1,t1,...,ρs k,t k then(ρx1,ρy1),...,(ρx n,ρy n)are valid pairs.” The ESPSP presents a formal problem de?nition of the idea outlined above.An example for the application of the ESPSP can be found in(?gure4). 10 2.2The actual Genome Reconstruction Problem The ESPSP approach works relatively well for sample data,but it also has several shortcom-ings,the most blatant one being the ambiguities in the results.Moreover,reality frequently tends to di?er from a simplistic model,so that some changes to the model have to be made in the following before the de?nition of the Genome Reconstruction Problem can be completed. 2.2.1Assigning orientations to the BES As already mentioned the reconstruction results obtained from the ESPSP are plagued by ambiguities.Applying the method to real data would lead to a number of possible arrange-ments for G’without any hint at which one could be the correct version.A solution to this dilemma is o?ered by assigning orientations to the BES.As we already know,the end sequences appear oppositely directed and convergent due to the nature of the BAC vectors and the ESP sequencing method.If the strands are being named by the signs”+”and”-”, an insert of non-composite DNA?anked by the end sequences x’and y’will appear in the orientation(+x’,-y’),which is to symbolise that x’lies on the”+”strand and”y”is located on the”-”strand.2 Consequently,after mapping x’and y’to G,the corresponding pairs(x,y)must possess the same properties concerning the orientation and this will remain una?ected by”legal”(-to avoid overloading the word valid-)reversals.This leads us to an updated de?nition of valid BES:Given the de?nition that a BES pair typically occurs in the orientation(+x,-y),let a pair of BES(x,y)in G be valid,if|y|?|x|≤L and x>0,y<0. The introduction of orientations requires to modify the de?nition of the reversal as well.In the same way a reversal?ips the signs of the blocks,it now?ips the orientation of an end sequence if it is located inside the interval[s,t].Apart from this the de?nition remains the same,except for the taking of the absolute value of x which has become necessary because x may be negative. For an interval[s,t]the reversal[0,1]→[0,1]is de?ned as ρs,t(x)= x if x ?sign(x)(t?(|x|?s))else (8) 2.2.2Multiple chromosomes The ESPSP approach is based on a unichromosomal genome.Since for a more realistic model multiple chromosomes should be allowed,the chromosomes of a multichromosomal genome are concatenated into a single virtual chromosome comprising a range from0to1.The fact 2The assignment of signs to strands can be deliberately chosen.In(Raphael et al.,2003)the mentioned orientation is the expected one for a typical BAC.Which sign belongs to which strand can,however,only be determined with further knowledge about the BACs employed. 11 that the distance between the last base in chromosome n and the?rst base in chromosome n+1should realistically not be1,because they are only neighbours in the concatenation and actually located at distinctly distant positions,is considered in the following de?nition of the distance function.The distance between x and y is denoted by d(x,y)and set to∞when x lies on a di?erent chromosome than y. d(x,y)= ||y|?|x||if x and y lie on the same chromosome ∞else (9) As described in the background section,sign?ipping on entire chromosomes is allowed in this model at no cost.Furthermore we assume that the concatenate is?xed. 2.2.3Events other than reversals Considering the background section(2.2),the restriction to the reversal as the single possible event on the genome is justi?ed.As all the other relevant events,namely fusions,?ssions and translocations can be modelled by reversals it is possible to regard only reversals.Other events,such as deletions,being neglectable for the large scale rearrangement scenario,no event de?nitions have to be added to the model. 2.2.4Coping with chimeric reads To complete the?nal problem de?nition,the issue of chimeric reads has to be addressed. Unaware of which invalid pairs of BES are actually indicators for rearrangements and which ones are artifacts,the method may come up with a wrong result because of faulty input data. Previously we have assumed chimeric BACs to be absent,which was a bit unrealistic.For the ESP Genome Reconstruction Problem we thus require that the”real”composite BES pairs have to be selected as input data. Fortunately,the parts of the genome combined in chimeric BACs are typically derived from distant genomic regions.As the genome is large,the probability of combining two relatively close regions is simply very small compared to the vast number of possibilities of combining a region with one of the many regions further away.On the other hand numerous BACs covering the same breakpoint will contain sequences from neighbouring genomic regions-otherwise they wouldn’t contain the same breakpoint inside their small insert lengths of about200kb. Consequently,a method to distinguish chimeric BACs from those that really contain break-points is given by a kind of cluster analysis:Plotting the distances of the BACs(i.e.the BES(x,y)that represent them)in G and marking the invalid pairs on the plot by di?erent colouring,the isolated invalid pairs are most likely the chimeric BACs and the invalid pairs occurring in clusters can with a high probability be associated with the actual composite 12 Figure5:Similar to the ESPSP the ESPGRP approach works by rearranging the reference genome G by reversals in way that the invalid pairs of BES become valid.This approach reconstructs G’and due to the introduction of orientations assigned to the BES the ambigu-ities of the ESPSP approach as shown in a)can be resolved.With orientations,as shown in b),the correct result can be distinguished from other possible arrangements when we consider that the BES of one pair of BES(x,y)must be convergent and oppositely directed.(Raphael et al.,2003) BACs.Based on these observations the step of choosing the correct input data can be re-alized as shown in the heuristic approach to the ESP Genome Reconstruction Problem in section4. 2.2.5The ESP Genome Reconstruction Problem With the amendments made in the previous subsections we are now able to de?ne the?nal Genome Reconstruction Problem.(Figure5)illustrates that the result in now unambiguous as it additionally the length restrictions requires valid pairs to have a correct orientation. (T2)ESP Genome Reconstruction Problem[ESPGRP](Raphael et al.,2003):”Given a set of oriented BES pairs generated by an ESP experiment,identify composite BES pairs(x1,y1),...,(x n,y n)in this set and?nd the minimum number of rearrangementsρs 1,t1, ...,ρs k,t k such that ifρ=ρs 1,t1 ,...,ρs k,t k then(ρx1,ρy1),...,(ρx n,ρy n)are valid pairs.” 13 3A heuristic approach to the Genome Reconstruction Problem In addition to the de?nition of the ESPGRP approach Raphael et al.(2003)present a heuristic solution to the problem which will be outlined here. 3.1Generating the input data With the observations made in section3.2.4the?rst step in tackling the ESPGRP is based on the assumption that chimeric BACs will most likely appear isolated and the actual composite BACs are expected to be found in clusters.A distance of two times the size of the maximum insert length L proved to be su?cient as a means of distinguishing the isolated pairs from the rest.Thus,let a pair of BES(x,y)be isolated,if its distance from other pairs is greater than2L.The distance is measured with the help of the distance function(formula9)as d(x,y)=d(x1,x2)+d(y1,y2). For the heuristic approach we compute for all pairs of BES(x,y)whether they are valid or not and afterwards plot their positions in G.The valid BES will be coloured black,the isolated invalid ones red and?nally,invalid pairs who are not isolated and form clusters will be painted blue.This is illustrated in(?gure6).The blue coloured clusters will be the input data for the ESPGRP.As the approach presented is a heuristic method it manual adjustments,e.g.in de?ning a range for microrearrangements or regarding the de?nition for isolation may become necessary.A decision based on the visualisation as given in(?gure6) may prove to be useful in this case. 3.2Finding a path of rearrangements Once the input data is speci?ed we take for granted that the clusters returned by the method described above correspond to composite BACs.The input consists of k clusters C1···C k of invalid BES pairs(x,y).The following approach is built on the assumption that the ESP data is sparse:we expect that there is only a single breakpoint contained in each BAC and, additionally,that each cluster is the result of a single rearrangement. Reconstructing G’from G means?nding a path of rearrangements(reversals)that transforms G into G’according to the requirements of the ESPGRP.The reversals we consider as the events which perform the rearrangements take an interval[s,t],that is to be inverted,as an argument.For the construction of G’we need to know what these intervals are for all clusters(i.e.under the assumption of sparse ESP data for the rearrangements,respectively) and the order in which the reversals have been executed.From the ESP data we only receive indicators for the occurrence of a breakpoint in a certain region.The sequences which are rearranged by the reversals,the synteny blocks,have yet to be determined. 14 Figure6:Example from the MCF7tumor data set used in Raphael et al.(2003).The points(x,y),each corresponding to a BES(x,y)are drawn according to the position of x and y-beginning and end of the BES-in the human genome.The original colouring has been altered to suit black and white prints:the black points are still the valid BES,clusters of invalid BES are drawn in light grey and the dark grey ones are the isolated invalid BES. Looked at with a higher resolution the elements of the clusters can be identi?ed,too.Some of the speckled points also consist of several clusters.The diagonal formed by the small range where|y?x|≤L comprises the black as well as some invalid points who miss the length requirements only slightly.Nonetheless they appear on the diagonal due to the scaling of the plot.Likewise some clusters,which are believed to be microrearrangements,can be seen on the diagonal,as the corresponding L lies in range between200kb and1.2mb-a small distance on the genome scale.On the full resolution plot30clusters assumed to represent composite BACs could be identi?ed,5of which were located close to the diagonal and therefore classi?ed as microrearrangements. 15 Accordingly,the heuristic approach aims at?nding the points s and t which separate the synteny blocks for the ith reversal(cluster i under the sparse ESP data assumption).The points s and t divide G into three parts:the region before s,the region after t and the section enclosed by s and t,thus A=[0,s),B=[s,t],C=(t,1].The goal is to?nd points s and t such that these three synteny blocks can be arranged in a manner that the respective invalid pairs of BES(x,y)become valid afterwards.As we have a cluster of BES,there are several invalid pairs all stemming from the same rearrangement which have to become valid. During this process care has to be taken not to render previously valid pairs,which might overlap with the BAC regarded,invalid. The heuristic now works by establishing intervals S and T denoting the approximate loca-tions of s and t,respectively.Inside the intervals all requirements for the pair of BES(x,y) considered to be valid are ful?lled,namely that x and y possess oppositely convergent ori-entations and that the length constraint is met for all x and y,i.e.for the most distant x and y: max x,y =|x?s|+|y?t|≤L(10) Now that intervals have been found where s and t could possibly lie in order for the rear-rangement to produce valid pairs,the next step delivers the precise location of s and t.All positions in the ranges S and T being equally well suited if only the validity is concerned,it is,however,a reasonable assumption to describe the optimal positions within these ranges as the ones which minimize the maximum length of BES pairs in G’: min s∈S,t∈T max x,y =|x?s|+|y?t|≤L(11) Finally,by implementing this approach,G is divided into2k+1synteny blocks derived from the k clusters.The division is the result of de?ning s and t for each cluster.The constraints we named above determine the arrangement of the blocks in G’and thus the architecture of the cancer genome.An example for the determination of arrangements and the reconstruction of a sample genome is given in(?gure7). 4Results for the MCF7breast cancer genome Applying the ESPGRP and the heuristic approach presented above to the MCF7breast cancer genome,a total of383invalid pairs could be identi?ed,127of which were arranged in30clusters interpreted as being indicative for rearrangements(Raphael et al.,2003).Five clusters were not considered in the?nal result as they were classi?ed as microrearrangements (|y?x|<1.2mb)not thought to have major e?ects on the overall layout of the genome.A further three clusters could not be resolved,i.e.no totally valid state could be reached and thus no reliable structure for the post-rearrangement genome G’.These three exceptions are possibly due to duplicated regions in the genome,where the assumption of having sparse ESP 16 Figure7:Examples from Raphael et al.(2003):a)Above intervals S and T comprising possible locations for s and t,respectively,are shown.These are later re?ned into the actual points s and t(below).Orientation constraints require that the end of block A be connected to the end of block B,which is performed by reversing B.In this way the constraints determine the arrangement of the synteny blocks in G’.b)Example for the reconstruction of a cancer genome G’from the original genome G by three reversals denoted by braces around the regions to be reversed.c)Which reversals are needed can be obtained from the ESP data showing six breakpoints symbolised by arcs connecting distant pairs.Following the arcs while writing down(block name)or-(block name),when the block is”entered”from the end,delivers the architecture of G’as shown in a) 17 data is violated.Duplicated regions could cause disturbances as breakpoints from several instances of a duplicated region will be associated with the same cluster. The di?culties connected with the duplicated regions could be overcome by using a di?erent data set,as the original one was especially enriched in certain duplicated regions due to its previous use.Also,a higher BAC coverage could help resolving previously unresolvable cases. For future directions an altered heuristic approach which does not rely on the sparse ESP data assumption is a possibility.The heuristic presented works quite well for this data set, but as we have seen disturbances may be caused by irregularities in the data.Furthermore a higher number of rearrangements than in the MCF7genome might cause the method to fail or require a very high ESP coverage in order to resolve all clusters. The complexity status of the ESPGRP is not yet known,but the existence of a heuristic approach proves the feasibility. Further insights still pending,as not all clusters could be resolved and the resolution of the genome map could still be improved,for the MCF7tumor genome the arrangement di?erences to G are shown in(?gure8).Some of the rearrangements were already con?rmed by biological data(Volik et al.,2003a)while other veri?cation e?orts are currently underway. Biological analysis based on ESP data usually involves sequencing the respective regions and a technique called chromosome painting or FISH(Fluorescent In Situ Hybridisation),which allows for a visual detection of rearrangements.FISH employs ampli?ed DNA fragments labeled with?uorescent dyes which hybridize with metaphase chromosomes.Rearrangements can easily be spotted when labeled DNA selected to target a speci?c locus hybridizes to other loci as well(?gure8b). Notably,the ESPGRP approach solely?nds an arrangement for G’,but not the actual path taken,which of course consists of di?erent kinds of events and not reversals only.Working on the preliminary result given in(?gure8)the most parsimonious scenario to explain the di?erences between G and the reconstructed G’consists of5reversals and15translocations as computed by GRIMM(Tesler,2001). 18 Figure8:a)The MCF7tumor genome as analysed by Raphael et al.(2003):The chromo-somes drawn are the ones where rearrangements occurred on the way from G(left)to the cancer genome G’(right).b)The table on the right hand side is an extract from the results section of Raphael et al.(2003)listing the breakpoints computed by the heuristic approach to the ESPGRP.On the left the rearrangement of sequences from chromosomes17and20 is con?rmed by FISH analysis(Volik,Zhao et al.,2003b).The same BAC(ampli?ed copies of it)hybridizes to chromosomes17and20. 19 5References Asakawa,S.et al.(1997),”Human BAC library:construction and rapid screening”,Gene 1997May20;191(1):69-79. Burgoyne,L.(2003)”Lecture Notes on Cancer”,Flinders University,Adelaide Caltech protocol(no publication date available)Caltech Genome Research Laboratory,Cal-ifornia Institute of Technology,”https://www.doczj.com/doc/d418396389.html,/protocol/321.htm” Hannenhalli,S.and Pevzner,P.(1995)“Transforming Men into Mice(polynomial algorithm for genomic distance problem)“,in:“Proceedings of the36th annual IEEE symposium on Foundations of Computer Science“,Milauwkee,Wisconsin,USA,pp.581-592 Pevzner,P.(2000)”Computational Molecular Biology-an algorithmic approach”,MIT Press,Cambridge,Massachusetts,USA,chapter10 Raphael,B.et al.(2003)”Reconstructing tumor genome architectures”,Bioinformatics, Vol.19,suppl.2,pp.ii162-ii171 Scully,R.(2000)”Role of BRCA gene dysfunction in breast and ovarian cancer predisposi-tion”,Breast Cancer Res,Vol2/02:234-330 Tesler,G.(2001)”GRIMM:genome rearrangements web server”,applications note,Bioin-formatics,Vol.18,no.3,pp.492-493 Volik,S.,Zhao,S.et al.(2003a)”End-sequence pro?ling:Sequence-based analysis of ab-berant genomes”,PNAS,Vol.100,pp.7696-7701 Volik,S.,Zhao,S.et al.(2003b),PNAS Supplementary online material for the previous article. 20 男：同学们大家好，欢迎收听英语沙龙。我是播音员··· 女：我是播音员··· 男：首先，先跟大家说一件好事儿。我们栏目现在准备正式改革了，如果你有好的建议，或者是希望本栏目设立的板块的话，那千万别浪费你的心思哟，赶紧写在站门口的本子上吧。 女：当然，我还得跟大家提个小醒，赶快参加话题发表吧，现在，可是有几个人的分数很高了呢。还不抓紧时间，礼物可就让别人拿走了呀。还有，下周四可能有活动哦。 男：好，步入正题。欧阳茜彤，你们家是不是常有喜事呢？ 女：当然了。大家说的祝福语可是一大套一大套的呢，基本没重的。 男：那你会说吗？ 女：你以为我那么2呀？连这点常识都没有，还算个中国人吗？ 男：唉，你弄错了。我说的是英语祝福语。 女：哦，那我可不会。 男：好吧，下面跟着好好学学。 关于友情的 From majestic mountains and valleys of green to crystal clear waters so blue, this wish is coming to you. 越过青翠的峻岭和山谷，直到晶莹湛蓝的水边，飞来了我对你的祝愿。 It's joy to know you, wishing the nicest things always for you, not only today, but all the year through because you are really a joy to know. 认识你是一种快慰，愿你永远拥有最美好的东西，不仅今天拥有，而且天天拥有，因为认识你真是一种慰藉。 A friend is a loving companion at all times. 朋友是永久的知心伴侣。 When I think of you the miles between us disappear. 当我想起你，相隔千里，如在咫尺。 You're wonderful friend, and I treasure you more with every year. 你是一位难得的挚友，我对你的珍重与岁俱增。 慰问 With warmth and understanding at this time of sorrow…and friendship that is yours for all the tomorrows. Uint2 II. Basic Listening Practice 3. Script W: Have you chosen your elective for next semester yet Are you taking French writing again M: Yes I am, but it’s compulsory for us next semester. So I think I’m gong to do marketing as an elective instead. Q: Which class will the man choose as his elective 4. Script M: Did you go to that businesses lecture on Friday I missed it and need to copy your notes. W: I’d say you could borrow my notes, but Sarah’s got them. Be careful not to miss Professor Brown’s seminar; he takes attendance in that. Q: What is the woman telling the man 3. Script W: Wow, Steven! In the library! What brings you here M: I’m enjoying the view. All the girls in fashion design are here are preparing for an exam on Monday. Q: Why is the man in the library 4. Script W: How’s your group doing with this statistics presentation Mine’s terrible. M: Yeah, mine too. David and Mike are OK, but Steven doesn’t pull his weight and Suzie’s never around. I don’t see how we can pass unless Steven and Suzie r ealize that this is their last chance. Q: What is the true of Steven and Suzie 5. Script W: You took an MBA at Harvard Business School, didn’t you What’s it like M: It’s expensive, about U.S. $ 40,000 a year, plus the costs of food and housing. But the tea ching is first class. The professors have a lot of practical experience. They use the case system of teaching; that is, you study how actual businesses grew or failed. Q: Why is he MBA teaching in Harvard Business School first class, according to the conversation Keys: 1.C 2.D 3.B 4.D 5.A III. Listening In Task1: On the First Day Script Harrison: I’m Harrison. Good to meet you. So you’ve bought the books for this biology class. Jenny: Sure, I think everyone had to before class started. Harrison: No. Usu ally no one does much on the first day because it’s still add-drop. Jenny: What’s that Harrison: Changing from class to class to find out which one is best. Hey, where are you from Jenny: Poland. Have you has this teacher before I’ve heard he is really goo d. Harrison: He’s good if you’re a hand-worker. He expects a lot. Jenny: Oh, I guess that’s good. I hope I can keep up with everyone else in the class. Maybe I need 新世纪大学英语视听说 教程答案 文档编制序号：[KK8UY-LL9IO69-TTO6M3-MTOL89-FTT688] Unit 1 Lesson A P2 A a8 b2 c1 d4 e145 f6789 g1367 h145 i67 P3 A B P3 B 1F 2T 3F 4T 5T 6F 7T P4 B Yes P6 A 1.Sigapore 2.Greenland 3.Angel Falls 4.Etna 5.the Andes 6.Mammonth Cave https://www.doczj.com/doc/d418396389.html,ke Baikal 8.the Sahara 9.Canada’s P8 A 2 P8 B 1257 P8 A The picture on the right. P9 B 1.very famous buildings 2.made of glass, steel,concrete 3.designed 4.style 5.1998 6.452 meters high 7.both the modern and the traditional side of my country P10 A 1.world-famous museum,Paris, 500 years,six million 2.ancient capital,big enough,millions of,shopping center P10 B 1.Louvre Museum 2.Kyoto,Japan 3.1989 4.Hiroshi Hara 5.To cope with the millions of visitors 6.brings new life into the city center 7.an ugly, modern mistake P11 B Answer 3 P11 C Answer 2467 Lesson B P13 A Sentences 1.3 and 5 are True. Whether English should be removed from the College Entrance Examination and give your reason? What is the aim of the reform? What are advantages and disadvantages? Are you agree with the reform that Samaritan 见义勇为[s?'m?ritn]will be taken into consideration in favorable scoring policy in college entrance examination and please give your reason. 教科书/作业答案 Unit1 Basic listening practice: CDABC Listening in: task 1: While the man is wondering why the woman is suddenly getting excellent marks, she says she read an article on studing and remembering. It talks about principles like “ Mental Visulization”, that is, creating a picture in one’s mind of what is to be remembered. This reminds the man of the principle “Association”, “Consolidation”new material into what one has already learned sixteen “Distributed Practice”shorter several days a muscle exercise Task2: FTFFF Task3: 1A 2.C3. D 4.B 5.D Further listening: task 1：ABDDC carrots, eggs, bananas, and milk (2) lost keys (3) a giant carrot and a banana hanging from it (4) a giant milk carton pouring milk over the carrot and banana (5) an egg-shaped UFO flying across the sky (6) The sound of the keys might remind you of having placed them in a drawer. The cold touch of the keys might remind Unit 1 Click Here for Language Learning Short Conversations 1.B 2. C 3. A 4.C 5.D 6.C 7. C 8.A 9.B 10.D Long Conversation 1.A 2. B 3.B 4. D 5. A Understanding a Passage 1. A 2.A 3. C 4. D 5. D Understanding a Movie Speech 1.honored, 2.interesting 3.invitation, 4.great, 5.wise, 6.reason, 7.key, 8.sense, 9.impressions, 10. importantly Homework Listening Task 1 1.D 2. D 3.A 4.B 5.D Task 2 1.B 2.B 3.D 4.A 5.C Task 3 1. added, 2.agreed, 3.create 4.increasingly important, 5.graduate, 6.expanded , 7.included, 8.the most commonly taught language, 9.followed, https://www.doczj.com/doc/d418396389.html,nguage program Unit 2 Chilling Out with the Folks Short Conversations 1.C 2. C 3. D 4.B 5.B 6.D 7. B 8.C 9.D 10.D Long Conversation 1.A 2. C 3.D 4.C 5.A Understanding a Passage 1.D 2.B 3. A 4. C 5. A Understanding a Movie Speech 第一单元 Outside view 1. 1 2 3 4 2.interview techniques lead in looking for boils down 3.professional job coach doing research practiced with present yourselfinterviewerpresentation and understanding in relationship to the jobsome examples dealing with problems 4. 1 3 5 7 8 9 10 11 5.what Samantha was doing wrong in her first job interview what she should do in future interviews Talk 1.B C A C 2.C A D Passage 1 1. 2.B D A D C Passage 2 Unite test DBACA BCDAD BDBC 第二单元 Outside view 1. 3 4 6 1.It's best suited to the download generation. Although the interview says "No one in the industry is suggesting the conventional bookshop is dead", the figures discussed indicate that the conventional bookshop will lose out to digital publishing. Downloadable books sell more than hardbacks and paperbacks. The e-book is delivered in a very convenient way. 2.undergoing a fundamental change600consuming content get it editedclick of a few buttons earning money for an author never really catch on Talk 1.B D A D 2.C B A Passage 1 1.20 once ten discussion travel writing 2.D A C B Passage 2 1. 1-B 2-G 3-E 4-A 5-D 6-F 7-C 2.literary specialist close linkswell-known writers the memorial literary historyrecent series of films the home ofsuccessful films English-speaking world The Differences Between a Seminar and a Workshop By Suzanne Patterson Many makeup artists often look to take short term training courses, such as seminars and workshops, to advance their skills instead of investing in costly makeup school classes. When looking for educational opportunities to advance your skill set it's important that you understand the structure and instructional differences between a seminar and workshop, and what to expect from each as outlined and explained below. The first thing you need to keep in mind about people coming together to participate in any kind of instructional situation is that each person possesses a learning dominance towards one of three distinct types of teaching styles, or modalities, that they lean towards the most: visual, auditory, or manipulative. These are the basics of information transfer, and it simply means some people learn better by seeing, some learn better by hearing, and some learn best with a hands-on experience added into the presentation. The good news is everyone has a combination of all three styles, but an individual will always favor one particular style over the others for learning because of inherent brain patterning. Class Structure: It is important to know your personal dominant style of learning so that you know how to get the most out of your investment. If you are a person that comprehends easily through hearing and watching others do things, such as a lecture with visual demonstration, then you will always do well with a seminar. If you are a person that not only has to hear about it and see it done, but experiment with hands-on in creating it yourself, then a workshop is always a good option. A good seminar or workshop will be constructed in such a way to provide the best learning curve to meet a combination of learning styles adequately in a timely fashion. However, you should always inquire beforehand how the class is actually planned out according to the percentage of lecture, demonstrations, and any hands-on work being presented so you can prepare to get the best possible benefit from the experience. This will help you decide if it meets your dominant style of learning. Functions of a Seminar: This class structure is based on the lecture and demonstration format. This means the instructor has prepared the concepts and techniques they will present and discuss through a combination of visual materials, interactive tools or equipment, and demonstrations. Note taking is always a good thing to do so that you have a form of review once the class is over. This format should always include some take home materials for the student that relates to the lecture and demonstration, and that gives them summary points to review along with their notes. A full laboratory phase (direct student hands-on working opportunity) is not a requirement for this kind of class structure, nor is it necessary in order for the seminar to be considered fully and successfully integrated into the student's educational and skill acquision process. As long as the lectures, 新一代大学英语视听说教程2答案新一代大学英语视听说教程2答案Unit 1 Personality and Development Listening and speaking 1, What does your taste in music reveal about your personality 1. creativity 2, favorite songs 3, 36,000 4, 104 different musical styles 5. their personality 6. conventional 7. more uneasy 8. outgoing 9. hard-working 10. stable 11. gentle 12.low 13. at ease with 14.intelligent 2. Does your career fit your personality? Listening and understanding I, imaginations 2, designs 3. patterns 4. graphic designer 5. landscape architect 6. procedures 7. instructions 8. details and data 9. routine and order 10. accountant II. cost estimator 12. see projects though 13. doers 14. big picture 15. sales agent 16. management analyst 17. logic 18. mysteries 19. detail 20. librarian 文档来源为:从网络收集整理.word版本可编辑.欢迎下载支持. B. Story dictation Listen to a story and fill in the missing words. The story will be read three times. The f the second time there will be a 10-second pause for each missing part. Now listen to the story: We met on a blind date. For me, it was love at first sight. He was perfect. He was my Prince Charming. (1)(I was a terrible romantic when I met him and I gave up m in a tuxedo. His skin was dark like brown sugar and his black hair was combed into carefully casual waves over he smiled at me, the smile spread across his whole face and (2)(lit his eyes so that they turned to golden hon He was strong, thoughtful, and quiet. When we were dating, he was such a gentleman, opening doors, letting sometimes, or he would get my favorite ice cream, and sometimes he would bring me lunch from my favorite Chin that's why we hit it off) . As they say, opposites attract! We dated for about a year before he proposed, and we together) . He was Mr. Right. Things were so great. We got a cute apartment, bought nice furniture, we entertained friends, we went out to e everyone said it would be) . It was exactly what I wanted: great guy, great apartment, great furniture, great ca B. Ask and answer the questions Ask and answer the following questions with a partner. 1. Do you think it is justified to succeed by making a victim of someone else? Your answer： Reference answer： No, no one has the right to sacrifice others for their own purposes. 2. Do you know anyone who often defames others? How do you deal with such people? Your answer： Reference answer： Yes, one of my classmates seems to enjoy defaming anyone that displeases him. I always s 2. in 3. in (the early 1770s) 英语角话题汇总(1) 一、变化的利与弊 每个人都必须面对社会的变迁、生活的变化，你喜欢变化吗？ed带你讨论变化的利与弊。 二、社会热点问题 人类正面临各种公共问题，经济、就业、房子等，richard带你讨论社会热点问题。 三、运动员成功的秘诀 ed带你观看比赛短片，讨论优秀的教练和队员成功的秘诀。 四、剑桥国际英语教学 来自美国的年青外教christian尝试剑桥国际英语的教学并讨论部分话题。 五、奥运会 a. 2008年北京奥林匹克奥运会。 b. 北京奥运会对中国会有怎样的影响？ 六、世界杯 2006年世界杯足球赛。 七、专题讲座 邀请澳大利亚艺术家lanm arr举办专题讲座。 八、交际英语 特邀广州英语频道主持人matthew与你分享如何通过交际英语实现有趣而愉快的谈话。 九、面试 在面试中如何赢得面试官的欣赏，你又需要做哪些准备，rich带你身临其境面试种种。 十、信仰 特殊人群有着有着不同的“奇怪”信仰，ed带你讨论佛教信徒、基督教信徒、无神论者观点。 十一、广告宣传 如何制作广告宣传自己，如何让产品与众不同？ed带你观看广告篇，发现其中奥秘。 十二、整形手术 整形手术上演着一幕幕人间悲喜剧，如何评判它的功过与是非？peter 整形手术在当今社会已司空见惯，你对此有何见解？ 十三、健康 你认为自己面临着食品卫生、健康等方面诸多问题吗？谈谈你的问题并讨论如何解决。 十四、文化差异 richard与讨论文化差异在宗教、婚姻等方面所引发的问题及爱情是否永恒的时尚话题。 十五、精力管理 知道你每天的最佳精力时段吗？matt将引导一场关于增强精力的讨论。 十六、毒品问题 一种新型类似可卡因的药物juicybake在社区蔓延，richard与你讨论是否非法。 十七、优先权 德国外教ed同你讨论优先权“priorities” 十八、 助人为乐、礼貌待人是否各有优缺点？德国外教ed设置不同情景带你进入热烈讨论。 十九、特殊职业 你想象过从事某些特殊职业吗？宇航员或者兽医这些已渐成为生活的一部分。 二十、乐观与悲观 关于乐观主义与悲观主义你有何见解？有多年英语教学经验，教育学专业背景德国外教与你一起探讨。 二十一、假如 如果你**世界，你将有何作为？喜爱中国功夫的英国外教理查德带给你新鲜时尚有趣的话题。 Unit 1 Access to success Listening to the world Sharing Scripts Part 1 V: Hi. There are a lot of things that I’d like to do but I’ve never done before. I’m not really a daredevil, so things like bungee jumping are not really my cup of tea, but I do know that trying new things makes you feel good. Today, I’m going to talk to people about trying new things and achievement. How do you feel about trying new things? Part 2 Ml: I’m up for trying new things. Depends what they are, obviously, um, some things I wouldn’t try, but I’d give most things a go. M2: I always enjoy trying new things. I like to meet new people and, try new food, see new places, see different things. It’s always nice to see that. Wl: I love to try new things. I love to travel. W2: If you try new things, you get more out of life, I think. M3: I’m always up for new things. I love traveling; I love trying exotic new foods, all that sort of stuff; seeing new cultures. Part 3 V: What stops you from trying new things? W2: Fear, probably. M2: Time and money complicates trying new things. It’s hard to find time to travel and it’s hard to afford enough money to travel as well. Ml: Er, I suppose, if it was dangerous and I could get injured. M3: Probably, at the moment, school - I don’t have a lot of time;I’ve got a lot of work. Money, as well. Wl: If it’s very dangerous; or …if my stomach just can’t handle it; or if I don’t have money. Part 4 V: What have you achieved in your life that makes you feel proud? M2: I’m very proud that I was able to go to Ethi opia and build houses for people who needed it. And, being a part of that team was really special to me. Ml: I’ve written plays and people come to see the plays and enjoyed them. And um, I’ve gone out and performed in front of, er, I suppose, thousands of people now and they’ve laughed. Wl: I guess um, finishing school, um, with a high level and - so far not a lot - but I’ve learnt French better than I thought I would, so, I’m proud of that, I guess. I’m still learning. M3: Probably proudest achievement is getting A grades in my end-of-year exams - helping me to get a place in university. W2: Well, I feel that I’ve become er, quite a good person and I guess I’m proud of that. Part 5 V: Who do you admire for their achievements and why? M3: Probably business leaders such as Richard Branson um, as he started off, you know as, as hardly 高级英语视听说课程介绍 一、课程的任务和教学目的 本课程是为英语专业班学生开设的指定必修公共课，是中级视听说的延续，起着进一步培养和提高学生英语听能和口头表达的作用。本课程的教学目的在于通过先进的多媒体教学手段，提高学生对语言真实度较高的各种视听材料的理解能力，从而提高用英语进行交际的能力。视、听、说相结合，相辅相成，以直观画面和情节内容为基础开展有针对性的训练。在听力方面，学生应能听懂英语国家人士所作的知识的讲座；在口头表达方面，能利用所掌握的英语表达自己的思想。 二、课程内容与基本要求 1.课程内容 英语视听说课程侧重语言能力的培养。教学内容丰富，包括人际关系沟通，电话会谈，信息交流，会议，接待客户，演示汇报等活动板块，每个板块中又分出六个主题，另外，为了全面发展学生的语言能力，本课程还根据需要增加了少量写作训练。为了让学生更好地掌握交际惯例，正确并恰当地用英语，让学生具备基本素质，提高跨文化沟通能力。 2.基本要求 视：本课程利用多媒体教学手段输入：给学生播放原汁原味的的英语教学短篇、英文听力材料和英、美原版电影，生动、活泼、真实、形象，以期充分调动学生视觉和听觉神经，产生较好的‘吸收’效果。 听：要求学生能听懂各种主题听力内容如日常生活对话、新闻、热门话题、科技等，在听完第一遍能听懂大意，第二遍能听懂细节，最后能用自己的语言复述听力内容，能听懂语速每分钟150wmp的一般性会话、报道和讲座以及英语国家的广播节目。 说：能就一主题或图片进行连续5分钟左右的较流利的陈述，对热门话题能进行一般性辩论。能较为详细地介绍自己的专业。日常会话不但语言正确，而且能注意语域的运用。在此基础上，组织学生‘模拟对话’、‘电影配音’和小组或全班‘课堂讨论’（Seminar）。教学重点在于发展两种口语能力：英语对话能力和英语连贯表达能力。前者通过模拟对话和学习电影片段配音来加强，后者通过课堂 大学英语视听说教程4答案 新视野大学英语视听说教程4第二版答案（最全最新版本含单元测试答案） Unit1 enjoy your feelings II C B D A D Listening In Task 1what a clumsy man! Keys: A C D C B Task 2causes of depression Keys: (1)families (2)chemicals (3)information (4) certain (5)self-esteen (6)thinking patterns (7)mood (8)divorce (9)physical abuse (10)financial difficulties (11)stress (12)anxiety Task 3 happiness index Keys: B D A A C Let’s Talk Keys: (1) shy (2) crying (3)scared (4) came down (5) fun (6) nice (7) two step (8) argue (9) touch (10) bad time (11) speak (12) fortable (13) brother (14) adults (15) children (16) secondary (17) growing (18) learn Further Listening and Speaking Task 1: Big John is ing! (S1) owner (S2) running (S3) drop (S4) run (S5) local (S6) yelling, (S7) lives!” (S8) As he’s picking himself up, he sees a large man, almost seven feet tall. (S9) The bartender nervously hands the big man a beer, hands shaking. (S10) “I got to get out of town! Don’t you hear Big John is ing?” Task 2 Reason and emotion Key : A B C C Dtt英语沙龙广播稿
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