Bio 7 Flashcards

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1
Q

7.1 List the four
nitrogen bases in
DNA and group
them based on the
number of rings.

A

Purine (double ring) consists of:
1. Adenine
2. Guanine
Pyrimidine (single ring) consists of:
3. Thymine
4. Cvtosine

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2
Q

7.1 Skill: Describe
and analyze the
process and the
results of the
Hershey and Chase
experiment.

A

*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!

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3
Q

7.1 Outline Rosalind
Franklin’s and
Maurice Wilkins’
investigation of
DNA structure by
X-ray diffraction.

A

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.

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4
Q

7.1 Outline the
features of DNA
structure that
suggested a
mechanism for DNA
replication.

A

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)

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5
Q

7.1 What is the
function of DNA
polymerases in the
process of DNA
replication?

A

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’.

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6
Q

7.1 Explain why
DNA replication is
continuous on the
leading strand and
discontinuous on
the lagging strand.

A

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.
****

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7
Q

7.1 List out all the
enzymes that carry
out DNA replication
and its function and
their functions.

A
  1. DNA gyrase: moves in advance of helicase and
    relieves strains in the DNA molecule that are
    created when the double helix is uncoiled.
  2. Helicase: uncoils the DNA double helix and
    splits it into two template strands
  3. Single-stranded binding proteins: keep the
    strands apart long enough to allow the template
    strand to be copied.
  4. 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.
  5. 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.
  6. 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.
  7. DNA polymerase I: removes the RNA primer
    and replaces it with DNA.
  8. 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).
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8
Q

7.1 How is DNA
replication different
in prokaryotic and
eukaryotic cell?

A

DNA replication initiates at a single site in
prokaryotes, and at multiple sites in eukaryotes.

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9
Q

7.1 Outline five
functions of non-
coding DNA
sequences found in
genomes. List
examples of non-
coding DNA
regions.

A

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

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10
Q

7.1 Application:
Outline how tandem
repeats are used in
DNA profiling.

A

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.

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11
Q

71 Define base
sequencing.

A

Base sequencing process in which the exact
sequence in a DNA fragment or molecule is
produced.

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12
Q

7.2 State which
direction
transcription occurs
in and why.

A

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).

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13
Q

7.2 Explain how
eukaryotic cells
modify mRNA after
transcription.

A

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.

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14
Q

7.2 Outline the
examples of non-
coding DNA in the
process of
transcription.

A

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.

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15
Q

7.2 Describe the
regulation of gene
expression by
proteins that bind to
specific base
sequences in DNA.

A

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.

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16
Q

7.2 Describe how
nucleosomes help
to regulate
transcription in
eukaryotes.

A

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

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17
Q

7.2 Outline the
relationship
between
methylation and
epigenetics.

A

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.

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18
Q

7.2 Explain how the
environment of a
cell and of an
organism has an
impact on gene
expression.

A

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

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19
Q

7.3 Explain the
process of
translation.

A

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

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20
Q

7.3 Distinguish
between bound
ribosomes and free
ribosomes.

A

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.

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21
Q

7.3 Outline the
structure of
ribosome

A

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):

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22
Q

7.3 Explain how the
speed of translation
is different in
prokaryotes and
eukaryotes.

A

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.

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23
Q

7.3 Describe the
secondary and
tertiary structure of
protein.

A

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.

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24
Q

7.3 Describe the
primary and
quaternary
structure of protein.

A

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

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25
Q

7.3 State the name
and function of the
binding sites on
ribosomes.

A

mRNA binding site on small ribosomal unit, binds
to mature mRNA strand to initiate translation
3 tRNA binding sites on large ribosomal unit:
-aminoacyl (A) site, where new amino acid to be
added to the chain binds to
-peptidyl (P) site, holds the growing polypeptide
chain
-exit (A) site, tRNA with no animo acid is, being
released into the cvtoplasm.

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26
Q

7.3 Explain how
tRNA-activating
enzymes illustrate
enzyme-substrate
specificity, and the
role of
phosphorylation.

A

20 amino acids, 20 different tRNA, because there
are 20 different enzymes needed to aid in the
attachment of each different amino acid to the
proper tRNA (on the 3’ end CCA acceptor stem)
To add amino acid to tRNA, the addition of a
phosphate and its accompanying energy from
ATP is also necessarv for the attachment of an amino acid to its proper tRNA (phospholyration)
Phosphorylation requires the hydrolysis of an ATP
molecule to provide the energy for the reaction:
ATP + H20 › AMP and PP (pyrophosphate).

27
Q

7.3 Outline the
structure of a tRNA
molecule

A

tRNA is composed of one chain of (RNA)
nucleotides
tRNA has a position/end/site attaching an amino
acid (reject tRNA contains an amino acid)
at the 3’ terminal / consisting of CCA/ACC
tRNA has an anticodon
anticodon of three bases which are not base
paired / single stranded / forming part of a loop
tRNA has double stranded sections formed by
base pairing
double stranded sections can be helical
tRNA has (three) loops (somethimes with an extra
small loop)
tRNA has a distinctive three dimensional / clover
leaf shape

28
Q

7.3 Explain the
existence of
polysomes in
electron
micrographs of
prokarvotes and
eukaryotes.

A

To allow the production of polypeptides and
hence proteins at a faster rate, multiple ribosomes
can attach to the same mRNA. The structure
formed is called a polysome.

29
Q

7.2 Distinguish
between enhancers,
promoters, and
silencers and
terminator.

A

An enhancer is a sequence of DNA that functions
to enhance transcription. A promoter is a
sequence of DNA that initiates the process of
transcription.
-transcription factors bind to enhancer/silencers
-RNA polymerase binds to promoter and
terminator region to start or end transcription
*promoters contain regulatory sequences where
transcription factors can bind
-the promoters of genes expressed in different
cell types will have different regulatory
sequences
-the binding of some trasncription factors helps
or blocks the binding of RNA polymerase
*enchancers and silencer are DNA sequences that
are NOT part of the gene.

30
Q

7.2 Explain the
process of
transcription
leading to the
formation of mRNA.

A

a. RNA polymerase; (polymerase number is not
required)
b. binds to a promoter on the DNA;
c. unwinding the DNA strands;
d. binding nucleoside triphosphates;
e. to the antisense strand of DNA;
f. as it moves along in a 5’>3’ direction;
g. Using complementary pairing/A-U and C-G;
h. losing two phosphates to gain the required
energy;
i. until a terminator signal is reached (in
prokaryotes);
j. RNA detaches from the template and DNA
rewinds;
k. RNA polymerase detaches from the DNA;
I. many RNA polymerases can follow each other:
m. introns have to be removed in eukarvotes to
form mature mRNA;

31
Q

7.1 Explain the
levels of
supercoiling.

A

DNA- nucleosome - beads on a string + 30nm
fiber > unreplicated interphase chromosome -
replicated metaphase chromosome
-DNA is packaged with histone proteins to create
a compacted structure called a nucleosome
-nucleosomes help to supercoil the DNA,
resulting in a greatly compacted structure that
allows for more efficient storage
-super coiling helps to protect the DNA from
damage and also allows chromosomes to be
mobile during mitosis and meiosis
-nucleosomes and their modifications control the
degree of DNA supercoiling.
Organisation of Eukaryotic DNA
-DNA is complexed with eight histone proteins
(an octamer) to form a complex called a
nucleosome
-nucleosomes are linked by an additional histone
protein (H1 histone) to form a string of
chromatosomes
-these then coil to form a solenoid structure (~6
chromatosomes per turn) which is condensed to
form a 30 nm fibre
-fibres then form loops, which are compressed
and folded around a protein scaffold to form
chromatin
-chromatin will then supercoil during cell division to form chromosomes that are visible (when
stained) under microscope

32
Q

7.1 Contrast
replication on the
the leading strand
and the lagging
strand of DNA

A

Because double-stranded DNA is antiparallel,
DNA polymerase must move in opposite
directions on the two strands
-leading strand, DNA polymerase is moving
towards the replication fork and so can copy
continuously
-lagging strand, DNA polymerase is moving away
from the replication fork, meaning copying is
discontinuous
-as DNA polymerase is moving away from
helicase, it must constantly return to copy newly
separated stretches of DNA
-lagging strand is copied as a series of short
fragments (Okazaki fragments), each preceded by
a primer
-primers are replaced with DNA bases and the
fragments joined together by a combination of
DNA pol I and DNA ligase

33
Q

7.1 Outline the need
for RNA primers in
DNA replication.

A

The RNA primer provides an initiation point for
DNA polymerase IlI, which can extend a
nucleotide chain but not start one.

34
Q

7.1 Define “coding
sequences” and
“repetitive
sequences” of DNA.

A

Coding sequences (regions) code for proteins
(They are transcribed and translated to produce a
protein). Repetitive sequences

35
Q

7.1 Outline the
process of X-rav
diffraction.

A

-X-ray crystallography is a technique to
determine the three dimensional structure of a
substance.
-samples of pure substance are crystalized. X-
rays pass through the crystal.
-some X-rays are scattered or diffracted by atoms
in the crystal.
-the diffraction pattern is recorded on
photographic paper.
-the diffraction pattern corresponds to the
structure of the substance.
-each substance has a unique diffraction pattern.
Rosalind Franklin and Maurice Wilkins produced
X-ray diffraction patterns of DNA.
(Franklin made careful observations of the
diffraction patterns to deduce the basic
dimensions of the molecule and that the
phosphates were on the outside. Franklin made a
crucial contribution towards solving the structure
of DNA.)

36
Q

7.1 Outline the
deductions about
DNA structure
made from the X-
ray diffraction
pattern.

A

Composition: DNA is a double stranded molecule
Orientation: Nitrogenous bases are closely
packed together on the inside and phosphates
form an outer backbone
Shape: The DNA molecule twists at regular
intervals (every 34 Angstrom) to form a helix (two
strands = double helix)
-the X indicated a helix.
-the regular diffraction pattern indicated the
dimensions of the molecule, such as its width,
were consistent.
-the distance between spots indicated the size of
one turn of the helix: this was 3.4 nm.
-the distance from the center to the outside of the
pattern indicated the distance between two bases
in the same strand; this was 0.34 nm.
-hence, there were ten bases per turn.
-the angle between a horizontal line through the
center and the X, showed the pitch of the helix.

37
Q

7.1 Define VNTR,
and tandem
repeats.

A

Variable Number of Tandem Repeat (VNTR) loci
are chromosomal regions in which a short DNA
sequence motif (such as GC or AGCT) is repeated
a variable number of times end-to-end at a single
location (tandem repeat).
Tandem repeats (or adiacent repeats) are units of
2 to 50 nucleotides that are repeated many times
(some repeats are longer).

38
Q

7.1 Explain why
VNTR are used in
DNA profiling

A

(Some regions of non-coding DNA contain highly
repetitive sequences.
-tandem repeats (or adjacent repeats) are units of
2 to 50 nucleotides that are repeated many times
(some repeats are longer).
-short tandem repeats are typically between two
and six nucleotides.)
-the number of repetitions of the tandem repeats
varies between individuals.
-PCR can be used to amplify the region
containing the tandem repeat from individuals

39
Q

7.1 Outline the
process of DNA
sequencing

A

-sequencing DNA determines the order of bases
in a DNA molecule.
-special nucleotides containing
dideoxyribonucleic acid are used to stop DNA
replication in preparation for base sequencing in
DNA profiling, because it lacks the OH group in
the 3’ Carbon.
-the -OH group on carbon 3 of the deoxyribose is
required for the condensation reaction that joins
the following nucleotide.
-there are four different manufactured
nucleotides with each containing a
dideoxyribonucleic acid.
-each of these four special nucleotides have a
different fluorescent marker attached to them.
-observing the position of these florescent
markers allows the sequencing of a segment of
DNA since they stop the DNA replication process
at the exact position they are added.
Gel electropheresis is used to separate mixtures
of DNA, RNA, or proteins according to molecular
size.
-used to determine the sequence of DNA as well,
as the longest fragment will move the least
-the shortest fragment will move the most after
electricity is applied to the gel.
-moving up the gel, each fragment is one nucleotide larger than the previous.

40
Q

7.2 Define gene
expression.

A

the process by which the instructions in our DNA
are converted into a functional product,
regulated by proteins that bind to specific base
sequences in DNA

41
Q

7.2 Outline two
examples of
environmental
influence on gene
expression.

A

Environmental factors can affect gene expression
such as:
-the production of skin melanin pigmentation
during exposure to sunlight in humans.
-in embryonic development, the embryo contains
an uneven distribution of chemicals called
morphogens.
-concentrations of the morphogens affect gene
expression contributing to different patterns of
gene expression
-and thus different fates of the embryonic cells
depending on their position in the embryo.

42
Q

7.2 Outline the
effect of acetylation
of nucleosome tails
on rates of gene
expression.

A

-residues of the amino acid lysine on histone tails
can have acetyl groups either removed or added.
-normally the lysine residues on histone tails bear
a positive charge that can bind to the negatively
charged DNA
-to form a condensed structure that inhibits
transcription.
-histone acetylation neutralizes these positive
charges allowing a less condensed structure with
higher levels of transcription.

43
Q

7.2 Describe the
three post-
transcriptional
modifications of
pre-mRNA in
eukaryotes.

A

Capping
-capping involves the addition of a methyl group
to the 5’-end of the transcribed RNA
-the methylated cap provides protection against
degradation by exonucleases
-also allows the transcript to be recognised by
the cell’s translational machinery (e.g. nuclear
export proteins and ribosome)

Polyadenylation
-polyadenylation describes the addition of a long
chain of adenine nucleotides (a poly-A tail) to the
3’-end of the transcript
-the poly-A tail improves the stability of the RNA
transcript and facilitates its export from the
nucleus

Splicing
-within eukaryotic genes are non-coding
sequences called introns, w hich must be removed
prior to forming mature mRNA
-the coding regions are called exons and these
are fused together when introns are removed to
form a continuous sequence
-introns are intruding sequences whereas exons
are expressing sequences
-the process by which introns are removed is
called splicing

44
Q

7.2 Describe the
process of
alternative RNA
splicing.

A

-alternative splicing is a process during gene
expression whereby a single gene codes for
multiple proteins.
-this occurs in genes with multiple exons.
-a particular exon may or may not be included in
the final messenger RNA.
-as a result, the proteins translated from
alternatively spliced mRNAs will differ in their
amino acid sequence and possibly in their
biological functions.

45
Q

7.2 Outline an
example of
alternative splicing
the results in
different protein
products.

A

In mammals, the protein tropomyosin is encoded
by a gene that has 1l exons. Tropomyosin pre-
mRNA is spliced differently in different tissues
resulting in five different forms of the protein. For
example, in skeletal muscle exon “2” is missing
from the mRNA and in smooth muscle, exons “3”
and “10” are not present.

46
Q

7.2 Describe the
effect of DNA
methylation on
gene expression.

A

-(the addition of methyl groups directly to DNA is
thought to play a role in gene expression.
-whereas methylation of histones can promote or
inhibit transcription, direct methylation of DNA
tends to decrease gene expression.
-(some cytosine bases in DNA can be chemically
modified by methylation, which involves adding a
methyl group.)
-methylation of DNA promotes the coiling of DNA by nucleosomes and stops transcription.
-(methylation of DNA is coordinated with
chemical modification of histones).
- (the amount of DNA methylation varies during a
lifetime and is affected by environmental factors)

47
Q

7.2 Define
epigenetic and
epigenome.

A

Epigenetics: via DNA methylation, the
environment can have an impact on gene
expression without causing changes to the DNA
base sequence.
Epigenome: the sum of all the epigenetic tags

48
Q

7.2 Discuss the role
of reprogramming
and imprinting on
epigenetic factors

A

-(different cells have their own methylation
pattern so that a unique set of proteins will be
produced in order for that cell to perform its
function.
-during cell division, the methylation pattern will
be passed over to the daughter cell.)
-sperm and eggs develop from cells with
epinegentic tags.
-when two reproductive cells meet, the
epigenome is erased through a process called
“reprogramming”,
-about 1% of the epigenome is not erased and
survives yielding a result called “imprinting”.
-for example, when a mammalian mother has
gestational diabetes, the high levels of glucose in
the fetal circulation trigger epigenetic changes in
the daughter’s DNA such that she is predisposed
to develop gestation diabetes herself.

49
Q

7.3 Outline the
process of
translation initiation.

A

-initiation involves the assembly of translation
machinery.
-(1) The small subunit of the ribosome binds to the
5’ end of the mRNA.
-(I)) The anticodon (UAC) of an initiator tRNA
(carrying methionine) binds to the start codon
(AUG) on the mRNA by hydrogen bonds.
-(Ill) Antiocdon to codon binding follows complementary base pair rules.
-(IV) The initiator tRNA assists in the binding of the
large subunit of the ribosome to the small subunit.
-(V) The initiator tRNA is in the P site of the large
subunit.

50
Q

7.3 Outline the
process of
translation
elongation,
including codon
recognition, bond
formation and
translocation.

A

Once the translation machinery has been
assembled, the polypeptide chain can be
synthesised which involves a repeated cycle of
events.
-(1) A charged tRNA binds at the A site.
-(ll) To bind to the A site, the anticodon of the
tRNA must be complementary to the codon on
the mRNA.
-(Ill) A peptide bond forms between the amino
acids at the P and A sites (the energy required is
provided by the charged tRNA).
-(IV) The tRNA at the P site detaches from its amino acid.
-(V) The ribosome moves one codon along the
mrna (5’ to 3)’.
-(VI) The tRNA with the growing polypeptide
chain is now in the P site.
-(VIl) The tRNA with no amino acid is now in the E
site and exits the ribosome (to be recharged by
tRNA activating enzyme).
-(VIII) A newly charged tRNA can now enter the A
site.

51
Q

7.3 Outline the
process the role of
the stop codon.

A

1) The ribosome moves down the mRNA until
there is a stop codon in the A site.
-(Il) No tRNA molecules can bind to stop codons.
-(Ill) Proteins called release factors bind to the A
site.
-(IV) This causes:
-the polypeptide chain to be released from the
tRNA in the P site.
-the ribosome seperates from the mRNA and
splits into large and small subunits

52
Q

7.3 List destinations
of proteins
synthesized on free
ribosomes.

A

The cvtoplasm, mitochondria and chloroplasts.

53
Q

7.3 List destinations
of proteins
synthesized on
bound ribosomes.

A

The ER, the Golgi apparatus, lysosomes, the
plasma membrane or outside the cell.

54
Q

7.3 Outline how a
ribosome becomes
bound to the
endoplasmic
reticulum

A

-(whether the ribosome is free in the cvtosol or
bound to the ER depends on the presence of a
signal sequence on the polypeptide being
translated.
-it is the first part of the polypeptide translated).
-as the signal sequence is created it becomes
bound to a signal recognition protein that stops
the translation until it can bind to a receptor on
the surface of the ER.
-once this happens, translation begins again with
the polypeptide moving into the lumen of the ER
as it is created.

55
Q

7.3 State the three
fates of proteins
synthesised by
ribosomes
associated with the
RER

A

1) To be secreted by exocytosis (e.g. antibodies.)
(2) To be plasma membrane proteins (e.g.
Voltage-gated potassium ion channels)
(3) To function in membrane-bound organelles
(e.g. digestive enzymes with lysosomes)

56
Q

7.3 Explain how the
chemical
characteristics of R
groups in the
polypeptide chain
affect protein
folding

A

If surrounded by water, amino acids with
hydrophobic (non-polar) R groups tend to be
located in the center of the protein and those with
hydrophilic (polar or charged) R groups tend to
be on the outside

57
Q

7.3 Describe the
structure of a
conjugated protein,
including the
prosthetic group.

A

-a prosthetic group is a non-protein molecule
tightly bound to the protein. -hemoglobin is
composed of four polypeptides (two alpha and
two beta chains).
-each polypeptide has a heme group.
-proteins containing a prosthetic group are
termed conjugated proteins.

58
Q

7.3 Outline the
process of
attaching an amino
acid to tRNA by the
tRNA activating
enzyme.

A

-ATP is required to supply energy to link the tRNA
with its specific amino acid via a covalent bond.
-the formation of this covalent bonds conserves
energy (originally from ATP), which is then used to
join amino acids via peptide bonds (at the
ribosome).
The reaction occurs in two steps:
-step number one is the activation of the amino
acid.
-ATP and the specific amino acid bind to the tRNA
activating enzyme.
-ATP is hydrolysed and the amino acid is
covalently linked to AMP.
-step number two is the attachment of amino acid
to tRNA.
-the specific tRNA molecule then binds to the
active site.
-the amino acid is covalently linked to the tRNA activating enzyme.
-ATP is hydrolysed and the amino acid is
covalently linked to AMP.
-step number two is the attachment of amino acid
to tRNA.
-the specific tRNA molecule then binds to the
active site.
-the amino acid is covalently linked to the tRNA
and AMP is released.
-the charged tRNA (tRNA with amino acid) is
released from the active site.

59
Q

7.3 Outline the
structure of tRNA
molecules.

A

-tRNA molecules have a stem-loop structure.
-the stem regions are double stranded due to
hydrogen bonding between complementary base
pairs.
-these are three loops, where there is no base
pairing.
-one loop is called the anticodon loop and
contains the three base anticodon sequence.
-there is an amino acid attachment site at the 3’
end of the tRNA molecule.

60
Q

7.3 Outline the
structure of a
polysome.

A

-polysomes are formed by several ribosomes
translating a single mRNA.
-in an electron microscope they appear as beads
on a string and represent multiple ribosomes
attached to a single mRNA molecule.
-(in prokaryotes, ribsoomes can attach to the 5’
end of the mRNA molecule as soon as
transcription starts.)

61
Q

7.1 Outline how X-
ray crystallography
is carried out.

A

X-ray crystallography is a technique used to determine the atomic and molecular structure of crystals. The general steps involved in carrying out X-ray crystallography are as follows:

Preparation of the sample: The first step involves preparing a suitable crystal sample of the molecule of interest. The crystal should be large enough and free from defects or impurities.
Data collection: The crystal is then placed in the X-ray beam, and the diffracted X-rays are collected by a detector. The crystal is rotated in the beam to collect data from different angles.
Data analysis: The collected data are then analyzed to determine the electron density of the crystal. The electron density is used to calculate the position and orientation of the atoms within the crystal lattice.
Model building: A model of the molecule is then built based on the calculated electron density. The model is refined by adjusting the positions of the atoms to improve the fit to the experimental data.
Validation: The final model is validated by analyzing the quality of the fit to the experimental data and checking for consistency with known chemical and physical properties.
Publication: The results of the crystallographic analysis are then published in a scientific journal, along with the final model and any relevant data or analysis

62
Q

7.1 The DNA-histone
complex is called:

A

chromatin

63
Q

7.1 Outline the
structure of the
nucleosomes in
eukaryotic
chromosomes.

A
  • contain histones
  • eight histone molecules form a cluster in a
    nucleosome
  • DNA strand is wound around the histones
    wound around twice in each nucleosome
  • (another) histone molecule holds the
    nucleosome (s) together