Allgemeine und molekulare Genetik 2

Descripción

Bachelor Mikrobiologie Test sobre Allgemeine und molekulare Genetik 2, creado por Anna Ullrich el 15/09/2016.
Anna Ullrich
Test por Anna Ullrich, actualizado hace más de 1 año
Anna Ullrich
Creado por Anna Ullrich hace alrededor de 8 años
17
1

Resumen del Recurso

Pregunta 1

Pregunta
What's right: Synthetic lethality ...
Respuesta
  • Combinatorial effects between alleles or substances
  • Effects in synthetic biology
  • An effect specific to a double mutation (not seen in the singles)
  • An effect specific to a mutation in combination with a SMI-small molecule inhibitor (not seen by either mutation or SMI alone)

Pregunta 2

Pregunta
What's right: Phage lambda lysogen ...
Respuesta
  • C1 represses lambda's lysogenic genes, c2 activates int- and c1 transcription, c3 stabilizes c2
  • C1 represses lambda's lysogenic genes, c2 represses int- and c1 transcription, c3 destabilizes c2
  • C1 represses lambda's lytic genes, c2 activates int- and c1 transcription, c3 stabilizes c2
  • C1 represses lambda's lytic genes, c2 represses int- and c1 transcription, c3 destabilizes c2

Pregunta 3

Pregunta
What's right: Bacterial transformation
Respuesta
  • The mechanism of gram+ and gram- DNA uptake is not related
  • Gram- bacteria possess an outer membrane requiring an additional pore for DNA uptake
  • Gram+ bacteria possess an outer membrane requiring an additional pore for DNA uptake
  • Gram+ and gram- bacteria have a pilus structure, a dsDNA receptor and an inner membrane pore in common

Pregunta 4

Pregunta
What's right: Intra- vs extragenic complementation
Respuesta
  • Genetic complementation between mutations in different genes does not require wild type alleles
  • Extragenic complementation is rare and requires mutations in different functional domains
  • Intragenic complementation is rare and depends on the presence of domains that function independently
  • Molecular complementation depends on addition of the wild gene

Pregunta 5

Pregunta
What's right: Dominant vs epistatic
Respuesta
  • a is epistatic to locus b , if the double mutant (a,a/b,b) has the same phenotype as the single mutant (a,a)
  • a is epistatic to b, means: Pht(a,b)=Pht(a,B) .. (although Pht(A,b)≠Pht(A,B))
  • dominant : Pht(A,A)=Pht(a,A)
  • dominant relates alleles of 1 gene, epistatic relates an allele to another gene
  • A is dominant over a: Pht(a,A) = Pht(a,a)

Pregunta 6

Pregunta
What's right: gene vs allele
Respuesta
  • a ncRNA is not encoded by a gene, because it is not translated
  • a centromere is not a gene, because it is not transcribed
  • a bacterial cistron is not a gene, because it contains several functional units
  • a gene is a functional unit in the genome, activated by transcription while an allele is a sequence variant of a gene
  • a bacterial operon is not a gene, because it contains several functional units

Pregunta 7

Pregunta
What's right: Falsifyability
Respuesta
  • According to Popper’s critical rationalism (P.c.r), a Sc.Hyp can’t be claimed true, without evidence
  • According to P.c.r. a single experiment, if well performed, can prove a hypothesis forever
  • According to P.c.r. a single experiment, if well performed, can disprove a hypothesis forever
  • According to P.c.r. a scientific hypothesis must make a prediction, which can be proven wrong

Pregunta 8

Pregunta
What's right: Suppressor mutation
Respuesta
  • A suppressor mutation blocks the expression of another gene
  • A suppressor mutation can’t occur in the same gene
  • A suppressor mutation must occur in tRNA
  • A suppressor blocks the appearance of a mutant phenotype

Pregunta 9

Pregunta
What's right: What is the phenotype?
Respuesta
  • Many genes do not cause a phenotype
  • The phenotype of a given mutation must be detectable under all conditions
  • The phenotype of a mutation can change according to the situation
  • The phenotype ist the observable impact of an allele

Pregunta 10

Pregunta
Connect the topic with the correct boxes/key words ... BACTERIAL RECOMBINEERING
Respuesta
  • attB/P/L/R
  • targetted DSB
  • RE enzyme free assembly of bacterial, genome-size DNA
  • Efficient, RE enzyme free transfer of inserts between plasmids
  • Inhibition of RecBCD
  • RE enzyme free sub cloning from large inserts
  • RE enzyme free correction of genetic defects in vivo
  • Bacterial immune system
  • Exo, Beta, Gam
  • Edition through DSB-repair

Pregunta 11

Pregunta
Connect the topic with the correct boxes/key words ... GATEWAY TECHNOLOGY
Respuesta
  • attB/P/L/R
  • targetted DSB
  • RE enzyme free assembly of bacterial, genome-sized DNA
  • Efficient, RE enzyme free transfer of inserts between plasmids
  • RE enzyme free sub cloning from large inserts
  • RE enzyme free correction of genetic defects in vivo
  • Bacterial immune system
  • Exo, Beta, Gam
  • Editing through DSB-repair
  • Cloning from donor to recipient

Pregunta 12

Pregunta
Connect the topic with the correct boxes/key words ... CRISPR
Respuesta
  • attB/P/L/R
  • targetted DSB
  • RE enzyme free assembly of bacterial, genome-size DNA
  • Efficient, RE enzyme free transfer of inserts between plasmids
  • Inhibition of RecBCD
  • RE enzyme free subcloning from large inserts
  • RE enzyme free correction of genetic defects in vivo
  • Bacterial immune system
  • Exo, Beta, Gam
  • Editing through DSB-repair

Pregunta 13

Pregunta
Connect the topic with the correct boxes/key words ... YEAST RECOMBINEERING
Respuesta
  • attB/P/L/R
  • targetted DSB
  • RE enzyme free assembly of bacterial, genome-size DNA
  • Efficient, RE enzyme free transfer of inserts between plasmids
  • Inhibition of RecBCD
  • RE enzyme free subcloning from large inserts
  • RE enzyme free correction of genetic defects in vivo
  • Bacterial immune system
  • Exo, Beta, Gam
  • Editing through DSB-repair
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