D108 普通生物學共筆
Chapter 20:
The Evolution of Genomes
學藝: 王恩昕、陳冠廷
組員: 阿念、ㄐㄐㄣ、陳品蓁、蘇花瓶、謝宜庭還是沒綽號
KMU D108 Biology CH20 The Evolution of Genomes
Ch20 Glossary
Archaea 古細菌
A diagram in the form of a tree 樹狀圖
α-lactakbumin α-乳清蛋白
Polyploidy 多倍體
Protein-Coding Genes 蛋白質表現基因
Proteome 蛋白質體
Proteomics 蛋白質體學
Pseudogene 偽基因
Banding pattern 條紋圖樣
Bioinformatics 生物資訊學
Rearrangement 重新排列
Regulatory sequence 調節序列
Repetitive DNA 重複 DNA
Centromere 中節
Copy-number variants(CNVs) 拷貝數變異
Repressor 抑制物
Retrotransposon 反轉錄轉位子
DNA intermediate DNA 中間物
Duplication 重複
Systems biology 系統生物學
Endothelium 內皮細胞
Exon shuffling 外顯子洗牌
Telomere 端粒
Tissue plasminogen activator (TPA) 組織血纖
維蛋白溶解酶原活化劑
Transposable element/Transposon 轉位子
(跳躍基因)
Transposition 轉位
Gene annotation 基因註解
Genetic (or linkage) mapping 基因(連鎖)圖譜
Genomics 基因體學
Globin 球蛋白
Tuberculosis 肺結核
Homeobox 同源箱
Homeotic gene 同源異型基因
Hot spot 熱點
Unequal crossing over 不對等互換
Vertebrate 脊椎動物
Insertion 插入
Inversion 倒位
Whole-Genome Shotgun 全基因霰彈槍法
Ligand 受體
Lysozyme 溶菌酶
Malaria 瘧疾
Metagenomics 總體基因體學
Microarray chip 微陣列晶片
Multigene Families 多基因家族
Noncoding DNA 非編碼型 DNA
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Overview
⚫ Complete genome sequences exist for a human, chimpanzee, E. coli, brewer’s
yeast, corn, fruit fly, house mouse, rhesus macaque, and other organisms
(人類、黑猩猩、大腸桿菌、啤酒酵母、玉米、果蠅、家鼠、獼猴、和其他生物體
皆有完整的基因組序列存在)
⚫ Genomics: investigate whole genes and their interactions
(基因體學: 研究整個基因序列和基因序列之間的相互關係)
⚫ Bioinformatics: a scientific field that develops methods for storing and analyzing
biological data
(生物資訊學: 一門發展生物資訊的儲存和分析的科學領域)
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KMU D108 Biology CH20 The Evolution of Genomes
Concept 20.1
The Human Genome Project fostered development of
faster, less expensive sequencing techniques
(人類基因組計畫促進快速且便宜的定序方法之發展)
⚫ Human Genome Project (人類基因組計畫) (1990-2003)
➢
The project had three stages (計畫分為三個階段)
1. Genetic (or linkage) mapping (基因(或連鎖)圖譜)
✓
Several thousand genetic markers(a gene or other identifiable DNA sequence)
(數千個遺傳標記(一個基因或是其他可作為辨識的DNA序列))
✓
The order and relative distance between genetic markers
(利用基因標記之間的順序和相對距離作出圖譜)
2. Physical mapping (物理圖譜)
✓
以DNA片段實際位置為依據
3. DNA sequencing (DNA 定序)
✓
以軟體分析並定序
**流程概述
⚫ Whole-Genome Shotgun Approach to Genome Sequencing (全基因散彈槍法)
➢ developed by J. Craig Venter in 1998
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ Next-generation sequencing (NGS) (次世代定序法) **請參考 CH19 p.5-8**
⚫ Metagenomics (總體基因體學)
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KMU D108 Biology CH20 The Evolution of Genomes
Concept 20.2
Scientists use bioinformatics to analyze genomes and
their functions
(科學家利用生物資訊學去分析基因組以及其功能)
1. National Center for Biotechnology Information (NCBI) (美國的 NCBI 資料庫)
➢ Genbank, the NCBI database of sequences, doubles its data approximately
every 18 months
(Genbank 是 NCBI 的序列資料庫，其資料量大約 18 個月就會加倍)
➢ Software is available that allows online visitors to search Genbank for
matches to
(線上造訪 Genbank 的人可以使用軟體去搜尋、比對以下資料)
2. Ensemble (http://asia.ensembl.org/index.html) A collection of DNA sequences
representing the genomes of 64 species
(集結了代表 64 種生物基因體的 DNA 序列)
3. European Molecular Biology Laboratory (http://www.embl-hamburg.de/)
(歐洲分子生物學實驗室 EMBL)
4. DNA Data Bank of Japan (http://www.ddbj.nig.ac.jp/)
(日本 DNA 資料庫 DDBJ)
5. BGI (Beijing Genome Institute)( http://www.genomics.cn/en/index)
(北京基因組學協會)
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ Identifying Protein-Coding Genes and Understanding Their Functions
(辨識蛋白質表現基因並且了解他們的功能)
➢ Using available DNA sequences, geneticists can study genes directly,
rather than taking the classical genetic approach, which requires
determining the genotype from the phenotype
(利用已知 DNA 序列，遺傳學家可以直接研究基因，而不是透過傳統的方
式，必須透過表現型才能決定基因型)
➢ Gene annotation (基因註解)
✓
Analysis of genomic sequences to identify protein-coding genes and
determine the function of their products
(分析基因組序列來辨識蛋白質表現基因，並且決定其產物的功能)
✓
Gene annotation is an largely automated process
(基因註解已經成為大量自動化的過程)
✓
Comparison of sequences of previously unknown genes with those of known
genes in other species may help provide clues about their function(將未知基因
與其他物種中已知的基因做比較，可以幫助我們找到有關此未知基因功能的線
索)
⚫ Understanding Genes and Gene Expression at the Systems Level
(於系統性的層次去了解基因及基因的表現)
➢ Proteomics (蛋白質體學)
systematic study of all proteins (系統性的蛋白質研究)
➢ Proteome (蛋白質體)
is the entire set of proteins expressed by a cell or a group of cells
(由一個細胞或一群細胞所表現出來的ㄧ整組蛋白質)
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KMU D108 Biology CH20 The Evolution of Genomes
➢ ENCODE (Encyclopedia of DNA Elements) (DNA 元件百科全書)
✓
Began in 2003
The aim of the project is to learn everything possible about functionally important
elements in human genome
(此計畫主要的目的為研究任何在人類基因組中可能具有重要功能的元素)
✓
Protein array→ IPA → pathway → gain-of-function Lose-of-function
⚫ How Systems Are Studied: An Example (系統如何被研究:一個實例)
➢ systems biology (系統生物學)
✓
aims to model the dynamic behavior of whole biological systems based on the study
of the interactions among the system’s parts(主要目的為研究系統當中交互影響的部
分，來模擬整個生物系統當中的行為機制)
✓
to map the protein interaction network in the yeast Saccharomyces
cerevisiae, researchers used sophisticated techniques to knock out (disable)
pairs of genes, one pair at a time, creating double mutant cells
(為了追蹤蛋白質在一種稱為 Saccharomyces cerevisiae 的酵母菌內的互動
模式，研究人員運用精細的科技讓成對的基因失去功能，而且一次使一對基
因喪失功能，來製造擁有突變基因的細胞)
✓
Computer software then mapped genes to produce a network-like “functional
map” of their interactions (然後利用電腦軟體去找出基因，並且依據基因之間
的交互作用，產生網狀的功能地圖)
✓
The systems biology approach is possible because of advances in bioinforma
(因為在生物資訊學上的進步，才使得系統生物學的方法成為可能)
The systems biology approach to protein
interactions
(由系統生物學方法繪出的蛋白質的交互作用網)
此圖裡面包含 Saccharomyces cerevisiae(一種出芽
中的酵母菌)的 4500 個基因的交互作用網格，每一
個白點都代表著一個尚未被歸類到任一方框的蛋白
質。每一個小框框代表了含有特定細胞功能的基因
群。右上的放大圖標出了一個大群組(代謝以及胺基
酸合成)中特定的一些功能小群組。
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⚫ Application of Systems Biology to Medicine (系統生物學在醫學上的應用)
➢ The Cancer Genome Atlas project (癌症基因組圖譜計畫)
✓
began from 2006
✓
200 forms of cancer & subtypes
(現存已知有約 200 種癌症或是其次型態)
✓
The Cancer Genome Atlas project is currently seeking all the common
mutations in three types of cancer by comparing gene sequences and
expression in cancer versus normal cells(癌症基因組圖譜計畫目前正藉比較癌
症細胞與正常細胞的基因序列和表現，尋找在三種癌症(肺癌、卵巢癌、腦部神
經膠質母細胞瘤)中的共同突變基因)
✓
In additional to Whole-genome sequencing, silicon and glass “chips” that hold
a microarray of most of the known human genes are now used to analyze
gene expression patterns in patients suffering from various cancers and other
diseases (除了將全部基因定序，帶有已知人類基因的矽和玻璃材質的微陣列
晶片，現正用於分析罹患癌症和其他疾病的病人之基因表現)
A human gene microarray chip
Tiny spots of DNA arranged in a grid on this
silicon wafer represent almost all of the genes
in the human genome. Using this chip,
researchers can analyze expression patterns
for all these genes at the same time
人類基因微陣列晶片
排列於此矽晶片上之柵格中的這些 DNA 小點
代表人類基因組中的全部基因。使用這種晶
片，研究人員可以同時分析人類所有基因的表
現模式，這可大幅減少試劑的使用量，並確保
所有的基因都處於相同的條件下
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KMU D108 Biology CH20 The Evolution of Genomes
About Microarray Chip (關於微陣列晶片)
以 DNA 晶片為例，製造出來的微陣列晶片含有整組基因的探針
1.
由A、B 兩組細胞分別分離出 mRNA，經由 RT-PCR(反轉錄 + PCR) 轉成 cDNA
2.
A 細胞的樣品加上紅色螢光，B 則是綠色，之後與 DNA 晶片進行雜合，帶有螢光的
cDNA 就會附著到晶片上
3.
分析時，A 細胞有表現的基因就會在晶片上顯示出紅色，B 細胞表現的基因則是綠色，
若是 A、B 皆會表現的基因則會顯示出黃色，沒有表現的則是黑色。透過此方法，可以
大規模的分析特定樣品中的基因表現狀況

除了 DNA 之外，exon、miRNA、蛋白質，等等其他的分子也都可以用微陣列晶片的方式
進行分析
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KMU D108 Biology CH20 The Evolution of Genomes
Concept 20.3
Genomes vary in size, number of genes, and gene
density(各個基因組的大小、基因數量、密度皆不相同)
⚫ Genome size (基因組大小)
➢ Eukaryotic genomes tend to be larger
(比起原核生物，真核生物的基因組通常比較大)
➢ Most bacterial genomes have 1 ~ 6 million base pairs(Mb), and most
plants and animals have genomes of at least 100 Mb; while humans
have 3,000 Mb
(大部分的細菌基因組為 1 ~ 6 Mb，且大部分的動物和植物的基因組至少 100
Mb；而人類有 3,000 Mb)

Mb：million base pairs(百萬個鹼基對)
➢ Genomes of archaea are, for most part, within the size of bacterial genomes
(在大部分的情況下，古細菌域的基因組大小介在細菌域的基因組大小範圍內)
➢ Aside from this general difference between prokaryotes and eukaryotes, a
comparison of genome sizes among eukaryotes fails to reveal any
systematic relationship between genome size and the organism’s
phenotype (除了原核和真核之間普遍的差異，真核生物間之基因組大小的
比較並未發現基因組大小及生物體的表現型之間有任何系統化的關係)
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KMU D108 Biology CH20 The Evolution of Genomes
Number of genes (基因數量)
➢ Free-living bacteria and archaea have from 1,500 to 7,500 genes, while
the number of genes in eukaryotes ranges from about 5,000 for
unicellular fungi to at least 40,000 for some multicellular eukaryotes
(自由生長的細菌和古細菌有 1,500 到 7,500 個基因，但是單細胞真菌
有大約 5,000個基因，而多細胞真核生物有 40,000 個基因)
➢ Number of genes is not correlated to genome size
(基因的數量和基因組的大小並無關聯)
➢ For example, the genome of the nematode C.elegans is 100 Mb in size
and contains roughly 20,000 genes. The Drosophila genome, in
comparison, is much bigger--only 13,700 genes
(線蟲 C.elegans 有 100Mb 大小的基因組，並且擁有大約 20,000 個基
因；而果蠅的基因組比線蟲大，卻只擁有 13,700 個基因)
➢ Vertebrate genomes “get more bang for the buck” from their coding
sequences because of extensive alternative splicing of RNA transcripts
(脊椎動物因可以進行 RNA 轉錄本的選擇性剪切，所以他們可以從基因
組中的編碼序列得到更多的組合)
➢ Gene density and noncoding DNA (基因密度與非編碼 DNA)
➢ Humans and other mammals have the lowest gene density, or number
of genes, in a given length of DNA
(人類和其他哺乳動物的基因密度極低，或者說在一段給定的 DNA 中，其基因
的數目很少)
➢ Multicellular eukaryotes have many introns within genes and
noncoding DNA between genes (多細胞真核生物的基因中，包含
許多內含子和非編碼DNA
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KMU D108 Biology CH20 The Evolution of Genomes
(基因組的大小和估計的基因數目)
(基因的數目)（每 Mb 所含的
(單倍數的基因組大小)
基因數）
(生物體)
(細菌)
(大腸桿菌)
(古細菌)
(紅麴菌)
(真核生物)
(釀酒酵母菌)
(優雅線蟲)
(阿拉伯芥)
(黑腹果蠅)
(智人、人類)
(川貝母(植物))
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KMU D108 Biology CH20 The Evolution of Genomes
Concept 20.4
Multicellular eukaryotes have much noncoding DNA
and many multigene families
(多細胞真核生物有許多非編碼基因和多基因家族)
(包含轉位子和相
關
序列的重複
Types of DNA sequences in the human genome(人類基因組內各種 DNA 序列)
The gene sequence that code for proteins or are transcribed into rRNA or tRNA molecules make up
only about 1.5% of the human genome
(大約只有 1.5%的人類基因組會形成蛋白質，或是轉錄成 rRNA 和 tRNA)
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ The bulk of many eukaryotic genomes consists of DNA sequences that neither
code for proteins nor are transcribed to produce RNAs with known functions
(大部分真核細胞的基因組不產生蛋白質，也不轉錄成有已知功能的 RNA)
⚫ Much evidence indicates that noncoding DNA (previously called junk DNA)
plays important roles in the cell
(許多證據顯示非編碼 DNA(以前稱垃圾 DNA)在細胞裡扮演很重要的角色)
⚫ For example, the genomes of humans, rats and mice contain almost 500
regions of noncoding DNA that are identical in sequence in all three species
(例如：人類和老鼠的基因組大約有 500 處的非編碼 DNA，而且這些序列，在
這幾種物種當中皆相同)
⚫ Pseudogene (偽基因)
A DNA segment that is very similar to a real gene but does not yield a functional
product; a DNA segment that formerly functioned as a gene but has become
inactivated in a particular species because of mutation
(為一與真的基因相似的 DNA 片段，但是不會產生有功能的產物；為一原本可
以有正常功能的基因，但因為突變，導致在特定物種當中無法被活化)
◆ Repetitive DNA (重複 DNA)
⚫ Nucleotide sequences, usually noncoding, that are present in many copies in a
eukaryotic genome. The repeated units may be short and arranged tandemly (in
series) or long and dispersed in the genome
(通常為非編碼的核苷酸序列，存在於真核生物基因組的複製體當中，這些重複
的片段可以很短且被排列為一系列，或是很長且散置於整個基因組中)
⚫ About 75% of the repetitive DNA is made up of units called transposable
elements and sequences related to them
(大約 75%的重複 DNA，是由稱為轉位子的元素和其相關的序列所組成)
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KMU D107 Biology CH20 The Evolution of Genomes
◆ Transposable Elements and Related Sequences
(轉位子與其相關序列)
⚫ Both prokaryotes and eukaryotes have stretches can move from one location
to another within the genome
(原核生物和真核生物皆擁有可以在基因組中轉移的基因)
⚫ The first evidence for wondering DNA segments came from American
geneticist Barbara McClintock’ s breeding experiments with Indian corn
(第一個有關神奇基因(轉位子)的證據是來自美國的基因學家 Barbara
McClintock 的印地安玉米繁殖實驗)
⚫ McClintock identified changes in the color of corn kernels that made sense
only if she postulated the existence of genetic elements capable of moving
from other locations in the genome into the genes for kernel color
(McClintock 認為玉米粒顏色的改變，必須在她的假設之下才能解釋；她假
設存在某種基因會在基因組中移動至與調控玉米粒顏色有關的基因上)
The effect of transposable elements on corn kernel color
(轉位子對於玉米粒顏色的影響)
Barbara McClintock first proposed the idea of mobile genetic elements after observing
variations in the color of the kernels on a corn cob
(觀察完玉米粒顏色的變異後，Barbara McClintock 首先提出可移動的基因的想法)
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◆ Movement of Transposons and Retrotransposons
(轉位子和反轉錄轉位子的移動)
⚫ Eukaryotic transposable elements are of two types
(真核細胞的轉位元有兩種)
➢ Transposon (轉位子)：
move within a genome by means of a DNA intermediate
(透過 DNA 中間物，在基因組中移動)
(可移動的轉位子複本 / DNA 中間物)
Transposon movement(轉位子的移動)
Movement of transposons by either the copy-and-paste mechanism or the cut-and-paste
mechanism involves a double-stranded DNA intermediate that is inserted into the genome
(不論是透過「複製和貼上」的機制，或是透過「剪下和貼上」的機制，都涉及被插入基因組內的
雙股 DNA 中間物)
(此圖為複製和貼上機制)
➢ Retrotransposon (反轉錄轉位子)
move by means of an RNA intermediate that is a transcript of the
retrotransposon DNA
(經反轉錄轉位子 DNA 轉錄出的單股 RNA 中間物在基因體中移動)
➢ Many retrotransposons have long terminal repeats (LTRs) at their ends
that may contain over 1,000 base pairs in each
(很多反轉錄轉位子在他們的末端擁有超過 1,000 個鹼基對的 LTRs)
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KMU D108 Biology CH20 The Evolution of Genomes
(反轉錄轉位子的複製本)
(合成單股的
RNA 中間
)
(插入)
Retrotransposon movement
(反轉錄轉位子的移動)
Movement begins with synthesis of a single-stranded RNA intermediate. The
remaining steps are essentially identical to part of the retrovirus replicative cycle
(此移動是從合成單股 RNA 中間物開始的，其餘的步驟與反轉錄病毒複製的循環很像：欲插
入至另一個位點，RNA 需先藉由反轉錄酶轉換回 DNA)
⚫ Sequences Related to Transposable Elements(和轉位子有關的序列)
➢ Multiple copies of transposable elements and sequences related to them
are scattered throughout eukaryotic genomes
(轉位子和與其相關序列的許多複本皆散佈在真核細胞的基因組中)
➢ LINE-1 (L1)
✓
17% of the human genome is made up of a type of retrotransposon
(人類基因組裡有 17%是由稱為 LIEN-1 的反轉錄轉位子所組成)
✓
L1 sequences have a very low rate of transposition
(L1 序列發生轉位的機率很低)
✓
These L1 elements are DNA that range in length from a few hundreds to 9000 bp.
(L1為長度數百到9000bp的DNA)
✓
The functional L1 are about 6500 bp and encode three proteins, including an
endonuclease that cuts DNA and a reverse transcriptase.
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(功能性L1長約6500bp編有三種蛋白質，包括一個核酸內切酶和反轉錄酶)
✓
L1 activity proceeds as follows:
•
RNA polymerase II (RNA聚合酶)transcribes the L1 DNA into RNA.
•
The RNA is translated by ribosomes in the cytoplasm(細胞質) into the
proteins.
•
The proteins and RNA join together and reenter the nucleus.
•
The endonuclease(核內切酶) cuts a strand of “target” DNA, often in the
intron of a gene.
•
The reverse transcriptase(反轉錄酶) copies the L1 RNA into L1 DNA which
is inserted into the target DNA forming a new L1 element there.
➢ SINEs
✓
SINEs 是短的non-coding DNA序列(100-400 base pair)，
✓
Represent reverse-transcribe RNA molecules originally transcribed by RNA
polymerase III.
(代表被RNA聚合酶反轉錄的分子)
✓
當中Alu elements含量最多，Alu為7S RNA(可組成signal recognition particle)的反
轉錄物。
✓
大部分Alu不含功能性分子，依靠L1去transposed (copied and pasted in new
locations)
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KMU D108 Biology CH20 The Evolution of Genomes
Other Repetitive DNA, Including Simple Sequence DNA
(其他的重複性 DNA，包含簡單序列 DNA)
⚫ Repetitive DNA that is not related to transposable elements has probably
arisen from mistakes during DNA replication or recombination
(與轉位子無關的重複性 DNA，可能是因 DNA 的複製過程或是 DNA 重組過
程發生錯誤而形成的)
➢ Such DNA accounts for about 14% of the human genome
(像這樣的 DNA 占了人類基因組的 14%)
⚫ Simple sequence DNA contains many copies of tandemly repeated short
sequences (簡單序列 DNA 包含很多一系列的短小且重複的序列)
…GTTACGTTACGTTACGTTACGTTACGTTAC…
⚫ In this case, the repeated unit (GTTAC) consists of 5 nucleotides. Repeated
units may contain as many as 500 nucleotides, but often contain fewer than 15
nucleotides
(在這個例子當中，重複的序列單元包含 5 個核苷酸；而重複的單元可以包含多
達 500 個核苷酸，但是通常都是少於 15 個核苷酸)
⚫ When the unit contains 2~5 nucleotides, the series of repeats is called short
tandem repeat (STRs)
(當此重複單元含有 2~5 個核苷酸時，我們稱為 STRs)
⚫ The repeats number can vary from person to person, and since humans are
diploid, each person has two alleles per site, which can differ
(重複的數目因人而異，而且因為人類為二倍體，所以有兩個等位基因，而每個
等位基因重複的數目也可以不同)
⚫ Simple sequence DNA makes up 3% of the human genome
(簡單序列 DNA 在人類基因組中占了 3%)
⚫ Much of a genome’s simple sequence DNA is located at chromosomal
telomeres and centromeres, suggesting that this DNA plays a structural role for
chromosomes
(很多基因組中的簡單序列 DNA 皆位於染色體的端粒或是中節上，暗示了此
DNA 在染色體中扮演了與結構有關的角色)
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ The DNA at centromeres is essential for the separation of chromatids in cell
division
(在中節處的 DNA 對於細胞分裂時，染色分體的分離很重要)
⚫ The simple sequence DNA located at telomeres prevents genes from being
lost as the DNA shortens with each round of replication
(位於端粒的簡單序列 DNA 可以在每次複製過程中，防止基因的遺失)
⚫ Short repetitive sequences provide a challenge for whole-genome shotgun
sequencing because the presence of many short repeats hinders accurate
reassembly of fragment sequences by computers
(短小的重複 DNA 為全基因霰彈槍法的定序方式帶來挑戰，因為許多短小的重
複片段會阻礙電腦對片段的精準排序)
◆ Genes and Multigene Families (基因和多基因家族)
⚫ Like the gene of bacteria, many eukaryotic genes are present as unique
sequences, with only one copy per haploid set of chromosomes
(就像細菌的基因，很多真核生物的基因以唯一序列的形式存在，即每一套單倍
數染色體中，都只有一個複本)
⚫ But in the human genome and the genomes of many other animals and plants,
solitary genes make up less than half of the total gene-related DNA
(但是在人類和很多其他動物、植物的基因組中，單獨存在的基因不到整個基因
相關 DAN 的一半)
⚫ The rest occur in multigene families, collections of two or more identical or very
similar genes
(其餘的皆由兩個或兩個以上的相同或相似的基因所組成，稱為多基因家族)
⚫ Some multigene families consist of identical DNA sequences, usually clustered
tandemly, such as those that code for rRNA products
(某些多基因家族由相同的 DNA 序列所組成，通常集結成一串，例如對應
rRNA 的基因)
⚫ The classic examples of multigene families of nonidentical genes are two
related families of genes that encode globins
(多基因家族裡基因不完全相同最經典的例子為，對應球蛋白的兩相關家族)
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ α-globins and β-globins are polypeptides of hemoglobin and are coded by
genes on different human chromosomes and are expressed at different times
in development
(α 球蛋白和 β 球蛋白是構成血紅素的多肽鏈，分別由位於不同染色體上的基因
所轉錄轉譯而成，並在人類發育的不同時期表現)
(Ψ：偽基因)
Embryo(胚胎)
Fetus and adult
(胎兒和成人)
Embryo
（胚胎）
21
Fetus(胎兒)
adult(成人)
KMU D108 Biology CH20 The Evolution of Genomes
Concept 20.5
Duplication, rearrangement, and mutation of DNA
contribute to genome evolution
(DNA 的複製、重新排列以及突變促成基因組的演化)
⚫ The basis of change at the genomic level is mutation, which underlies much of
genome evolution
(突變是造成基因組階層改變的原因，佔了多數的基因組演化)
⚫ It seems likely that the earliest forms of life had a minimal number of genes —
those necessary for survival and reproduction
(似乎在最早的生命形式當中，只含有極少量生存與繁殖所必須的基因)
⚫ The size of the genome has increased over evolutionary time, with the extra
genetic material providing the raw material for gene diversification
(基因組的大小隨著時間的演進，已經增加許多，藉由增加額外的基因物質，提
供原料使得基因呈現多樣化)
◆ Duplication of Entire Chromosome Sets(複製整個基因組)
⚫ An accident in meiosis can result in one or more extra sets of chromosomes, a
condition known a polyploidy
(發生在減數分裂的意外，可能導致多出一或多個染色體對，造成所謂的多倍體
現象)
➢ Although such accidents would most often be lethal, in rare cases they could
facilitate the evolution of genes
(雖然這種意外通常是致命的，但是在少數的案例中，它可以加速基因的演化)
⚫ The genes in the one or more extra sets can diverge by accumulating
mutations
(位於多出的ㄧ或多個基因組對當中的基因，可以透過累積變異的方式來產生分
歧)
➢ The outcome of this accumulation of mutations may be the branching off of a
new species
(這些累積後的變異可能造成新的分支，並且形成新物種)
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Alterations of Chromosome Structure (染色體結構的改變)
⚫ Sometime in the last 6 million years, when the ancestors of humans and
chimpanzees diverged as species, the fusion of two ancestral chromosomes in
the human line led to different haploid numbers for humans(n=23) and
chimpanzees (n=24)
(6百萬年前的某個時候，人類和黑猩猩的祖先產生分歧，形成不同的物種；祖
先的其二染色體在人類的支線當中發生了融合，導致人類(n=23)和黑猩猩
(n=24)各有不同數目的染色體)
⚫ The banding patterns in stained chromosomes suggested that the ancestral
versions of current chimp chromosomes 12 and 13 fused end to end, forming
chromosomes 2 in an ancestor of the human lineage
(在已經染色的染色體上可以看到條紋的圖樣，這些圖樣顯示黑猩猩祖先的第12
號和第13號染色體曾在末端進行融合，在後代人類的祖先形成第2號染色體)
Human and chimpanzee chromosomes
人類第 2 條染色體上像是中節與端粒的序
列，與黑猩猩第 12.13 條的端粒、第 13 條
的中節相符。這顯示人類祖先的 12.13 對
染色體透過末端相接，形成現今人類的第 2
條染色體。而其中第 12 對染色體的終結仍
具其功能，但 13 的中節已失去功能。
⚫ Researchers also compared the DNA sequence of each human chromosome
with the whole-genome sequence of the mouse
(研究者也比對了人類和老鼠的每一個染色體裡的基因組序列)
人類的 16 號染色體裡有很大一部份主
要是來自老鼠的 7、8、16、17 號染色
體。這代表自從老鼠和人類分家以
來，這些基因就一直待在老鼠世系和
人類世系中
⚫ Performing the same comparative
⚫
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ Performing the same comparative analysis between chromosomes of humans
and other mammalian species allowed the researchers to reconstruct the
evolutionary history of chromosomal rearrangement
（在人類和其他哺乳類物種間，進行相同的比較分析可以讓研究人員重建染色
體重新排列的演化史）
⚫ The rate of duplications and inversion seems to have begun accelerating about
100 million years ago, around 35 million years before large dinosaurs became
extinct and the number of mammalian species began rapidly increasing
(大約在一億年前，重複和倒位的機率似乎開始上升；而大約三千五百萬年前左
右，大量的恐龍開始滅絕，取而代之的是哺乳類的物種)
➢ The apparent coincidence is interesting because chromosomal
rearrangements are thought to contribute to the generation of new species
(這個明顯的巧合很有趣，因為染色體的重新排列被視為與促成新物種有關）
⚫ Analysis of the chromosomal breakage points associated with the
rearrangements showed that specific sites were used over and over again
(分析染色體斷裂處和重新排列之間的關係，顯示某些特定的位置不斷被重複使
用)
⚫ A number of these recombination “hot spots” correspond to locations of
chromosomal rearrangements within the human genome that are associated
with congenital diseases
(很多「熱點」(即不斷被重複使用的地方)，正好對應到人類基因組中，與先天
疾病有關之染色體重新排列的位置)
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Duplication and Divergence of Gene-sized Regions of DNA
（基因大小的DNA之重複與分歧）
⚫ Unequal crossing over during prophase I of meiosis can results in one
chromosome with a deletion and another with a duplication of a particular
region
(第一次減數分裂前期，染色體不對等的互換，將導致其中一個染色體有部分基
因的缺失現象，而另外一個染色體則會有部分基因的重複現象）
⚫ Transposable element can provide homologous sites where nonsister
chromatids can cross over, even when other chromatid regions are not
correctly aligned (轉位子(跳躍基因)可以提供相對應的位置給非姊妹染色分體進
行互換，即使其他染色分體上的區域並沒有正確的排列）
Gene duplication due to unequal crossing over
Such recombination between misaligned nonsister chromatids of homologous chromosomes produces
in chromatid with two copies of the gene and one chromatid with no copy (當減數分裂聯會時，染色體發
生了彎曲，可能會造成基因在互換後，染色體有增加或缺失的現象。如圖中一條染色體的染色分體具有
重複的基因複本，另一個染色分體則沒有該基因複本)
◆ Evolution of Genes with Related Functions: The Human Globin
Genes(具有相關功能之基因的演化，以人類的球蛋白基因為例）
➢ A comparison of gene sequences within a multigene family can suggest the
order in which the genes arose
(在多基因家族裡比較基因的序列，可以知道基因起源的順序)
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KMU D108 Biology CH20 The Evolution of Genomes
➢ Re-creating the evolutionary history of the globin genes using this approach
indicates that they all evolves from one common ancestral globin gene that
underwent duplication and divergence into the α–globin and β– globin
ancestral genes about 450~500 million years ago
(運用此方法(比較序列)重新建立球蛋白基因的演化史，可以得知大約在四億五
千萬到五億年前，同一種祖先的球蛋白基因經過重複和分歧的現象，形成α球
蛋白基因和β球蛋白基因)
A model for the evolution of the human α–globin and β–globin gene families
from a single ancestral globin gene
(α球蛋白及β球蛋白基因家族自同一個先祖基因的演化模型)
➢ After the duplication events, differences between the genes in the globin
family undoubtedly arose from mutations that accumulated in the gene
copies over many generations
(在多次的重複之後，多世代所累積的變異所導致了基因在球蛋白家族之間的
差異)
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Evolution of Genes with Novel Functions
（新功能基因的演化）
⚫ In the evolution of the globin families, genes duplication and subsequent
divergence produced family members whose protein products performed
functions similar to each other
(在球蛋白基因家族的演化中，基因重複和隨後的分歧，產生了其蛋白產物可執行
類似功能(氧運輸)的家族成員)
⚫ However, an alternative scenario is that one copy of a duplicated gene can
undergo alterations that lead to a completely new function for the protein
product
(另外，重複基因的一份複本可能產生改變，導致蛋白質產物具有另外一種新的功
能)
⚫ The genes for lysozyme and α–lactakbumin are good examples of this type of
situation (溶菌酶和 α-乳清蛋白的基因，是很好的例子)
⚫ The two proteins are quite similar in their amino acid sequences and three
dimensional structures (兩者在胺基酸序列和 3D 的結構上很相似)
✓ Lysozyme (溶菌酶):An enzyme that helps protect animals against
bacterial infection by hydrolyzing bacterial cell walls
(藉由水解細菌的細胞壁來幫助動物抵抗細菌)
✓ α–lactakbumin (α-乳清蛋白):a nonenzymatic protein that plays a role
in milk production in mammals
(非酵素的蛋白質，在哺乳類動物的產乳過程中扮演重要的角色)
➢ both genes are found in mammals, whereas only the lysozyme gene is
present in birds
(兩種基因均可在哺乳類動物中被發現，但是鳥類卻只有溶菌酶的基因)
➢ these findings suggest that at some time after the lineages leading to
mammals and bird had separated, the lysozyme gene was duplicated in the
mammalian lineage but not in the avian lineage
(這個發現顯示，在導致哺乳動物世系和鳥類世系分歧的時間點後，溶菌酶基
因在哺乳動物世系經歷了重複事件，但在鳥類世系則否)
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Rearrangements of Parts of Genes : Exon Duplication and Exon
Shuffling (基因部分重排:外顯子重複與重組)
⚫ The presence of introns may have promoted the evolution of new proteins
by facilitating the duplication or shuffling of exons (內含子的存在可能會藉由
加速外顯子的重複與重組，促進新蛋白質的演化)
⚫ A particular exon within a gene could be duplicated on one chromosome
and deleted from the other(在基因中的ㄧ特定外顯子，可以在染色體上重
複，或是從其他染色體上刪除)
⚫ The gene with the duplicated exon would code for a protein containing a
second copy of the encoded domain (ㄧ個擁有重複外顯子的基因，可以轉
錄轉譯出該蛋白質的第二個複本)
⚫ This change might augment its function by
1. Increasing its stability (增加穩定性)
2. Enhancing its ability to bind a particular ligand
(增加其和特定配體的連接能力)
3. Altering some other property(改變一些特性)
◆ Exon shuffling (外顯子的重組)
⚫ the occasional mixing and matching of different exons either within a gene or
between two different (non-allelic) genes owing to errors in meiotic
recombination
(由於減數分裂重組中的錯誤，不同的外顯子在基因內或兩個不同的（非等位）基
因之間隨意地混合和配對)
⚫ This process could lead to new proteins with novel combinations of functions
(此過程可產生擁有新功能組合的蛋白)
⚫ Take tissue plasminogen activator (TPA) for example
(以組織血纖維蛋白溶解酶原活化劑為例)
➢ The TPA protein is an extracellular protein that help control blood clotting
(TPA 為細胞外的蛋白質，負責控制血液的凝結)
➢ TPA 基因在血管內皮細胞(endothelium)上被表現，可以產生能溶解血栓(blood
clot)的蛋白
➢ TPA is thought to have arisen by several instances of exon shuffling and
duplication (TPA 基因被認為是藉由數次外顯子重組和重複而衍生出來的)
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KMU D108 Biology CH20 The Evolution of Genomes
(表皮成長因子基
因)因)
(纖維連接蛋白基
因)
(血纖維蛋白溶解酶原基
Evolution of a new gene by exon shuffling
(藉由外顯子的重新排列(洗牌)可以使基因演化，並產生新的基因)
◆ How transposable Elements Contribute to Genome Evolution
（轉位子(跳躍基因)如何促成基因組的演化）
⚫ Multiple copies of similar transposable elements may facilitate recombination, or
crossing over, between different chromosomes
(許多相似的轉位子可能加速不同染色體間的重組或是互換)
⚫ If a transposable element inserts within a regulatory sequence, the transposition
may lead to increased or decreased production of one or more proteins
(如果轉位子插入一調節序列中，則轉位的現象可能導致蛋白質產量的增減)
➢ Regulatory sequence(調節序列):sequence that codes for a protein, such as
a repressor, that controls the transcription of another genes or a group of
genes
(為一段可以產生蛋白質的序列，像是抑制物，可以控制其他基因的轉錄)
➢ Transposable elements may carry a gene or groups of genes to a new
position (在轉位的過程中，轉位子可能攜帶ㄧ個或一群基因到另ㄧ個新位
置)
➢ Transposable elements may also create new sites for alternative splicing
in an RNA transcript (轉位子也可能在 RNA 轉錄產生新的選擇性剪接位)
➢ In all cases, changes are usually detrimental but may on occasion prove
advantageous to an organism(改變通常有害但有時對生物會有好處)
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KMU D108 Biology CH20 The Evolution of Genomes
Concept 20.6
Comparing genome sequences provides clues to
evolution and development
(比較基因體的序列，可以提供生物演化與發展之線索)
⚫ Comparisons of genome sequences from different species reveal a lot about
the evolutionary history of life, from very ancient to more recent
(比較不同物種之間的基因體，可以揭露很多從很久以前到現在、關於生命的演化
史)
◆ Comparing Genomes(基因體之間的比較)
⚫ Comparing genomes of closely related species sheds light on more recent
evolutionary events, whereas comparing genomes of very distantly related
species helps us understand ancient evolutionary history
(親緣關係較近的物種之間的基因組比較可以幫助我們了解近期的生物演化事件，
而親緣關係較遠的物種之間的基因組比較可以幫助我們了解古代的演化歷史)
⚫ The evolutionary relationships between species can be represented by a
diagram in the form of a tree, where each branch point marks the divergence
of two lineages (物種間的演化關係可由樹狀圖表示，每個分支點表示兩譜系的
分歧)
(細菌)
(真核生物)
(共同祖先)
(古細菌)
(哺乳類的分化)
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Comparing Distantly Related Species(親緣關係較遠之物種的比較)
⚫ Highly conserved genes have changed very little over time
(高度保留基因幾乎不隨時間改變)
➢ 在親緣關係很遠的物種間卻有相似度很高的基因，稱為高度保留基因
⚫ These help clarify relationships among species that diverged from each other
long ago(這些基因有助於釐清很久以前就分化的物種之間的關係)
⚫ Bacteria, archaea, and eukaryotes diverged from each other between 2 and 4
billion years ago(細菌、真細菌和真核生物約在 20~40 億年前就已分化)
⚫ Highly conserved genes can be studied in one model organism, and the results
applied to other organisms(可以選擇某物種來研究高度保留基因，並將結果應用
到其他物種)
◆ Comparing Closely Related Species(親緣關係較近之物種的比較)
⚫ Genetic differences between closely related species can be correlated with
phenotypic differences(親緣關係接近的物種間的基因差異，與他們表現型的差
異有關)
⚫ For example, genetic comparison of several mammals with non-mammals helps
to identify what it takes to make a mammal(例如，比較數種哺乳類和非哺乳類動
物的基因就可以幫助我們了解要成為哺乳類需要什麼基因)
⚫ Human and chimpanzee genomes differ by 1.2%, at single base-pairs, and by
2.7% because of insertions and deletions(人類和黑猩猩的基因只有 1.2%的差異
(以單一核苷酸來看)，或 2.7%的差異(以長片段的 DNA 來看，這是因為有基因
序列的插入與缺失))
⚫ Several genes are evolving faster in humans than chimpanzees. These include
genes involved in defense against malaria and tuberculosis and in regulation of
brain size, and genes that code for transcription factors (許多基因像是抵抗瘧
疾、肺結核、控制腦的大小、負責製造轉錄因子等基因，在人類的演化速度比黑
猩猩快)
⚫ Differences in the FOXP2 gene may explain why humans but not chimpanzees
communicate by speech(兩者在 FOXP2 基因表現的差異，或許可以解釋為什麼
人類可以藉由語言溝通而黑猩猩卻無法)
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KMU D108 Biology CH20 The Evolution of Genomes
⚫ Humans and chimpanzees differ in the expression of the FOXP2 gene, whose
product turns on genes involved in vocalization(人類和黑猩猩對於 FOXP2 的基
因表現不同，FOXP2 基因的產物會啟動其他和發聲有關的基因)
◆ FOXP2 gene functions in vocalization in vertebrate
(FOXP2 基因作用於脊椎動物的發聲)
⚫ FOXP2 一種製造轉錄因子且演化相當迅速的基因，許多證據顯示它與脊椎動物發聲有關
⚫ 科學家利用基因工程，製造出 FO XP2 基因的突變種來進行研究
⚫
Experiment 1: Researchers cut thin sections of brain and
stained them with reagents that allow visualization of brain
anatomy in a UV fluorescence microscope.
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KMU D108 Biology CH20 The Evolution of Genomes
實驗一：
研究者將老鼠大腦的一部份切成薄片，並以試劑染色，使腦內構造(brain anatomy)可以被 UV
螢光顯微鏡觀察
結果一：FOXP2 基因異常會導致大腦內的細胞排列混亂
嚴重程度：homozygote（同型合子）> heterozygote（異型合子）>wild type
實驗二：研究者將把剛出生的幼鼠與母鼠分開，並記錄幼鼠所發出的叫聲次數
結果二：
homozygote：沒有叫聲
heterozygote：極少叫聲
wild type：正常
結論：FOXP2 基因在老鼠的溝通功能上扮演著重要的角色，實驗結果也增強其他研究結果(對
⚫
鳥類以及人類的研究)的可信度，並支持以下假說：FOXP2 在不同的生物體中或許也以類似的
方式運作(控制發聲)
Comparing
Genomes Within a Species (比較同一物種的基因體)
⚫ Variation within humans is due to single nucleotide polymorphisms, inversions, deletions,
and duplications(人類基因變異大多是因為：單一核苷酸變異(SNPs)，基因倒位、缺失和
重複)
⚫ Most surprising is the large number of copy-number variants(CNVs)(其中最驚人的是
數量龐大的拷貝數變異)
⚫ These variations are useful for studying human evolution and human health(這些變異
(SNPs、CNVs、STRs)對研究人類演化史和人體健康有很大的幫助) (相關內容請詳見
CH19)
Copy-number variants (CNVs)：一種結構變異，為基因組
上某些區域有缺失或重複，而造成個體擁有一個或多個的某基
因或是某其區段，例如：正常染色體基因順序為ABCD
產生變異→ABCCD(重複) or ABD(缺失)→CNVs產生
如右圖，因為基因重複造成CNVs，現在染色體內有兩段一樣
的基因。
由於CNVs是影響一段區域，而SNPs只影響單一個核苷酸，
故CNVs更有機會影響表現型，且與複雜的疾病與失調有關。
研究顯示，CNVs 在人類基因組裡總量超過 8 千個，且佔基因體
的 13%，如此高密度的 CNVs 也使人開始懷疑什麼才是「一個正
常的人類基因體」
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KMU D108 Biology CH20 The Evolution of Genomes
◆ Widespread Conservation of Developmental Genes Among Animals
⚫ Molecular analysis of the homeotic genes in Drosophila has shown that they all include
a 180-nucleotide sequence called a homeobox, which codes for a 60-amino-acid
homeodomain in the encoded proteins (在果蠅體內同源異型基因的分子分析顯示，這些
基因全都包含一段相同的核苷酸序列，稱為同源箱，約由 180 個核苷酸所組成，而這些核
苷酸序列可以轉錄出 60 個胺基酸所組成的蛋白質)
⚫ An identical or very similar nucleotide sequence has been discovered in the homeotic
genes of many invertebrates and vertebrates (在脊椎動物和無脊椎動物的同源異型基因
中，也可以找到相同或非常相似的核苷酸序列)
⚫ Homeotic genes in animals were named Hox genes(homeobox-containing genes),
because homeotic genes were the first genes found to have this sequence (動物中的同
源異型基因稱為 Hox 基因(含同源箱基因的縮寫)，因為同源異型基因是最早被發現具有
這種序列的基因)
⚫ Homeobox genes code for a domain that allows a protein to bind to
DNA and to function as a transcription regulator (同源箱的基因負責製
造一個區域，使蛋白質和 DNA 結合，並成為可以調節轉錄的轉錄因
子)
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KMU D108 Biology CH20 The Evolution of Genomes
(胸部)
(生殖部位)
(腹部)
Effect of differences in Hox gene
expression in crustaceans and
insects
(Hox 基因影響甲殼綱和昆蟲綱在身
體結構上不同的表現)
Conservation of homeotic genes in a
fruit fly and a mouse
Homeotic genes that control the form of
anterior and posterior structures of the
body occur in the same linear sequence
on chromosomes in Drosophila and
mice
(果蠅和老鼠之同源異型基因的保留性)
調控身體前端和後端結構的同源異型基
因，以相同的順序出現在果蠅和老鼠的
染色體上
35
KMU D108 Biology CH20 The Evolution of Genomes
Campbell's Biology, 9e (Reece etal.) Chapter 20
The Evolution of Genomes
Although the major overhaul of this chapter occurred during the 8th edition, questions have been
added for every concept in this new edition. The new questions test on a variety of levels, from
basic knowledge and understanding to synthesis, and some present images from the textbook that
require interpretation.
Multiple-Choice Questions
1) For mapping studies of genomes, most of which were far along before 2000, the
three-stage method was often used. Which of the following is the usual order in which the stages
were performed, assuming some overlap of the three?
A)genetic map, sequencing of fragments, physical map
B)linkage map, physical map, sequencing of fragments
C)sequencing of entire genome, physical map, genetic map
D)cytogenetic linkage, sequencing, physical map
E)physical map, linkage map, sequencing
Topic: Concept 20.1
Skill:Application/Analysis
2) What is the difference between a linkage map and a physical map?
A)For a linkage map, markers are spaced by recombination frequency, whereas for a physical map
they are spaced by numbers of base pairs (bp).
B)For a physical map, the ATCG order and sequence must be achieved; however, it does not for the
linkage map.
C)For a linkage map, it is shown how each gene is linked to every other gene.
D)For a physical map, the distances must be calculable in units such as nanometers.
E)There is no difference between the two except in the type of pictorial representation.
Topic: Concept 20.1
Skill:Knowledge/Comprehension
3) How is a physical map of the genome of an organism achieved?
A)using recombination frequency
B)using very high-powered microscopy
C)using restriction enzyme cutting sites
D)using sequencing of nucleotides
E)using DNA fingerprinting via electrophoresis
Topic:
Concept 20.1
Skill:Knowledge/Comprehension
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KMU D108 Biology CH20 The Evolution of Genomes
4) Which of the following most correctly describes a shotgun technique for sequencing a genome?
A)genetic mapping followed immediately by sequencing
B)physical mapping followed immediately by sequencing
C)cloning large genome fragments into very large vectors such as YACs, followed by sequencing
D)cloning several sizes of fragments into various size vectors, ordering the clones, and then
sequencing them
E)cloning the whole genome directly, from one end to the other
Topic:
Concept 20.1
Skill:Knowledge/Comprehension
5) The biggest problem with the shotgun technique is its tendency to underestimate the size of the
genome. Which of the following might best account for this?
A)skipping some of the clones to be sequenced
B)missing some of the overlapping regions of the clones
C)counting some of the overlapping regions of the clones twice
D)having some of the clones die during the experiment and therefore not be represented
E)missing some duplicated sequences
Topic:
Concept 20.1
Skill:Synthesis/Evaluation
6) What is metagenomics?
A)genomics as applied to a species that most typifies the average phenotype of its genus
B)the sequence of one or two representative genes from several species
C)the sequencing of only the most highly conserved genes in a lineage
D)sequencing DNA from a group of species from the same ecosystem
E)genomics as applied to an entire phylum
Topic:
Concept 20.1
Skill:Knowledge/Comprehension
7) Which procedure is not required when the shotgun approach to sequencing is modified as
sequencing by synthesis, in which many small fragments are sequenced simultaneously?
A)use of restriction enzymes
B)sequencing each fragment
C)cloning each fragment into a plasmid
D)ordering the sequences
E)PCR amplification
Topic:
Concept 20.1
Skill:Application/Analysis
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KMU D108 Biology CH20 The Evolution of Genomes
8) What is bioinformatics?
A)a technique using 3-D images of genes in order to predict how and when they will be expressed
B)a method that uses very large national and international databases to access and work with
sequence information
C)a software program available from NIH to design genes
D)a series of search programs that allow a student to identify who in the world is trying to sequence
a given species
E)a procedure that uses software to order DNA sequences in a variety of comparable ways
Topic:Concept 20.2
Skill:Knowledge/Comprehension
9) What is proteomics?
A)the linkage of each gene to a particular protein
B)the study of the full protein set encoded by a genome
C)the totality of the functional possibilities of a single protein
D)the study of how amino acids are ordered in a protein
E)the study of how a single gene activates many proteins
Topic:
Concept 20.2
Skill:Knowledge/Comprehension
10) Bioinformatics can be used to scan sequences for probable genes looking for start and stop sites for
transcription and for translation, for probable splice sites, and for sequences known to be found in
other known genes. Such sequences containing these elements are called
A)expressed sequence tags.
B)cDNA.
C)multigene families.
D)proteomes.
E)short tandem repeats.
Topic:
Concept 20.2
Skill:Knowledge/Comprehension
11) A microarray known as a GeneChip, with most now-known human protein coding sequences, has
been developed to aid in the study of human cancer by first comparing two to three subsets of
cancer subtypes. What kind of information might be gleaned from this GeneChip to aid in cancer
prevention?
information about whether or not a patient has this type of cancer prior to treatment
A)evidence that might suggest how best to treat a person's cancer with chemotherapy
B)data that could alert patients to what kind of cancer they were likely to acquire
C)information about which parent might have provided a patient with cancer-causing genes
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KMU D108 Biology CH20 The Evolution of Genomes
D)information on cancer epidemiology in the United States or elsewhere
Topic:
Concept 20.2
Skill:Application/Analysis
12) What is gene annotation in bioinformatics?
A)finding transcriptional start and stop sites, RNA splice sites, and ESTs
B)describing the functions of protein-coding genes
C)describing the functions of noncoding regions of the genome
D)matching the corresponding phenotypes of different species
E)comparing the protein sequences within a single phylum
Topic:
Concept 20.2
Skill:Knowledge/Comprehension
13) Why is it unwise to try to relate an organism's complexity with its size or number of cells?
A)A very large organism may be composed of very few cells or very few cell types.
B)A single-celled organism, such as a bacterium or a protist, still has to conduct all the complex life
functions of a large multicellular organism.
C)A single-celled organism that is also eukaryotic, such as a yeast, still reproduces mitotically.
D)A simple organism can have a much larger genome.
E)A complex organism can have a very small and simple genome.
Topic:
Concept 20.3
Skill:Synthesis/Evaluation
14) Fragments of DNA have been extracted from the remnants of extinct woolly mammoths, amplified,
and sequenced. These can now be used to
A)introduce into relatives, such as elephants, certain mammoth traits.
B)clone live woolly mammoths.
C)study the relationships among woolly mammoths and other wool-producers.
D)understand the evolutionary relationships among members of related taxa.
E)appreciate the reasons why mammoths went extinct.
Topic: Concept 20.3
Skill:Synthesis/Evaluation
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KMU D108 Biology CH20 The Evolution of Genomes
15) If humans have 2,900 Mb, a specific member of the lily family has 120,000 Mb, and a yeast has ~13
Mb, why can't this data allow us to order their evolutionary significance?
A)Size matters less than gene density.
B)Size does not compare to gene density.
C)Size does not vary with gene complexity.
D)Size is mostly due to "junk" DNA.
E)Size is comparable only within phyla.
Topic: Concept 20.3
Skill:Synthesis/Evaluation
16) Which of the following is a representation of gene density?
A)Humans have 2,900 Mb per genome.
B)C. elegans has ~20,000 genes.
C)Humans have ~20,000 genes in 2,900 Mb.
D)Humans have 27,000 bp in introns.
E)Fritillaria has a genome 40 times the size of a human.
Topic: Concept 20.3
Skill:Application/Analysis
17) Why might the cricket genome have 11 times as many base pairs as that of Drosophila
melanogaster?
A)The two insect species evolved at very different geologic eras.
B)Crickets have higher gene density.
C)Drosophila are more complex organisms.
D)Crickets must have more noncoding DNA.
E)Crickets must make many more proteins.
Topic: Concept 20.3
Skill:Synthesis/Evaluation
18) The comparison between the number of human genes and those of other animal species has led to
many conclusions, including
A)the density of the human genome is far higher than in most other animals.
B)the number of proteins expressed by the human genome is far more than the number of its
genes.
C)most human DNA consists of genes for protein, tRNA, rRNA, and miRNA.
D)the genomes of other organisms are most significantly smaller than the human genome.
Topic: Concept 20.3
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KMU D108 Biology CH20 The Evolution of Genomes
19) Barbara McClintock, who achieved fame for discovering that genes could move within genomes, had
her meticulous work ignored for nearly four decades, but eventually won the Nobel Prize. Why was
her work so distrusted?
A)The work of women scientists was still not allowed to be published.
B)Geneticists did not want to lose their cherished notions of DNA stability.
C)There were too many alternative explanations for transposition.
D)She allowed no one else to duplicate her work.
E)She worked only with maize, which was considered "merely" a plant.
Topic: Concept 20.4
Skill:Application/Analysis
20) What is the most probable explanation for the continued presence of pseudogenes in a genome
such as our own?
A)They are genes that had a function at one time, but that have lost their function because they
have been translocated to a new location.
B)They are genes that have accumulated mutations to such a degree that they would code for
different functional products if activated.
C)They are duplicates or near duplicates of functional genes but cannot function because they
would provide inappropriate dosage of protein products.
D)They are genes with significant inverted sequences.
E)They are genes that are not expressed, even though they have nearly identical sequences to
expressed genes.
Topic: Concept 20.4
Skill:Synthesis/Evaluation
21) What characteristic of short tandem repeat DNA makes it useful for DNA fingerprinting?
A)The number of repeats varies widely from person to person or animal to animal.
B)The sequence of DNA that is repeated varies significantly from individual to individual.
C)The sequence variation is acted upon differently by natural selection in different environments.
D)Every racial and ethnic group has inherited different short tandem repeats.
Topic: Concept 20.4
Skill:Knowledge/Comprehension
22) Alu elements account for about 10% of the human genome. What does this mean?
A)Alu elements cannot be transcribed into RNA.
B)Alu elements evolved in very ancient times, before mammalian radiation.
C)Alu elements represent the result of transposition.
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KMU D108 Biology CH20 The Evolution of Genomes
D)No Alu elements are found within individual genes.
E)Alu elements are cDNA and therefore related to retrotransposons.
Topic: Concept 20.4
Skill:Synthesis/Evaluation
23) A multigene family is composed of
A)multiple genes whose products must be coordinately expressed.
B)genes whose sequences are very similar and that probably arose by duplication.
C)the many tandem repeats such as those found in centromeres and telomeres.
D)a gene whose exons can be spliced in a number of different ways.
E)a highly conserved gene found in a number of different species.
Topic: Concept 20.4
Skill:Knowledge/Comprehension
24) Which of the following can be duplicated in a genome?
A)DNA sequences above a minimum size only
B)DNA sequences below a minimum size only
C)entire chromosomes only
D)entire sets of chromosomes only
E)sequences, chromosomes, or sets of chromosomes
Topic: Concept 20.5
Skill:Application/Analysis
25) In comparing the genomes of humans and those of other higher primates, it is seen that humans
have a large metacentric pair we call chromosome 2 among our 46 chromosomes, whereas the
other primates of this group have 48 chromosomes and any pair like the human chromosome 2 pair
is not present; instead, the primate groups each have two pairs of midsize acrocentric
chromosomes. What is the most likely explanation?
A)The ancestral organism had 48 chromosomes and at some point a centric fusion event occurred
and provided some selective advantage.
B)The ancestral organism had 46 chromosomes, but primates evolved when one of the pairs broke
in half.
C)At some point in evolution, human ancestors and primate ancestors were able to mate and
produce fertile offspring, making a new species.
D)Chromosome breakage results in additional centromeres being made in order for meiosisto
proceed successfully.
E)Transposable elements transferred significantly large segments of the chromosomes to new
locations.
Topic: Concept 20.5
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KMU D108 Biology CH20 The Evolution of Genomes
Skill:Application/Analysis
26) Unequal crossing over during prophase I can result in one sister chromosome with a deletion and
another with a duplication. A mutated form of hemoglobin, known as hemoglobin Lepore, is known in
the human population. Hemoglobin Lepore has a deleted set of amino acids. If it was caused by
unequal crossing over, what would be an expected consequence?
A)If it is still maintained in the human population, hemoglobin Lepore must be selected for in
evolution.
B)There should also be persons born with, if not living long lives with, an anti-Lepore mutation or
duplication.
C)Each of the genes in the hemoglobin gene family must show the same deletion.
D)The deleted gene must have undergone exon shuffling.
E)The deleted region must be located in a different area of the individual's genome.
Topic: Concept 20.5
Skill:Synthesis/Evaluation
27) When does exon shuffling occur?
A)during splicing of DNA
B)during mitotic recombination
C)as an alternative splicing pattern in post-transcriptional processing
D)as an alternative cleavage or modification post-translationally
E)as the result of faulty DNA repair
Topic: Concept 20.5
Skill:Knowledge/Comprehension
28) What are genomic "hot spots"?
A)the locations that correspond to most genetic diseases
B)the areas of a genome that most often mutate due to environmental effects
C)the locations that most often correspond with chromosomal breakpoints
D)the locations that correspond to most genetic diseases and the locations that most often
correspond with chromosomal breakpoints
E)the locations that correspond to most genetic diseases, the areas of a genome that most often
mutate due to environmental effects, and the locations that most often correspond with chromosomal
breakpoints
Topic: Concept 20.5
Skill:Knowledge/Comprehension
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KMU D108 Biology CH20 The Evolution of Genomes
29) In order to determine the probable function of a particular sequence of DNA in humans, what might
be the most reasonable approach?
A)Prepare a knockout mouse without a copy of this sequence and examine the mouse phenotype.
B)Genetically engineer a mouse with a copy of this sequence and examine its phenotype.
C)Look for a reasonably identical sequence in another species, prepare a knockout of this
sequence in that species, and look for the consequences.
D)Prepare a genetically engineered bacterial culture with the sequence inserted and assess which
new protein is synthesized.
E)Mate two individuals heterozygous for the normal and mutated sequences.
Topic: Concept 20.6
Skill:Synthesis/Evaluation
30) Homeotic genes contain a homeobox sequence that is highly conserved among very diverse
species. The homeobox is the code for that domain of a protein that binds to DNA in a regulatory
developmental process. Which of the following would you then expect?
A)that homeotic genes are selectively expressed over developmental time
B)that a homeobox-containing gene has to be a developmental regulator
C)that homeoboxes cannot be expressed in nonhomeotic genes
D)that all organisms must have homeotic genes
E)that all organisms must have homeobox-containing genes
Topic: Concept 20.6
Skill:Synthesis/Evaluation
31) Which of the following studies would not likely be characterized as eco-devo?
A)the study of a particular species to see whether or not it has developmental regulation
B)a study of the assortment of homeotic genes in the zebra
C)a comparison of the functions of a particular homeotic gene among four species of reptiles
D)studying the environmental pressures on developmental stages such as the tadpole
E)a fossil comparison of organisms from the Devonian era
Topic: Concept 20.6
Skill:Application/Analysis
32) A recent report has indicated several conclusions about comparisons of our genome with that of
Neanderthals. This report concludes, in part, that, at some period in evolutionary history, there was
an admixture of the two genomes. This is evidenced by
A)some Neanderthal sequences not found in humans.
B)a small number of modern H. sapiens with Neanderthal sequences.
C)Neanderthal Y chromosomes preserved in the modern population of males.
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KMU D108 Biology CH20 The Evolution of Genomes
D)mitochondrial sequences common to both groups.
Topic: Concept 20.6
Skill:Synthesis/Evaluation
Art Questions
Use the following figure to answer the next few questions.
Figure 20.1 Types of DNA sequences in the human genome.
The pie chart in Figure 20.1 represents the relative frequencies of the following in the human genome:
I. repetitive DNA unrelated to transposons
II. repetitive DNA that includes transposons
III. unique noncoding DNA
IV. introns and regulatory sequences
V. exons
33) Which region is occupied by exons only (V)?
A)A
B)B
C)C
D)D
E)E
Topic: Concept 20.4
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KMU D108 Biology CH20 The Evolution of Genomes
Skill: Application/Analysis
34) Which region includes Alu elements and LI sequences?
A)A
B)B
C)C
D)D
E)E
Topic: Concept 20.4
Skill:Application/Analysis
Use the following figure to answer the next few questions.
Figure 20.2
Figure 21.2 shows a diagram of blocks of genes on human chromosome 16 and the locations of blocks
of similar genes on four chromosomes of the mouse.
35) The movement of these blocks suggests that
A)during evolutionary time, these sequences have separated and have returned to their original
positions.
B)DNA sequences within these blocks have become increasingly divergent.
C)sequences represented have duplicated at least three times.
D)chromosomal translocations have moved blocks of sequences to other chromosomes.
E)higher mammals have more convergence of gene sequences related in function.
Topic: Concept 20.5
Skill:Synthesis/Evaluation
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KMU D108 Biology CH20 The Evolution of Genomes
36) Which of the following represents another example of the same phenomenon as that shown in
Figure 20.2?
A)the apparent centric fusion between two chromosome pairs of primates such as chimps to form
the ancestor of human chromosome 2
B)the difference in the numbers of chromosomes in five species of one genus of birds
C)the formation of several pseudogenes in the globin gene family subsequent to human divergence
from other primates
D)the high frequency of polyploidy in many species of angiosperm
Topic: Concept 20.5
Skill:Synthesis/Evaluation
Scenario Questions
Use the following information to help you answer the next few questions.
Multigene families include two or more nearly identical genes or genes sharing nearly identical
sequences. A classical example is the set of genes for globin molecules, including genes on human
chromosomes 11 and 16.
37) How might identical and obviously duplicated gene sequences have gotten from one chromosome to
another?
A)by normal meiotic recombination
B)by normal mitotic recombination between sister chromatids
C)by transcription followed by recombination
D)by chromosomal translocation
E)by deletion followed by insertion
Topic: Concept 20.4
Skill:Application/Analysis
38) Several of the different globin genes are expressed in humans, but at different times in development.
What mechanism could allow for this?
A)exon shuffling
B)intron activation
C)pseudogene activation
D)differential translation of mRNAs
E)differential gene regulation over time
Topic: Concept 20.4
Skill:Synthesis/Evaluation
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KMU D108 Biology CH20 The Evolution of Genomes
End-of-Chapter Questions
The following questions are from the end-of-chapter “Test Your Understanding” section in Chapter 20 of
the textbook.
39) Bioinformatics includes all of the following except
A)using computer programs to align DNA sequences.
B)analyzing protein interactions in a species.
C)using molecular biology to combine DNA from two different sources in a test tube.
D)developing computer-based tools for genome analysis.
E)using mathematical tools to make sense of biological systems.
Topic: End-of-Chapter Questions
Skill:Knowledge/Comprehension
40) One of the characteristics of retrotransposons is that
A)they code for an enzyme that synthesizes DNA using an RNA template.
B)they are found only in animal cells.
C)they generally move by a cut-and-paste mechanism.
D)they contribute a significant portion of the genetic variability seen within a population of gametes.
E)their amplification is dependent on a retrovirus.
Topic: End-of-Chapter Questions
Skill:Knowledge/Comprehension
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KMU D108 Biology CH20 The Evolution of Genomes
41) Homeotic genes
A)encode transcription factors that control the expression of genes responsible for specific
anatomical structures.
B)are found only in Drosophila and other arthropods.
C)are the only genes that contain the homeobox domain.
D)encode proteins that form anatomical structures in the fly.
E)are responsible for patterning during plant development.
Topic: End-of-Chapter Questions
Skill:Knowledge/Comprehension
42) Two eukaryotic proteins have one domain in common but are otherwise very different. Which of the
following processes is most likely to have contributed to this similarity?
A)gene duplication
B)RNA splicing
C)exon shuffling
D)histone modification
E)random point mutations
Topic:
End-of-Chapter Questions
Skill:Application/Analysis
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KMU D108 Biology CH20 The Evolution of Genomes
1~5
BACDE
26-30
BCECA
6~10
DCBBA
31-35
CBAED
11~15
CABDC
36-40
ADECA
16~20
CDBBE
41-42
AC
21~25
ACBEA
50