Lecture 18: DNA Replication
components necessary for DNA synthesis: substrates, template, primer and
Substrates: A,T, G, C base pairs
Template: the original strand of DNA
Primers: RNA primers base paired with the template to allow DNA polymerase to
Enzymes: all enzymes and proteins required to make the replisome. (to synthesize
structure of DNA with respect to polarity and complementarity: DNA has a sugar-
phosphate backbone. It has 4 nitrogenous bases: adenine, guanine, thymine and
cytosine. Adenine and guanine are purines (double-ring structures). Thymine and
cytosine are pyrimidines (single-ring structures). A base pairs with T, C base pairs
with G. DNA is polar because it has a 5’ end and a 3’ end.
5’ and 3’ end of a DNA strand: The 5’ end of DNA has a phosphate group. The 3’ end
has a hydroxyl group. DNA synthesis occurs in the 5’ to 3’ direction.
relationship between DNA synthesis and chromosome duplication during S phase:
DNA is synthesized during S phase, and therefore all the chromosomes (bundles of
DNA) get duplicated.
structure of replication fork and replication bubble:
problem that arises during replication of the ends of linear chromosomes: DNA
polymerase can only extend DNA from a free 3’ OH group. Because the lagging
strand ends on a 3’ end, there is no way to finish the DNA even if a primer were
there, as we would not be able to replace the primer because there would be no free
OH group. This means that the chromosome would get progressively shorter and
shorter if there were no mechanism to fix this.
how telomerase addresses the above problem: the ends of chromosomes contain a G
rich series of repeats, called a telomere. Telomerase recognizes the tip of an existing
repeat series. Using an RNA template, telomerase elongates the parental strand in
the 5’ to 3’ direction and adds additional repeats as it moves down the parental
strand. The lagging strand is then completed by DNA polymerase alpha, which
carries a DNA primase as one of its subunits. In this way the original information at
the ends of linear chromosomes is completely copied into the new DNA, and the
only thing lost is a bit of the telomere.
Lecture 19: Gene Mutation
characteristics of the Ames Test as a screen for chemical mutagens: the bacterium
Salmonella typhimurium is used because there is a mutant that is unable to synthesize the amino acid histidine. Therefore when this mutant is grown on
minimal medium, it will not grow. The Ames test takes possible chemical mutagens
and puts them on minimal medium with the mutant bacteria. If the bacteria start to
grow, it is assumed that the chemical caused a back mutation, allowing the bacteria
to start to synthesize histidine. This means that the chemical causes mutations in
bacteria and possibly other organisms and is a positive Ames Test.
characteristics of mutation vs. DNA damage: a mutation is a heritable change in
double-stranded DNA sequence such as base substitutions,
insertions/deletions/reorganizations. DNA damage is a physical abnormality in the
DNA. Damage can be recognized by enzymes and be correctly repaired. If it cannot
be repaired then transcription, translation and replication will be blocked and/or
the cell with go through apoptosis.
mechanisms of spontaneous substitution mutation arising from errors in
replication: Sometimes a base pair can undergo a tautomeric shift, which means that
the electrons within the base pair shift and then it pairs with a different base pair.
An example is thymine, which normally has 2 hydrogen bonds with adenine.
Sometimes it undergoes a tautomeric shift and will base pair with guanine and have
3 hydrogen bonds.
mechanisms of spontaneous indel mutation arising from errors in replication:
Because of the sequence (same base pairs all in a row), one or the other of the
strands can loop out. Therefore when the DNA is replicated, one of the sides is too
mechanism by which base analogues, such as 5-bromouracil, increase mutation
frequency: 5-bromouracil is indistinguishable from thymine and will be put in DNA
instead of thymine. The bromine atom leads to a molecule that is tautomerically
unstable, which is bad as it will always change which base it wants to pair with.
likely effects of silent, missense and nonsense mutations on protein function: (all are
caused by a change of a single base pair)
Silent: the changed codon will specify for the same amino acid, due to the
degeneracy of the genetic code.
Missence: the changed codon will specify for a different amino acid. This may or may
not alter the protein function in a big or small way.
Nonsense: the changed codon will specify for a stop. This causes a premature
termination of the protein during translation. This will almost certainly affect the
protein in a big way, it will likely be partially functional at best.
mechanism of frameshift mutation and likely effect on protein function: A frameshift
is cause by an insertion or deletion of a base pair. This alters the whole reading
frame from there on. The resulting polypeptide is usually non-functional because of
the significantly altered amino acid sequence. likely effects of mutations outside of coding regions: Some of the non-coding regions
are used for transcriptional, translational, post-transcriptional regulations. These
mutations would then affect the regulations. It really depends on which part is being
Lecture 20: Chromosome Mutation
mechanism of action of DNA damage by non-ionizing radiation (UV): pyrimidines, in
particular thymines, that are side by side can absorb photons of light and this causes
a reorganization of electrons which causes covalent bonds and creates a thymine
dimer. This distorts the helix and blocks replication and transcription.
mechanism of action of DNA repair enzymes:
Proofreading: if a newly added nucleotide is mismatched, DNA polymerase will
reverse and use a a built-in deoxyribonuclease to remove the incorrect nucleotide.
The enzyme then resumes working forward. (RNA polymerases cannot proofread).
Reverse transcriptases can’t really proofread either.
Excision repair: the part where the bad base pair is is cut out and then it is filled in.
Photoreactivation: In most organisms thymine dimers can be repaired by
photoreactivation. Photoreactivation is a repair process in which photolyase
enzymes directly reverse thymine dimers via photochemical reactions. Lesions on
the DNA strand are recognized by these enzymes, followed by the absorption of light
wavelengths (particularly blue light). This absorption enables the photochemical
reactions to occur, which results in the elimination of the pyrimidine dimer,
returning it to its original state.
mechanism of action of DNA damage by ionizing mutagenesis: When ionizing
radiation passes through your cells, it creates reactive oxygen. This then damages
everything, from proteins to DNA.
mechanism of production of aneuploid (abnormal number of chromosomes)
gametes during meiosis: This happens by nondisjunction, which is the failure of
homologous pairs to separate during the first meiotic division or of chromatids to
separate during the second meiotic division.
difficulties of studying effects of radiation exposure on large populations: its hard to
isolate large populations, and there are infinite amounts of other factors affecting
the population, because it is so large and therefore more diverse.
Lecture 21: Immunogenetics
role of phagocytosis in immune response: This is the non-specific part of the
immune system. It goes after bacteria and other antigens and engulfs them and
breaks them down.
characteristics of innate vs. adaptive immune response: The innate (non-specific) immune response goes after any sort of antigen it sees as abnormal. The adaptive
immune response involves inherited mechanisms that lead to the synthesis of
molecules such as antibodies that target pathogens in a specific way.
mechanism of generation of diverse antibody genes: The receptors and the
antibodies the cell will eventually produce are tetramers. There are two heavy
chains and two light chains that make up this protein. On the light chain gene, there
are V segments and J segments. V means variable, J means joining. There are several,
sometimes hundreds of V segments and 4-5 J segments. As the B-cell matures, it
purposefully causes mutations in order to build an immunoglobulin gene. The
segments are like lego inherited from parents to make your own genes. This is called
somatic recombination. On the heavy chain gene, it is similar but they also have D
segments. D stands for diversity. Ds are put with Js, the DJs with Vs.
structure of antibodies and corresponding B cell receptor (BCR): Each antibody
molecule consists of 4 polypeptide chains: two identical light chains and two
identical heavy chains. These form a Y structure, called a B-cell receptor (BCR). The
BCR is different in every type of antibody because of the reasons stated above.
role of chromosomal rearrangements of antibody sequences in activation of
oncogenes: the DNA in your lymphocytes gets rearranged and sometimes it can
mess up and you can have antibody genes (promoters, etc) controlling oncogenes.
mechanism giving rise to immunological memory: some of the activated
lymphocytes differentiate into memory cells that circulate in the body, ready to
initiate a rapid immune system response on subsequent exposure to the same
plausible explanations for why different cell lines in chimeras do not mount immune
responses against each other: the sheep can tolerate the goat because the immune
system forms after the embryogenesis of the geep. Therefore the immune system
treats both the sheep and the goat as "self".
Lecture 22/23: DNA Technologies I and II (ELYSIA)
Recall that the surprising ability of the “solar slug” Elysia to maintain functional
chloroplasts stolen while feeding on Vaucheria algal cells (kleptoplasty) is an
unsolved mystery in science. One possibility is that genes required for plastid
maintenance have undergone evolutionary lateral gene transfer from the nucleus of
the algae to the nucleus of the slug. These two lectures were designed to analyze a
primary research article and showcase how the authors used selected DNA
technologies to generate data in support of their hypothesis that a particular
photosynthetic gene (psbO) from the nucleus of Vaucheria is also present in Elysia.
Overall, you should be able to I) identify the various types of data shown by this
paper; II) explain how these data were obtained; and III) conclude whether each
type of data does, or does not, support the hypothesis. I) 1. Picture of the amplified portion of the cDNA of the psbO transcript on agarose
gel in Vaucheria and Elysia 5 months after algal feeding.
2. Northern blot analysis of Vaucheria and Elysia psbO transcript (RNA)
3. RACE-amplified sequence of psbO gene in DNA of Vaucheria, egg Elysia, and adult
4. Translation of the MSP amino acid sequences in Vaucheria and Elysia. They are
identical between the two and have a tripartite targeting signal.
II)1. Total RNA was isolated in Vaucheria and then was reverse transcribed into
cDNA (complementary DNA, half RNA, half DNA). Degenerate primers were created
for the psbO gene from other published sequences and the cDNA was amplified
using PCR. The amplified cDNA was then electrophoresed.
2. a Northern blot analysis was done on the cloned psbO transcript. This is done by
gel electrophoresis on the RNA and then detected with a hybridized probe.
3. Homologous primers were designed from the RACE-amplified sequence of the
Vaucheria psbO cDNA. By PCR, these primers amplified the genomic DNA of
Vaucheria, Elysia eggs and adult Elysia tissue as well as adult Elysia cDNA (created
4. The amino acid sequences were analyzed by phylogenetic methods, based on
amino acid sequences of 25 mature proteins and carried out using maximum
III) 1. Supports the hypothesis because we would not normally expect psbO
transcription 5 months after feeding as the plastids cannot produce it themselves.
2. Somewhat supports the hypothesis as the probe reacted with both species but
there was not picture of a negative control to really prove the hypothesis
3. Supports the hypothesis because it shows the gene in Elysia egg, proving Elysia
doesn’t need to eat algae to get the gene.
4. Does not support the hypothesis because although the sequences are identical,
the tripartite targeting sequence can only be used in chloroplasts with 4
membranes, like Vaucheria, and the chloroplasts Elysia retains only have 2
membranes (the chloroplast endoplasmic reticulum isn’t retained in the engulfed
location of PsbO gene in photoautotrophic organisms: It is found in the nuclear
location and role of PsbO gene product in photosynthetic electron transport: PsbO
gene product is found on the lumenal side of the thylakoid membrane. Found in
complex 2 of the electron transport chain.
purpose of molecular size markers in electrophoresis: A molecular weight size
marker is used to identify the approximate size of a molecule run on a gel, using the
principle that molecular weight is inversely proportional to migration rate through
a gel matrix. Therefore when used in gel electrophoresis, markers effectively
provide a logarithmic scale by which to estimate the size of the other fragments
(providing the fragment sizes of the marker are known). These markers can be
composed either of different proteins of known size, or nucleic acids of different sizes. Markers are loaded in lanes adjacent to samples lanes before the
commencement of the run.
interpretation of agarose gel data: Because a larger molecular weight means a
slower migration rate through the gel, it is easy to compare many things this way.
For example, seeing if two samples have the same gene (Vaucheria and Elysia)
mechanism of polymerase chain reaction: PCR uses repeated cycles of heating and
cooling to make many copies of a specific region of DNA. First the temperature is
raised to near boiling (94℃), causing the double stranded DNA to separate, or
denature into two single strands. Then the temperature is decreased (45-65℃), and
short DNA sequences called primers bind/anneal to complementary matches on the
target DNA sequence. The primers bracket the target sequence to be copied. At a
slightly higher temperature (72℃), the enzyme Taq polymerase binds to the primed
sequences and adds nucleotides to create the second strand on each individual
strand, thus doubling the amount of DNA. This cycle is repeated many times to make
multiple copies of the sequence.
role of thermal cycling in polymerase chain reaction: The changes in temperature
allows the DNA to separate, the primers to anneal and the Taq polymerase to add
role of primers in polymerase chain reaction: The primers (one for the bottom and
top strand) are about 16-25 base pairs in length. They anneal to certain parts on the
DNA, thus specifying the region that will be multiplied.
role of Taq polymerase in