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Life Science 4 > Lecture 21 > Flashcards

Lecture 21 Flashcards

1
Q

the complete set of DNA within a single cell of an organism

A

genomes

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2
Q

is a discipline in genetics that applies recombinant DNA, DNA sequencing methods, and bioinformatics to sequence, assemble, and analyze the function and structure of genomes

A

genomics

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3
Q

Human genome project sequenced human genome, model system genomes, and –

A

EST (expressed sequence tags) = cDNAs

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4
Q

map the fragments, sequence the fragments, and piece it together = slow and careful (public)

A

clone-by-clone sequencing approach

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5
Q

Take a lot of copies of the genome
Break it up into little fragments
Sequenced all the little fragments
Computer puts it all together

A

shotgun sequencing

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6
Q

next generation sequencing: after each nucleotide is added, its identity is determined by –

A

fluorescence

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7
Q

Once a sequence is obtained, it must be –

A

annotated

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8
Q

annotating a sequence includes – the genes, promoters, exons/introns, and function of genes

A

locating

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9
Q

process of attaching biological functions to DNA sequences

  • describes both structural and functional features of a gene
  • goal = identify known genes, regulatory sequences, etc as well as to identify sequences that are likely to be genes through their function
A

annotation

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10
Q

– cDNAs allow designation of all introns and exons

A

complete

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11
Q

– cDNAs (referred to as expressed sequence tags (EST)) allow expressed genes to be identified

A

incomplete

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12
Q

same info from all cell types from the same individual
includes: cis-elements (promoter, enhancer, etc.)
introns, repetitive sequences, etc.

A

genomic DNA

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13
Q

– is expressed genes that is derived from mRNA contains coding regions (no introns or regulatory sequences)

A

cDNA

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14
Q

What type of library do you need to piece the genome together?

A

genomic

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15
Q

What type of library do you need to determine the exon/intron complete sequence and structure of a gene?

A

genomic and cDNA

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16
Q

expression levels are based on frequency of –

A

cDNA clones

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17
Q

exon/intron structure are based on

A

cDNA clones

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18
Q

alternative splicing are based on –

A

cDNA clones

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19
Q

databases contain info on function which is based on –

A

homology to known genes

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20
Q

Where experimental data is not available, – is used to predict gene structure from ORFs, splice sites, promoter sequences

A

bioinformatics

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21
Q

Bioinformatics: the use of – approaches to decipher DNA-sequence information

A

computational

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22
Q

sequences without a stop codon, appear to possibly code for polypeptides

A

open reading frames

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23
Q

DNA – analysis can analyze mRNA concentrations – a cheaper alternative to sequencing.

A

microarray

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24
Q

Sequence analysis of all living organisms in an environment

A

metagenomics

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25
Q

in metagenomics – are over represented

A

most common species

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26
Q

Addition of foreign DNA into the genome.

A

transgenic

27
Q

Removal of a specific piece of DNA from the genome.

A

knock-out

28
Q

Replacement of a specific piece of DNA from the genome with a different piece of DNA.

A

gene replacement

29
Q

transgenes from different species

A

heterologous

30
Q

human cells can properly translate bacterial genes because the genetic code is –

A

universal

31
Q

– sequences are not universal: promoters, enhancers are species specific so need to use sequences specific for host

A

regulatory

32
Q

– vectors are vectors that have been furnished with sequences capable of directing efficient transcription and translation of transgenes

A

expression

33
Q

All vectors need

– and Selectable marker so that you can transform it and select for things

A

Origin of replication

34
Q

Series of restriction sites into which the gene to be expressed is inserted in recombinant clones

A

MCS (multi-cloning site)

35
Q

You can put a human gene in the MCS
If you have the bacterial promoter and transcription terminator you can express a human gene
But need to get rid of human introns since bacteria cannot splice introns therefore you need to use –

A

cDNA

36
Q

Eukaryotic expression vector must contain –, polyadenylation signal, adn MCS

A

TATA box

37
Q

different species may prefer different codons, reflected in levels of particular tRNA

A

codon bias

38
Q

factors that affect heterologous protein synthesis:

codon bias, – and high levels of synthesis can lead to aggregation and precipitation of inactive protein

A

post-translational modifications (rarely occur in bacterial cells)

39
Q

some proteins produced in E. coli for humans include insulin
human growth hormone
erythropoietin (induces red blood cell formation)
–

A

proteases used in detergents

40
Q

– is also commonly used to express heterologous genes

A

Yeast

41
Q

replicates in both bacteria and yeast

A

shuttle vector

42
Q

With this shuttle vector, DNA sequences can be manipulated in – where manipulation is easier, after which the modified plasmids can be shuttled into yeast for heterologous protein expression

A

E. coli (bacteria)

43
Q

an advantage of using S. cerevisiae as a protein factory is that it has the ability to splice introns and – modify proteins

A

post-translationally

44
Q

S. cerevisiae can – proteins out of the cell

A

secrete

45
Q

stable transgenic yeast can be made by integration of a plasmids without –

A

an origin of replication

46
Q

stable transgenic yeast can be made by double crossover through -

A

circular or linear plasmid with homology to genome

47
Q

integrates introduced DNA at a random, nonhomologous location

A

illegitimate recombination

48
Q

integration of entire plasmid is useful for – or making transgenic yeast

A

gene disruption

49
Q

double crossover is useful for gene disruption or –

A

gene replacement

50
Q

In homologous recombination with circular DNA, a single crossover results in integration of introduced DNA without the – of target gene

A

replacement

51
Q

In homologous recombination with circular DNA, a double crossover results in the – of target gene

A

replacement

52
Q

In homologous recombination with linear DNA, a single crossover results in integration of introduced DNA and –

A

terminal deletion

53
Q

In homologous recombination with linear DNA, a double crossover results in the – of target gene

A

replacement (or knockout)

54
Q

linearized DNA molecules recombine at a – frequency than circular ones so in yeast a double crossover at linearized plasmid is the best choice

A

higher

55
Q

Common technique used to determine the function of a gene is to disrupt it and examine the phenotype

A

gene knockout

56
Q

Using a –, it is a simple process in yeast to disrupt any gene

A

selectable marker

57
Q

Drosophila genome contains a predicted POU-domain gene, pdm-3 but nothing is known about its function.
How can you figure out where in the body pdm-3 is functioning?

A

transgenic fly

58
Q

Advantages of transgenic animals: can examine the – of a gene of interest

A

expression pattern

59
Q

Advantages of transgenic animals: can – particular cells or cell types with a reporter

A

label

60
Q

Advantages of transgenic animals: can – a normal or mutant allele of a gene to look for associated phenotypes

A

overexpress

61
Q

Disadvantages of transgenic animals: integration site is – so expression can vary depending on where the foreign DNA integrates

A

random

62
Q

Disadvantages of transgenic animals: Random integration can sometimes disrupt genes and cause –.

A

unrelated mutant phenotypes

63
Q

Disadvantages of transgenic animals: Often, multiple copies of the transgene are integrated, so the expression level cannot be controlled and is often much – than normal.

A

higher

Life Science 4 (11 decks)

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