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Genetics > Chapter 8 > Flashcards

Chapter 8 Flashcards

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Biology 101 flashcards
1
Q

Transcription

A

The biological process where information from DNA is converted to RNA
- involves synthesis of RNA from DNA template by a protein called RNA polymerase

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2
Q

Major types of RNA

A
  • Messenger RNA: encode the sequence of amino acids in a polypeptide
  • Ribosomal RNA: large and small subunits
  • Transfer RNA: Carries amino acids to ribosomes
  • Small nuclear RNA: form spliceosomes
  • MicroRNA: base pairing with certain mRNAs, altering their stability and efficiency of translation
  • Small interfering RNA: Regulate mRNA stability and translation
  • Telomerase RNA: acts as a tempolate
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3
Q

What can RNA do

A
  • Intermediate for information exchange between DNA and protein
  • A modifier of gene expression
  • An enzyme that may work with protein
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4
Q

Small interfering RNA

A
  1. Encorporated into RISC complex
  2. scans and finds complementary mRNA to dsRNA
  3. binds to target mRNA and induces cleavage
  4. mRNA is now cut and recognized as abnormal to cell
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5
Q

Transcription in bacteria

A
  • Translation starts before transcription finishes
  • one mRNA is being translated by multiple RNA polymerase complexes
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6
Q

What are the four steps of transcription in bacteria

A
  1. Promoter recognition
  2. Transcription initiation
  3. Chain elongation
  4. Chain termination
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7
Q

Promoter in transcription

A

Not transcribed - involved with regulating transcription by controlling access of RNA polymerase and may be bound by transcription factors

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8
Q

RNA coding region

A

Region that is transcribed from start to termination
- includes untranslated regions and translated regions

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9
Q

Bacterial Promotors

A
  • 2 consensus sequences (10 and 35)
  • recognized by the RNA polymerase holoenzyme
  • Spacing relative to the start of transcription is important but sequences in between can be anything
  • Can vary between different genes
  • RNA polymerase can recognize multiple consensus sequences because different versions of protein called sigma subunit change its conformation and DNA-binding specificity
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10
Q

What is a promoter

A
  • A promoter is a region of DNA lies upstream of a gene
  • Region is recognized and bound to by RNA polymerase which then transcribes that gene
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11
Q

Promotor in bacteria

A

RNA polymerase typically binds to one promoter and then transcribes several genes in an operon

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12
Q

Promoters in eukaryotes

A

Each gene has its own promoter; binding of RNA polymerase requires the help of other proteins called general transcription factors

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13
Q

Start codon RNA code

A

The start codon of RNA is complementary to the 3’ end of the template strand and corresponds to an 5’ ATG 3’ on the 5’ end of the non-template stand and a 5’ CAT 3’ on the template strand

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14
Q

Sigma subunits

A

Influence what genes are transcribed in bacteria by interacting with RNA polymerase

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15
Q

Termination of Bacterial Transcription

A
  • Two ways: intrinsic and rho-dependent
  • Both involved the formation of secondary structures between complementary nucleotides in the RNA molecule
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16
Q

Transcription in eukaryotes

A
  • Transcription in the nucleus; translation in the cytoplasm
  • Much more diversity in promotor sequences
  • Transcription makes mRNA which is processed before translation
  • RNA has exons and introns
  • DNA is associated with chromatin which influences transcription
  • RNA is more stable in eukaryotes
17
Q

3 RNA polymerases in eukaryotes

A
  • RNA polymerase I: several ribosomal RNA genes
  • RNA polymerase II: protein coding genes and most small nuclear RNA genes
  • RNA polymerase II: tRNA genes, one small nuclear RNA gene and one ribosomal RNA gene
18
Q

How does one identify where a promoter is

A
  • Introduce mutations in upstream sequence and examine effect on transcription
  • If the mutations are in promoter region, the mutation is associated with lower expression
  • Another approach is to check whether protein binds to upstream regions - only if promotor region in sequence (will be heavier)
  • DNA footprint protection assay
  • Analysis of sequence conversion
19
Q

DNA footprint protection Assay

A
  1. Label end DNA with 32p
  2. Transcriptional protein complex added to experimental DNA strand
  3. DNase I added to experimental and control: cleaves unprotected DNA
  4. Use gel electrophoresis to organize fragments by length
  5. potential promoter region will have no strands cut at that length
20
Q

Regulatory sequences of RNA polymerase II

A
  • Many different developmental stages and tissue types that have transcriptomes that require precise regulation
  • Enhancer sequences increase the level of transcription of specific genes - can be up or downstream and close or far away
  • silencer sequences decrease the level of transcription of specific genes
  • chromatin also influences transcription in a tissue-specific and development-stage specific way
21
Q

Half-life of transcripts in bacteria

A

seconds to minutes

22
Q

Half-life of transcripts in eukaryotes

A

hours to days

23
Q

Posttranscriptional processing in eukaryotes

A
  • 5’ capping: after the first 20 to 30 nucleotides of mRNA are synthesized, a modified nucleotide is added to the 5’ end
  • 3’ polyadenylation: a section of the 3’ end of the pre-mRNA is replaced with a string of adenines
  • Intron splicing
24
Q

What is the purpose of 5’ capping

A
  • Prevents degradation
  • facilitates transfer to the cytoplasm
  • splicing of introns
  • increases translation efficiency
25
Q

What is the purpose of 3’ polyadenylation

A

Thought to be associated with the termination of transcription

26
Q

What is the purpose of intron splicing

A

Gets rid of bits of the transcript that are not translated

27
Q

Types of introns

A
  • Group I: self-splicing; located in eukaryotes, bacteria and bacteriophages
  • Group II: self-splicing; located in eukaryotic organelles, bacteria and archaea
  • Pre-mRNA: Spliceosomes; located in eukaryotic nuclear genes
  • rRNA and tRNA: enzymatic splicing; located in eukaryotes, bacteria, archaea
28
Q

Intron Recognition sequences

A
  • 5’ splice site: at the beginning of the intron, immediately following the 5’ exon, most introns begin with GU nucleotides
  • 3’ splice site: at the end of the intron, just before the nest exon, most introns end with AG
  • Branch sites: near the end of the intron, a conserved sequence exist where the lariat forms, connecting the 5’ end of the intron to a branch point A within the branch site recognition sequence
29
Q

Intron Splicing

A
  • small nuclear ribonucleoproteins or snRNPs bind to these recognition sequences and form a lariat structure
  • snRNPs make up the spliceosomes
30
Q

Lariat formation

A
  • First the 5’ end of the intron is cleaved from the 3’ end of the preceding exon
  • then, formation of the lariat involves a type of covalent bond called 2’-5’ phosphodiester bond
31
Q

Exon ligation

A
  • The lariat is cleaved at the 3’ end of the intron and degraded
  • flanking exons are ligated
32
Q

What are three mechanisms can explain alternative RNA transcription and splicing

A
  1. Alternative promoters: different transcription start points in different cell types
  2. Alterative polyadenylation: different mature mRNAs
  3. Alternative pre-mRNA splicing: Pre-mRNA spliced in alternative patterns in different cell types
33
Q

Why do eukaryotes have introns whereas bacteria do not

A

As genomes size increases there are more and bigger introns in the genome

34
Q

RNA and translation

A
  • Translation requires RNA enzymes (transfer RNAs and ribosomal RNAs) that interact with proteins
  • These tRNA and rRNAs must be transcribed
35
Q

Processing of ribosomal and transfer RNA in bacteria

A
  1. Transcription produces a 30s per-RNA
  2. RNA cleavage releases rRNAs and tRNAs
36
Q

Processing of ribosomal and transfer RNA in eukaryotes

A
  1. Transcription synthesizes a 45S per-rRNA transcript
  2. Pre-RNA cleavage produces 3 rRNAs
  3. rRNA processing and ribosome assembly takes place in a compartment in the nucleus called nucleolus
37
Q

Transfer RNA structure

A

Four double stranded stems, three of them with single-stranded loops, form the secondary structure of tRNA
- Arm opposite anticondon binds amino acid
- Anticodon lines up with mRNA sequence

Genetics (18 decks)

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