sequencing：使用二代测序原因：高通量，短序列 不用长序列原因： 1.算法错误率高 2.长序列测序将嵌合体基因错误积累。嵌合体基因：通过重组由来源与功能不同的基因序列剪接而形成的杂合基因
sequencing：
增多的total length>N>gap>missing in genome The reads with a frequency >
1 were called duplicated reads, and we defined the duplication rate as the count of duplicated reads / the count of total reads. 利用further reads mapping gap后仍旧存在的gap，原因是：此区域大部分是串联重复区域和转座子 串联重复区域是什么？ the non-interspersed repeat sequences using RepeatMasker with the “-noint” option, including Simple_repeat, Satellites, and Low_complexity repeats panda装配过程中失掉了一部分串联重复 Loose tandem repeats, requiring that percent of matches larger than 90% and percent of Indels less than 10%; Exact tandem repeats, requiring that percent of matches equal to 100% and percent of Indels equal to 0%
evaluate-1：
1.realigned all the usable sequencing reads onto the scaffolds using SOAPaligner
2.CG content:
3. Comparison of assembled scaffolds and 26 panda mRNA gene sequences in GenBank: The known panda mRNA gene sequences were downloaded from GenBank
4. sequenced and assembled nine BACs independently using Sanger sequencing technology
5.
evaluate-2：
1.annotation 2.mannual check
estimated the genome size:
L为reads平均长度，K为k-mer长度；knum为所有的k-mer总个数，kdepth为k-mer频数的期望深度（即k-mer曲线中主峰对应的横坐标位置）；bnum为测序reads覆盖碱基的总个数，bdepth为覆盖碱基的期望深度。 L=average_reads_length= 52 bp K=17 kdepth=k-mer曲线中主峰对应的横坐标位置=15 bdepth为覆盖碱基的期望深度 bnum为测序reads覆盖碱基的总个数 为什么用C值？ We therefore used C-values (haploid DNA content in picograms), as this is proportional to genome size
repetitive sequences：
1.the Repbase transposable-element consensus sequences were annotated using mammalian genomes other than the panda.
2.about 70Mb of transposable-element sequences (3% of the genome) had a ,10% divergence rate from the consensus (Supplementary Fig. 9), which are likely to be active transposable elements of recent origin.
Comparative genomics-Sequence comparison：
1.segmental duplication：也称为节段重复，是真核基因组内高度同源的序列元件。 whole-genome assembly comparison (WGAC)受 assembly whole-genome shotgun sequence detection or WSSD：Considering the average depth was about 2.47 times that of the whole genome, we estimated that the total length of the duplicated copies was about 34.3 Mb 2.Conserved sequences among the panda, dog and human genomes.Each of these genomes contained ,1.4 Gb of non-repetitive sequence, 3.The <target> dog or human and <query>panda are usually just the names of files containing the sequences to be aligned, in either FASTA, Nib, or 2Bit format. However they can be HSX index files that refer to the sequences indirectly, and they also can specify pre-processing actions such as selecting a subsequence from the file (see Sequence Specifiers for details). With certain options such as ??self the <query> is not needed; otherwise if it is left unspecified the query sequences are read from stdin (though this does not work with random-access formats like 2Bit). As a special case, the <target> is omitted when the ??targetcapsule option is used, since the target sequence is embedded within the capsule file. 4.there were 4–5 times more rearrangements in dog than in panda, which provided evidence that the panda has a lower divergence rate than the dog
annotation：
1.predict/annotation “With synteny” means genes predicted on regions with synteny to the human or dog, and the fragmental genes were conjoined by building gene-scaffolds. “Out of synteny” means genes predicted on regions without synteny evidence to the other species; Pseudo-genes, are those containing more frame errors than a specified threshold. 2.验证： 3.The panda was similar to the human with respect to all of these key parameters 4.Figure S11 | Analysis of sequence completeness of the predicted genes. a.Alignment rate comparison between panda (annotation后的)and dog using single-copy genes, both panda and dog genes were aligned to human genes, and the alignment rate was calculated for each pair of orthologous genes. On average, the dog and panda genes covered 96.2% and 93.5% of the human gene sequences, b. The ratio of missing exons（未annotation的）. The annotated panda genes were compared with the human genes, and the ratio of missing length was calculated on 5’-end, 3-end, and middle of genes unannotated exons were at the 5‘ or 3’ ends of genes; these exons were usually very small conclusion：quality of the predicted panda genes was comparable to that of other well-annotated mammalian genomes.
Comparative genomics-PSGs prediction：
1.one specific for the panda lineage, one specific for the dog lineage, and one combining evidence from all five species included in the alignmen 2. 3.The panda and the dog lineage only share six PSGs： MWU&PE consistent with the results from previous genome-wide positive selection scans in mammalian genomes， 4.
Panda-specific characteristics：
diet:
1.在遗传学中，Ka/Ks或者dN/dS表示的是异意替换（Ka）和同意替换（Ks）之间的比例。这个比例可以判断是否有选择压力作用于这个蛋白质编码基因。 不导致氨基酸改变的核苷酸变异我们称为同义突变，反之则称为非同义突变。一般认为，同义突变不受自然选择，而非同义突变则受到自然选择作用。在进化分析中，了解同义突变和非同义突变发生的速率是很有意义的。常用的参数有以下几种：同义突变频率(Ks)、非同义突变频率(Ka)、非同义突变率与同义突变率的比值(Ka/Ks)。如果Ka/Ks>1，则认为有正选择效应。如果Ka/Ks=1，则认为存在中性选择。如果Ka/Ks<1，则认为有纯化选择作用。 移码突变是由DNA序列中的许多核苷酸的indel引起的遗传突变，其不能被3整除。由于密码子基因表达的三重特性，插入或缺失可以改变阅读框，导致与原始翻译完全不同。序列中较早的缺失或插入发生，蛋白质的改变越多。
2.The third exon contained a 2-bp (‘GG’) insertion; the sixth exon contained a 4-bp (‘GTGT’) deletion. 3.coding not protein:purify selection: 如果某DNA突变对于生物是有害的，但是却不是致命的（立即被消灭），那么这个突变就将处于纯化选择作用之下。 weak selection can lead to different evolutionary fates due to different population sizes and reduced population size can increase the fixation of slightly deleterious mutations. low fecundity rate: a.Alignment b.Phylogenies of FSHB genes in different mammalian species
Population genetics：
1.杂合率：a. panda vs human;b.常 vs X 2.Substitution matrix of the panda heterozygous SNPs in the whole genome. The ratio of transition / transversion is 2.1.
原文地址：https://www.cnblogs.com/yuanjingnan/p/11215479.html