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AP Bio S2 > Unit 6: gene expression and regulation > Flashcards

Unit 6: gene expression and regulation Flashcards

1
Q

conservative model

A

the parental strands direct synthesis of an entirely new double stranded molecule
- the parental stranded are fully conserved

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2
Q

semi-conservative

A

the two parental strands: each make a copy of itself
- after one round of replication, the two daughter molecules each have one parental and one new strand

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3
Q

dispersive model

A

the material in the two parental strands is dispersed randomly between the two daughter molecules.
- After one round of replication, the daughter molecules contain a random mix of parental and new DNA

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4
Q

DNA replication begins in sites called

A

origins of replication
- various proteins attach to the origin of replication and open to the DNA to form a replication fork

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5
Q

Helicase

A

unzips DNA strands at each replication fork

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6
Q

topoisomerase

A

help prevent strain of the replication fork by relaxing supercoiling

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7
Q

Primase

A

initiates replication by adding short segments of RNA, called primers to, the parental DNA strand
Primers serve as the: foundation for DNA synthesis

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8
Q

the enzymes that synthesize DNA can only attach new DNA nucleotides to

A

to an existing strand of nucleotides

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9
Q

DNA polymerase III

A

attaches to each primer on the parental strand and moves in the 3’ and 5’ direction
- as it moves; it adds nucleotides to the new strand in the 5’ to 3’ direction

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10
Q

DNAP III follow helicase known as the

A

leading strand and requires one primer

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11
Q

DNAP III on the other parental strand moves away from the helicase is known as the

A

lagging strand and requires many primers

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12
Q

leading vs lagging strand

A

leading is synthesized in one continuous segment
-lagging strand moves away from the replication fork it is synthesized in chunks

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13
Q

okazaki fragments

A

segments of the lagging strand

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14
Q

after DNAP III form an ______, ______ replaces ____ nucleotides with ___ nucleotides

A

okazaki fragments, DNAP I, RNA, DNA

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15
Q

dna ligase

A

joins a okazaki fragment forming a continous DNA strand

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16
Q

how are genes on DNA protected from dna becoming shorter and shorter?

A

telomeres: repeating units of short nucleotides sequences that do not code for genes
- help form a cap at the end of DNA to help postpone erosion
- the enzyme telomerase adds telomeres to DNA

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17
Q

as _____ adds nucleotides to the new DNA strand, it proofreads the bases

A

-DNA polymerase

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18
Q

if segments of DNA are damaged in DNA replication…

A

nuclease can remove segments of nucleotides and DNA polymerase and ligase can replace the segments

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19
Q

protein are ____ made up of ____

A

polypeptides, amino acids (linked by peptide bonds)

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20
Q

transcription

A

the synthesis of RNA info from DNA
- allows for: “message” of the DNA to be transcribed
- occurs in the nucleus

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21
Q

translation

A

the synthesis of polypeptide using info from RNA
- occurs at the ribosome
- a nucleotide sequence becomes an amino acid sequence

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22
Q

what does mRNA do?

A

mRNA is synthesized during transcription using a DNA template
- mRNA carries info from the DNA to the ribosomes in the cytoplasm

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23
Q

what does tRNA do?

A

each tRNA carry a specific amino acid
- can attach to mRNA via their anticodon (a complementary codon to mRNA)

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24
Q

what does rRNA do?

A

-helps form ribosomes
- helps link amino acids together

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25
Transcription-initiation
1. RNA polymerase molecules attach to a promoter region of DNA - promoter regions are upstream of the desired gene to transcribe
26
Eukaryote vs Prokaryote initiation transcription
eukaryote: promoter region is called TATA box - transcription factors help RNA polymerase bind Prokaryote: RNA polymerase can bind directly to promoter
27
transcription-elongation
- RNA polymerase opens the DNA and reads the triplet code of the template strand - 3' to 5' direction but mRNA transcript elongates 5' to 3' - RNA polymerase moves downstream
28
transcription- termination
prokaryotes: - RNA polymerase detaches - mRNA transcript is released and proceeds to translation -mRNA DOES NOT need modifications eukaryotes: - RNA polymerase transcribes a sequence of DNA called the polyadenylation signal sequence - codes for polyadenylation signal - releases the pre-mRNA from the DNA - must undergo modifications before going to translation
29
5' cap modification
the 5' end of the pre-mRNA receives a modified guanine nucleotide
30
poly A-tail
the 3' end of the pre-mRNA receives adenine nucleotides
31
both the 5' cap and poly a tail functions
- help mature mRNA leave the nucleus - help protect the mRNA from degradation - help ribosomes attach to the 5' end of the mRNA
32
RNA splicing
the spliceosome remove introns (do not code for AA) and joined the exons together
33
prokaryote vs eukaryote ribosomes subunits
pro: small subunit (30s) large subunit (40s) euk: small subunit (40s) large subunit (60s)
34
what are the large subunit three sites?
A site: amino acid site (holds the tRNA carrying an amino acid) P site: polypeptide site (holds the tRNA carrying the growing polypeptide chain) E site: exit site
35
Translation: initiation
- small ribosomal subunit binds to the mRNA and a charged tRNA binds to start codon, AUG, on the mRNA - next the large subunits binds
36
translation: elongation
- the next tRNA comes into the A site -mRNA is moved through its ribosome and its codons are read - all organisms use the same genetic code, it supports the idea of common ancestry
37
translation elongation 3 steps
1. codon recognition: the appropriate anticodon of the next tRNA goes to the A site 2. peptide bond formation: peptide bonds are formed that transfer the polypeptide to the A site tRNA 3. translocation: the tRNA in the A site moves to the P site, then it goes to the E site. The next A site is open for the next tRNA
38
translation: termination
- a stop codon in the mRNA reaches the A site of the ribosome - the stop codon signals for a release factor - hydrolyzes the bond that holds the polypeptide to the p-site - polypeptide releases - all translational units dissemble
39
protein structures
Primary: chain of amino acids secondary: coils and folds due to hydrogen bonds forming tertiary: side chain interaction quaternary: 2+ polypeptide chains interacting
40
some polypeptides require __ proteins to __ correctly and some require ___ before it can be ___ in the cell
chaperone, fold, modifications, functional
41
Explain how trp operon is a repressible operon
- the trp operon controls the synthesis of tryptophen - transcription is active - allosteric enzyme is only active when tryptophan binds to it - when too much tryptophan builds up in bacteria, it is most likely to bind to the repressor turning it active, which will then temporarily shut off transcription for tryptophan
42
explain how lac operon is an inducible operon
the lac operon control synthesis of lactase - transcription is off - a lac repressor is bound to the operator (allosterically active) - inducer=allolactose - when present it will bind to the lac repressor and turn the lac repressor off (allosterically inactive) - genes now can be transcribed
43
How can DNA be modified?
1. histone acetylation: adds acetyl groups to histones which looses the DNA 2. DNA methylation: adds methyl groups to DNA which causes chromatin to condense
44
transcription initiation
- once chromatin modifications allow the DNA to be more accessible, specific transcription factors bind to control elements - sections of non coding DNA that serve as binding sites - gene expression can be increased or decreased by binding it activators or repressors to control elements
45
translation initiation
- translation can be activated or repressed by initiation factors - microRNA and small interfering RNA's can bind to mRNA and degrade it or block translation
46
morphogenesis
the physical process that gives an organism its shape
47
mutations
changes in the genetic material of a cell which can alter phenotypes - primary source of genetic variation - any disruption can cause new phenotypes
48
substitution
the replacement of one nucleotide and its partner with another pair of nucleotide
49
silent
change still codes for the same AA (redundancy in the genetic code)
50
missense
changes results in a different AA
51
nonsense
change results in a stop codon
52
frameshift mutations
when the reading frame of the genetic info is altered - insertion: a nucleotide is inserted - deletion- a nucleotide is removed
53
large scale mutations
1. nondisjunction: chromosomes do not separate properly during meiosis 2. translocation: a segment of one chromosome moves to another 3. inversion: a segment is reversed 4. duplication: a segment is repeated 5. deletion: a segment is lost
54
transformation
uptaking of DNA from nearby cell
55
transduction
viral transmission of genetic material
56
conjugation
cell to cell transfer of DNA
57
transposition
movement of DNA segments within and between DNA molecules
58
gel electrophoresis
- separate DNA fragments - dna is loaded into wells on one end of a gel and an electric current is applied - DNA fragments are negatively charged so they moved towards the + electron
59
polymerase chain reaction
- make several copies of a specific DNA - segments of DNA are amplified - results can be analyzed using gel electrophoresis
60
dna sequencing
determine the order of nucleotide in DNA

AP Bio S2 (7 decks)

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