Bio 7 Flashcards
(63 cards)
7.1 List the four
nitrogen bases in
DNA and group
them based on the
number of rings.
Purine (double ring) consists of:
1. Adenine
2. Guanine
Pyrimidine (single ring) consists of:
3. Thymine
4. Cvtosine
7.1 Skill: Describe
and analyze the
process and the
results of the
Hershey and Chase
experiment.
*It was known that some virus consists of solely
DNA and a protein coat.
-studied viruses that specifically infect bacteria:
phages
-phages have a protein coat with DNA inside
They coated one virus with radioactive
phosphorous coating (radiolabled DNA with
phosphorous 32), and the other one with
radioactive sulphur 35 (radiolabled proteins)
Allowed infection on bacteria (E Coli) and
centrifuged the bacteria, in which the bacteria will
turn into a pellet, and the virus will remain as
supernatant.
Testing DNA:
Supernatant is non-radioactive (proteins remained
outside)
Pellet is radioactive (Phosphorous has entered)
Testing proteins:
Supernatant is radioactive (where there is leftover proteins)
Pellet is non-radioactive (bacteria)
Results: DNA consists of genetic material!
7.1 Outline Rosalind
Franklin’s and
Maurice Wilkins’
investigation of
DNA structure by
X-ray diffraction.
Rosalind Franklin’s experiment made a discovery
on the structure of DNA, even though the credit
was taken by Crick and Watson.
She used the method of X-ray diffraction to
investigate in the structure of DNA, by posing a X
ray beam onto a crystallized DNA molecule,
which the diffracted rays were used to elucidate
details of the molecular structure of DNA.
Their results suggested the following:
1. DNA forms a helix.
2. DNA twists every 34 Armstrong, 10 bases (3.4A
each)
3. There is a second helical structure (double
helix)
4. Nitrogenous bases are closely packed together
on the inside, and phosphorous is outside of the
DNA, where P is clinging.
7.1 Outline the
features of DNA
structure that
suggested a
mechanism for DNA
replication.
Structures:
-DNA helix is both tightly packed and regular in
structure
-phosphates and sugars form an outer backbone
in the structure of DNA.
-composed of an equal number of purines (A + G)
and pyrimidines (C + T); suggesting that two
strands must run in antiparallel directions
-these nitrogenous bases are paired (purine +
pyrimidine) within the double helix
-for this pairing between purines and pyrimidines
to occur, the two strands must run in antiparallel
directions
-adenine and thymine paired via two hydrogen
bonds, whereas guanine and cytosine paired via
three hydrogen bonds
The 2 mechanisms suggested are:
1. Replication occurs via complementary base
pairing (adenine pairs with thymine, guanine pairs
with cytosine)
2. Replication is bidirectional (proceeds in
opposite directions on the two strands) due to the
antiparallel nature of the strands
3. If the bases were always paired this way, then
this would describe the regular structure of the
DNA helix (shown by Franklin)
7.1 What is the
function of DNA
polymerases in the
process of DNA
replication?
DNA polymerases can only add nucleotides to
the 3’ end of a primer; responsible for covalently
linking nucleotides to the end of a DNA strand
during replication.
The direction of DNA replication is therefore 5’ to
3’.
7.1 Explain why
DNA replication is
continuous on the
leading strand and
discontinuous on
the lagging strand.
DNA replication is continuous on the leading
strand and discontinuous on the lagging strand.
Leading strand moves towards the replication
fork from 3 prime to 5 prime.
Lagging strand move in the opposite direction
(form 5 prime to 3 prime) to the replication fork,
so replication is discontinuous.
****
7.1 List out all the
enzymes that carry
out DNA replication
and its function and
their functions.
- DNA gyrase: moves in advance of helicase and
relieves strains in the DNA molecule that are
created when the double helix is uncoiled. - Helicase: uncoils the DNA double helix and
splits it into two template strands - Single-stranded binding proteins: keep the
strands apart long enough to allow the template
strand to be copied. - DNA polymerase Ill: adds nucleotides in a 5’ to
3’ direction. It is on the leading strand it moves in
the same direction as the replication fork, close to helicase. - DNA primase: adds a short length of RNA
attached by base pairing to the template strand of
DNA. This acts as a primer, allowing DNA
polymerase to bind and begin replication. - DNA polymerase Ill: starts replication next to
the RNA primer and adds nucleotides in a 5’ to 3’
direction. It therefore moves away from the
replication fork on the lagging strand. Short
lengths of DNA are formed between RNA primers
on the lagging strand, called Okazaki fragments. - DNA polymerase I: removes the RNA primer
and replaces it with DNA. - DNA ligase seals up the nick by making another
sugar-phosphate bond.
**Polymerases catalyse the formation of covalent
bonds between the sugar of one nucleotide and the phosphate group of the following nucleotide
(a condensation reaction).
7.1 How is DNA
replication different
in prokaryotic and
eukaryotic cell?
DNA replication initiates at a single site in
prokaryotes, and at multiple sites in eukaryotes.
7.1 Outline five
functions of non-
coding DNA
sequences found in
genomes. List
examples of non-
coding DNA
regions.
Some examples of non-coding DNA:
1) telomeres
-occur on the ends of chromosomes and have a
protective function
-present to ensure that no coding DNA are lost,
as the helicase do not unwind to the end of DNA.
-in this case non-coding DNA are lost instead of
useful, coding DNA.
2) introns
-important in RNA transcription to ensure that
useful coding DNA (exons) are not lost
3) genes for tRNAs
-they do not code for the creation of proteins
-folds to form tRNA molecules instead
4) gene regulating sequences
-promoter, terminator, enhancers, silencers
-sequences that are involved in the regulation of
translation process
7.1 Application:
Outline how tandem
repeats are used in
DNA profiling.
These short tandem repeats are unique for each
individual. Analysing regions of short tandem
repeats allows identification of family
relationships, possible criminal activity, and
identification of disaster victims.
DNA analysis involves the use of restriction
enzymes and gel electrophoresis. DNA fragment
bands are produced when restriction enzymes
and gel electrophoresis are used. The unique
position of these fragment bands on the gel is
then analysed to determine relationships.
71 Define base
sequencing.
Base sequencing process in which the exact
sequence in a DNA fragment or molecule is
produced.
7.2 State which
direction
transcription occurs
in and why.
Transcription occurs in a 5’ to 3’ direction,
because nucleotides (NTPs) can only be added to
the 3’ end of the growing mRNA sequence
(added to the OH group to form a
phosphodiester bond).
7.2 Explain how
eukaryotic cells
modify mRNA after
transcription.
The mRNA produced by transcription in
eukaryotic cells goes through several changes
before it leaves the nucleus to enter the
cytoplasm.
I the introns on the mRNA is removed via
splicing, and leaves the exons behind. Small
nuclear RNAs (snRNAs) known as spliceosomes
bring about this splicing, and can increase the
number of proteins one gene can code for.
The remaining exons are chemically connected to
each other.
Different proteins are produced when introns are
removed in the splicing process. It is also possible
for exons to change position in the splicing
process. This will allow the production of different
proteins.
2) then, a cap is added to the 5’ end of the mRNA,
and a poly-A-tail is added to the 3’ end, both for protective measures.
3) a mature mRNA is formed, and exits the
nucleus.
7.2 Outline the
examples of non-
coding DNA in the
process of
transcription.
The enhancer and silencer as well as the
promotor and terminator region in the DNA are
examples of non-coding DNA, as they do not
code for the synthesis of specific proteins.
However, they are important in the process of
regulation transcription, so they are regulatory
sequences.
Enhancers are sections of DNA which proteins
may combine with to increase the rate of
transcription of a particular gene. Silencers are
sections of DNA to which proteins may attach.
This decreases the rate of transcription of a gene.
The binding of RNA polymerase to the promoter region initiated transcription, while the the
terminator stops transcription and dissociates the
RNA polymerase, mRNA strand, and DNA strand
from each other.
7.2 Describe the
regulation of gene
expression by
proteins that bind to
specific base
sequences in DNA.
Proteins which bind to DNA have a controlling
factor in gene expression. They are called
transcription factors, which bind to different non-
coding regions of the DNA to regulate
transcription rate.
Transcription activators are proteins which cause
looping of DNA. The looping of DNA may result
in a shorter distance between the activator and
the promoter regions of a gene. This will increase
the expression of that gene.
Repressor proteins may bind to segments of DNA
known as silencers. This prevents transcription
and gene expression.
7.2 Describe how
nucleosomes help
to regulate
transcription in
eukaryotes.
The packaging of DNA that occurs at
nucleosomes serves as a regulator of
transcription in eukaryotic cells.
DNA wrapped around histones in nucleosomes is
inaccessible to transcription enzymes, but
chemical modification of the tails of histones is an
important factor in determining whether a gene
will be expressed or not.
A number of different types of modification can
occur to the tails of histones including the addition of an acetyl group, the addition of a
methyl group or the addition of a phosphate
group.
Chemical modification of histone tails can either
activate or deactivate C genes by decreasing or
increasing the accessibility of the gene to M
transcription factors
neutralization (through acetylation) = less
condensed structure = higher levels of
transcription
7.2 Outline the
relationship
between
methylation and
epigenetics.
Cytosine in DNA can be converted to
methylcytosine by the addition of a methyl group
(-CH3).
Methylation inhibits transcription, so is a means of
switching off expression o certain genes. The cells
in a tissue can be expected to have the same
pattern o methylation and this pattern can be
inherited in daughter cells produced by mitosis.
Environmental factors can inluence the pattern o
methylation and gene expression
Many cancer cells have either a larger amount of
methvlation or a lower amount of methvlation
than non-cancerous cells. The presence of methyl
groups also seems to play a role in the maternal
or paternal expression of a gene.
Fluorescent markers can be used to detect
patterns of methylation in the chromosomes.
Analysis of the patterns has revealed some trends:
1. Patterns o methylation are established during
embryo development and the percentage o C-G
sites that are methylated reaches a maximum at
birth in humans but then decreases during the rest
o an individuals lie.
2. At birth identical twins have a very similar
pattern of methulation, but differences accumulate during their lifetimes, presumably due
to environmental differences. This is reflected in
the decreasing similarity between identical twins
as the grow older.
7.2 Explain how the
environment of a
cell and of an
organism has an
impact on gene
expression.
Organisms with the same genotypes often
express different phenotypes when in different
environments. Much effort has gone into twin
studies especially for twins raised apart.
The pattern o chemical markers established in the
DNA o a cell is the epigenome and research into
it is epigenetics
7.3 Explain the
process of
translation.
translation involves initiation, elongation/
translocation and termination:
Initiation:
mRNA binds to the small sub-unit of the
ribosome:
ribosome slides along mRNA to the start codon;
anticodon of tRNA pairs with codon on
mRNA:complementary base pairing (between
codon and anticodon);
(anticodon of) tRNA with methionine pairs with
start codon / AUG is the start codon;
Elongation:
Synthesis of the polypeptide involves a repeated
cycle of events.
second tRNA pairs with next codon; peptide bond
forms between amino acids;
ribosome moves along the mRNA by one
codon;movement in 5’ to 3’ direction;
tRNA that has lost its amino acid detaches;
another tRNA pairs with the next codon/moves
into A site;
tRNA activating enzymes;
link amino acids to specific tRNA;
Termination:
stop codon (eventually) reached;
ston codon doesn’t code for any amino acid
7.3 Distinguish
between bound
ribosomes and free
ribosomes.
Free ribosomes are not attached to the ER, and
synthesizes proteins that are primarily used within
the cell.
Bound ribosome are attached to the ER, and
synthesizes proteins to be secreted or used in the
lysosome.
Whether a ribosome is attached to the
endoplasmic reticulum or not seems to be
determined by a signal sequence of specific
amino acids called for on the mRNA strand
coming from the nucleus.
7.3 Outline the
structure of
ribosome
small subunit and large subunit;
mRNA binding site on small subunit;
three tRNA binding sites / A, P and E tRNA
binding sites;
protein and RNA composition (in both subunits):
7.3 Explain how the
speed of translation
is different in
prokaryotes and
eukaryotes.
Translation can occur immediately after
transcription in prokaryotes due to the absence of
a nuclear membrane.
The speed of protein synthesis in prokaryotic cells
is faster than in eukaryotic cells. This is due to two
factors:
• Non-coding sequences do not exist in
prokaryotic DNA. Therefore, there is no need to
process the mRNA produced by transcription to
remove introns.
• Prokaryotic cells do not have a nucleus.
Therefore, in these types of cells mRNA does not
have to move through the nuclear membrane to
attach to a ribosome.
7.3 Describe the
secondary and
tertiary structure of
protein.
Secondarv:
-alpha helix: coiled form
-beta pleated sheets: folded form
-held by hydrogen bonds between the NH and
C=O between 2 amino acids (Hydrogen bonds
occur between oppositely charged polar regions
of the carboxyl and amino groups of amino acids
in a polypeptide)
Tertiary:
The folding of a protein to create the tertiary
structure is quite specific based on interactions of
the amino acid -groups present.
-hydrogen bonds
-disulfide bridges
-ionic interactions
-hydrophobic or hydrophilic interactions
all between R groups.
Polar and non-polar amino acids are important in
determining the tertiary structure of a protein. The
tertiary structure of proteins is especially
important when the protein is an enzyme (enzyme
substrate specificity)
A protein’s primary and secondary structures do
not change when it folds to form the tertiary
structure.
7.3 Describe the
primary and
quaternary
structure of protein.
Primary:
-a sequence of amino acids held by peptide
bonds (only covalent b. is present)
The primary structure of a group
protein determines the secondary, tertiary, and
quaternary levels of protein structure.
Quatenary:
-have multiple polypeptide chains combined to
form a single structure.
-involves all the bonds present in the primary,
secondary, and tertiary structures.
-may include prosthetic groups (non-polypeptide
components) and form a conjugated protein
-haemoglobin is an example of a conjugated
protein
-it has haem groups which contain iron atoms for
oxygen attachment
