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Chapter 17: Meiosis
Meiosis is a process of generating haploid cells (gametic cells), cells with only one set
of chromosomes.
In humans, meiosis occurs only in the ovaries or testes.
Meiosis allows for sexual reproduction, that is the exchange of chromosomes from
two organisms.
Various types of sexual reproductive life cycles:
1. Animals
2. Most fungi and some alga
3. Plants and some alga
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Interphase I: DNA synthesis occurs. Each chromosome becomes replicated.
Meiosis I:
1. Prophase I: During synapsis, homologous chromosomes pair together in tetrads.
Portions where the homologous chromosomes are held close together (chiasmata,
pl. chiasma) can switch places.
o A phenomenon called crossing over takes place. Non-sister chromatids are
crossed to hold homologous pairs together, but sometimes, portions of the
chromosomes swap places.
2. Metaphase I: Homologous chromosome pairs align along the metaphase plate.
Kinetochore microtubules attach, extending from the centrosomes.
3. Anaphase I: Homologous chromosomes separate and move towards opposite
poles.
4. Telophase I and cytokinesis: Cleavage furrows or cell-wall formation occur to form
two daughter cells.
`Meiosis II
1. Prophase II: spindle apparatus forms; chromosome pairs move towards metaphase
plate
2. Metaphase II: Centromeres align on metaphase plate; kinetochore fibers attach
3. Anaphase II: sister chromatids separate; individual chromosomes move towards
outer pole
4. Telophase II and cytokinesis: cleavage furrow or cell-wall created; four daughter
haploid cells result
How does meiosis contribute to genetic variation?
1. Independent assortment: The homologous chromosomes can align in either
direction at metaphase I.
2. Crossing over: Non-sister chromatids “exchange” DNA information.
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Chapter 18: Genetics
Gregor Mendel, an Austrian monk, had too much time on his hands.
In 1843, he entered an Augustinian monastery, where after years of
studying religion, he studied peas. From 1851-1853. Mendel studied
at the University of Vienna with the physicist Doppler, who encouraged
that mathematics be used to study natural phenomena. Mendel also
studied with Unger, a famous botanist.
Mendel chose peas for their ease of use, rapid growth, and distinct
traits.
His experiments began with true-breeding plants. He hybridized two true-breeding
plants and charted the results of each character.
Typical monohybrid cross: looked at one character and checked for trait variations in
offspring
P-generation: true-breeding plants
F1-generation: first filial generation or hybrid generation
F2-generation: second filial generation or offspring of hybrid generation
The initial idea was that hybrids should be a “mix” of the true-breeding parents. This
notion was termed blending.
Guess for F1:
Purple flowers (x) white flowers  pale purple flowers
Result for F1:
Purple flowers (x) white flowers  purple flowers
Result for F2
F1 purple flowers (x) F1 purple flowers  705 purple & 224 white
3:1 ratio
What Mendel termed “heritable factor” we now call genes.
Mendelian genetics:
1. Alternative versions of genes (different alleles) account for variations in inherited
characters.
2. For each character, an organism inherits two alleles, one from each parent.
3. If the two alleles differ, then one, the dominant allele, is fully expressed in the
organism’s appearance; the other, the recessive allele, has no noticeable effect on
the organism’s appearance.
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4. The two alleles fro each character segregate during gamete production. “law of
segregation”
Homozygous: having identical alleles
 Homozygous recessive (rr)
 Homozygous dominant (RR)
Heterozygous: having two different alleles (Rr)
Dihybrid cross-tests: observing heritable patterns of two characters
Regular Mendelian genetics follows basic rules of inheritance.
 The presence of at least 1 dominant allele always produces the dominant trait.
 The presence of 2 recessive alleles always produces the recessive trait.
 Autosomal conditions: genes found on non-sex chromosomes (humans #1-44)
 Sex-linked conditions: genes found on sex-chromosomes (humans #45-46)
Incomplete dominance: The heterozygous genotype has a phenotype that appears to
be the blend of the two allelic forms.
Codominance: The heterozygous genotype has a phenotype of both allelic forms.
1. MN blood groups
 M and N are molecules found on red blood cells.
 can be MM, MN, or NN
2. Tay-Sachs disease: a disease that prevents production of enzymes to metabolize
gangliosides (brain lipids) and accumulation leads to brain degeneration
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

A person with Tt genotype can detect the enzyme deficiency and be fully
function. This means a person who is heterozygous also shows incomplete
dominance biochemically.
A person with Tt genotype can produce half-functioning enzymes and half nonfunctioning enzymes. This means a person who is heterozygous also shows
codominance molecularly.
Multiple alleles: More than 2 alleles exist to create the phenotype.
Pleiotropy: genes that affect multiple phenotypic
traits
Epistasis: gene at one locus that alters the
phenotype of another gene at a neighboring
locus
Quantitative characters: characters that vary
on a continuum, rather than an either/or situation
(as Mendel found in peas)
o Usually results from polygenic
inheritance (multiple genes affect one
phenotype)
o Example: skin coloration, height
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Characters are often multifactorial: requires many proteins to create one phenotype
* * * Pedigrees * * *
Pedigrees: a way to show genetic history in blood relatives
Widow’s peak = dominant
Straight hair line = recessive
* * * Disorders * * *
Disorders are any non-productive phenotypes. Disorders are not evenly distributed
among the various ethnic groups because of the geography (thus genetic) isolation.
Recessive disorders: harmful phenotypes from recessive alleles
 Affected individuals: homozygous recessive (rr)
 Carriers: heterozygous (Rr)
 Normal: homozygous dominant and heterozygous (RR and Rr)
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Example: cystic fibrosis (1:2500 European descendants affected; 1:25 carrier)
 The chloride ion channels are not fully functional in homozygous recessive
individuals. They are not capable of transporting chloride ions between cells or into
the extracellular fluid. Because of increased (chloride ion), the mucus coat on
certain cells becomes unusually thick and sticky. Mucus buildup leads to increase
infections because increased mucus levels disable some antiboties in the body.
Example: Tay-Sachs disease
Example: Sickle-cell anemia (1:10 African descendants carriers)
 Affected: homozygous recessive individuals
 Carriers with some evolutionary advantage: heterozygous
 Normal: homozygous dominant
Dominant disorders: harmful phenotypes from dominant allele
Lethal dominant alleles are less common than recessive alleles in the population. Why?
Example: achondroplasia (type of dwarfism)
Example: Huntington’s disease
 Affects individuals ages 35-45
 Degenerates nervous system
 Chromosome #4 may carry Huntington’s disease on a locus near the tipe.
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Ways to detect disorders:
1. using Mendelian crosses to show possible outcomes of diseases
2. using pedigrees
3. carrier recognition: determining
whether parents are
heterozygous carriers
4. fetal testing
 amniocentsis: Around the
14-16th week of pregnancy,
doctors remove about 10
mL of amniotic fluid. They
can culture cells and
karyotype the
chromosomes or detect for
the presence of any
chemicals.
 Chorionic villi sampling:
Within 24 hours of extracting
fetal tissue, cells are
cultured. Karyotypes can be
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

made of the chromosomes.
Ultrasound: uses sound waves to produce images
Fetoscopy: scope inserted into the uterus that transmits light
Karyotypes can reveal sex of offspring and any possibility of nondisjunction.
* * * The Mathematics of Genetic Probability * * *
Punnett squares: brute-force way of solving genetics problems
The purpose of using a Punnett square is to figure out the probability of two organisms
interbreeding to produce ____ trait.
Monohybrid test crosses can reveal the parental genotypes.
Purple flower (x) White flower
PP or Pp
pp
Two possible outcomes:
p
p
PP x pp
P
Pp
Pp
P
Pp
Pp
100% purple, Pp
p
p
Pp x pp
P
Pp
Pp
p
pp
pp
50% purple (Pp); 50% (pp)
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Common monohybrid ratios:
Parental Genotypes
AA x aa
Aa x Aa
Aa x aa
aa x aa
Offspring outcome
100% heterozygous genotype
100% dominant phenotype
25% AA, 50% Aa, 25% aa
1:2:1 genotype
3:1 phenotype
50% Aa, 50% aa
1:1 genotype
1:1 phenotype
100% recessive genotype
100% recessive phenotype
MULTIPLICATION RULE: determines the probability (chance) that two
independent events will occur together
Example 1: What is the probability that is I toss two coins, both will be heads if the odds for heads are 1
out of 2 (½)?
½x½=¼
Example 2: A plant that is Yy genotype is allowed to self-pollinate (Yy x Yy). What chance does the
offpsring have for being yy?

Because the law of independent assortment Yy will segregate into separate
gametes. Each parent has a ½ chance of donating a y gamete.
½x½=¼
Example 3: A plant that is Yy is crossed with a yy plant. What chances does the offspring have for
being Yy?


Yy parent has a ½ chance of donating y gamete.
yy parent has 2/2 chances of donating y gamete.
½x1=½
Example 4: A plant has a genotype YyRr. If this plant were allowed to self-pollinate,
what is the probability that the offspring will be YYRR?
 The parent has a ¼ chance of donating any of these gametes: YR, Yr, yR, yr.
 ¼ chance of YR from sperm
 ¼ chance of YR from ovum
¼ x ¼ = 1/16
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ADDITION RULE: determines the probability of events that occur in two of more ways
Example: Yy sperm is fertilized with a Yy egg. What is the probability that the resulting
offspring will be Yy?


½ Y from sperm and ½ y from egg  ¼ probability
½ Y from egg and ½ y from sperm  ¼ probability
¼+¼=½
Solving trihybrid crosses: PpYyRr x Ppyyrr
What is the probability that the offspring would have at least two recessive alleles?
 ppyyRr, ppYyrr, Ppyyrr, PPyyrr, ppyyrr
ppyyRr
 ¼ pp (x) ½ yy (x) ½ Rr = 1/16
Pp x Pp  ¼ pp
Yy x yy  ½ yy
Rr x rr  ½ Rr
ppYyrr
 ?
Ppyyrr
 ?
PPyyrr
 ?
ppyyrr
 ?
Add all the probabilities together. 3/8 offspring will have at least two recessive
phenotypes.
Practice problem: AaBbRr x AaBbRr
What are the probabilities that the offspring will display all dominant phenotypes?
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Chapter 19: Gene expression
* * * Transcription * * *
Transcription is the process of creating an RNA
(ribonucleic acid) strand from a DNA template.
DNA is a double helical structure, but only one of
the strands “codes” for the RNA.
Transcription is slightly different in eukaryotes and
prokaryotes.
Phases of transcription:
1. Initiation
2. Elongation (propagation)
3. Termination
PROKARYOTIC TRANSCRIPTION:
1. Initiation: process of “beginning” transcription
This is the rate limiting step. If initiation does
not take place, transcription cannot take place.
RNA polymerase is the enzyme that must bind
to DNA
RNA polymerase is a holoenzyme – made of
up the core enzyme (two alpha and two beta
subunits) and the sigma factor.
holoenzyme
The core enzyme has high affinity for DNA, but the sigma factor is not added to the
core enzyme until transcription is ready to occur.
RNA polymerase must bind to the promoter region. The promoter region is usually
a sequence of many thymine and adenine nucleotides (TATAAT) about 10
nucleotides before the start of transcription.
 TATA box or Pribnov box - The TATA box is said to have consensus (tightly
bound to RNA polymerase).
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 In front of the TATA box, about 35 nucleotides from the start of transcription, is
another consensus nucleotide sequence.
Together the TATA box and the consensus sequence form the promoter region.
Promoter
Operator
downstream
What happens if you mutate the promoter region?
o DNA topoisomerase I “unwinds” the DNA by introducing negative supercoils.
DNA gyrase also introduces negative supercoils. By this, they relax DNA to
form an open complex.
Once the DNA is opened, the core enzyme latches onto the DNA. The core enzyme
slides upstream and down stream of the DNA but cannot transcribe the DNA. This
is called promoter scanning.
Not until the sigma factor binds to the core enzyme does transcription begins.
Depending on how tightly the holoenzyme binds to the promoter region (strength)
directly correlates with how efficient transcription will be.
2. Elongation: process of adding nucleotides to make an RNA strand (only single
stranded)
RNA polymerase (holoenzyme) begins transcription adding A-U, C-G.
A = adenine
U = uracil
C = cytosine
G = guanine
RNA polymerase unwinds the DNA to create an RNA replication fork.
Problems:
 RNA polymerase cannot “fall off” of the DNA strand in the middle of transcription.
To ensure tight binding, NusA protein is used to help lock-in the tight binding.
 RNA polymerase must continually unwind at the head of transcription and rewind
the DNA after nucleotides are wound.
3. Termination: a process of ending transcription
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 Protein-independent: hair pin loops (stem loops, inverted repeats of G and
C)
 Protein-dependent: Rho-factor
Rho is a protein that binds to the RNA and travels behind the RNA
polymerase. Once rho catches up to the RNA polymerase, termination takes
place.
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EUKARYOTIC TRANSCRIPTION:
Eukaryotes have three different RNA polymerases:
1. RNA pol I: rRNA
2. RNA pol II: mRNA
3. RNA pol III cytoplasmic and small nuclear RNA
Eukaryotes also have more than 4 subunits in the core enzyme.
Transcription takes place in the nucleus.
Because DNA in eukaryotes is bound around protein histones (DNA + histone =
nucleosomes), for transcription to take place, the protein histones must be
disassembled or denatured.
Transcription in the nucleus produces precursor RNA.
Not until RNA processing is mRNA (messenger RNA) created and exported through
nuclear pores to the cytoplasm.
1. Initiation:
The TATA box is about 30 nucleotides upstream of the start.
There is a consensus GC rich nucleotide sequence at 40 and 110 nucleotides
upstream of start on DNA.
If the consensus is CT rich, then the sequence is fungi.
There are sequences called enhancers further upstream of the GC or CT rich
sequences.
The RNA polymerase holoenzyme binds to the promoter region (TATA box). A
TATA box binding protein (TBP) is often present.
2. Elongation: similar to prokaryotes, nucleotides added
3. Termination: similar to prokaryotes
RNA processing: The RNA is long and has a non-coding sequence on the 3’ end
called the trailer. There are introns which are noncoding sequences interspersed with
the coding RNA. The RNA made is very unstable.
The initial RNA transcript is called heterogeneous nuclear RNA (hnRNA). It is easily
degraded.
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A 5’-cap is added, and also a 3’-polyA tail is added.
CAP: guanyltransferase adds GTP molecules to add a guanine to the 5’ end of the
RNA. The guanine becomes methylated (add –CH3). The mRNA becomes heavily
methylated to stabilize the RNA.
Poly-A tail: polyadenylation (adding
adenine nucleotides to the 3’ end)
Small nuclear ribonucleoprotein
particles (snRNP or SNURP) add the
adenine nucleotides.
RNA can self-splice. It will remove the
RNA introns automatically. snRNP are
involved in intron removal in complexes
called spliceosome.
In eukaryotes, the RNA is “processed.”




Introns are removed
Exons are joined.
5’-cap is added.
Poly-A tail is added.
There are two genes that encode two
specific RNA strands that fold to become
ribosomal RNA and transfer RNA.
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* * * Translation: Protein Synthesis * * *
Translation is the process of
ribosomes forming peptide bonds
between amino acids, using
mRNA as the template. The
result is the production of
proteins.
Ribosomes are created from rRNA, ribosomal RNA. Two strands of rRNA fold to make
the large and the small subunit of the ribosomal complex.
Prokaryote: 70S = 30S + 50S
Eukaryote: 80S = 40S + 60S
Codon: sequence of three RNA nucleotides that correspond with an anticodon of a
transfer RNA molecule, carrying an amino acid
1. initiation:
 Transfer RNA activation: tRNA strands are “charged” with a corresponding amino
acid. GTP dephosphorylates to GDP, supply the necessary “energy” to fuel this
reaction.
 The small and large ribosomal subunits are separated by initiation factors
(proteins).
 f-met: formylmethionyl-tRNA binds to the smaller ribosomal subunit and together
bind to the initiator codon AUG.
 The larger subunit binds to the initiator complex, and translation begins. The
larger subunit has two “pockets” or regions named for their functions: P
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(transpeptidation) and A (acceptor) sites. The f-met-tRNA is positioned in the Psite.
2. Elongation:
 Elongation factors
locate the appropriate
t-RNA with the
matching anticodon of
the codon near the Asite.
 An enzyme called
peptidyl transferase to
the A-site tRNA
transfers the amino
acid of the t-RNA in the
P site. Peptide bonds
link the two amino
acids.
 Translocation: The
ribosome shifts such
that the A-site tRNA is
now in the P-site.
 The process repeats,
and the amino acid
chain grows.
 The process stops when the ribosome encounters a codon that indicates “stop.”
3. Termination:
 A water molecule dissociates the polypeptide chain from the t-RNA.
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* * * Operons * * *
Central dogma: Genes (DNA) code for the production of amino acids, which fold into
proteins and are expressed as the phenotype of an individual. Genes also code for the
production of protein enzymes.
Assuming no mutations, how is gene expression regulated?
1. Feedback inhibition: If too much gene is expressed and there is ample supply of
the necessary protein, the product of the metabolic pathway will inhibit its own
production.
Example: The cell produces cyclin at the G-checkpoints. If cyclin levels are high, the
cell will initiate the breakdown of its cyclin to slow down (regulate) the release of
MPF.
Example: All bacteria require tryptophan (an amino acid –trp) to survive. If the
bacteria does not find trp in the environment, it must synthesize its own trp. If too
much trp has been made, to conserve energy, the high [trp] signals the pathway to
stop.
2. Enzymatic inhibition: Cells can adjust for its own enzyme catalytic levels by
introducing allosteric or non-allosteric inhibition.
How organisms control gene expression:
Operons: transcription units that can consist of multiple genes (polycistronic) or a
single gene (monocistronic)
 Polycistronic operons (bacteria) are made up of a promoter region, an operator, and
the genes, which they control.
 They control genes by producing the necessary enzymes that allow or prevent
transcription.
Trp operon: a pathway for regulating
tryptophan production




The operator is always set to
“go.”. It allows for the RNA
polymerase holoenzyme to
precede transcription.
The operator is always on.
To turn the operator off, the
operon must have a
repressor protein.
The repressor protein is
made with a regulatory gene
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that is found upstream of the operator.
 Tryptophan acts as a corepressor to turn on the repressor protein.
 The trp operon is repressible
Lac operon: a pathway for breaking down lactose (disaccharide sugar) into galactose
and glucose for energy




The operator is always set to “stop.” It does not allow for the RNA polymerase
holoenzyme to proceed transcription.
A repressor is always “on” the operator.
An inducer molecule binds to the repressor to inactivate the repressor. The
inducer is allolactose. In the presence of lactose, allolactose binds to the
operator to turn it on.
The lac operon is inducible, which means that it must be induced to begin
transcription.
Positive regulation:
cAMP (cyclic AMP) builds
up in concentration when
glucose is absent. CRP
(cAMP regulatory
protein) binds to the
allosteric site of CRP.
This activates
transcription. The CRP+
cAMP binds to the
promoter region of the
DNA, this helps the RNA
polymerase to bind to the
promoter.
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* * * Mutation * * *
Mutations are any chances in the genetic makeup of an organism.
 Mutations do not lead to disorders in every situation.
Point mutations: changes to any
one nucleotide
 Substitutions – Polymerase
may add one nucleotide in place
of the correct one.
-Due to redundancy (wobble), the
mutation may not lead to a change
in the amino acid sequence. These
are called silent mutations.
 Insertions – frameshift
mutation; Because of the
addition of a nucleotide, the
ribosomes will read a different
sequence of codons than
intended.
 Deletions – frameshift mutation;
Because of the subtraction of a
nucleotide, the ribosomes will
read a different sequence of
codons than intended.
How do mutations occur?
1. Mistakes during DNA replication
2. Mistakes during recombination
(crossing-over)
3. Mistakes during transcription
4. Mistakes during translation
5. Mutations: chemical or physical substances that can cause DNA mutations
 X-rays
 UV  thymine dimers
 High energy radiation
 Base analogues – chemicals with similar chemical structure as DNA nucleotides
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Ames test: simple testing method developed by Bruce Ames to test for the mutagenic
activity of certain chemicals
How are DNA mutations corrected?
 Excision repair
 Removal of lesions
 Postreplication repair
 Recombinant repair
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Chapter 20: Biotechnology
Somatic nuclear transfer (nuclear
transplantation):
The nucleus is removed from a
differentiated somatic cell and transplanted
into an empty egg cell.
UV light is used to destroy the nucleus in
the egg cell.
The transplanted egg cell is forced (with
additions of hormones, etc) to undergo
embryonic development.
Dolly and Polly the sheep were both cloned
with this technology.
Recombinant plasmids: altering the
genetic makeup of plasmids (circular
pieces of bacterial DNA)
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Transformation: inserting pieces of bacterial DNA into a bacterium, altering its genetic
makeup
Phage libraries
Plasmid libraries:
cDNA libraries:
complementary DNA
libraries
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Polymerase chain reaction (PCR): technique for making DNA libraries (multiple
copies) using polymerase enzyme and thermal cycling
Gel electrophoresis in RFLP technology:
RFLP: Restriction fragment length polymorphism
Most of the genome consists of non-coding (“junk”) DNA. These junk sequences are
heritable in the same ways that coding DNA (genes) are. It is fair to say that two related
organisms should share similar sequences of junk DNA.
Restriction enzymes were isolated from bacteria, which had them as defense against
viruses. Viruses cleave their genetic material into the bacterial genome, and restriction
enzymes are used to cut out the viral DNA.
Restriction enzymes only recognize specific DNA sequences called palindromes.
Ex: 3’ – AGGCCT – 5’
5’ – TCCGGA – 3’
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The sequences read the same in a 3’ to 5’ direction. The enzyme would break the
phosphodiester bonds between the A and G, making a zig-zag cut, leaving “sticky
ends.”
If the palindrome sequence is CCGG the cut would be made between C and G.
GGCC
The resulting cut would leave “blunt ends.”
If two pieces of DNA are treated to the same restriction enzyme, chances are that they
would have palindrome sequences in nearly the same places in their non-coding DNA if
the DNA are related. The result of treating with restriction enzymes is the creation of
DNA fragments of various lengths (RFLPs). The DNA fragments of each sample are
loaded into the wells of an agarose gel. The gel is placed into an electrophoresis
chamber with a running buffer solution, and plugged into an electric source to generate
a voltage differential gradient. The electric charge separates the DNA bands in a
process called gel electrophoresis.
RFLP analysis:
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Sanger method:
Southern blotting:
What are ways in which these technologies can help human societies? (gene therapy?)
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