基因组学复习题终极版

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 genome

1.2 Better understanding the knowledge of heritable genetic variations in human genome

1.3 Apply new knowledge of gene and metabolic pathway to develop new effecive methods for human disease treatment

1.4 To explain the mechanism of evolution and variation among different species

1.5 To clone and characterize the key functional genes in crop

1.6 Using genomics tools to increase the crop yield and resolve the food crisis in the world

2.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 water

3.How dose DNA function advantage over RNA?

Both can serve as genetic material; many viruses use RNA as their genetic material. DNA probably evolved as the genetic material for cells so that RNA could be used as messenger

RNA 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 so

RNA had to 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 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 its original order to help maintain the original sequence on the opposite strand when a correction enzyme comes along to clip out the mismatch

4.Short descriptions the two important experiments which proved that genes are made of DNA

(a)Inject mouse with harmless bacteria, the mouse survive.

Inject mouse 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 mouse

Inject mouse with harmless 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 the 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 caused by chemical mutagens

6.What are the possible effects of DNA mutations on organism?

①lethal ②loss or gain some functions ③alter genotype without changing the phenotype

④contributive to the survival of species, promote evolution

1. The effects of mutations on genomes -Synonymous change -Non-synonymous change

2. The effects of mutations on multi-cellular organisms -Loss-of-function -Gain-of-function

3. …on microorganisms ①Auxotrophs ②Conditional-lethal mutants ③Inhibitor-resistant mutants ④Regulatory mutants

7.What is DNA repair? What is the function of DNA repair?

DNA repairing refers to a collection 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 inactivated by DNA damage.

Direct repair、Excision repair、Mismatch repair、Nonhomologous end-joining：used to mand double-strand breaks、Recombinational repair、SOS repair

8.What is the central dogma?

DNA replication with DNA polymerase.DNA transcript to RNA with RNA polymerase.RNA reverse transcript to DNA with reverse transcriptase. translation,ribosome

9.What is the constitutive expression?

1. Concept

Expression 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, GAPDH and Ubiquitin

2. Features

①genes are expressed continuously and won’t be influenced by cell environmental conditions

②relatively constant expression level

3. Function

To maintaining the basic cell processes or structure

10.What are the two main gene expression patterns?

1、Temporal specificity

The gene expression occurs during a specific time and obey strict order. Temporal specificity of multicellular organisms is also called stage specificity

2、Spatial specificity

In multi-cellular organisms, the expression level of a certain gene is different among different tissues or organs in a certain developmental 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: the 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 chain

12.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 encoded in genetic material (DNA or mRNA sequences) is translated into proteins (amino acid sequences) by living cells.

STOP codons:UAA,UAG,UGA Start codon:AUG

13.How many types of enzymes involved in gene cloning?

①DNA ligases: which join DNA molecules together by synthesizing phosphodiester bonds between nucleotides at the ends of two different molecules, or at the two ends of a single molecule.②DNA polymerases: which are enzymes that synthesize new polynucleotides complementary to an existing DNA

or RNA template.③Alkaline phosphatase：which removes the phosphate group present at the 5′terminus of a DNA molecule.④Other 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 sequence

TypesⅠ：less useful because they cut DNA randomly and sequences of the resulting fragments are not precisely known.Such as Eco B、Eco K

TypesⅡ：over 2300 typeⅡenzymes, cutting DNA is always at the same place, either within the recognition sequence or very close to it, have palindromic structure

Types Ⅲ: recognize two separate non-palindromic sequences that are inversely oriented (e.g. EcoP15). They do not cut DNA at the same points

15.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 single restriction enzyme cutting sites

（4）several selectable marker genes

16.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. ?Negative 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 protein 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 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 double-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 shortened 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 for 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 invading 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 λDNA into the E.eoli genome?

The att sites each contain an identical 15 base pair core sequence. The 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 recombination.

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 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 results 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 replication 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 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 difficulties 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 repetitive 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 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 provide 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 for 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 homozygote 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 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 position 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 intact 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, producing 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, enabling 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 the 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 of 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 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 could 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 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 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 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 not 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 conditions, 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 noncoding sequences.

19. What are the three primary modifications that can be used to improve the location of ORFs by computer analysis?

Computer programs can be modified to screen for codon bias, exon-intron boundaries, and upstream regulatory sequences of genes.

20. What are the two major problems that occur when attempting to use computer analysis to identify ORFs in the genomes of higher eukaryotes?

Simple ORF scan is less effective with DNA of higher eukaryotes.This is partly because there is substantially more space between the real genes in a eukaryotic genome, increasing chances of

finding spurious ORFs. But main problems with genomes of higher eukaryotes in general is that their genes are often split by introns and so do not appear as continuous ORF in DNA sequence.

21. What structural features of functional RNA molecules, such as tRNA and rRNA, can be searched for in a genome sequence to identify the genes encoding these RNA molecules?

Functional RNA molecules have their own distinctive features, which can be used to aid their discovery in a genome sequence

The most important of these features is ability to fold into secondary structure. These structures include intermolecular base pairing and stem-loops or hairpins.?Functional RNA genes also have distinctive regulatory sequences.? Functional RNA genes often locate in empty spaces in a genome sequence

22. What are two limitations that arise when northern analysis is used to determine the number of genes present in a DNA fragment?

Some genes contain optional exons and may encode different-sized mRNA molecules. Also, it is possible that not all of the genes present in a DNA fragment will be expressed in the cells from which the RNA is isolated.

23. What is the differences between orthologus and paralogous genes?

Orthologous genes are homologous genes present in different organisms and paralogous genes are homologous genes present in the same organism.

24. What is cDNA and how to obtain an individual cDNA?

A cDNA is a copy of an mRNA and so corresponds to coding region of a gene, plus any leader or trailer sequences that are also transcribed.

To obtain an individual cDNA, a cDNA library must first be prepared from all of mRNA in tissue being studied.