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Flashcards in Module 3 Review and Practice Exam Deck (125):
1

You determine that you have only 3 copies left of an important DNA fragment, so you decide to amplify it. Using flanking primers, how many PCR cycles would you have to run to generate over one billion (10^9) copies of the fragment?

?? "This question has not been graded."


29

2

What enzyme is required for PCR? State the specific name of the enzyme, not just the type of enzyme.

?? "This question has not been graded."

Taq polymerase

3

Apart from the enzyme, what 3 DNA molecules are required (give the technical names by which they are called)? What must be true about the sequences of these molecules? Note that dNTPs are individual nucleotides, not DNA molecules.

??

The complement, single DNA strand, and 2 short oligonucleotide primers are required to flank each side of the target sequence.

The sequence of the primers and complement DNA strand must be complementary to one another, so the primers can bind to the DNA.

Single stranded template DNA; two specific primer molecules; the two primer molecules have to be complementary to part of the template DNA molecule.

4

Name the 3 basic steps of PCR and describe the molecular processes that occur in each. How is each step induced?

??
The first step of denaturization is induced by a high temperature, which facilitates the denaturing of double stranded DNA into single stranded DNA.
The second step of hybridization/annealing is induced by a lower temp, which allows the primers to bind or anneal to the complement DNA, on either side of the target sequence.
The final, third step of elongation is induced by a medium temperature, slightly higher than step 2, which allows the thermodynamically stable Taq polymerase enzyme to elongate the primers by synthesising new DNA.


- Denature target DNA, forming a single strand template DNA molecule. This process is induced by raising the temperature to separate two DNA strands.
- Anneal specific primers to flank DNA target of interest. These primers are complementary to the template DNA. This process is induced by cooling down the environmental temperature and let the primer molecule form hydrogen bonds with the template DNA molecule.
- Extension of primer by Taq DNA polymerase, creating a complementary strand of the template DNA molecule.

5

How does this system work to amplify DNA?

??
Each run of PCR results in double the amount of initial DNA, thus after several runs, the initial target DNA is significantly amplified.

The system works to amplify DNA, by the addition of enzymes to the targeted region which allow it to be flanked, then synthesized by the Taq polymerase. After each trial the taregt region is synthesized at an exponential rate.

6

Cloning reactions are done with DNA that has been cloned by restriction digestion and not by PCR. Using what you know about the way PCR works, why would you not want to use DNA from PCR to create DNA for cloning?

Normally in DNA replication, polymerase makes errors one out of every 1010 nucleotides inserted.

(In addition, Taq polymerase used in PCR is less faithful because it does not have a proofreading subunit).

Because PCR amplifies from previous sequences if an error is made early on it will be proliferated in the sequence.

This answer it true (feedback)

7

Two genes that evolved from the same common ancestral gene, but are now found as homologs in different organisms are called ____________ .

orthologs

8

PCR

Taq polymerase

9

This term refers to the work undertaken by large teams of researchers who, through a concerted effort, clone and sequence the DNA of a particular organism.

genome project

10

During gel electrophoresis, __ will migrate more rapidly than __.

During gel electrophoresis, small DNA fragments will migrate more rapidly than large DNA fragments.

11

In the previous list of cloned fragments, the fragments needed to make the longest possible contig, with the least amount of overlap, are...

Cloned fragment, Markers present
A, M4 M5
B, M1 M2 M3
C, M6 M7 M8
D, M3 M4 M5 M6
E, M3 M4 M5
F, M8 M9 M10

b,d,c,f

B, M1 M2 M3
D, M3 M4 M5 M6
C, M6 M7 M8
F, M8 M9 M10

12

Which of the following enzymes is used to make complementary DNA (cDNA) from RNA?

reverse transcriptase

13

Electrophoresis separates DNA fragments of different sizes, but this technique does not indicate which of the fragments contains the DNA piece of interest. This problem is solved by

Removing the bands from the gel and hybridizing them with a known strand of DNA complementary to the gene of interest.

14

Which technique would NOT be used to find a gene for a functional protein in a sequenced region of a genome?

See if a SNP database contains sequences in the region.

15

Which techniques would be used to find a gene for a functional protein in a sequenced region of a genome?

Scan the region for promoter sequences.

Scan the region for ORFs.

Scan the region for intron splice sites.

See if an EST database contains sequences in the region.

16

List two especially useful characteristics of cloning vectors.

high copy number and antibiotic resistance gene(s)

17

0.1% frequency of recombination is observed ....

....in genes located very close to one another on the same chromosome.

18

Some vectors such as pUC18 and others of the pUC series contain a large number of restriction enzyme sites clustered in one region. What term is given to this advantageous arrangement of restriction sites?

polylinker

19

The set of all proteins encoded by the genome is called the _______ .

proteome

20

The difference between a genetic screening experiment and a selection experiment is that a screening experiment involves ________, whereas a selection experiment creates conditions that ________ irrelevant organisms.

The difference between a genetic screening experiment and a selection experiment is that a screening experiment involves visual examination, whereas a selection experiment creates conditions that eliminate irrelevant organisms.

21

You are handling a paternity lawsuit brought against five potential fathers by a woman. You isolated DNA from the mother, the child, and all the potential fathers. After using PCR to amplify specific polymorphic loci from each individual, you fractionate the amplified products on an agarose gel and stain with ethidium bromide to visualize the DNA fingerprints (shown below). Mo = mother; Ch = child; M1–M5 = potential fathers. Do these results confirm that any of the men are the child's biological father? Explain your answer.

??

Any band that the child has that the mother has as well is considered uninformative. (this doesn't mean that that particular one actually came from the mother - it could have come from the father if he has it as well). Any band that the child has that is absent in the mother must have come from its father. Based, on this, M4 is the child's father.

22

Compare the fields of structural, functional, and comparative genomics. What is the purpose of each?

??

structural genomics is the organization and sequece of geneic info in a genome. functional genomics is tells what the sequences do and compartive genomics compares and contrast differences in gene sequences

Structural genomics is the organization and sequence of the genetic information within a genome. The purpose of this is to create maps that provide information on locations of genes, molecular markers, and chromosome segments. Functional genomics characterizes what the sequences do. This identifies their function. Comparative genomics compares similarities and differences in gene content, function, and organization among genomes of different organisms.

23

In the genetic map of the human genome, one map unit is approximately 850,000 bp. For the genome of the eukaryotic yeast Saccharomyces cerevisiae, one map unit is approximately 3000 bp. What is a map unit, and why is it so different in these two different types of organisms?

??

A map unit is the distance on genetic maps based on percentage of recombination. This is the rate that the chromosome crosses over. The numbers are different because in general, multicellular eukaryotes have more DNA than simple, single-celled eukaryotes.

A map unit is the amount of measured recombination b/w two link points. One map unit in a human is many more bp than in yeast due to the repitition in human genes. The amount of homologous pairs in humans must be lower than yeast.

24

A PCR technique that fills in small gaps by using the end of a cloned sequence as a primer to amplify into adjacent uncloned fragments.

Primer walking

25

real-time PCR

DNA quantification

26

The smallest number of clones that represents the entirety of the genome are called what?

minimum tiling path

27

A _______________ family is a group of evolutionarily related genes that arose through repeated evolution of an ancestral gene.

multigene

28

Which of the below are NOT steps in the production of genome sequence maps:

Isolate whole chromosomes.

29

A fragment of DNA is cloned into a plasmid with a sequencing primer binding site. After dideoxy sequencing, the gel pattern shown in this diagram is obtained. What was the sequence of the DNA strand that acted as the template in the sequencing reaction?

5' GCTAGCA 3'

30

Before sequencing, the DNA fragment was cloned into a plasmid. On the strand that acted as the template in the sequencing reaction, what base of the cloned fragment was closest to the primer?

A

31

A principal problem with inserting an unmodified mammalian gene into a bacterial plasmid, and then getting that gene expressed in bacteria, is that

bacteria cannot remove eukaryotic introns.

32

Restriction endonucleases are especially useful if they generate "sticky" ends. What makes an end sticky?

single stranded complementary tails

33

If a restriction enzyme cuts a circular DNA into 3 fragments, how many restriction sites are there in the DNA?

3

34

A section of a genome is cut with three enzymes: A, B, and C. Cutting with A and B yields a 10-kb fragment. Cutting with B and C yields a 2-kb fragment. What is the expected result from a digest with A and C, if the C site lies in between the A and B sites?

An 8-kb fragment

35

For a physical map of a chromosome, distances are measured in units of

base pairs.

36

Which are steps in the production of genome sequence maps:

When sequences are obtained, assemble and organize the sequences in order.

Identify molecular markers on specific chromosomes.

Read the sequence of individual piece of the genome.

37

2 of the 7 different pea traits examined by Mendel involved genes that we now know are linked. Knowing this, can you explain why Mendel was still able to use results from his crossing experiments to develop the principle of independent assortment?

The genes are linked on the same chromosome, but are far enough apart ( > 50 map units) from each other so that they still sort independently.

38

2 of the 7 different pea traits examined by Mendel involved genes that we now know are linked. Knowing this, can you explain why Mendel was still able to use results from his crossing experiments to develop the principle of independent assortment?

??

The genes are linked on the same chromosome, but are far enough apart ( > 50 map units) from each other so that they still sort independently.

39

Compare the transcriptome of an organism with the proteome. What is described by each? Which one will generally have more macromolecules, and why?

????????????????

The transcriptome of an organism is a set of every RNA produced in that organism.

The Proteome is the set of every protein produced in that organism.

Transcriptome - set of all RNA transcripts produced by the genome at any one time. Bigger.

Proteome - complete set of proteins expressed during a cell's lifetime.

It takes 3 RNA codons to code for one protein, so it is natural to have more RNA molecules (transcriptome) than proteins (proteome).

The transcriptome is all of the RNA transcribed from the DNA in an organism, the proteome is all of the proteins in an organism. The transcriptome will generally have more macro molecules, because not all of the transcripts produce (e.g. introns) will code for something.

The transcriptome identifies all of the RNA transcripts in a genome. The proteome identifies all of the proteins that are encoded by that genome. The transcriptome will generally have more macromolecules because it also includes all non-coding RNA.

40

Explain why the greatest diversity of human SNPs is found among African people.

???

The greatest diversity of human SNPs is found in African people because studies suggest that humans first evolved in Africa. Since Africans have been around the longest they have had the greatest amount of time to create SNPs.

Humans are thought to have first evolved in Africa. These populations are the oldest with the greatest amount of time to accumulate polymorphisms

41

The "distance" between two linked gene pairs can be expressed as a percentage. Name the unit based on percent recombination that was created in honor of the scientist who pioneered the use of fruit flies for genetic research.

centimorgan

42

YAC

centromere

43

This is the study of "all genes in an organism in their entirety.”

genomics

44

The following statements are NOT true.

Antibodies are used for Northern blot analysis.

VNTRs are highly conserved in human populations.

45

The following statements are TRUE.

Coding sequences for gene products can be isolated from cDNA libraries.

PCR amplification generates large numbers of linear DNA fragments.

RNA molecules can be used as hybridization probes in Southern blot analysis.

46

ß-galactosidase

lacZ

47

Between which 2 genes would you expect the highest frequency of recombination?

A____W__E_________G

A and G

48

Plasmids are important in biotechnology because they are

a vehicle for the insertion of foreign genes into bacteria

49

A human gene with a disease phenotype is going to be mapped by positional cloning.

Which would be the most useful for this task?

Data about the inheritance of SNP markers in families with the disease

50

You are doing an experiment to characterize a 3000bp clone using two different restriction enzymes. Enzyme 1 (E1) produces 2 fragments in a single digest of 1400 and 1600bp. Enzyme 2 (E2) also produces 2 fragments of 1400 and 1600bp. When a double digest using both of these enzymes is done, it results in 2 fragments of 1400bp and 200bp. Based on this data, choose the correct restriction map from the choices given below.

1400 | 200 | 1400

51

Of the DNA sequences below, which would probably be the harder to determine?

CGATATATATATATATACGAT

GGCATCACGAGCTGCATTCGCA

CGATATATATATATATACGAT

The repetitive region in “A” would make it harder to determine with certainty, even though it is shorter than “B.”

52

A set of overlapping DNA fragments that form a contiguous stretch of DNA is called a _________.

contig

53

What is the function of dideoxynucleotides in Sanger DNA sequencing?

They stop synthesis at a specific site, so the base at that site can be determined.

54

One of the primary reasons for the necessity of generating a large number of clones in a eukaryotic genomic library is that

each vector can take up only a relatively small fraction of the eukaryotic DNA.

55

Sanger sequencing

DNA sequencing by synthesis of DNA chains that are randomly terminated by incorporation of a nucleotide analog (dideoxynucleotides) followed by sequence determination by analysis of resulting fragment lengths in each reaction.

56

Explain why the genetic map distance between two genes on the same chromosome may be inconsistent with the physical map distance.

E.g., for three loci A, B, and C, on the same chromosome, explain why the genetic distance might be A-[20 centimorgans]-B-[20 cM]-C, while the physical distance might be A-[200 kilobases]-B-[100 kb]-C.

??

The genetic map distance is measured by using the recombinant frequency. It shows the relative distance between different genes on a chromosome in the unit of centimorgan. The physical map distance is measured objectively and directly in the unit of nucleotide base.

57

The lungfish Protopterus aethiopicus has a genome 38 times larger than that of humans. Most of the DNA in this species is noncoding repetitive DNA. How could you create a library of clones that would let you compare just the genes in the lungfish to the genes in humans?

??

You could generate cDNA libraries and compare the transcribed regions of the genome

Creating a cDNA library of each organism would allow this. This can be done by using reverse transcriptase to create DNA transcripts of mRNA with small hairpin loops at the end. DNA polymerase will then synthesize a complement strand. This method gives a library of all expressed genes.

58

Describe the 3 basic components of a typical plasmid cloning vector and the reason/use for those plasmid vector components.

??

-Origin of replication: ensures that the vector is replicated within the cells

-Selectable markers: enable all cells containing the vector to be identified

-Restriction sites: needed so a DNA fragment can be inserted

59

A linear DNA fragment is cut with a restriction enzyme to yield two fragments. There is/are __ site(s) for this enzyme in this fragment.

??

1

60

You have cut DNA from source A with restriction enzyme #1 and you have cut DNA from source B with restriction enzyme #2. Both of these restriction enzymes leave a 4 base single stranded overhang. You want to ligate these restricted fragments together. What must be true for this to be successful?

??!!!!!

The single stranded overhangs must be complementary to one another and DNA ligase must be present to seal the fragments together.

The RE must have complementary sticky ends so that they will anneal together

61

A map of the order, overlap, and orientation of physically isolated pieces of the genome.

physical map

62

cloning vector

plasmid

63

Crossing over (genetic recombination) occurs in...

Meiosis I prophase

64

There are different challenges that exist for sequencing prokaryotic and eukaryotic genomes.

Which challenge is correctly paired with the type of genome to which it relates?

Eukaryotic: repetitive DNA

65

What is bioinformatics?

a method that uses very large national and international databases to access and work with sequence information

66

PCR is....

a technique for amplifying DNA sequences in vitro

67

A typical prokaryotic genome has

1 million base pairs of DNA, containing 1000 genes.

68

Nucleic acid blotting is widely used in recombinant DNA technology. In a Southern blot, one generally

hybridizes filter-bound DNA with a DNA probe

69

What is the enzymatic function of restriction enzymes?

to cleave nucleic acids at specific sites

70

All of the following are characteristics of the genomics revolution EXCEPT_____________

Inability to understand single genes

71

All of the following are characteristics of the genomics revolution

Enabled reverse genetics approach to genetics research

Facilitated collaborative research networks

Ability to conduct discovery-based research

Large scale acquisition of DNA sequences

72

A BLAST search is done to

find similar gene or protein sequences.

73

As a model system, what are three of the advantages of the mouse as a model system?

??

1. Mice have similar body plans and stages of development as humans
2. Similar genome size and number of chromosomes to humans
3. Many human genes have homologs in mice.

74

Is it possible for two different genes located on the same chromosome to assort independently? Explain your answer.

??

yes, if the two genes are located very far apart from each other.

75

What are Northern analyses used for? Describe the steps involved in performing a Northern analysis, and describe how levels of gene expression are determined.

??

Northern blotting screens mRNA to determine expression characteristics of a gene. mRNA is extracted, isolated and then separated by electrophoresis. After electrophoresis, the sample is transfered to a nylon membrane and transfered via capillary action. Levels of gene expression can be determined using microarrays.

Northern blotting is a procedure used to transfer RNA from a gel to a solid support.
1) RNA isolation
2) Probe generation
3) Denaturing agarose gel electrophoresis
4) Transfer to solid support
5) Hybridization with probe
6) Washing
7) Detection
Levels of gene expression are determined by running a sample RNA on same gel to provide accurate sizing ladder or by the use of microarrays

76

transgene

foreign DNA

77

shuttle vector

multiple hosts

78

A map of the distribution of cloned genomic DNA from genomic clone libraries.

Physical map

79

What is a transgenic organism?

??!!!

An organism that has been permanently altered by the addition of a DNA sequence to its genome.

An organism that stably carries a foreign gene within is genome

80

What is the purpose of an antibiotic resistance gene in a plasmid cloning vector?

??!!!

The LacZ gene is a selectable marker. Acts as a reporter gene which encodes beta-galctosidase. Expression of the lacz gene causes bacterial host cells carrying pUC18 to produce blue colonies when grown on medium containing a compound Xgal.

To determine if the vector is present in host cell, Good selection marker

it is a selectable marker to distinguish hosts with or without the vector

81

Which of the following are the important proteins needed for cloning a eukaryotic gene into a bacterial plasmid?

restriction enzymes specific for the target genes

DNA ligase

82

What do PCR, reverse transcription, and dideoxy DNA sequencing all have in common?

All produce DNA chains as a product.

83

10% recombination is equal to how many map units?

10mu

84

A human gene with a disease phenotype is going to be mapped by positional cloning.

Which would be the most useful for this task?

Data about the inheritance of SNP markers in families with the disease

85

One of the primary reasons for the necessity of generating a large number of clones in a eukaryotic genomic library is that .....

each vector can take up only a relatively small fraction of the eukaryotic DNA.

86

Describe one major difference in the organization or content of prokaryotic and eukaryotic genomes.

??

Prokaryotes with larger genomes will have more genes. In eukaryotes, there is little association between genome size and number of genes.

Eukaryotic genomes contain repetitive DNA that is largely absent in prokaryotic genomes

Genes are more densely spaced in prokaryotes verses eukaryotes

Prokaryotic genomes typically encode fewer genes than eukaryotic genomes

87

An individual has the following genotype. Gene loci (A) and (B) are 15 cM apart.

Indicate all the possible gametes that this individual can produce, and the proportions of expected progeny genotypes if a testcross is performed on this individual.

??

What do we know?

–15% recombination frequency –Means 15% recombinant progeny, 85% parental progeny
–Testcross is Ab/aB X abab

–Results
Ab = 42.5% (half of 85% parentals)

aB = 42.5% (half of 85% parentals)

AB = 7.5% (half of 15% recombinants)

ab = 7.5% (half of 15% recombinants)

88

What are 3 key differences between a genomic and a cDNA library?

??

Genomic contains both noncoded and coded DNA. Whereas cDNA only has transcribed regions. cDNA only contains sequences found in mature mRNA where the introns have been removed.

A genomic library contains all the DNA sequences found in a genome. A genomic library must contain a large number of clones to ensure that all DNA sequences are represented.
A cDNA library contains all the DNA sequences that are transcribed into mRNA.
A cDNA library is enriched with fragments from actively transcribed genes and introns do not interrupt the cloned sequences

A genomic library contains genomic DNA sequences and represents the entire genome of the organism. A cDNA library contains sequences found in a mRNA population from a particular tissue and represents only those genes expressed in that tissue.

89

Another word for a “DNA chip” (microscopic spots of oligonucleotides bound to glass that can be fluorescently labelled to identify levels of expression).

microarray

90

in situ hybridization

chromosome spread

91

expression vector

protein

92

A gene construct that indicates when transcription occurs because the protein is easily identified (often GUS or GFP).

reporter gene

93

Name 2 methods that have been used to produce mutations in a forward genetics approach.

UV light, ethyl methane sulfonate (EMS), and nitrosoguanidine cause single base-pair changes or small deletions and insertions, usually resulting in conditional mutations.
Transposons are also used for mutagenesis.



UV light, EMS, transposons, Nitrosoguandandine,

UV light, EMS and nitrosoguanine
mutagenic agents, x-rays, radiation, chemical mutagens (EMS), transposable elements, UV light.

94

You are doing an experiment to characterize a 3000bp clone using two different restriction enzymes. Enzyme 1 (E1) produces 2 fragments in a single digest of 1400 and 1600bp. Enzyme 2 (E2) also produces 2 fragments of 1400 and 1600bp. When a double digest using both of these enzymes is done, it results in 2 fragments of 1400bp and 200bp. Based on this data, choose the correct restriction map from the choices given below.

1400 l 200 l 1400

95

A section of a genome is cut with 3 enzymes: A, B, and C.

Cutting with A and B yields a 10-kb fragment. Cutting with B and C yields a 2-kb fragment. What is the expected result from a digest with A and C, if the C site lies in between the A and B sites?

An 8-kb fragment

96

For a physical map of a chromosome, distances are measured in units of

base pairs

97

Plasmids are important in biotechnology because they are

a vehicle for the insertion of foreign genes into bacteria

98

Which of the following enzymes is used to make complementary DNA (cDNA) from RNA?

reverse transcriptase

99

List 4 applications of PCR technology. Do not describe what PCR *does* (well, you can if you want, but you won't get points for it). Instead, list activities or fields in which PCR is useful.

1) Identification of RE variants
2) Screening for genetic disorders
3) Diagnostic screening for infectious organisms
4) Forensics
5) Paleobiology
6) Microsatellite analysis

100

The transcriptome of a genome contains more components than the proteome. Explain why this is true.

??

The transcriptome of a genome contains more components than the proteome because not all RNA encodes proteins. RNA can also encode tRNA and rRNA. Also, not all of mRNA is ever translated. On the other hand, all proteins come from RNA.

Transcription of some genes does not result in a protein-coding mRNA. For example, some genes are transcribed to yield tRNAs, rRNAs, miRNAs, etc. The point is that not every component of the transcriptome codes for a component of the proteome, but every component of the proteome is coded for by a component of the transcriptome.

101

Name and describe 3 different kinds of bacterial cloning vectors.

??

Bacterial artificial chromosomes (BAC), Plasmid, Cosmid

Plasmid, Phage, Bacterial artificial chromosomes (BAC), Expression Vectors

-Cosmids
-Bacterial artificial chromosomes (BAC)
-Expression vector

- Plasmid cloning vector: It generally can load fragments of about 5-10kb long. It is derived from bacterial plasmids. It has polylinker regions, a selection system, an insert discrimination system, and an origin of replication.

- Phage cloning vector: it can insert about 10-15kb; it is derived from bacterial virus (phage). To insert DNA fragments, the central thrid portion of the phage DNA, resulting in a left arm and a right arm and a centrol region. The fragments of interest then are iinserted between the two arms, ligated by DNA ligase.

- Cosmid cloning vector: it can carry up to 50kb inserts; it is made of part phage and part plasmid, which makes it to be a engineered vector

- Bacterial Artificial chromosomes cloning vector

102

What is the purpose of the LacZ gene in a plasmid cloning vector?

??

The LacZ gene is a selectable marker. Acts as a reporter gene which encodes beta-galctosidase. Expression of the lacz gene causes bacterial host cells carrying pUC18 to produce blue colonies when grown on medium containing a compound Xgal.

103

Rank from “roughest” to “fine detail” for the amount of resolution allowed by the following methods of mapping (1=roughest, 4=finest):

Linkage map- 1
Cytogenetic map- 2
Restriction map- 3
Sequence map- 4

104

Match the following terms with their definitions.

Ortholog, Syntheny, Paralog, Homolog.

Homolog: Closely related genes based on sequence and function.

Ortholog: Homologus genes of the same locus inherited from a common ancestor.

Paralog: Genes related by gene duplication in the genome.

Syntheny: Conservation of the same groups of genes in the chromosomes of 2 or more species

105

Paralog

Genes related by gene duplication in the genome.

106

Homolog

Closely related genes based on sequence and function.

107

Ortholog

Homologus genes of the same locus inherited from a common ancestor.

108

Synteny

Conservation of the same groups of genes in the chromosomes of 2 or more species.

109

transgene

Foreign DNA

110

A set of overlapping DNA fragments that form a contiguous stretch of DNA is called a _________.

contig

111

Some vectors such as pUC18 and others of the pUC series contain a large number of restriction enzyme sites clustered in one region. What term is given to this advantageous arrangement of restriction sites?

polylinker

112

A principal problem with inserting an unmodified mammalian gene into a bacterial plasmid, and then getting that gene expressed in bacteria, is that

bacteria cannot remove eukaryotic introns.

113

A BLAST search is done to

find similar gene or protein sequences.

114

Figure A below shows a restriction map of a rare prokaryotic gene with its direction of transcription indicated by the arrow. Figure B shows the unique cloning region (i.e., multiple cloning site) contained within a plasmid-cloning vector. The blackened region in Figure A represents the amino acid coding sequence of a protein that can be used in humans as a vaccine. The stippled region in Figure B is a highly active, constitutive (unregulated) prokaryotic promoter region. Letters indicate the cleavage sites for different restriction enzymes. Known gene sequences are indicated by short thick lines. Explain how you would isolate and fuse the coding region (Figure A) behind the indicated promoter in the cloning vector (Figure B) to produce large amounts of the protein in bacterial cells. Assume that the cloning vector carries the gene for tetracycline (an antibiotic) resistance. Letters represent different restriction enzymes.

would use restriction enzyme (RE) C with either RE D or RE B to clone the gene in the expression vector so that the beginning of this gene is directly 3' of the strong promoter.

By using restriction enzyme (RE) C with RE D, the coding region will be directly 3' of the strong promoter and then seal the sequences together with ligase.

I would cleave the gene with RE C and D (or B) to isolate the coding region. I would then use RE C again on the cloning vector in order to fuse the gene to plasmid (using the same RE would create sticky ends that can be annealed and ligated with DNA ligase). At this point PCR can be carried out on the new structure containing the plasmid and the gene of interest in order to produce large amounts of the protein.

115

In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

a. Identify progeny classes representing the double-crossover (DCO) gametes (e.g., the least frequent class; v ct+ s+), and parental gametes. (This is already known from the description, but confirm that the parental type is the most frequent offspring class; e.g., v+ ct+ s+.) Next, identify the locus whose alleles, if switched, will convert from DCO to parental class, and vice versa. For example, switching the mutant allele (v) in the DCO class to the parental allele v+ converts the DCO phenotype to the parental phenotype. The (v) locus therefore lies between the (ct) and (s) loci, and the relative order of this linkage group is (ct+ v+ s+) or (s+ v+ ct+). Both orders are correct because, from this experiment alone, we have no way of orienting (ct) or (v) relative to either end of the chromosome containing this linkage group.

b. s - v: (14 + 20 + 1+1)/(1200) = 0.030 = 3.0% = 3.0 cM
v - ct: (81 + 73 + 1+1)/(1200) = 0.13 = 13.0% = 13.0 cM

C. The expected number of double crossovers (DCO EXP) equals the product of the number of observed single crossovers (SCO). Use the calculated map distances for each of the two SCO regions, and apply the multiplicative rule to calculate the expected number of double crossovers: DCO EXP = (0.03)(0.13) = 0.004 (0.4%). Therefore, 0.4% of the 1200 total progeny, or (0.004)(1200) = 4.8 progeny, are expected to be DCO progeny (DCO EXP). The observed number of double-crossover progeny (DCO OBS) = 2. The coefficient of coincidence (CC) = DCOOBS /DCO EXP = 2/4.8 = 0.427; and the interference (I) = 1 - CC, therefore, I = 1 − 0.417 = 0.573.

116

Discuss the differences, and at least one similarity, between recombination and independent assortment.

??

Recombination - the sorting of alleles into new combinations

Independent assortment - in meiosis, each pair of homologous chromosomes assort independently of other homologous pairs.

Recombination only shuffles already existing genetic variation and does not create new variation at the involved loci.

Similarity - both result in an increase in genetic variation

117

Mario Capecchi, Sir Martin Evans, and Oliver Smithies recently won a Nobel Prize for gene targeting (gene knockouts) in mice.

Describe the steps involved in creating a knockout mouse.

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Insert the knockout gene into embryonic stem cells.
Grow transformed ES cells on medium.
Select surviving ES cell colonies, test for presence of knockout gene, grow up culture of knockout cells.
Insert ES cells into blastocyst. Mate chimera to black mice.
Test agouti progeny for presence of knockout gene.
Mate siblings to establish homozygous lines.

118

Why are telomeres and centromeres particularly difficult to sequence?

They consist of highly repetitive DNA, and so strand slippage issues can confuse the determination of a consensus sequence.

119

We have looked at the cloning experiments involved in producing Snuppy. Describe the specific technique that was used and how the results demonstrated that Snuppy was in fact a clone of the donor Afghan hound.

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Use microsatellite analysis of canine-specific microsatellites for 8 different canine markers between the donor-dog and snuppy. Both reached the same peak for each loci, therefore, they must be clones.

Microsatellite analysis was used to show that snupppy was a clone. Microsatellite loci are highly variable loci that contain a large number of DNA repeats (eg. 2, 3, or 4 nucleotides in length) at the population level. However, an individual can only have 2 of these alleles at any one microsatellite locus. By comparing the alleles that snuppy had at 8 different microsatellite loci with those allele that the donor afghan hound and the surrogate mother had, it was shown that snuppy had exactly all of the same alleles as the afghan hound, providing that snuppy was a clone of the donor afghan

120

Tony, who is not diseased, has a sister with cystic fibrosis (CF). Neither of his parents have CF.Tony is expecting a child with Tina. Tina's family history is unknown.

If the frequency of heterozygotes in the general population is 1/50, what is the probability that Tony and Tina's child will have CF? Explain each factor in your calculation.

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(2/3)(1/50)(1/4) = 1/300

121

The haploid human genome contains about 3 × 10^9 nucleotides.

On average, how many DNA fragments would be produced if this DNA was digested with restriction enzyme PstI (a 6-base cutter)?

RsaI (a 4-base cutter)? How often would an 8-base cutter cleave?

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PstI would result in an average of 4.8 DNA fragments

RsaI would result in an average of 76.9 DNA fragments.

An 8-base cutter would cleave every 8 65,536 bases, on average.

122

A family expresses the previously-unrecorded blue tongue trait. Genome analysis of a few affected individuals reveals that they possess 2 mutant homologs of a gene not previously characterized. In order to determine if and how these mutations result in blue tongue, construct a pedigree for the family and run Southern blot analysis on each of the family members. The pedigree and Southern blots are shown below, and organized as follows:
-The Southern blot results for each individual are located directly below that individual's symbol on the pedigree (for example, the results for the male in the 1st generation are shown in Lane 1).
-Shaded regions within the homologs represent sequences that hybridize to the probe used for the Southern analysis.
-Arrows indicate cleavage sites used for the Southern analysis.
Which homologue(s) is (are) linked to blue tongue? What is the relationship between these homologues (simple dominance, incomplete dominance, etc.)?

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Only individuals with the restriction sites on homolog B producing a 1.0 kb fragment have blue tongue. Hence it would seem that this is simple dominance, when there is a mutation that causes a second restriction site to appear on homolog B the blue tounge phenotype is expressed.



homolog B is linked to blue tongue and is dominant to homolog A

123

The full-length (i.e., containing the entire protein-coding region) cDNA for a specific eukaryotic gene in humans is 1500 nucleotides long. You screen a pig genomic library with this cDNA and isolate two genomic clones of different lengths. Both clones are sequenced and found to be 1900 and 2100 nucleotides long from start codon to stop codon. Screening of genomic libraries of several other organisms reveals that all of them contain only one genomic clone -- pigs seem to be the exception to the rule here. What evolutionary events might have led to the presence of two genomic clones in pigs, and the discrepancies in their length compared to the cDNA probe? How is this representative of a general type of occurrence in molecular genetic evolution?

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There was likely a duplication of this gene in pigs then the gene diverged from the pigs and humans or simply the humans lost a copy of the genes

This gene was most likely present in a common ancestor, the presence of two copies could be from gene duplication and the increase in length could be due to elongation of repeats included in the sequence.

124

If the recombination frequency between genes (A) and (B) is 5.3%, what is the distance between the genes in map units on the linkage map?

??

125

cDNA library

mRNA