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Molecular Genetics Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Nucleotide

A
  • ribose sugar, nitrogenous base, and
    phosphate group.
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2
Q

Nucleoside

A
  • ribose sugar and nitrogenous base.
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3
Q

DNA is a polymer of ______ that have
hydrogen on the ribose sugar’s 2’ carbon. RNA is a
polymer of nucleotides that have ______ groups
on the ribose sugar’s 2’ carbon. This is the reason
DNA is called deoxyribonucleic acid, while RNA is
called ribonucleic acid.

A

1) nucleotides
2) OH-

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4
Q

Since ______ have more hydrogen bonds, a
______ temperature is needed to break DNA
strands.

A

1) G-C bonds
2) higher

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5
Q

Nucleosomes

A
  • are complexes of DNA wrapped
    around histone proteins. Each nucleosome has
    nine histones total. The central core contains two
    of each histone H2A, H2B, H3 and H4. On the
    outside, a single histone, H1, holds the DNA in
    place.
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6
Q

Chromatin

A
  • refers to the overall packaging of DNA
    and histones.
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7
Q

2 Types of Chromatin include

A

1) Euchromatin
2) Heterchromatin

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8
Q

Euchromatin

A
  • nucleosomes are “loosely
    packed”, so DNA is readily accessible for
    transcription.
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9
Q

Heterochromatin

A
  • nucleosomes are “tightly
    packed”, so DNA is mostly inactive.
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10
Q

Histones are _____ charged while DNA is
_______ charged, allowing proper binding.

A

1) positively
2) negatively

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11
Q

Acetylation

A

-of histones removes positive charges,
relaxing DNA-histone attractions and allowing for
more transcription to happen.

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12
Q

Deacetylation

A
  • of histones increases positive
    charges, tightening DNA-histone attractions and
    decreasing transcription.
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13
Q

Methylation

A
  • of histones adds methyl groups, either increasing or decreasing transcription.
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14
Q

An _______ is required to initiate DNA replication where the DNA strands first separate. Organisms with ______ DNA such as bacteria have a single origin of replication while organisms with ______ DNA such as humans have multiple origins of replication.

A

1) origin of replication
2) circular
3) linear

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15
Q

DNA undergoes _______,
where each new double helix produced by
replication has one “new” strand and one “old”
strand.

A

1) semiconservative replication

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16
Q

DNA is ______, meaning that the _______
(terminal phosphate group) of one strand is
always next to the _____ (terminal hydroxyl
group) of the other strand and vice versa.

A

1) antiparallel
2) 5’ end
3) 3’ end

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17
Q

Steps of Replication

A

1) Initiation
2) Elongation
3) Termination

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18
Q

Initiation

A
  • creating origins of replication at
    A-T rich segments of DNA because A-T bonds
    only have two hydrogen bonds and are easier
    to split apart.
  • a promoter sequence (aka
    promoter) next to the gene attracts RNA
    polymerase to transcribe the gene.
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19
Q

Elongation

A
  • producing new DNA strands using different types of enzymes.
  • transcription bubble forms and RNA polymerase travels in the 3’ → 5’ direction
    on the template strand. However, it extends
    RNA in the 5’ → 3’ direction.

Involves:
Helicase
Single-strand binding proteins
topoisomerase
Primase
sliding clamp proteins
DNA polymerase(s)
leading strand
lagging strand
DNA ligase

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20
Q

Helicase

A

-unzips DNA by breaking hydrogen bonds between strands, creating a replication fork. As it unzips the strands, helicase leads to supercoiling (tension ahead of the replication fork).

-Separates complementary strands at the replication fork.

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21
Q

Single-strand binding proteins

A
  • bind to uncoiled DNA strands, preventing
    reattachment of the strands to each other.
  • Proteins that prevent
    the two strands from
    coming back together
    after separate.
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22
Q

Topoisomerase

A
  • nicks the DNA double
    helix ahead of helicase to relieve built-up
    tension and supercoiling.
  • Relaxes the DNA double
    helix from the tension
    and supercoiling the
    opening helix is creating.
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23
Q

Primase

A
  • Primase places RNA primers at the origin
    of replication to create 3’ ends for nucleotide addition.
  • Provides a 3’ hydroxyl
    group for DNA
    polymerase to attach
    new nucleotides to.
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24
Q

Sliding clamp proteins

A
  • hold DNA polymerase onto the template strand.
  • Helps to hold DNA
    polymerase to the
    template strand.
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25
Q

DNA polymerase

A
  • adds free nucleoside
    triphosphates to 3’ ends. DNA polymerase can only add nucleotides onto an
    preexisting 3’ hydroxyl group provided by primase.
  • The class of enzymes
    that extends DNA in the
    5’ to the 3’ direction.
    Several have
    proofreading
    capabilities that allow
    them to catch synthesis
    errors.
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26
Q

The leading strand

A

-is produced
continuously because it has a 3’ end that
faces the replication fork.

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27
Q

The lagging strand

A
  • is produced
    discontinuously because its 3’ end is facing
    away from the replication fork. Thus, many
    RNA primers are needed to produce short
    DNA fragments called Okazaki fragments.
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28
Q

A different _______ replaces RNA primers with DNA.

A

1) DNA polymerase

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29
Q

DNA ligase

A

-glues separated fragments of DNA together.
- GLues together separate
pieces of DNA.

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30
Q

Termination

A
  • replication fork cannot
    continue, ending DNA
    replication.
  • a termination sequence (aka
    terminator) signals to RNA polymerase to stop transcribing the gene.
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31
Q

Telomeres

A
  • are noncoding, repeated
    nucleotide sequences at the ends of linear chromosomes. They are necessary in
    eukaryotes because when the replication fork reaches the end of a chromosome, a
    small segment of DNA from the telomere is not replicated and lost (no RNA primer is
    present to help produce another Okazaki fragment).
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32
Q

To review, the _____ checkpoint regulates cell cycle
transition from the _____ into the S phase,
checking for favorable conditions to grow. If
_______, the cell will remain in _______ and
will not enter the S phase for DNA replication.

A

1) G1/S
2) G1 phase
3) unfavorable
4) G0

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33
Q

Telomerase

A
  • Adds repetitive DNA to
    the ends of eukaryotic
    chromosomes, which
    prevents critical
    information from being
    lost.

-Telomerase is an enzyme that extends telomeres to prevent DNA loss.

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34
Q

Genes

A
  • are instructions within DNA that code for proteins. However, they must first be transcribed into RNA before being translated into proteins. In a gene, the promoter region comes first, then the
    gene operator, then the gene.
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35
Q

Specifically, DNA undergoes transcription to
produce single-stranded __________.

A

1) mRNA

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36
Q

Because ______ do not have membrane-enclosed nuclei, both transcription and
translation occur simultaneously in the cytosol. RNA polymerase opens up DNA, forming a transcription bubble.

A

1) Prokaryotes

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37
Q

Before transcription can occur, a _______
combines with _______ to form RNA polymerase
holoenzyme, giving it the ability to target specific
DNA promoter regions.

A

1) sigma factor
2) prokaryotic core RNA
polymerase

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38
Q

There are two types of termination in bacteria:

A

1) Rho independent termination
2) Rho-dependent termination

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39
Q

Rho independent termination

A

— a termination sequence is reached and the
RNA transcript folds into a hairpin loop →
RNA polymerase falls off and transcription
ends.

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40
Q

Rho-dependent termination

A

— A Rho protein binds to the RNA transcript,
moving 5’ → 3’ to catch up and displace
RNA polymerase, ending transcription.

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41
Q

operon

A

is a group of genes that function as a
single unit that is controlled by one promoter. The
operator region is present near the operon’s
promoter.

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42
Q

To regulate the promoter, _______ bind to the operator regions, while _______
bind to the promoter sites. (prokaryotes)

A

1) repressor
2) activator

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43
Q

The lac operon

A

is an inducible operon (it must beinduced to become active). LacZ, lacY, and lacA
are the three genes contained within the lac
operon that encode proteins required for lactose
metabolism. The lac operon will only be induced
when glucose is not available as an energy source,
so lactose must be used.

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44
Q

The lac repressor protein

A

is the first way that the
lac operon is controlled. This protein is constitutively expressed (always on). Thus, the lac repressor protein is always bound to the
operator, blocking transcription. However, when
lactose is present it is converted to allolactose.
Allolactose binds directly to the repressor and
removes it from the operator, allowing transcription to occur.

45
Q

cAMP levels and catabolite activator protein
(CAP)

A
  • are the second level of lac operon regulation.
    cAMP levels are inversely related to glucose levels,
    so when glucose is low, cAMP is high. cAMP binds
    to catabolite activator protein (CAP), which then
    attaches near the lac operon promoter to help
    attract RNA polymerase, promoting transcription.

Glucose and lactose present: - Moderate transcription.
Repressor not bound and CAP not bound.

Glucose and lactose absent:
- No transcription. Repressor is bound and CAP is bound.

Glucose present and lactose absent:
- No transcription. Repressor is bound and CAP is not bound.

Glucose absent and lactose present:
- High transcription. Repressor not bound and CAP is bound.

46
Q

trp operon

A
  • is responsible for producing the amino acid tryptophan. It is known as a repressible operon because it codes for
    tryptophan synthetase and is always active unless the presence of tryptophan in the
    environment represses the operon.
47
Q

______ binds to the trp repressor protein,
which then attaches to the ______ on the trp
operon to prevent tryptophan production. Thus,
this is the first level of trp operon regulation. When
tryptophan is not present in the environment, the
trp operon will undergo transcription because the
trp repressor protein will be _______. (prokaryotes)

A

1) Tryptophan
2) operator
3) inactive

48
Q

Unlike in prokaryotes, eukaryotic transcription
occurs in the ______ and uses ________ to transcribe most genes.

A

1) nucleus
2) RNA polymerase II

49
Q

Transcription factors

A
  • are needed in eukaryotes
    to help RNA polymerase bind to promoters. The
    TATA box is a sequence in many promoters that
    transcription factors can recognize and bind to.
50
Q

Enhancers

A
  • are DNA sites that activator
    proteins can bind to; they help increase transcription of a gene.
51
Q

Silencers

A
  • are DNA sites that repressor
    proteins can bind to; they decrease transcription of a gene.
52
Q

Enhancers and silencers can be far ______ or
_______ from the gene, so DNA from these
sites are thought to loop around to colocalize
with RNA polymerase.

A

1) upstream
2) downstream

53
Q

The poly A signal

A
  • is located within the terminator
    sequence and stimulates polyadenylation
    (addition of adenine nucleotides to the 3’ end of
    the mRNA).
54
Q

Exonucleases

A
  • are enzymes that cleave nucleotides
    from the polynucleotide chain at the ends of the
    chain. Exonuclease activity only results in sticky
    ends.
55
Q

Endonucleases

A
  • are enzymes that cleave
    nucleotides from the polynucleotide chain from
    the inside of the polynucleotide chain.
    Endonuclease activity can result in either sticky or
    blunt ends.
56
Q

Post-transcriptional modification (PTM) (Eukaryotes)

A
  • describes the
    conversion of pre-mRNA into processed mRNA,
    which leaves the nucleus.
57
Q

There are 3 main types of PTM

A

1) 5’ capping
2) Polyadenylation of 3’ end
3) Splicing out introns

58
Q

5’ capping

A
  • 7-methylguanosine cap is added to the 5’ end of the mRNA during elongation,
    protecting the mRNA from degradation.
59
Q

Polyadenylation of the 3’ end

A
  • addition of
    the poly A tail to the 3’ end to prevent
    degradation by exonucleases.
60
Q

Splicing out introns

A
  • introns are stretches of
    noncoding DNA that lie between regions of
    coding DNA (exons). Splicing refers to removing introns from pre-mRNA using
    spliceosomes. “Splice signals” present within
    introns signal to the spliceosome where to cut.
61
Q

Alternative splicing

A
  • describes a single pre-mRNA
    having multiple possible spliced mRNA products.
    Thus, the same pre-mRNA can produce many
    different proteins.
62
Q

miRNAs (micro RNA)

A

-are small RNA molecules that silence mRNA expression as a method of
post-transcriptional gene regulation by base-pairing with parts of sequences on the mRNA transcript that inhibits their translation.

63
Q

snRNAs (small nuclear RNA) and proteins

A
  • make up the functional part of a spliceosome and are
    collectively referred to snRNPs (small nuclear
    RiboNucleic Proteins).
64
Q

_____ and _____ are
important players in translation, the process of
converting mRNA into protein products.

A

1) ribosomes
2) tRNA

65
Q

Difference in ribosome makeup: Eukaryotes vs Prokaryotes

A

1) Eukaryotes - small (40S) and large (60S)
subunits form a 80S ribosome. They are
composed of rRNA (ribosomal RNA) and
proteins. The subunits are made in the
nucleolus and assembled once they are
exported to the cytosol.

2) Prokaryotes - small (30S) and large (50S)
subunits form a 70S ribosome. They are also
composed of rRNA and proteins, but are
assembled together in the nucleoid.

66
Q

A codon

A
  • is a group of three mRNA bases (A, U, G,
    or C) that code for an amino acid or terminate
    translation. There are 64 codon combinations
    total but only 20 amino acids, so degeneracy is
    present (multiple codons code for the same amino
    acid).
67
Q

Start Codon

A
  • AUG (Methione)
68
Q

Stop codons

A
  • UAA, UAG, UGA (end translation, do
    not code for any amino acid)
69
Q

An anticodon

A
  • is a group of three tRNA bases (A, U, G, or C) that base pairs with a codon. Each tRNA carries an amino acid to be added to the growing protein.
70
Q

Aminoacyl-tRNA

A
  • refers to a tRNA bound to an amino acid.
71
Q

Aminoacyl-tRNA synthetase

A
  • is the enzyme that
    attaches an amino acid to a specific tRNA using the
    energy from ATP.
72
Q

Ribosomal Binding sites for tRNA

A
  1. A site - A for aminoacyl-tRNA, which first
    enters at this site.
  2. P site - P for peptidyl-tRNA, which carries the
    growing polypeptide.
  3. E site - E for exit site. The tRNA from the P site
    is sent here and released from the ribosome.
73
Q

The ribosome catalyzes the formation of a _____ between the polypeptide in the P site and the newly added amino acid in the A site. Afterwards, the polypeptide is transferred to the _______ and the ribosome shifts one codon
down the mRNA. The _____ will now be empty and
ready to accept another aminoacyl-tRNA. The tRNA
from the ______ will be transferred to the _____ and
will leave the ribosome. During ______ (occurs in which the tRNA molecule
at the A site moves to the P sit), and the tRNA at
the P site moves to the E site (A → P → E)

A

1) peptide bond
2) A site’s tRNA
3) A site
4) P site
5) E site
6) translocation

74
Q

Chaperonins

A
  • Specialized proteins found in both eukaryotic and prokaryotic organisms and function in assisting newly
    synthesized polypeptides to fold into their correct
    shape.
75
Q

A DNA mutation

A
  • is a heritable change in the DNA nucleotide sequence that can be passed down to
    daughter cells.
76
Q

2 main types of DNA mutations:

A

1) Base substitution (point mutation)
2) Frameshift mutation

77
Q

Base substitutions (point mutations)

A
  • one nucleotide is replaced by another.
  • includes: silent mutation, missense mutation, and nonsense mutation
78
Q

Silent mutations

A
  • no change in amino
    acid sequence. Due to “third base wobble”,
    mutations in the DNA sequence that affect
    the third base of a codon can still result in
    the same amino acid being added to the
    protein. Relies on the degeneracy
    (redundancy) of translation.
79
Q

Missense mutations

A
  • single change in
    amino acid sequence. Can either be conservative (mutated amino acid similar
    to unmutated) or non-conservative (mutated amino acid different from unmutated).
80
Q

Nonsense mutations

A
  • single change in
    amino acid sequence that results in a stop
    codon. Results in early termination of
    protein
81
Q

Frameshift mutations

A
  • are mutations that
    result in a shift in the reading frame, changing the
    way the mRNA transcript is read.
  • there are 2 types: insertions and deletions
82
Q

Insertions

A
  • adding nucleotides into the
    DNA sequence - can shift the reading frame.
83
Q

Deletions

A
  • removing nucleotides from the DNA sequence - can shift the reading frame.
84
Q

A null mutation

A
  • can also occur, a null
    (non-functional) allele is produced that lacks the
    function of the normal, wild-type allele.
85
Q

Factors that contribute to DNA mutations:

A

● DNA polymerase errors during DNA replication.
● Loss of DNA during meiosis crossing over.
● Chemical damage from drugs.
● Radiation
● Transposons (jumping genes) - DNA sequences in prokaryotes and eukaryotes that can move and integrate into different places in
the genome and cause mutations.

86
Q

Factors that prevent DNA mutations:

A

● DNA polymerase proofreading by DNA
polymerase.
● Mismatch repair machinery that checks uncaught errors.
● Nucleotide excision repair that cuts out damaged DNA and replaces it with correct
DNA using complementary base pairing.

87
Q

Chromosomal mutations

A
  • occur and affect the entire chromosome rather than individual nucleotides.
  • There are four types:
    1) duplication
    2) translocation
    3) deletion
    4) inversion
88
Q

Duplication

A
  • A region of DNA is
    duplicated, resulting in a larger chromosomal arm and an atypical banding
    pattern.
89
Q

Translocation

A
  • A piece of one chromosome breaks off and attaches to
    another chromosome. Translocation increases chromosomal arm length and
    results in an abnormal banding pattern. This is the only mutation that affects both chromosomes.
90
Q

Deletion

A
  • A portion of the chromosome is
    deleted, resulting in a shorter
    chromosomal arm.
91
Q

Inversion

A
  • A portion of the chromosome becomes inverted on the arm of the
    chromosome. Results in an abnormal banding pattern, but does not affect the
    length of the chromosome.
92
Q

______ are not living because they must infect
living cells to multiply.

A

1) Viruses

93
Q

The ______ is a viral protein coat that is made of
subunits called _______. Some viruses also
have a phospholipid envelope that they pick up
from the host cell membrane.

A

1) capsids
2) capsomere

94
Q

In order for an infection to continue to spread,
viruses undergo ______ to create new
viruses that can further infect other cells/hosts.

A

1) Viral replication

95
Q

The viral replication cycle is as follows:

A
  1. Attachment - binding of a virus to host cell
  2. Penetration - virus crosses through the
    host’s cell membrane.
  3. Uncoating - viral capsid is removed and
    degraded by host enzymes.
  4. Synthesis - components of viral capsid are
    manufactured.
  5. Assembly - viral capsid components assemble to form the viral capsid.
  6. Release - last step of viral replication, fully assembled viruses are released.
96
Q

Two viral life cycle types:

A

1) Lysogenic Cycle
2) Lytic Cycle

97
Q

Lysogenic cycle

A
  • virus is considered dormant
    because it inserts its own genome into the host’s genome and does not harm the host. Each time the host genome undergoes
    replication, so does the viral genome.
98
Q

Lytic cycle

A
  • virus takes over host to replicate and does cause harm to the host. The viral
    particles produced can lyse the host cell to find other hosts to infect.
99
Q

Viruses can ____ between the lysogenic and lytic
cycles. For example, _____ can stimulate a virus in the lysogenic cycle to replicate
and enter the lytic cycle.

A

1) switch
2) favorable conditions

100
Q

Retroviruses (eg. HIV)

A
  • have an RNA genome that
    infects host cells. They contain an enzyme called
    reverse transcriptase, which converts their RNA
    into cDNA (complementary DNA). The cDNA can
    integrate into the host genome and enter the
    lysogenic cycle.
101
Q

_____ are asexual and divide by ______,
so they only receive genes from one parent cell
and do not increase genetic diversity through
reproduction. ______ are bacteriophage
genomes that have been integrated into the host
genome.

A

1) asexual
2) binary fission
3) prophages

102
Q

Bacteria must increase genetic diversity through
______, which describes the
transfer of genes between individual organisms.

A

1) horizontal gene transfer

103
Q

There three methods of horizontal gene
transfer:

A

1) conjugation
2) transformation
3) transduction

104
Q

Conjugation

A
  • bacteria use a cytoplasmic
    bridge called a pili to copy and transfer a special plasmid known as the F plasmid (fertility factor). If a bacteria contains an F
    plasmid, it is referred to as F+. If not, it is referred to as F-. To review, plasmids are
    circular DNA pieces that are independent from
    a bacteria’s single circular chromosome.
105
Q

Transformation

A
    • bacteria take up
      extracellular DNA. Bacteria are referred to as
      competent if they can perform transformation.

Electroporation is the process of using electrical impulses to force bacteria to become
competent.

106
Q

Transduction

A
  • viruses transfer bacterial DNA between different bacterial hosts. This occurs
    when a bacteriophage enters the lysogenic cycle in its host and carries bacterial DNA
    along with its own genome upon re-entering
    the lytic cycle.
107
Q

Quorum sensing

A
  • mechanism of communication
    by cells, regulating the release of signaling
    molecules that affect microbial metabolism and
    gene expression. Dependent on cell density.
108
Q

Bacteria can also contain ______, which are
extrachromosomal pieces of DNA in the form of a
plasmid that contains antibiotic resistance genes.

A

1) R-factors

Biology (15 decks)

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