AP Bio
Ch.18 “Genetics of Viruses and Bacteria”
The Genetics of Viruses
Objectives:
1. Explain why viruses are obligate intracellular
parasites.
They can reproduce only within a host cell
Contain no:

Metabolic enzymes

ribosomes

Other equipment for making proteins
5.
Explain how a virus identifies its host cell.

Recognition systems evolved in the virus

Lock-and-Key” fit between proteins on the outside of
the virus and specific receptor molecules on the
surface of cells.

Some viruses have broad host range
o West Nile virus: mosquitoes, birds and
humans

Some have only one species
Poliovirus, measles infect humans only
6.
Describe bacterial defenses against phages.

Natural selection favors bacterial mutants with
receptor sites that are no longer recognized by a
particular type of phage

Restriction enzymes (restriction endonucleases) chop
up foreign DNA (bacterial DNA is chemically modified)

Lysogenic cycle: coexist with host
7.
Distinguish between the lytic and lysogenic reproductive
cycles, using phage lambda as an example.
Lytic: (T4) fig. 18.6

attachment

phage DNA injected

Phage gene makes enzyme to degrade host DNA

Viral genome transcribed using host enzymes

Self-Assembly of viral proteins & DNA

Last phage gene makes enzyme to damage bacterial
cell wall, fluid enters, cell bursts, releasing 100-200
new viruses.
Lysogenic: (Phage - lambda) fig. 18.7

Attachment

Phage DNA injected





 DNA circularizes
Phage DNA integrates into host DNA (prophage)
Phage gene makes protein, keeps other phage genes
silent.
Bacterium reproduces by binary fission (prophage
copied)
Prophage might exit bacterial chromosome, initiating
lytic cycle.
Temperate Viruses: Phages using both cycles
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8.
Describe the reproductive cycle of an enveloped virus.
Fig. 18.8
Nearly all animal viruses with RNA genomes have an
envelope, few bacteriophages do.

Glycoproteins on the viral envelope bind to specific
receptor molecules on the host cell, promoting viral
entry into the cell.

The capsid and viral genome enter the cell. Digestion
of the capsid by cellular enzymes releases the viral
genome.

The viral genome functions as a template for
synthesis of complementary RNA strands by a viral
enzyme.

New copies of viral genome RNA are made using
complementary RNA strands as templates.

Complementary RNA strands also function as mRNA,
which is translated into both capsid proteins (cytosol)
and glycoproteins (ER) for the viral envelope.

Vesicles transport envelope glycoproteins to the
plasma membrane

A capsid assembles around each viral genome
molecule.

Each new virus buds from the cell, its envelope
studded with viral glycoproteins embedded in
membrane derived from the ER.
Explain the reproductive cycle of the herpesvirus.
Fig. 18.9, 10

Envelope derived from the nuclear membrane of the
host

Have double-stranded DNA genome

Reproduce within the host cell nucleus using combo
of viral and cellular enzymes to replicate and
transcribe their DNA.

Minichromosomes in nuclei of nerve cells are copies
of herpesvirus DNA

Remain latent until physical or emotional stress
triggers new round of active viruses (blisters).
9.
Describe the reproductive cycle of retroviruses.
Fig. 18.10
Come equipped with enzyme reverse transcriptase
transcribes RNA into DNA.
Opposite of central paradigm (DNARNA Protein)
Retroviruses: HIV (human immunodeficiency virus causes
AIDS: acquired immunodeficiency syndrome)

All retroviruses contain two identical single-stranded RNA
and two molecules of reverse transcriptase.

After entry, reverse transcriptase synthesizes two strands
of complementary DNA.

Viral DNA enters cell’s nucleus and integrates into the
DNA of a chromosome.

Now called a provirus, permanent.

Viral proteins made: capsid, rev. transc.(cytoplasm),
envelope glycoproteins (ER).

Vesicles transport the glycoproteins from the ER to the
cell’s plasma membrane.

Capsids are assembled around viral genomes and rev.
trnsc.

New viruses bud off from host cell.
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10. List some characteristics that viruses share with
living organisms and explain why viruses do not fit
our usual definition of life.

Biologically inert

unable to replicate its genes (no ribosomes, enzymes)

unable to regenerate its own supply of ATP

unable to reproduce independently
11. Describe the evidence that viruses probably evolved
from fragments of cellular nucleic acids.

Depend on cells so evolved after cells

May have evolved several times

Viral genomes have more in common with host
genome than other viruses

Some viral genes are essentially identical to genes of
the host
12. Define and describe mobile genetic elements.
Possible Original sources of viral genomes
Plasmids

Small circular DNA molecules

Found in bacteria and yeasts

Exist apart from cell’s genome

Can replicate independently of genome

Occasionally transferred between cells
Transposons

DNA segments that can move from one location to
another within a cell’s genome
Mobile genetic elements

Plasmids

Transposons

Viruses
13. Describe the best current medical defenses against
viruses.

Antibiotics kill bacteria by inhibiting enzyme-catalyzed
processes specific to the pathogen; viruses don’t have
their own enzymes.

Vaccines are harmless variants of pathogenic microbes
that stimulate the immune system to mount defenses
against the actual pathogen or virus. (Viral flu vaccine,
smallpox, chicken pox.)

Although vaccines can prevent certain viral illnesses,
medical technology can do little, at present, to cure most
viral infections once they occur.

Antiviral drugs resemble nucleosides (pentose sugar plus
nitrogen base, like adenosine in ATP) and as a result,
interfere with viral nucleic acid synthesis.

Currently, multidrug treatments, or cocktails, have been
found to be most effective. These are combinations of two
nucleoside mimics and a protease inhibitor, which
interferes with an enzyme required for assembly of virus
particles.
Explain how AZT helps to fight HIV infections.

AZT (azidothymidine) curbs HIV reproduction by
interfering with the synthesis of DNA by reverse
transcriptase.
 Acyclovir impedes herpesvirus reproduction by
inhibiting viral polymerase that synthesizes viral DNA
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The Genetics of Bacteria
24. Describe the significance of R plasmids.

R plasmids confer resistance to antibiotics for the bacteria
that carry them. They can be transmitted thru sex pili to
other bacteria.
24. Explain how the widespread use of antibiotics
contributes to R plasmid-related disease.

Exposure of bacteria to antibiotics will kill all susceptible
bacteria but not those that happen to have R plasmids
with genes that counter the antibiotic.

Natural selection predicts that the small fraction of
resistant bacteria that survive will go on to repopulate the
organism or environment.

Many R plasmids have genes that encode sex pili and
enable plasmid transfer from one bacterial cell to another
by conjugation.

Some R plasmids carry as many as ten genes for
resistance to that many antibiotics.
30. Using the trp operon as an example, explain the
concept of an operon and the function of the operator,
repressor, and corepressor. Fig. 18.21
Operon: A unit of genetic function common in bacteria
and phages, consisting of coordinately regulated clusters
of genes with related functions.
Operator: a master switch that controls a cluster of
functionally related genes.
Repressor: A protein that suppresses the transcription of
a gene. Binds the operator to turn off the operon.
Corepressor: A small molecule that cooperates with a
repressor protein to switch an operon off.

The trp operon contains 5 genes needed to
synthesize (anabolic) tryptophan.

The trp operon: regulated synthesis of repressible
enzymes.

In the presence of tryptophan, it acts as a
corepressor, binding the repressor and turning the
operon off.

In the absence of tryptophan (the corepressor), the
repressor is inactive, the operon turns on and the
enzymes to synthesize tryptophan are made.
32. Describe how the lac operon functions and explain the
role of the inducer, allolactose. Fig. 18.22

The lac operon contains 3 genes needed to
metabolize (catabolic) lactose.

The lac operon: regulated synthesis of inducible
enzymes.

Lactose absent, repressor active, binds operator,
turns operon off.

The lac repressor is innately active, and in the
absence of lactose it switches off the operon by
binding to the operator.

Lactose present, repressor inactive, operon on.

Allolactose, an isomer of lactose, derepresses the
operon by inactivating the repressor, The enzymes
for lactose utilization are induced. Allolactose is the
inducer.
In summation:
Trp (anabolic): repressor + corepressor (tryptophan) =
active repressor
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Lac (catabolic): repressor + inducer (allolactose) =
inactive repressor
34. Distinguish between positive control by CAP (fig.
18.23) and negative control by the lac repressor (fig.
18.22a) and give examples of each from the lac
operon.




Bacteria preferentially use glucose for glycolysis
instead of lactose. How can they detect glucose
concentration?
Mechanism depends of the interaction of an allosteric
regulatory protein with a small organic molecule
Small organic molecule: cyclic AMP (cAMP) which
accumulates when glucose is scarce.
The regulatory protein, called catabolite activator
protein (CAP), is an activator of transcription.
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Ch. 19 “Eukaryotic Genomes:
Organization, Regulation, and
Evolution.
1. Compare the structure and organization of
prokaryotic and eukaryotic genomes.
Genomes:
shape
Challenges expressing
genes
Prokaryotic
circular
same
Size of genome
small
Specialization of cells
NA
Noncoding DNA
none
DNA organization
Some
organization
using DNAProtein
complexes
Higher structural levels
none
Must alter patterns of
gene expression in
response to changes in
environmental
conditions.
Different cell types
5x greater (25,000
genes)
Crucial in
multicellular
organisms
Enormous amount
Elaborately
organized using
DNA-Protein
(histones)
complexes called
chromatin
Several levels of
supercoiling
same
One type
Regulation of genome
expression occurs at …
Post-transcriptional
control?
Eukaryotic
linear
Several types with one
genome, different
subset of genome
expressed
transcription
No
Yes, nuclear envelope
separates nucleolus
from cytoplasm
2. Describe the current model for progressive levels of
DNA packing in eukaryotes. Fig. 19.2
a. DNA double helix
b. Nucleosomes: DNA (-) + histone proteins (+) = “beads
on a string.”
c. 30-nm fiber: nucleosomes coil to form chromatin fiber
d. 300-nm fiber: looped domains: loops attached to nonhistone protein scaffold. May attach to nuclear lamina
for organization.
e. metaphase chromosome: 1,400-nm fiber
Activity 19.1 “DNA Packing”
3. Explain how histones influence folding in eukaryotic
DNA.
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Association of DNA and histones remains essentially intact
throughout the cell cycle.
Histones:

leave transiently during replication

stay during transcription

changes in shapes and positions of nucleosomes can
allow RNA-synthesizing polymerases to move along
the DNA.

Next level of packing

Interactions between the histone tails of one
nucleosome and linker DNA and nucleosomes to either
side.

Causes extended DNA to fold

Chromatin of each chromosome occupies a specific
restricted area within the interphase nucleus, attach to
nuclear lamina, and the chromatin fibers of different
chromosomes do not become entangled.
2.
Distinguish between heterochromatin and
euchromatin.
Interphase chromatin:

is much less condensed than mitotic chromatin.

Shows several of same levels of higher-order
packing.

Centromeres and telomeres exist in a highly
condensed state similar to a metaphase
chromosome.

Heterochromatin: visible as irregular clumps with a
light microscope.

Euchromatin (“true chromatin”) less compacted for
transcription.

Dense packing of DNA in chromosomes prevents
gene expression by not leaving room for enzymes.
THE CONTROL OF GENE EXPRESSION
Activity: Overview: Control of Gene Expression”
5. Explain the relationship between differentiation and
differential gene expression.
Cell differentiation: during development of a multicellular
organism, its cells undergo a process of specialization in form
and function.

Typical cell expresses about 20% of its genes at any
given time.

Muscle cells, highly differentiated, express even less.

All cells have same genome. (Immune cells are an
exception)

The subsets of genes expressed in the cells of each
type is unique, allowing these cells to carry out their
specific function.

Only 1.5% of DNA in humans codes for protein.

Small fraction codes for RNA

Most of the rest is noncoding (RNA of unknown
function?)
Differential gene expression: The differences between cell
types are due not to different genes being present, but to the
expression of different genes by cells with the same genome.
6. Describe at what level gene expression is generally
controlled. Fig. 19.3 Other examples are places where
expression can be turned on or off, accelerated, or slowed
down. Activity: “Control of Transcription”, “PostTranscriptional Control”

Most important level: transcription initiation (Prok &
Euk)
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





Chromatin modification incl. histone acetylation and
DNA demethylation
Translation: DNA to RNA
RNA processing
Transport of RNA to cytoplasm
Peptide Cleavage: chemical modification, transport to
cellular destination
Degradation of protein
Important control points in eukaryotic gene
expression Obj. 7-9.
Chemical modifications of histones and DNA of chromatin
influences both chromatin structure and gene expression.
7. Explain how histone acetylation and DNA methylation
affect chromatin structure and the regulation of
transcription. Fig. 19.4
Histone acetylation: acetyl groups (-COCH3) are attached to
+ charged lysines in histone tails; deacetylation is the removal
of acetyl groups.
Positive charges on lysines are neutralized, they no longer
bind to neighboring nucleosomes. Therefore, less compaction.

Transcription proteins have easier access to genes in
acetylated regions.

Enzymes that acetylate or deacetylate histones are closely
associated with, or components of, the transcription
factors that bind to promoters.

Acetylation enzymes:
o Modify chromatin structure
o Recruit components of transcription
machinery.
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Histone Code Hypothesis: specific combinations of
modifications, rather than the overall level of histone
acetylation, help determine the chromatin configuration, which
in turn influences transcription.
DNA Methylation: turns genes off.
Examples:

Most plants and animals have methylated bases
(usually cytosine)

Barr bodies or X inactivation highly methylated

Cells in which genes are not expressed (due to
specialization)
Other interesting facts:

Proteins that bind methylated DNA recruit histone
deacetylation enzymes. Dual mechanism represses
transcription.

Essential during development to turn off certain genes
for cell specialization.

Once methylated, genes stay that way through
successive cell divisions.
During replication, where one strand is methylated, enzymes
correctly methylate the daughter strand after each round of
replication.

Therefore, methylation patterns are inherited.

Genomic Imprinting in mammals, where methylation
permanently regulates expression of either the
maternal or paternal allele of certain genes at the start
of development.
8. Define epigenetic inheritance.
Epigenetic Inheritance: Inheritance of traits transmitted by
mechanisms not directly involving the nucleotide sequence.
Enzymes that modify chromatin structure appear to be integral
parts of the cell’s machinery for regulating transcription.
SKIP, SKIP, SKIP to my lou…
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14. Describe the process and significance of alternative
RNA splicing.
Alternative RNA splicing: different mRNA molecules are
produced from the same primary transcript, depending on
which RNA segments are treated as exons and which as
introns.
Regulatory proteins specific to a cell type control intron-exon
choices by binding to regulatory sequences within the primary
transcript. Fig. 19.8
15. Describe factors that influence the life span of mRNA
in the cytoplasm. Compare the longevity of mRNA in
prokaryotes and in eukaryotes.

Life span of mRNA important in determining the pattern of
protein synthesis in a cell.

Prok. mRNA usually degrades within a few minutes of
their synthesis. Allows for quick changes to environmental
conditions. (adaptive)

Euk. mRNA hours, days, even weeks.

Process (at least in yeast):
1. Shorten poly-A tail
2. Triggers enzymes that remove 5’cap
3. Removal regulated by nucleotide sequences in the
mRNA (found in 3’ UTR of molecule)
4. Nuclease enzymes rapidly chew up the mRNA
Factors:

nucleotide sequences in the mRNA (found in 3’ UTR of
molecule)

microRNAs
o precursor folds back on itself (hairpin)
o Dicer cuts it into shorter segments
o One strand of each short double-stranded
RNA is degraded; the other strand (miRNA)
then associates with a cmplex of proteins.
o Protein-bound miRNA can base-pair with
any target mRNA that contains the
complementary sequence.
o Prevents gene expression either by
degrading target mRNA or by blocking its
translation.

siRNAs: small interfering RNAs.
o Similar to miRNA
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16. Explain how gene expression may be controlled at the
translational and post-translational level.
Activity: “Review: Control of Gene Expression”
THE MOLECULAR BIOLOGY OF CANCER
17. Distinguish between proto-oncogenes and
oncogenes. Describe three genetic changes that can
convert proto-oncogenes into oncogenes.
Proto-oncogenes: normal genes that code for proteins that
stimulate normal cell growth and division.
Oncogene: cancer-causing genes that are created by:

Movement of DNA within the genome

Amplification of a proto-oncogene (multiple copies)

Point mutations in a control element or in the protooncogen itself. Fig. 19.11
18. Explain how mutations in tumor-suppressor genes
can contribute to cancer.
Tumor-suppressor genes: normal products inhibit cell
division.

Functions of tumor-suppressor proteins:

Repair damaged DNA

Control the adhesion of cells to each other or to the
extracellular matrix. Normal anchorage is crucial to
normal cells and often absent in cancers.

Components of cell-signaling pathways that inhibit the
cell cycle.
Cancer can be categorized into two groups based on their
normal effects on the cell cycle:
“Accelerators”: ras stimulates the cell cycle (i.e. cell
division).
“Brakes”: p53 inhibits the cell cycle.
19. Explain how excessive cell division can result from
mutations in the ras proto-oncogenes.
“ACCELERATORS”
Mutations in ras occur in about 30% of human cancers.
Normal pathway:



Ras gene
ras protein
messenger of signal from plasma membrane protein
to a cascade of protein kinases

stimulates cell growth.
Hyperactive ras gene (product of oncogene) issues signals on
its own. Overstimulates cell growth.
CELL-CYCLE STIMULATING PATHWAY: RAS
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20. Explain why a mutation knocking out the p53 gene can
lead to excessive cell growth and cancer. “BRAKES”
Mutations in p53 occur in about 50% of human cancers.
P53 a.k.a. “The Guardian Angel”
Normal Pathway:

DNA damage signals 

protein kinases

leads to activation of p53 gene

protein that inhibits cell division.
Errors in p53 will result in a loss of “brakes” and the cell can
replicate errors, like overactive ras errors.
21. Describe three ways that p53 prevents a cell from
passing on mutations caused by DNA damage.
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
Allows time for DNA repair.

Can turn on proteins that repair DNA.

May trigger apoptosis (cell death)
Ensures mistakes are not replicated.
22. Describe the set of genetic factors typically
associated with the development of cancer.
Changes in a tumor parallel a series of genetic
changes:
Colorectal cancer, best understood.

Loss of tumor-suppressor gene (APC or other: small
benign growth (polyp)

Activation of ras oncogen

Loss of tumor-suppressor gene (ex. DCC: larger
benign tumor: adenoma)

Loss of tumor-suppressor gene p53

Additional mutations (malignant tumor: carcinoma)
23. Explain how viruses can cause cancer. Describe
several examples.
Viruses play a role in about 15% of human cancer cases
worldwide.
Viruses integrate their DNA into host DNA. Therefore:

A retrovirus may donate an oncogene to the cell.

May disrupts a tumor-suppressor gene, inactivating it.

Convert a proto-oncogene to an oncogene by adding
a promoter.

Some viruses produce proteins that inactivate p53
and other tumor-suppressor proteins.
24. Explain how inherited cancer alleles can lead to a
predisposition to certain cancers.
Multiple genetic changes are required to produce a cancer cell.
This helps explain the observation that certain cancers run
in some families.
An individual inheriting an oncogene or a mutant allele of a
tumor-suppressor gene is one step closer to accumulating
the necessary mutations for cancer to develop than is an
individual without any such mutations.
15% of colorectal cancers are inherited.
5-10% of breast cancer patients had a strong inherited
predisposition.
BRCA-1, BRCA-2 (BReast CAncer) Mutations in these genes
are found in at least half of inherited breast cancers.
Inheriting one faulty allele increases risk for breast cancer from
2% to 60%.
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GENOME ORGANIZATION AT THE DNA
LEVEL
25. Describe the structure and functions of the portions of
eukaryotic DNA that do not encode protein or RNA.
Genome organization tells us much about how genomes have
evolved and continue to evolve.
Exons: expressed as protein, rRNA, or tRNA.
Introns and regulatory sequences:
Unique noncoding DNA:
Repetitive DNA:
Repetitive DNA (incl. transposable elements and Alu):
Transposable elements, “jumping genes” discovered by
Barbara McClintock, in maize (Indian corn).

Transposons: move within genome by means of
a DNA intermediate

Retrotransposons: use RNA intermediate
Many code for protein but do not carry out normal cellular
functions. Therefore, described as “non-coding DNA.”
Alu Elements:
Family of similar sequences
10% of genome
Much shorter than functional transposable elements (300bp)
Do not code for any protein.
Many Alu elements are transcribed into RNA molecules;
function unknown.
Repetitive DNA

Mistakes in replication or recombination

Large-segment duplications: long stretch copied from
one chromosomal location to another, on the same or
a different chromosome
Simple sequence DNA (aka: satellite DNA)

Contains many copies of tandemly repeated short
sequences.

Example: GTTAC repeated 3-several hundred
thousand times.

Centrifuges into its own band, hence: satellite DNA

May be structural: heavy in telomeres and
centromeres
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29. Using the genes for rRNA as an example, explain how
multigene families of identical genes can be
advantageous for a cell.
Collections of identical or very similar genes.
Usually clustered in tandem
Most code for RNA (except histone genes)
Example:

rRNA genes (3) encoded in a single transcription unit that
is repeated tandemly hundreds to thousands of times

Helps cells to quickly make the millions of ribosomes
needed for active protein synthesis.

Primary transcript is cleaved to yield the three rRNA
molecules.

Combine with protein and (5srRNA) to form ribosomal
subunits.
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Ch. 20 DNA Technology
1. Explain how advances in recombinant DNA technology
have helped scientists study the eukaryotic genome.
Genetic Engineering: The direct manipulation of genes for
practical purposes.
Examples:

Manufacture of hundreds of protein products
(hormones and blood-clotting factors)

Recombinant DNA can be made then introduced into
cultured cells that replicate the DNA and express its
genes, yielding a desired protein.
Biotechnology: the manipulation of organisms or their
components to make useful products.
Examples:

Using microbes to make wine and cheese

Selective breeding of livestock, crops, flowers.

Both use naturally occurring mutations and genetic
recombinations.

In vitro manipulation is modern and enables
modification of specific genes and to move them
between organisms as distinct as bacteria, plants and
animals.
2. Describe the natural function of restriction enzymes
and explain how they are used in recombinant DNA
technology.
Restriction Endonucleases: In nature, bacteria protect
themselves from invasion by viral bacteriophages (“bacteria
eaters”) or other bacteria by cutting up foreign DNA, a process
called “restriction.” These enzymes cut DNA molecules at a
limited number of locations.
Gene cloning and genetic engineering were made possible by
discovery of restriction enzymes. They gave scientists a pair of
microscopic scissors that cut at specific sites consistently.
3. Explain how the creation of sticky ends
by restriction enzymes is useful in
producing a recombinant DNA molecule.

Example: EcoR1, cuts DNA at a
specific 6-base sequence.

Creates “sticky ends” that will anneal
easily with complementary sticky ends.

Use same restriction enzyme with
different DNA.

Mix together.

Any complementary bases across
sticky ends will join

Forms recombinant DNA.
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4. Outline the procedures for cloning a eukaryotic gene in
a bacterial plasmid.
Cloned genes are useful for two main purposes:

to make many copies of a particular gene

to produce a protein product
Examples of uses:

sequencing the gene

producing the protein

gene therapy

basic research
5. Describe techniques that allow identification of
recombinant cells that have taken up a gene of interest.

ampR

restriction enzyme cuts lacZ gene, if gene is taken in
there, (recombined) the X-gal will not be hydrolyzed,
white colonies.

Non-recombined plasmids, colonies will be blue; they
hydrolyze X-gal.
6. Define and distinguish between genomic libraries using
plasmids, phages, and cDNA.
Genomic Libraries: A set of thousands of
DNA segments from a genome, each carried by
a plasmid, or other cloning vectors. (Each
plasmid or virus is analogous to a book in the
library.)
“Other vectors” include

viral DNA

cDNA
cDNA: contains only a subset of genes that
were transcribed into mRNA in the original cell.
One advantage of a plasmid or viral vector over
cDNA is that they contain regulatory sequences and introns
which mRNA and therefore, cDNA, would not.
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cDNA or complementary DNA
Procedure:

extract mRNA from cell of interest

reverse transcriptase used in vitro to make singlestranded DNA (transcripts of mRNA).

Get rid of mRNA by enzyme degradation

Add DNA polymerase to add the second strand to
single-stranded DNA.

Restriction Enzyme recognition sequence at each end

Insert cDNA into vector DNA to create library
Creates “stripped-down” versions of genes.
Good for:

When you don’t want whole genome, just mRNA or
protein.

studying genes from specific cell types

following changes in pattern of gene expression thru
development.
7. Describe the role of an expression vector.

To overcome differences between prok/euk gene
expression (DNA control sequences like promoters)
expression vectors are used.

Contains a highly active prokaryotic promoter just
upstream of a restriction site where the eukaryotic
gene can be inserted in the correct reading frame.

Euk have introns, prok don’t have introns and do not
have RNA-splicing machinery. Use cDNA, includes
only exons. Just add bacterial promoter and other
control elements.
8. Describe two advantages of using yeast cells instead of
bacteria as hosts for cloning or expressing eukaryotic
genes.
1. They are just as easy to grow as bacteria

Have plasmids (rarity in euk). Will accept
recombinant plasmids and express the genes.

YAC: Yeast Artificial Chromosomes will behave
normally during mitosis, replicating foreign DNA.

Can carry a much longer DNA segment than
plasmids

More likely to contain an entire gene.
2. Eukaryotic proteins will not function unless they are
modified after translation like adding carbs and lipids
9. Describe two techniques to introduce recombinant DNA
into eukaryotic cells.

Electroporation: brief electrical pulse creates
temporary holes in plasma membrane

Injection

Plants: Agrobacterium (soil bacterium)
10. Describe the polymerase chain reaction (PCR) and
explain the advantages and limitations of this procedure.

ADVANTAGES: A small quantity of DNA can be
amplified, in vitro, to a useable amount. (To be
sequenced or introduced into a vector: yeast, plasmid,
bacterium)

LIMITATIONS: Occasional errors during PCR replication
imposes limits of the # of good copies that can be made.
Procedure for PCR:
Starting materials:

double-stranded DNA containing the target nucleotide
sequence to be copied
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


Heat resistant DNA polymerase
All four nucleotides
Two short, single-stranded DNA molecules (primers)
One primer is complementary to one end, the other
primer is complementary to the other end.
During each PCR cycle the samples are warmed, then cooled.

Each cycle will double the target DNA sequence.

After about 20 cycles, target sequence DNA
increased a billion-fold.
11. Explain how gel electrophoresis is used to analyze
nucleic acids and to distinguish between two alleles of a
gene. P.393
12. Describe the process of nucleic acid hybridization.
Once you have made a genomic library with bacteria, how can
you tell which are for the gene of interest?
The key is the radiolabeled probe DNA that will anneal with its
complementary DNA (a segment of the target gene).
13. Describe the Southern blotting procedure and explain
how it can be used to detect and analyze instances of
restriction fragment length polymorphism (RFLP).
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The Southern blot was named after the inventor’s name of
Southern. Subsequent procedures for similar blotting
techniques were named: Northern, Western, Eastern.
Fig. 20.10 Shows how to detect specific nucleotide sequences
within different samples of genomic DNA.
The example given compares three individual with different
genotypes:

homozygous normal hemoglobin

homozygous sickle-cell hemoglobin

heterozygous (one normal allele, one sickle-cell
allele.)
14. Explain how RFLP analysis facilitated the process of
genomic mapping.
Cytogenetic mapping: created starting point for genomic
mapping.
Karyotyping determines # of chromosomes and banding
pattern
FISH: Fluorescence in situ hybridization of whole
chromosomes
Linkage Mapping: Ordering of genetic markers such as
RFLPs, simple sequence DNA, and other polymorphisms
(about 200 per chromosome. Based on recombination
frequencies. Resulted in map with 5,000 markers.
Physical Mapping: Made by cutting the DNA of each
chromosome into a number of restriction fragments and then
determining the original order of the fragments.

The key is to make fragments that overlap and then
use probes to find the overlaps.
First round uses:

YAC (yeast artificial chromosome) and

BAC (bacterial artificial chromosome)

because they can be so large.
After these long fragments are ordered,
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



each fragment is cut into smaller pieces
which are cloned in plasmids or phages
ordered in turn, and finally
sequenced.
DNA Analysis and Genomics
15. Explain the goals of the Human Genome Project.
1990-2003

Read out and make freely available the complete
DNA sequence of the human being.

To gain a comprehensive picture of how the genes
are turned on and off in different tissues and at
different times, so we can see how genes collaborate
in modules or circuits.

It’s an ensemble picture that we’ve never had before.
Genomes Sequenced:
E. coli:
Saccharomyces cerevisiae:
Caenorhabditis elegans:
Drosophila melanogaster:
Mus musculus:
Hemophilus influenzae:
Homo sapiens:
enteric bacteria
yeast
nematode
fruit fly
mouse
flu
humans
16. Explain how linkage mapping, physical mapping, and
DNA sequencing each contributed to the genome mapping
project.
These were the three successive stages of the HGP.
Linkage Mapping: Ordering of genetic markers such as
RFLPs, simple sequence DNA, and other polymorphisms
(about 200 per chromosome. Based on recombination
frequencies. Resulted in map with 5,000 markers.
Physical Mapping: Made by cutting the DNA of each
chromosome into a number of restriction fragments and then
determining the original order of the fragments.

The key is to make fragments that overlap and then
use probes to find the overlaps.
First round uses:

YAC (yeast artificial chromosome) and

BAC (bacterial artificial chromosome)

because they can be so large.
After these long fragments are ordered,

each fragment is cut into smaller
pieces

which are cloned in plasmids or
phages

ordered in turn, and finally

sequenced.
17. Describe the alternate approach to
whole-genome sequencing pursued by J.
Craig Venter and the Celera Genomics
Company.
Skip genetic mapping and physical mapping
stages and start directly with the sequencing of
random DNA fragments.
Powerful computer programs would then
assemble the resulting very large number of
overlapping short sequences into a single
continuous sequence.
18. Explain how researchers recognize
protein-coding genes within DNA sequences.

DNA sequences are collected in computer banks
available on the internet.
Software scans sequences for:

transcriptional and translational start and stop signals

RNA-splicing sites,
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


Other telltale signs of protein-coding genes
Short coding sequences similar to those present in
known genes. (ESTs: expressed sequence tags) are
cataloged in computer databases.
Identifies gene candidates (putative genes)
19. Describe the surprising results of the Human Genome
Project.

Genome size generally increases from prok to euk, it
does always correlate with biological complexity
among eukaryotes.

The number of genes an organism has is often lower
than expected from the size of its genome.

Human genome is approx. 25,000 genes, much lower
than the 50,000 – 100,000 expected

Human genome is only 1.5x the number found in the
fruit fly and the nematode worm.

Relative to other organisms studied, human genes
account for a much smaller fraction of the human
genome.
20. Explain how the vertebrate genome, including that of
humans, generates greater diversity than the genomes of
invertebrate organisms.

Gene expression is regulated in more subtle and
complicated ways in vertebrates than in other
organism.

Non-coding DNA may function in these regulatory
mechanisms.

Alternative splicing of RNA transcripts

Nearly all human genes contain multiple exons.
o About 75% of these multi-exon genes are
alternatively spliced.
o Assume each alternatively spliced gene
specifies 3 different polypeptides
o Total # of different polypeptides = 75,000
added to

post-transcriptional cleavage or addition of carb
groups to increase variation of polypeptides.

Much larger number of possible interactions between
gene products that result from greater polypeptide
diversity
23. Define and compare the fields of proteomics and
genomics.
Genomics: The systematic study of full sets of genes and
their interactions.
Proteomics: the systematic study of full protein sets
(proteomes) encoded by genomes.

Because proteins, not genes, actually carry out the
activities of the cell, we must study when and where they
are produced in an organism, and also how they interact, if
we are to understand the functioning of cells and
organisms.

A first step in systems-biology approach is defining gene
circuits and protein interaction networks.

With the use of computer science and math to process
and integrate vast amounts of biological data, researchers
can detect and quantify the many combinations of
interactions.
24. Explain the significance of single nucleotide
polymorphisms in the study of human evolution.

Much of human diversity is in single nucleotide
polymorphisms (SNP’s or snips).

These occur about once in 1,000 bp.
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




Comparing your personal gene sequence with someone of
same gender you would be 99.9% identical.
SNP’s make useful genetic markers in studying human
evolution
Such as migratory routes
ID disease genes and genes that affect our health in more
subtle ways.
Likely to change practice of medicine in 21st century.
PRACTICAL APPLICATIONS OF DNA
TECHNOLOGY
25. Describe how DNA technology can have medical
applications in such areas as the diagnosis of genetic
disease, the development of gene therapy, vaccine
production, and the development of pharmaceutical
products.
diagnosis of genetic disease:

Use of PCR and labeled nucleic acid probes to detect
RNA of disease in blood using RT-PCR (take RNA +
o
reverse transcriptasecDNA then PCR and probe for
HIV sequence.
Examples of diseases that can be diagnosed this way:
o Sickle-cell disease
o Hemophilia
o Cystic fibrosis
o Huntington’s disease
o Duchennne muscular dystrophy
o
RFLP analysis: to find genetic markers with diseasecausing alleles that are different than normal allele.
gene therapy:
The alteration of an afflicted individual’s genes with normal
genes.

Cells receiving normal allele would need to reproduce
throughout lifetime of individual. Bone marrow cells
(include stem cells for blood and immune system) are
prime candidate. Fig. 20.16 p. 403

Challenge: DNA receiving vector DNA may insert near
gene involved in proliferation and development of blood
cells, causing leukemia.

Worry: eugenics, some researchers see no difference
between gene therapy and organ transplant.
vaccine production:

Two types: Inactivated, Attenuated

Both present viral surface proteins to immune system
to prime it for the next time it comes in contact.

The protein alone could be created using DNA
techniques.
pharmaceutical products:

Human insulin: for diabetics

Human growth hormone treats people with too little
HGH caused by a form of dwarfism.
26. Explain how DNA technology is used in the forensic
sciences.

DNA Fingerprinting: DNA testing reveals unique DNA
sequence from body fluids of tissue left at crime
scene.

Presented in court as autoradiograph.
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27. Describe how gene manipulation has practical
applications for environmental and agricultural work.
Environmental Cleanup:

Certain microorganisms’ ability to transform chemicals
are being engineered into other microorganisms

Then used to treat environmental problems

Example: bacteria can extract heavy metals from
environment turn into compounds that are easily
recoverable. May become important in mining
minerals and cleaning toxic mining wastes.
Agricultural:

Transgenic animals and plants to create better
products.

Pharm animals: producers of large amounts of an
otherwise rare biological substance for medical use,
like, human hormone or blood clotting factor.
o Cows secrete chemical in milk
o Chickens secrete chemical in eggs
o Product can be purified more easily than
from a cell culture.
28. Describe how plant genes can be manipulated using
Ti plasmid carried by Agrobacterium as a vector.
See fig. 20.19
29. Explain how DNA technology can be used to improve
the nutritional value of crops and to develop plants that
can produce pharmaceutical products.
Increased Nutritional Value:

Transgenic rice plants that produce yellow rice grains
containing beta-carotene, which our body uses to
make vitamin A.

“Golden Rice” could help prevent vit A deficiency in
the half of the world’s population that depends on rice
as a staple food.

Currently, large numbers of young children in SE Asia
suffer from vit A deficiency, which leads to

vision impairment and

increases susceptibility to disease.
Pharm Crops:

Human proteins for medical use

Viral proteins for use as vaccines

Vaccines for hepatitis B

antibody produced by transgenic tobacco plants that
interferes with bacteria that cause tooth decay.
30. Discuss the safety and ethical questions related to
recombinant DNA studies and the biotechnology industry.
Hazardous new pathogens might be created
Cancer cell genes transferred into bacteria or viruses

Strict laboratory procedures designed to:

protect researchers from infection by engineered
microbes and

to prevent the microbes from accidentally leaving the
lab.

Strains of microorganisms to be used in recombinant
DNA experiments are genetically crippled to ensure
that they cannot survive outside the lab.

Obviously dangerous experiments have been banned.
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GM organisms:
A Genetically Modified organism is one that has acquired by
artificial means one or more genes from the same or another
species.
Most are crops
Fears:

Transgeneic plants might pass their new genes to
close relatives in nearby wild areas.

Crop plants that carry resistance to herbicide,
disease, or insect pests cross-pollinated with wild
relatives, it may create “super weeds.”

Protein products of transgenes might lead to allergic
reactions
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