Molecular Genetics Flashcards
Nucleotide
- ribose sugar, nitrogenous base, and
phosphate group.
Nucleoside
- ribose sugar and nitrogenous base.
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.
1) nucleotides
2) OH-
Since ______ have more hydrogen bonds, a
______ temperature is needed to break DNA
strands.
1) G-C bonds
2) higher
Nucleosomes
- 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.
Chromatin
- refers to the overall packaging of DNA
and histones.
2 Types of Chromatin include
1) Euchromatin
2) Heterchromatin
Euchromatin
- nucleosomes are “loosely
packed”, so DNA is readily accessible for
transcription.
Heterochromatin
- nucleosomes are “tightly
packed”, so DNA is mostly inactive.
Histones are _____ charged while DNA is
_______ charged, allowing proper binding.
1) positively
2) negatively
Acetylation
-of histones removes positive charges,
relaxing DNA-histone attractions and allowing for
more transcription to happen.
Deacetylation
- of histones increases positive
charges, tightening DNA-histone attractions and
decreasing transcription.
Methylation
- of histones adds methyl groups, either increasing or decreasing transcription.
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.
1) origin of replication
2) circular
3) linear
DNA undergoes _______,
where each new double helix produced by
replication has one “new” strand and one “old”
strand.
1) semiconservative replication
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.
1) antiparallel
2) 5’ end
3) 3’ end
Steps of Replication
1) Initiation
2) Elongation
3) Termination
Initiation
- 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.
Elongation
- 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
Helicase
-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.
Single-strand binding proteins
- 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.
Topoisomerase
- 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.
Primase
- 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.
Sliding clamp proteins
- hold DNA polymerase onto the template strand.
- Helps to hold DNA
polymerase to the
template strand.
DNA polymerase
- 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.
The leading strand
-is produced
continuously because it has a 3’ end that
faces the replication fork.
The lagging strand
- 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.
A different _______ replaces RNA primers with DNA.
1) DNA polymerase
DNA ligase
-glues separated fragments of DNA together.
- GLues together separate
pieces of DNA.
Termination
- replication fork cannot
continue, ending DNA
replication. - a termination sequence (aka
terminator) signals to RNA polymerase to stop transcribing the gene.
Telomeres
- 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).
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.
1) G1/S
2) G1 phase
3) unfavorable
4) G0
Telomerase
- 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.
Genes
- 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.
Specifically, DNA undergoes transcription to
produce single-stranded __________.
1) mRNA
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.
1) Prokaryotes
Before transcription can occur, a _______
combines with _______ to form RNA polymerase
holoenzyme, giving it the ability to target specific
DNA promoter regions.
1) sigma factor
2) prokaryotic core RNA
polymerase
There are two types of termination in bacteria:
1) Rho independent termination
2) Rho-dependent termination
Rho independent termination
— a termination sequence is reached and the
RNA transcript folds into a hairpin loop →
RNA polymerase falls off and transcription
ends.
Rho-dependent termination
— A Rho protein binds to the RNA transcript,
moving 5’ → 3’ to catch up and displace
RNA polymerase, ending transcription.
operon
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.
To regulate the promoter, _______ bind to the operator regions, while _______
bind to the promoter sites. (prokaryotes)
1) repressor
2) activator
The lac operon
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.
The lac repressor protein
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.
cAMP levels and catabolite activator protein
(CAP)
- 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.
trp operon
- 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.
______ 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)
1) Tryptophan
2) operator
3) inactive
Unlike in prokaryotes, eukaryotic transcription
occurs in the ______ and uses ________ to transcribe most genes.
1) nucleus
2) RNA polymerase II
Transcription factors
- 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.
Enhancers
- are DNA sites that activator
proteins can bind to; they help increase transcription of a gene.
Silencers
- are DNA sites that repressor
proteins can bind to; they decrease transcription of a gene.
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.
1) upstream
2) downstream
The poly A signal
- is located within the terminator
sequence and stimulates polyadenylation
(addition of adenine nucleotides to the 3’ end of
the mRNA).
Exonucleases
- are enzymes that cleave nucleotides
from the polynucleotide chain at the ends of the
chain. Exonuclease activity only results in sticky
ends.
Endonucleases
- 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.
Post-transcriptional modification (PTM) (Eukaryotes)
- describes the
conversion of pre-mRNA into processed mRNA,
which leaves the nucleus.
There are 3 main types of PTM
1) 5’ capping
2) Polyadenylation of 3’ end
3) Splicing out introns
5’ capping
- 7-methylguanosine cap is added to the 5’ end of the mRNA during elongation,
protecting the mRNA from degradation.
Polyadenylation of the 3’ end
- addition of
the poly A tail to the 3’ end to prevent
degradation by exonucleases.
Splicing out introns
- 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.
Alternative splicing
- describes a single pre-mRNA
having multiple possible spliced mRNA products.
Thus, the same pre-mRNA can produce many
different proteins.
miRNAs (micro RNA)
-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.
snRNAs (small nuclear RNA) and proteins
- make up the functional part of a spliceosome and are
collectively referred to snRNPs (small nuclear
RiboNucleic Proteins).
_____ and _____ are
important players in translation, the process of
converting mRNA into protein products.
1) ribosomes
2) tRNA
Difference in ribosome makeup: Eukaryotes vs Prokaryotes
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.
A codon
- 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).
Start Codon
- AUG (Methione)
Stop codons
- UAA, UAG, UGA (end translation, do
not code for any amino acid)
An anticodon
- 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.
Aminoacyl-tRNA
- refers to a tRNA bound to an amino acid.
Aminoacyl-tRNA synthetase
- is the enzyme that
attaches an amino acid to a specific tRNA using the
energy from ATP.
Ribosomal Binding sites for tRNA
- A site - A for aminoacyl-tRNA, which first
enters at this site. - P site - P for peptidyl-tRNA, which carries the
growing polypeptide. - E site - E for exit site. The tRNA from the P site
is sent here and released from the ribosome.
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)
1) peptide bond
2) A site’s tRNA
3) A site
4) P site
5) E site
6) translocation
Chaperonins
- Specialized proteins found in both eukaryotic and prokaryotic organisms and function in assisting newly
synthesized polypeptides to fold into their correct
shape.
A DNA mutation
- is a heritable change in the DNA nucleotide sequence that can be passed down to
daughter cells.
2 main types of DNA mutations:
1) Base substitution (point mutation)
2) Frameshift mutation
Base substitutions (point mutations)
- one nucleotide is replaced by another.
- includes: silent mutation, missense mutation, and nonsense mutation
Silent mutations
- 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.
Missense mutations
- 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).
Nonsense mutations
- single change in
amino acid sequence that results in a stop
codon. Results in early termination of
protein
Frameshift mutations
- 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
Insertions
- adding nucleotides into the
DNA sequence - can shift the reading frame.
Deletions
- removing nucleotides from the DNA sequence - can shift the reading frame.
A null mutation
- can also occur, a null
(non-functional) allele is produced that lacks the
function of the normal, wild-type allele.
Factors that contribute to DNA mutations:
● 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.
Factors that prevent DNA mutations:
● 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.
Chromosomal mutations
- occur and affect the entire chromosome rather than individual nucleotides.
- There are four types:
1) duplication
2) translocation
3) deletion
4) inversion
Duplication
- A region of DNA is
duplicated, resulting in a larger chromosomal arm and an atypical banding
pattern.
Translocation
- 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.
Deletion
- A portion of the chromosome is
deleted, resulting in a shorter
chromosomal arm.
Inversion
- 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.
______ are not living because they must infect
living cells to multiply.
1) Viruses
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.
1) capsids
2) capsomere
In order for an infection to continue to spread,
viruses undergo ______ to create new
viruses that can further infect other cells/hosts.
1) Viral replication
The viral replication cycle is as follows:
- Attachment - binding of a virus to host cell
- Penetration - virus crosses through the
host’s cell membrane. - Uncoating - viral capsid is removed and
degraded by host enzymes. - Synthesis - components of viral capsid are
manufactured. - Assembly - viral capsid components assemble to form the viral capsid.
- Release - last step of viral replication, fully assembled viruses are released.
Two viral life cycle types:
1) Lysogenic Cycle
2) Lytic Cycle
Lysogenic cycle
- 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.
Lytic cycle
- 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.
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.
1) switch
2) favorable conditions
Retroviruses (eg. HIV)
- 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.
_____ 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.
1) asexual
2) binary fission
3) prophages
Bacteria must increase genetic diversity through
______, which describes the
transfer of genes between individual organisms.
1) horizontal gene transfer
There three methods of horizontal gene
transfer:
1) conjugation
2) transformation
3) transduction
Conjugation
- 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.
Transformation
- bacteria take up
extracellular DNA. Bacteria are referred to as
competent if they can perform transformation.
- bacteria take up
Electroporation is the process of using electrical impulses to force bacteria to become
competent.
Transduction
- 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.
Quorum sensing
- mechanism of communication
by cells, regulating the release of signaling
molecules that affect microbial metabolism and
gene expression. Dependent on cell density.
Bacteria can also contain ______, which are
extrachromosomal pieces of DNA in the form of a
plasmid that contains antibiotic resistance genes.
1) R-factors
