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genetics > ch15: genes and how they work > Flashcards

ch15: genes and how they work Flashcards

(64 cards)

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1
Q

the neurospora will only grow if

A

it can make arginine

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2
Q

If the gene mutated was arg H, would the cells grow if
you supplemented the media with Arginosuccinate?

A

no

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3
Q

If the gene mutated was arg E, would the cells grow if
you supplemented the media with Arginosuccinate?

A

yes

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4
Q

one gene/one polypeptide hypothesis

A

many proteins are made from multiple polypeptide stands (ex. hemoglobin)

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5
Q

central dogma

A

information only flows from DNA to RNA to protein
transcription: DNA to RNA
translation: RNA to protein

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6
Q

transcription (3)

A
  • DNA directed synthesis of RNA
  • only one strand of the DNA is used (template strand)
  • T in DNA replaced by U in RNA
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7
Q

translation (3)

A
  • mRNA produced from transcription used to direct synthesis of polypeptides
  • takes place at ribosome (RNA protein complex)
  • required several kinds of RNA
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8
Q

stop codons

A

three codons (UUA, UGA, UAG) used to terminate translation (dont code for an amino acid)

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9
Q

start codon

A

codon (AUG) used to signify the start of translation; codes for the amino acid methionine as well

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10
Q

code is degenerate, meaning that

A

some amino acids are specified by more than one codon (64 codons, only 20 amino acids)

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11
Q

DNA template strand

A

the template by which the RNA sequence is made

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12
Q

DNA coding strand

A

has the same nucleotide sequence as the RNA sequence (except I instead of T)

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13
Q

upstream

A

closer to the 5’ end of the coding strand than the start site (numbered -1, -2, -3, etc)

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14
Q

downstream

A

closer to the 3’ end of the coding strand than the start site (numbered +2, +3, +4, etc)

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15
Q

core enzyme

A

drives RNA synthesis

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16
Q

sigma factor

A

helps RNA polymerase recognize the beginning of genes

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17
Q

only complete core enzyme and sigma factor can

A

properly initiate RNA synthesis

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18
Q

initiation

A

RNA polymerase finds and binds to promotor region of gene

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19
Q

elongation

A

the RNA transcript is synthesized from the DNA template

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20
Q

termination

A

specific termination sequences cause the RNA polymerase to stall and then stop transcription, releasing the RNA transcription

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21
Q

promotor

A

short sequence upstream of the start site

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22
Q

start site

A

first base transcribed into RNA

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23
Q

promotor forms a

A

recognition and binding site for the RNA polymerase

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24
Q

1) prokaryotic initiation of transcription (5)

A
  • -35 sequence binds to sigma subunit
  • DNA helix unwinds at TATAAT box (AT rich region)
  • RNA polymerase binds to the unwound DNA
  • RNA transcription begins at the start site (+1)
  • sigma factor released (no longer needed)
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25
2. prokaryotic elongation of transcription (3)
- grows in the 5' to 3' direction as ribonucleotides are added onto the 3' end - transcription bubble: unwound section of DNA template, where RNA transcription occurs - after transcription bubble passes, the now-transcribed DNA is rewound as it leaves the bubble
26
3) prokaryotic termination of transcription (4)
- signal "stop" to the RNA polymerase - RNA transcript is released - RNA polymerase releases DNA - DNA rewinds back into helix
27
prokaryotic transcription (3)
- single RNA polymerase - initiation of mRNA synthesis does not require a primer - requires a transcription unit (promotor, start site, termination site)
28
how is eukaryotic transcription different (3)
1. initiation: requires a series of transcription factors 2. elongation: no difference from prokaryotes 3. termination: termination sites not as well defined; RNA transcript is modified after transcription
29
in eukaryotes, how is the primary transcript modified to become mature mRNA (3)
1. addition of a 5' cap: protects from degradation; involved in translation initiation 2. addition of a 3' poly-A tail: created by poly-A polymerase; protection from degradation 3. removal of non-coding sequences (introns): pre-mRNA splicing done by spliceosome
30
introns
non-coding sequences
31
exons
sequences that will be translated
32
primary mRNA need to be processed to get mature mRNA, how is the pre-mRNA processed
by splicing out the introns, making sure that only the exons (coding sequences) will be translated
33
small ribonucleoprotein particles (snRNPs)
a complex of proteins and snRNA that recognizes the intron-exon boundaries
34
introns always begin and end with
the same 2-base sequence (recognition)
35
snRNPs bind to the
beginning of the intron and a branch site within the intron
36
spliceosome
the splicing organelle responsible for removing introns
37
most introns are labeled with a GU at the beginning and an AG at the end. what would that mature mRNA look like if one of these bases were mutated
wherever that mutation is on the mRNA, that intron stays in along with the exons that are supposed to be there
38
alternative splicing
a single primary transcript can be spliced into different mature mRNAs ex: exons 1,2,3 can splice together and skip 4 ex: exons 1,2,4 can splice together and skip 3
39
tRNA (transfer RNA)
carries amino acids to the ribosome for incorporation into a polypeptide (protein); is an adaptor protein
40
what does aminoacyl-tRNA synthetases do
an amino acid that adds other amino acids to the acceptor stem of tRNA
41
how does tRNA get charged (3)
1) animo acid is activated: ATP provides energy to bring amino acid to enzyme 2) tRNA binds to the enzyme 3) amino acid attaches to tRNA, charged tRNA is then released
42
charged tRNA
has an amino acid added using the energy from ATP
43
ribosomes do not verify amino acid attached to tRNA, so it's up to aminoacyl-tRNA synthetase to
put the right amino acid on
44
the ribosome has two primary functions:
1) decode the mRNA code: small subunit of ribosome 2) form peptide bonds: peptidyl transferase- enzymatic component of the large subunit of ribosome
45
the ribosome has multiple tRNA binding sites (3):
1) E site: binds the tRNA that carried the last amino acid 2) P site: binds the tRNA attached to the growing peptide chain 3) A site: binds the tRNA carrying the next amino acid
46
three steps of translation
initiation, elongation, termination
47
initiation of prokaryotic translation
- recognizes start codon (AUG) - initiation complex includes charged tRNA, small ribosomal subunit and mRNA strand - large subunit now added - initiator tRNA bound to P site - A site is empty
48
initiation of eukaryotic translation (similar to prokaryotes except...)
- initiating amino acid is different - more complicated initiation complex - lack of an RBS: small subunit recognizes and binds to 5' cap of mRNA
49
elongation of prokaryotic translation
- elongation factor binds to a charged tRNA and brings it to the empty A site - peptidyl transferase forms peptide bond - translocation of ribosome - addition of successive amino acids occurs as a cycle
50
wobble pairing during translation
wobble pairing allows less stringent pairing between the 3' base of the codon and the 5' base of the anticodon so one tRNA can recognize multiple codons (this is why 3rd base of codon can be variable)
51
termination of prokaryotic translation
- elongation continues until the ribosome encounters a stop codon - stop codons are recognized by release factors which release the polypeptide from the ribosome
52
protein targeting
the mechanism by which a cell transports proteins to the appropriate positions in the cell or outside of it
53
point mutations
alter a single base
54
ase substitution
substitute one base for another
55
silent mutation
one base changes, same amino acid is coded for
56
missense mutation
one base changes, changes amino acid inserted
57
nonsense mutations
changed to stop codon
58
what kind of mutation is sickle cell anemia
missense mutation
59
frameshift mutations (3)
- addition or deletion of a single base - much more serious consequences - alter reading frame downstream
60
triplet repeat expansion mutation
- repeat unit is expanded un the disease allele relative to the normal - caused by "slippage" during DNA replication; replicates the same sequence over and over ex: huntingtons disease, fragile X syndrome
61
chromosomal mutations: deletion
part of the chromosome is lost; changes the structure of a chromosome
62
chromosomal mutations: duplication
part of the chromosome is copied; change the structure of a chromosome
63
chromosomal mutation: inversion
part of the chromosome is in reverse base-pair order; changes the structure of a chromosome
64
chromosomal mutations: translocation
part of the chromosome is moved to a new location; changes the structure of a chromosome

genetics (11 decks)

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