基因组学复习题终极版

1、1. From your understand, what are the main challenges in genomics that we will confront with in the future?1.1 To characterize the structures and functions of human genome1.2 Better understanding the knowledge of heritable genetic variations in human genome1.3 Apply new knowledge of gene and metabol

2、ic pathway to develop new effecive methods for human disease treatment1.4 To explain the mechanism of evolution and variation among different species1.5 To clone and characterize the key functional genes in crop1.6 Using genomics tools to increase the crop yield and resolve the food crisis in the wo

3、rld2. What are the advantages of DNA as the genetic material？2.1. DNA contains a large amount of information.2.2. Complementary base pairs ensure accurate replication.2.3. High stability in water3. How dose DNA function advantage over RNA? Both can serve as genetic material; many viruses use RNA as

4、their genetic material. DNA probably evolved as the genetic material for cells so that RNA could be used as messengerRNA which carries the information for protein synthesis to the ribosomes. The mRNA is metabolically unstable because it is rapidly broken down by RNAse. DNA must be stable soRNA had t

5、o evolve to fulfill this function. Both DNA and RNA have the same coding capacity. They both are polymers with similar potential length. DNA that commonly exists in all living organisms but is not really common with RNA. Some viruses are exceptional because they exist with a single strand of DNA or

6、with a double strand of RNA. The most important by far is that DNA is typically double stranded. This has a number of advantages, the most immediate being when one strand breaks the entire molecule does not fall apart if an error is made in one strand on one base the other strand is still there in i

7、ts original order to help maintain the original sequence on the opposite strand when a correction enzyme comes along to clip out the mismatch4. Short descriptions the two important experiments which proved that genes are made of DNA (a)Inject mouse with harmless bacteria, the mouse survive.Inject mo

8、use with harmless bacteria plus transforming principle, the mouse dies, and extract virulent bacteria from mouse Inject mouse with harmless bacteria plus transforming principle and treated with protease ribonuclease, the mouse dies, and extract virulent bacteria from mouseInject mouse with harmless

9、bacteria plus transforming principle and treated with protease deoxyribonuclease, the mouse survives, and can not extract virulent bacteria from mouse(b) 35S is only in DNA, 32P is only in protein. Make 35S and 32P marked phage attached to the bacteria, the phage was marked with 32P and 35S, after t

10、he phage detached,centrifuge,we can find 32P in the bottom pellet of bacteria,and 35S in the top phage protein capsid. 5. Please list the names of three main the types of DNA mutations.(1) Some mutations are spontaneous errors(2) Some mutations are caused by physical mutagens(3) Some mutations are c

11、aused by chemical mutagens6. What are the possible effects of DNA mutations on organism? lethal loss or gain some functions alter genotype without changing the phenotype contributive to the survival of species, promote evolution1. The effects of mutations on genomes -Synonymous change -Non-synonymou

12、s change2. The effects of mutations on multi-cellular organisms -Loss-of-function -Gain-of-function3. on microorganisms Auxotrophs Conditional-lethal mutants Inhibitor-resistant mutants Regulatory mutants7. What is DNA repair? What is the function of DNA repair? DNA repairing refers to a collection

13、of processes by which a cell identifies and corrects damage to the DNA molecules. It is essential that cells possess efficient repair systems, without these repair systems a genome would not be able to maintain its essential cellular functions for more than a few hours before key genes became inacti

14、vated by DNA damage.Direct repair、Excision repair、Mismatch repair、Nonhomologous end-joining：used to mand double-strand breaks、Recombinational repair、SOS repair8. What is the central dogma?DNA replication with DNA polymerase.DNA transcript to RNA with RNA polymerase.RNA reverse transcript to DNA with

15、 reverse transcriptase. translation,ribosome9. What is the constitutive expression?1. ConceptExpression of a gene that is transcribed at a relatively constant level regardless of the cell environmental conditions, for example, the expression of housekeeping genes to produce proteins such as Actin, G

16、APDH and Ubiquitin2. Features genes are expressed continuously and wont be influenced by cell environmental conditions relatively constant expression level3. FunctionTo maintaining the basic cell processes or structure10. What are the two main gene expression patterns?1、Temporal specificityThe gene

17、expression occurs during a specific time and obey strict order. Temporal specificity of multicellular organisms is also called stage specificity2、Spatial specificityIn multi-cellular organisms, the expression level of a certain gene is different among different tissues or organs in a certain develop

18、mental stage ， then resulting in different distribution of specific protein in tissues and organs. Spatial specificity is also called cell or tissue specificity.11. What are the three key stages of transcription? Initiation: describes the synthesis of the first nucleotide bonds in RNA Elongation: th

19、e enzyme moves along the DNA and extends the growing RNA chain Termination: involves recognition of the point at which no further bases should be added to the chain12. What is the genetic code and what are the stop codons and start code?The genetic code is the set of rules by which information encod

20、ed in genetic material (DNA or mRNA sequences) is translated into proteins (amino acid sequences) by living cells.STOP codons:UAA,UAG,UGA Start codon:AUG13. How many types of enzymes involved in gene cloning?DNA ligases: which join DNA molecules together by synthesizing phosphodiester bonds between

21、nucleotides at the ends of two different molecules, or at the two ends of a single molecule.DNA polymerases: which are enzymes that synthesize new polynucleotides complementary to an existing DNA or RNA template.Alkaline phosphatase：which removes the phosphate group present at the 5 terminus of a DN

22、A molecule.Other enzymes:Taq DNA polymerases、nucleases、end-modification enzymes etc.14. What is restriction enzyme and how many types does it have?A restriction endonuclease is an enzyme that binds to a DNA molecule at a specific sequence and makes a double-stranded cut at or near that sequenceTypes

23、：less useful because they cut DNA randomly and sequences of the resulting fragments are not precisely known.Such as EcoB、EcoKTypes：over 2300 type enzymes, cutting DNA is always at the same place, either within the recognition sequence or very close to it, have palindromic structureTypes : recognize

24、two separate non-palindromic sequences that are inversely oriented (e.g. EcoP15). They do not cut DNA at the same points15. What are the ideal requirements for a plasmid?（1）small size, high copy number, easy for manipulation（2）can replicate independently in the host cell chromosome（3）one or more sin

25、gle restriction enzyme cutting sites（4）several selectable marker genes16. Please outline the four regulatory patterns forming by the interaction of transcriptional regulatory protein and small effect factor.Dependent on regulatory protein : Positive control: active protein, promote gene expression.N

26、egative control: inactive protein, prevents or inhibits a gene from being expressed.Dependent on small molecule : Induction: inducer, a small molecule can trigger gene transcription by binding to a protein activator. Repression: repressor, it refer to inhibition of transcription by binding of protei

27、n repressor.1. What is the role of recombination in genome evaluation? Recombination allows for major changes and extensive restructuring of genomes. In the absence of recombination, genomes would undergo little change and be fairly static structures.2. How can the resolution of a holliday structure

28、 yield two different results?Resolution of a Holliday junction can generate parental or recombinant duplexes, depending on which strands are nicked. Both types of product have a region of heteroduplex DNA.3. Describe how the double-strand break model explains how gene conversion occurs? In the doubl

29、e-strand break model, homologous recombination initiates with a double-strand break in one of the molecules. One strand in each half of the broken chromosome is shortened, leaving 3' overhangs. One of the overhangs invades the other intact DNA molecule to set up a Holliday junction. The shortene

30、d DNA strands are extended by DNA polymerase, with the DNA synthesis in the region being converted using the DNA molecule that did not undergo the original double-strand break as a template.4. Some E.eoli strains that are used for propagating recombinant plasmids contain recA defects to be useful fo

31、r researchers working with recombinant plasmids?（1）Establishment of hetero-duplex during HR by forming a protein RecA-coated DNA filament that is able to invade intact double helix and set up D-loop（2）An intermediate in formation of D-loop is a triplex structure, a three-strand DNA helix in which in

32、vading poly-nucleotide lies within major groove of intact helix and forms hydrogen bound with base pairs it encounters.5. What are the properties of the attP and aftB sites that mediate integration of DNA into the E.eoli genome? The att sites each contain an identical 15 base pair core sequence. The

33、 core sequences are flanked by variable sequences: B and B' (each just 4 hp in length) in the bacterial genome, and P and P' in the phage DNA. P and P' are both over 100 by in length. Mutations in the core sequence inactivate the att site so that it can also no longer participate in reco

34、mbination.6. How is the new copy of a retroelement inserted into a genome? a. Insertion of new copy of retroelement into genome occurs preferentially at certain position.b. Insertion involves removal of two nucleotides from 3'end of double-stranded retroelement by integrase enzyme. c. Integrase

35、also makes a staggered cut (交错切口）in genomic DNA so that both retroelement and integration site now have the 5' overhang.d. These overhang might not have complementary sequences but they still appear to interact in some way so that retroelement becomes inserted into genomic DNA.e. Interaction res

36、ults in loss of retroelement overhangs and filling in of gaps that are left, which means that integration site becomes duplicated into a pair of direct repeats, one at either end of inserted retroelement.7. What is the role of tRNA molecules in the replication of retroelements? The first step in rep

37、lication of a retroelement is synthesis of an RNA copy. This RNA molecule is then converted into double-stranded DNA. The first stage in this conversion is synthesis, by reverse transcription, of a single-stranded DNA copy of the RNA molecule. This strand synthesis reaction is primed by a tRNA that

38、anneals to a site within the 5' long terminal repeat of the RNA copy of the retroelement.8. Give examples of the harmful effects that transposons can have on a genome. 9. Why are maps required for the sequencing of genomes? If a map of a genome was unavailable, what would be the major difficulti

39、es in obtaining a genome sequence?A genome map provides a guide for the sequencing experiments by showing the positions of genes and other distinctive features. If a map is unavailable then it is likely that errors will be made in assembling the genome sequence, especially in regions that contain re

40、petitive DNA.10. How has PCR made the analysis of RFLPs much faster and easier? What was required to map RFLPs prior to the utilization of PCR? The primers for the PCR are designed so that they anneal either side of the polymorphic site, and the RFLP is typed by treating the amplified fragment with

41、the restriction enzyme and then running a sample in an agarose gel. Before the invention of PCR, RFLPs were typed by Southern hybridization, which is time-consuming.11. How does the linkage between genes provide a critical component to genetic mapping? Discuss how genetic markers can be linked to pr

42、ovide maps of individual chromosomes.If a pair of genes display linkage then they must be on the same chromosome. If crossing-over is a random event then the recombination frequency between a pair of linked genes is a measure of their distance apart on the chromosome. The recombination frequencies f

43、or different pairs of genes can be used to construct a map of their relative positions on the chromosome.12. Why is a double homozygote used for test crosses in linkage analysis experiments? Why is it preferable that the homozygote alleles be recessive for the traits being tested? The double homozyg

44、ote will produce gametes that are all the same genetically and if they are recessive then this parent will not contribute to the phenotype of the offspring.13. Genetic mapping techniques require at least two alleles for a given marker, while physical mapping techniques do not rely on the presence of

45、 alleles to map genomes. Discuss how the technique of fluorescent in situ hybridization can be used to map genome locations even if there is no genetic variation present at a given position. FISH uses a fluorescently labeled DNA fragment as a probe to bind to all intact chromosome. The binding posit

46、ion can be determined and this information used to create a physical map of the chromosome.14. How does a scientist prepare a clone library of DNA from just a single chromosome? Individual chromosomes can be separated by flow cytometly. Dividing cells are carefully broken open so that a mixture of i

47、ntact chromosome is obtained. The chromosomes are then stained with a fluorescent dye. The amount of dye that a chromosome binds depends on its size, so larger chromosomes bind more dye and fluoresce more brightly than smaller ones. The chromosome preparation is diluted and passed through a fine, pr

48、oducing a stream of droplets, each one containing a single chromosome. The droplets pass through a detector that measures the amount of fluorescence, and hence identifies which droplets contain the particular chromosome being sought. An electric charge is applied to these drops, and no others, enabl

49、ing the droplets containing the desired chromosome to be deflected and separated from the rest.15. What are the ideal features of a DNA marker that will be used to construct a genetic map? To what extent can RFLPs, SSLPs, or SNPs be considered "ideal" DNA markers?The text indicates that th

50、e ideal features include high frequency in the genome being studied, ease of typing, and the presence of multiple alleles. This implies that SSLPs should be the "ideal" markers, but in reality SNPs are more popular. A discussion of this apparent paradox of the relative importance of each o

51、f the three criteria, demands consideration and in particular a realization that the critical feature of an "ideal" marker is high density.16. Which is more useful-a genetic or a physical map? what purpose the map is intended to fulfill. A map designed to aid a sequencing project might not

52、 be the same as one designed to enable individual genes to be cloned. If it is concluded that for sequencing purposes a physical map is more useful, and in fact a genetic map has little or no direct value (which is a reasonable inference to make from a reading of Chapter 4), then the discussion coul

53、d turn to how easy or otherwise it would be to locate the genes in a genome sequence, and to assign functions to those genes without any prior knowledge of where the genes are. These issues are discussed in Chapter 5.From above description we know that the map provide the framework fro carrying out

54、the sequencing phase of the project with both method. If the map indicates the positions of genes, then it can also be used to direct the initial part of a clone contig project to the interesting regions of a genome, so that the sequences of important genes are obtained as quick as possible.Genetic

55、mapping: it is based on use of genetic techniques to construct maps showing the positions of genes and other sequence features on a genome. Techniques include cross-breeding experiments or, in case of humans, the examination of family histories.Physical mapping: it uses molecular biology techniques

56、to examine DNA molecules directly on order to construct maps showing the positions of sequence features, including genes.17. How does a dye-quenching experiment determine if an oligonucleotide has hybridized to a DNA molecule containing a single nucleotide polymorphism? When the oligonucleotide is n

57、ot hybridized to the target sequence, the fluorescent label and quenching molecule are next to each other and the fluorescence is quenched. When the oligonucleotide binds to a target sequence, the fluorescent label is located away from the quenching molecule. By controlling the hybridization conditi

58、ons, the oligonucleotide will only bind to the target sequence if all the nucleotides are complementary.18. Why is it relatively easy to identify ORFs in prokaryotic genomes by computer analysis?Computers can readily scan all six reading frames of a DNA sequence for ORFs. In addition, as a random DNA sequence would possess a stop codon at least every 100-200 by and most genes contain more than this number of codons, it is fairly straightforward to identify coding sequences in bacterial genomes that lack introns and other significant noncodi