Genetics Exam 1 Study Guide
Lecture 1
a. Identify central dogma of molecular biology and examples of exceptions to this dogma.
a. General:
i. DNA  DNA = replication
ii. DNA  RNA = Transcription
iii. RNA  Protein = Translation
b. Exception
i. RNA  DNA = reverse transcription
ii. RNA  RNA = RNA replication
c. Special test tube
i. DNA  Protein = Direct translation from DNA to protein
b. Identify the organization of the human genome (approx. # of genes) number of chromosomes
and how DNA is packaged into chromatin.
a. Chromosome number in human genome is 22 autosomal and 1 sex chromosome
b. There are approximately 21000-26000 protein coding genes in the human body
i. Exons  code for pieces of proteins
ii. Introns  spacers between exons, contain information but are NOT coding
sequences
c. Major associated proteins with DNA are histones
d. DNA is tightly packed in the nucleus of every cell. DNA wraps around special proteins
called histones, which form loops of DNA called nucleosomes. These nucleosomes coil
and stack together to form fibers called chromatin.
i. Telomeres  sequences at the ends of chromosomes  it protects ends from
degradation or fusion
e. DNA is anti – parallel, the strands twist to form major groove and a minor groove
f. Most protein interactions with the DNA strand occur at the major groove
g. There are three forms of DNA  B form, A form, Z form  MOST COMMON is B form
c. Identify the structure and function of genes.
a.
d. Identify the major pyrimidine and purine bases in DNA and RNA.
a. DNA
i. Pyrimidines
1. Thymine, Cytosine
ii. Purines
1. Adenine, Guanine
b. RNA
i. Pyrimidines
1. Uracil, Cytosine
ii. Purines
1. Adenine, Guanine
e. Distinguish between nucleoside and nucleotide structures.
a. Nucleoside  Sugar + Base
b. Nucleotide  Sugar + Base + Phosphate
f. Identify how nucleotides are linked in strands of RNA and DNA.
a. Nucleotides are linked through phosphodiester bonds
b. 5’3’ directionality
g. Elucidate the concept of base pairing in DNA.
a. Base pairing occurs through hydrogen bonding
i. A hydrogen bonds with T through TWO hydrogen bonds
ii. C hydrogen bonds with G through THREE hydrogen bonds
iii. For RNA U hydrogen bonds with A through TWO hydrogen bonds
h. Compare/contrast DNA packaging in prokaryotic and eukaryotic cells.
a. Prokaryotes and eukaryotes package their DNA molecules with protein in structures
called chromosomes. A prokaryotic chromosome is circular and resides in a cell region
called the nucleoid. The types of proteins found in prokaryotic chromosomes, known as
the nucleoid-associated proteins, differ from the histone proteins that appear in
eukaryotic chromosomes and cause the prokaryotic chromosomes to form looped
structures. Unique eukaryotic chromosome packaging features include tight coiling,
dense packing, enclosure within a nuclear membrane and linear rather than circular
structures.
i. Identify the structure of the nucleosome and describe its function in eukaryotic cells.
a. Histone 1  Acts as a linker and stabilizes the nucleosome
b. Nucleosome
i. Fundamental subunit of chromatin
ii. Each nucleosome in composed of two turns of DNA wrapped around 8 histones
total. Each set of histone (4 histone in 1 set) is H2A, H2B, H3, and H4.
j. Distinguish between poly-nucleosome and solenoid structure of chromatin.
a. Poly – nucleosome  one octamer + H1 histone, multiple of those create beads on the
string and that is the poly – nucleosome
b. Solenoid  poly – nucleosome looped around very tight
k. Identify differences between DNA and RNA.
a. DNA has Thymine base, RNA has Uracil in place of thymine
b. RNA has a hydroxyl group on carbon 2, DNA does not (that is why it is called
deoxyribose)
c. DNA is double stranded, RNA is single stranded
d. Various types of RNA  mRNA, tRNA, rRNA
i. Side note  A mRNA molecule is said to be monocistronic when it contains the
genetic information to translate only a single protein chain (polypeptide). This is
the case for most of the eukaryotic mRNAs.
l. Identify the life cycle of a retroviruses.
a. Infectious virus binds to the envelope receptor. The bound virus enters the cell through
Fusion. Once in the cell, the virus releases its RNA and reverse transcription occurs 
The DNA now integrates with our own DNA forming a viral genomic RNA which then
undergoes translation. The viral RNA then integrates with the cell membrane and leaves
the cell enveloped in the cell membrane making it undetectable by our immune system.
m. Identify the clinical uses of 5-fluorouracil, doxorubicin and azidothymidine.
a. 5 – fluorouracil
a. Pyrimidine base
b. Inhibits synthesis of thymidine nucleotides used for DNA replication
c. Used in chemotherapy to “knockdown” DNA synthesis  cell replication
b. Doxorubicin
a. Anthracycline antibiotic  works by intercalating into the DNA interior and
disrupting the helix  used as a chemotherapy agent, disrupts DNA replication
 cell replication (mitosis)
c. Azidothymidine
a. First drug approved for treatment of AID and HIV infection  anti-retroviral
b. Reverse transcriptase inhibitor
c. Also inhibits DNA polymerase y
Key Concepts
•
The human genome is subdivided into a large nuclear genome with more than 26,000 genes,
and a very small circular mitochondrial genome with only 37 genes
•
Human genes are usually not discrete entities: their transcripts frequently overlap those from
other genes, sometimes on both strands.
•
Duplication of single genes, subchromosomal regions, or whole genomes has given rise to
families of related genes.
•
Genes are traditionally viewed as encoding RNA for the eventual synthesis of proteins, but many
thousands of RNA genes make functional noncoding RNAs that can be involved in diverse
functions.
•
Noncoding RNAs often regulate the expression of specific target genes by base pairing with
their RNA transcripts.
•
Some copies of a functional gene come to acquire mutations that prevent their expression.
These pseudogenes originate either by copying genomic DNA or by copying a processed RNA
transcript into a cDNA sequence that reintegrates into the genome (retrotransposition).
•
Occasionally, gene copies that originate by retrotransposition retain their function because of
selection pressure. These are known as retrogenes.
•
Transposons are sequences that move from one genomic location to another by a cut-andpaste or copy-and-paste mechanism. Retrotransposons make a cDNA copy of an RNA transcript
that then integrates into a new genomic location.
•
Very large arrays of high-copy-number tandem repeats, known as satellite DNA, are associated
with highly condensed, transcriptionally inactive heterochromatin in human chromosomes.
Lecture 2
•
List and define the various players and steps in DNA synthesis and replication
Proteins
DNA helicase
DNA polymerase
DNA clamp
Single stranded binding proteins (SSB)
Topoisomerase
DNA Gyrase
DNA ligase
Primase
Telomerase
•
Step 1  Origin of replication
•
•
Function
Unwinds DNA helix at replication fork
Builds daughter DNA strand by adding dNTPs in
5’3’ direction, error – correction and proof
reading
Proteins that DNA Pol III from leaving DNa parent
strand
Binds to unwound single strand and prevents re –
annealing to helix  supports new DNA
formation
Relaxes DNA from supercoiled structure
Relieves strain on unwinding strands of DNA
Re – anneals strands and joins Okazaki fragments
Provides RNA primers for DNA Pol to build new
DNA
Eukaryotes – lengths chromosome ends by
adding repetitive sequence to “cap”
The double helix “unwinds”
•
Step 2  Nucleotides are added in the correct order to create a complementary strand
•
Step 3  New strand is sealed
Explain the importance of semiconservative replication of DNA
•
•
Semiconservative replication allows for us to keep a “mother” strand while replicating
the daughter strand, so we always have an “original” copy of the DNA strand.
List the enzymes and proteins involved in unwinding DNA for replication
Enzyme
Primase
DNA Pol III
Function
Synthesizes RNA primer
Synthesizes and proofreads new strand of
DNA
Replaces RNA primer with DNA and
proofreads new DNA
Joins DNA all together
DNA Pol I
DNA Ligase
•
Explain the function of DNA Polymerase I-IV
E. coli polymerase
DNA Pol I
DNA Pol II
DNA Pol III
DNA Pol IV
•
•
Function
RNA primer removal
3’5’ exonuclease
5’3’ exonuclease
DNA repair
3’5’ exonuclease
DNA replication/synthesis
3’5’ exonuclease
DNA Repair
DNA Damage
DNA repair
Bulky hydrocarbons
attached to bases
Bulky hydrocarbons
attached to bases
Explain the nucleotidyl transfer reaction for DNA chain elongation.
•
This process extends the DNA daughter strand
•
It requires free 3’ hydroxyl for new phosphodiester bond formation
•
The DNA base at 3’ end, will undergo SN2 – like reaction adding the next nucleotide to
the daughter DNA
List the five best known eukaryotic DNA polymerases and describe their function.
DNA Polymerase
Alpha
Beta
Gamma
Delta
Epsilon
Enzyme activity
Polymerase primase
3’5’ exonuclease
Polymerase
Polymerase
3’5’ exonuclease
Polymerase
3’5’ exonuclease
Polymerase
3’5’ exonuclease
5’3’ exonuclease
Functions
Primer synthesis
Repair
Repair
Mitochondrial DNA
replication
Lagging strand synthesis
Repair
Leading strand synthesis
Gap filling in lagging strand
•
For a DNA replication fork, describe the following: leading strand synthesis, lagging strand
synthesis, continuous and discontinuous synthesis.
•
•
•
DNA replication moves along the parent strand in the 5’ to 3’ direction
•
DNA replication occurs very easy on the leading strand
•
RNA primase adds the first nucleotide to the nascent chain, while DNA
polymerase just chills near the replication fork moving with the fork and adding
nucleotides preserving the proper anti-parallel orientation
•
This process is the continuous DNA synthesis
Lagging strand
•
DNA polymerase can only move in the 5’3’ direction, so an enzyme does the
work in the lagging strand
•
The lagging strand replicates in small fragments called Okazaki fragments 
100-200 nucleotides in humans that are synthesized in the 5’3’ direction
AWAY from the fork
•
The Okazaki fragments are then put together by DNA ligase creating the
continuous strand.
•
This type of replication is known as discontinuous
Okazaki fragments are the fragments that are synthesized during the lagging strand
synthesis because the enzyme in the lagging strand moves away from the fork, when the
fork is unwinding in the opposite direction.
Describe DNA proof-reading and explain its importance.
•
•
•
Describe Okazaki fragments and explain why they are needed for DNA synthesis.
•
•
Leading strand synthesis
In bacteria, all three DNA polymerases (I, II and III) have the ability to proofread, using 3'
→ 5' exonuclease activity. When an incorrect base pair is recognized, DNA polymerase
reverses its direction by one base pair of DNA and excises the mismatched base.
Describe the cell cycle and its regulation in eukaryotes
•
•
•
Define telomeres and describe telomerase activity and its role in DNA replication
•
Telomere is the end sequence at the end of the chromosome that serves as a “cap” to
protect the chromosome. Over time telomeres shorten, and eventually cell division
stops.
•
Telomeres allow cells to divide without losing genes. Without telomeres, chromosome
ends could fuse together and corrupt the cell’s genetic blueprint causing malfunction,
cancer, or cell death.
Compare and contrast DNA synthesis in eukaryotes and prokaryotes
Eukaryotic DNA replication
Quantity of DNA large (46 chromosome)
 multiple points of origin
Replication proceeds bi – directionally
from point of origin
All origins of replication start
simultaneously and meet at ends
Small Okazaki fragments (250nts)
Prokaryotic Replication
Quantity of DNA small (1 chromosome)
 single point of origin
Replication proceeds bi – directionally
from point of origin
Large Okazaki fragments (1000-2000 nts)
•
•
Identify defects/diseases of DNA replication (give examples).
Lecture 3
a. Identify mechanisms DNA proof-reading and distinguish the importance of these
mechanisms.
i. DNA Polymerase I, II, III in prokaryote can proofread DNA from 3’5’
exonuclease
ii. DNA polymerase delta and epsilon can proofread DNA from 3’5’ exonuclease
in eukaryotes
b. Identify the sources and types of DNA damage and health consequences.
i. Deamination: (CU base change, A  hypoxanthine )
ii. Depurination: loss of purine base (A or G)
iii. Dimer formation: (T-T and/or T-C) ; base become cross-linked T-T is more
common caused by UV light
iv. Alkylation: an “alkyl group” (-CH3) is added to bases can be harmless
v. Chemical induced : cause mutations by base mis-pairing during replication or
stopping polymerase entirely
vi. Oxidative Damage (ROS): G oxidizes to 8-oxo-guanine can cause SSB and DSB
important in cell organelles
vii. DNA replication errors: wrong or modified nucleotide inserted
viii. Double Strand Breaks (DSBs): induced by ionizing radiation, transposons,
topoisomerase, homing endonucleases, mechanical stress on chromosomes, or
single strand nick during replication and transcription
c. Identify the mechanisms of mutagenesis.
i. Mutagenesis is the term used to describe a mutation in DNA that CAN be
replicated. This leads to several diseases and cancers
ii. Electrophilic form find a nucleophile in the purine and or pyrimidine rings
1. Causes:
a. Base analogs, DNA intercalators, DNA adducts
d. Identify and distinguish the four types of DNA repair mechanisms.
i. Methyl – directed mismatch repair
1. If mismatch occurs, it will cause a distortion of the helix
2. Repair enzyme must be able to recognize new strand from template
strand
ii. Base excision repair (BER)
1. Correct specific chemical damage to DNA
2. Enzyme needed  glycosylases(specific) and AP endonucleases
a. Mechanism:
i. Glycosylase recognizes specific base damage
ii. Cleaves glycosyl bond and removes base
iii. AP endonuclease cleaves DNA backbone
iv. DNA polymerase trims gap and DNA ligase repairs gap
v. Correct base replaced using template strand
iii. Nucleotide Excision Repair (NER)
1. Defects in this pathway causes
a. Xeroderma Pigmentosum, Cockayne Syndrome,
Thrichotridystrophy
2. Mechanism
a. Damage is recognized, the helix is partially opened upon binding
of several proteins (protein complex)  Structure specific
endonucleases cut the DNA (nicking)  Exonucleases now take
that “cut” part out (digestion)  DNA polymerase I
resynthesizes the “gap” (resynthesis)  DNA ligase now closes
the helix (ligation)
iv. Direct Repair
1. Demethylation of DNA by a methyltransferase
e. Distinguish between homologous and non-homologous genetic rearrangements.
i. THESE ARE DOUBLE STRANDED BREAKS, the first four types of repair mechanism
were single stranded breaks
ii. Homologous  using an existing strand of DNA to repair the damaged DNA
iii. Non – homologous  Breaks at the end are directly ligated without the need
for the homologous strand. This process restores the DNA integrity but the DNA
sequence at the junction may the altered
1. RAD51, P53, RAD54, BRCA2 are oncogenes (tumor suppressor genes).
Mutation in these genes cause DNA damage (cancer).
f. Identify the functions of transposons (LINE, SINE and ALU) and their role in
pathology/disease
i. Transposon  DNA sequence that can change its position within the genome
1. LINE  Long interspersed elements (1000 base pairs)
2. SINE  Short interspersed elements (100 base pairs)
a. ALU  a short stretch of DNA important in human evolution
g. Identify the six hereditary DNA repair disorders:
i.
ii. Blooms syndrome  autosomal recessive, mutation in BLM gene, stalled
holiday junction, male and female are infertile
iii. Werner Syndrome  WRN gene, RECQL2 chromosomal instability, autosomal
recessive
iv. Nijmegen syndrome  mutation in NBS1 gene, autosomal recessive.
h. Identify the Biochemical basis of:
i. hereditary non-polyposis colorectal cancer (HNPCC)
1. MSH2 and MLH1 are the genes for HNPCC
ii. Burkitt’s lymphoma.
1. Chromosomal Translocation (8:14)
2. C-myc proto – oncogene
3. If translocations are passed on to the next generation through germ
cells, the offspring could have partial trisomy and partial monosome
due to the juxtaposition of chromosome DNA
Lecture 4
1. Compare and contrast eukaryotic with prokaryotic transcription
1. Prokaryote

Initiation
1) RNA polymerase (RNAP) binds to one of several specificity factors,
sigma, to form a holoenzyme. At this stage the DNA is double –
stranded.
2) The DNA is unwound and becomes single – stranded near the initiation
site. This form the open complex.
3) The sigma factor eventually dissociates from the core enzyme, and
elongation proceeds.

Elongation
1) Process through which nucleotides are added to the growing RNA chain.

Termination
1) Intrinsic terminator
a. Terminator sequences within the RNA that signal the RNA
polymerase to stop. It forms a hairpin structure that leads the
dissociation of the RNAP from the DNA template.
2) Rho – dependent termination
a. Terminator factor called rho factor which is a protein to stop
RNA synthesis at specific sites.
2. Eukaryotic

Initiation
1) Promoter region  The promoter regulates where, when and to what
level a gene is expressed
a. RNA polymerase binds to the promoter and initiates gene
transcription, leading to the production of mRNA
2) The binding cascade to initiate transcription
a. TFIID binds to TATA box  TFIIA and TFIIB bind next  RNA Pol
binds to this complex that now also includes TFIIFinitiation
complex completed by addition of TFIIE, TFIIJ and TFIIH  RNA
polymerase is activated by an ATP dependent phosphorylation
step  transcription initiated at the start point

Elongation
1) Process through which nucleotides are added to the growing RNA chain.

Termination
2. Name the main sites of non-coding DNA segments
1. Introns are non – coding regions of DNA segments
2. Telomeres, Enhances, Promoters, Termination
3. Identify steps involved in eukaryotic processing of primary mRNA (capping, intron removal)
1. mRNA made as a precursor (hnRNA) then post-transcriptionally processed to mature
form
1. STEPS:
1. Removal in introns – splicing of exons
2. Addition of 5’ cap for stability (added as transcribed)
3. Addition of poly A tail for termination signal
2. CAP structure: unsual 5’ – 5’ pairing of methylated GMP at 5’ end of mRNA
1. NOTE  This process requires folate and B12 and is one of the processes
affected by vitamin deficiencies
5. Distinguish between introns and exons and know how they are processed.
1. Introns are removed to mature RNA
2. Exons are spliced together with accuracy by spliceosome
6. Identify steps to final spliced mRNA product
1. 5’ GU, 3’ AG that is your spliced mRNA and then you ligase the two exons.
7. Define structural elements of mature RNA that allow it to be stable in the cell
1. 5’ CAP and poly A tail protect the mature RNA from degradation
8. Identify different types of DNA-binding molecules that are used to regulate transcription
1. TFIID binds to TATA Box  TFIIA and TFIIB bind next (see objective 1)
9. Distinguish the roles of three eukaryotic RNA polymerases
1) RNA polymerase I  rRNA (localized to nucleolus)
2) RNA polymerase II  mRNA (polypeptide coding)
3) RNA Polymerase III  tRNA and other small RNAs
10. Identify three types of RNA and their function include components of ribosomal structure
1) Messenger RNA (mRNA)  carries the message from the DNA to the ribosome for
translation
2) Ribosomal RNA (rRNA)  Forms major structural complex of the ribosome + proteins
3) Transfer RNA (tRNA)  acts as an adapter molecule aligning amino acids with mRNA
sequence help in the ribosome
11. Delineate function of tRNA
1) tRNA ensures that mRNA containing code for amino acids in a polypeptide is translated
correctly. Minimum of 20 tRNAs one for each amino acid
2) enzyme = nucleotidyl transferase
12. Identify the biochemical basis of β-Thalassemia
1) Microcytic anemia  not enough HbA to fill RBC
a. Autosomal recessive  beta-globin gene on chromosome 11
b. Deficiency in synthesis of beta – globin chains (hemoglobin); excess alpha –
chains
13. Identify and distinguish the site of action of α-amantin and Diptheria toxin
1) Alpha – amantin  acts as a potent and specific inhibitor of RNA polymerase II and
mRNA synthesis in eukaryotes (animal and plants)
2) Diphteria toxin  blocks protein translation. ADP – ribosylation of EF2 inhibits protein
synthesis leading to cell death
Lecture 5 and 6
1. Identify the features of the genetic code (you do not need to memorize the code!). Include the
concept of degeneracy and how it affects the code.
i.
Genetic code is a triplet  each word in the code is composed of three nucleotide bases
called a codon
ii.
Nucleotide sequences are always written from the 5’ end to the 3’ end
iii.
64 different combinations of bases
iv.
Termination codons  UAG, UGA, UAA (THEY ARE STOP CODONS)
v.
Initiation codon  AUG (codes for methionine)
2. Identify the wobble hypothesis and its importance.
i.
Third position in codon is often called the “wobble position” due to the degeneracy
being most common at that site
ii.
The degeneracy fold depends on how many bases can substitute in a spot and not
change the meaning of the code
a) Example  the third position in the codon GC_ is a four – fold degenerate site
because any of the four bases will encode alanine
iii.
The hypothesis proposes four relationships:
a) First 2 bases of a codon always form strong Watson Crick base – pair with the
anticodon and confer most of the coding specificity
b) The first base of anticodon (5’3’) or the 3rd base (3’5’) called the wobble
base. The wobble permits tRNA to read more than one codon with the
maximum limits of three codons
c) When the first base in the 5’ end in the anticodon is U then the third base in
codon can be either A or G. The U forms strong Watson-crick base pair with A
and wobble pairing with G.
d) When I or some other modified base present at 5’ end of anticodon then tRNA
can recognize three different codons. The bases that can pair in this case are
A,U, and C.
3. Identify a reading frame and the concepts of being “in frame” versus a “frame shift”.
i.
Reading frame has 3 possible frames within one strand
a) Since the nucleotides are triples, each nucleotide of the triple can be the start
for your reading frame (AUG start codon)
ii.
Frame shift  Is a mutation that is either an insertion or deletion of nucleotide which
disrupts the triplet.
4. Identify the six types of mutations that can occur during translation.
i.
Point mutation  a single base change
a) Silent  a change that specifies the same amino acid
b) Missense  a change that specifies a different amino acid
c) Nonsense  a change that produces a stop codon
ii.
Insertion  an addition of one or more bases
5.
6.
7.
8.
iii.
Deletion  a loss of one or more bases
Identify the steps involved in translational mechanism.
i.
Protein synthesis in Eukaryotes
a) Initiation of translation involved formation of a complex composed of
I.
Methionyl – tRNA  initially forms a complex with the protein
eukaryotic initiation factor 2 which binds GTP
II.
mRNA
1. in this rections hydrolysis of ATP is required because a helicase
is needed to unwind the hairpin loop
III.
ribosome
b) Elongation
I.
Three main steps
1. Binding of incoming charged tRNA
2. Peptide bond formation
3. Translocation of the ribosome
II.
When Met-tRNA is bound to the P site, the mRNA codon in the A site
determines which aminoacyl –tRNA will bind to that site
III.
In eukaryotes, the incoming aminoacyl-tRNA first combines with
elongation factor EF1a containing bound GTP before binding to the
mRNA ribosome complex
IV.
When the aminoacyl-tRNA –EF1-GTP complex binds to the A site, GTP
is hydrolyzed to GDP.
V.
Contrast: Elongation in prokaryotes is similar – except that the
corresponding factor for EF1 is named EF-Tu and the associating
elongation factors are called EF-Ts instead of EF1.
c) Termination  STOP CODON is found
Distinguish the activity of aminoacyl tRNA synthetase.
i.
Each family member enzyme recognizes a specific amino acid and all the tRNAs that
correspond to that amino acid
ii.
Catalyze a two-step reaction that results in the covalent attachment of the carboxy
group of an amino acid to the 3’ end of its corresponding tRNA
iii.
They have a proof-reading or editing activity that can remove amino acids from the
enzyme or the tRNA molecule
Distinguish the structure and function of ribosomal RNA compared to other major RNAs. Identify
all components of ribosomal structure.
i.
There are two ribosomal subunits
a) Eukaryotes are 60S and 40S for a total of an 80S subunit
b) The small ribosomal subunit
I.
Bind mRNA and is responsible for the accuracy of translation by
facilitating correct base pairing between codon and mRNA
c) Large subunit
I.
Catalyzes formation of peptide bonds that link the amino acids in a
growing polypeptide chain
Distinguish in general terms how tRNAs become charged with amino acids.
tRNAs become charged with amino acids when interacting with the ribosome  this
happens at three sites
a) A – site is the aminoacyl site where incoming charged (aminoacyl-tRNAs) attach
b) P – site is the peptidyl site where the peptide bond is formed with the incoming
amino acid adding on to the peptide to form a new carboxy – terminus
c) E – site is the exist site where the now depleted tRNA is released from the
ribosome
9. Identify the components of the process of translation.
i.
Messenger RNA (mRNA – single strand 5’3’ read)
ii.
Ribosome
iii.
Charged tRNA (tRNA + specific amino acids)
iv.
Various accessory factors
v.
GTP to provide energy
10. Identify the clinical use of streptomycin, erythromycin, tetracyclines, chloroamphenicol.
i.
Steptomycin  bind to the 30S ribosomal subunit of prokaryotes, thereby preventing
formation of the initiation complex
ii.
Erythromycin  Binds to the 50S ribosomal subunit and prevents translocation
iii.
Tetracyclines  Binds to the 30S ribosomal subunit and inhibits binding of aminoacyl –
tRNA to the A site
iv.
Chloroamphenicol  Binds to the 50S ribosomal subunit and inhibits
peptidyltransferase
i.
Lecture 7
a. Identify the steps and processes involved in the mechanism of diauxic growth.
a. Key features
i. Initial phase same as normal growth curve
ii. Stationary phase is a “lag” phase, synthesis of enzymes to utilize a 2nd carbon
source
iii. 2nd exponential growth phase to utilize lactose
iv. Exhaust 2nd carbon source culture (lactose) declines and dies
v.
b. Distinguish between various regulatory mechanisms used in eukaryotes.
a. Modification of DNA
b. Transciptional modification
c. Posttranscriptional
d. Translational modification
e. Posttranslational modification
c. Define and distinguish between the following alterations in genes: loss, amplification,
rearrangement
a. Gene amplification  duplication of a region of DNA that contains a gene
i. Common sources of gene amplification are: homologous recombination,
retrotransposition, aneuploidy, polyploidy.
b. Gene loss  loss of heterozygosity results in the loss of the entire gene and the
surrounding chromosomal region
c. Gene rearrangement  usually followed after DNA double helices breaks and non –
homologous recombination occurs. When ends rejoin together, you produce a new
chromosomal arrangement of genes.
d. Identify the steps and processes involved in alternative splicing –expansion of the transcriptome
a. Exon skipping  exon is spliced out of the primary transcript or retained
b. Mutually exclusive exons  one of two exons is retained in mRNAs after splicing, but
not both.
c. Alternative donor site  An alternative 5’ splice junction is used, changing the 3’
boundary of the upstream exon
d. Alternative acceptor site  an alternative 3’ splice junction is used, changing the 5’
boundary of the downstream exon
e. Intron retention  a sequence may be spliced out as an intron or simply retained.
e. Identify the steps and processes involved in post-translational regulatory mechanisms
a. Phosphorylation  adds a phosphate to serine, threonine or tyrosine
b. Lipidation  attaches a lipid, such as a fatty acid to a protein chain
c. Ubiquination  adds ubiquitin to a lysine residue of a target protein marking it for
destruction
d. Disulfide bond  covalently links the “S” atoms of two different cysteine residues
e. Acetylation  Adds an acetyl group to the N – terminus of a protein to increase stability
f. Glycosylation  attaches a sugar, usually to an “N” or “O” atom in an amino acid side
chain
f. Identify the steps and processes involved in the function of chaperones in protein folding
a. Protein folding, maintenance of proteome integrity, and protein homeostasis critically
depend on a complex network of molecular chaperones.
b. Chaperons are proteins that mediate post-translation protein folding
i. Two functions:
1. Foldases  support folding of proteins in ATP – dependent manner
2. Holdases  bind folding intermediates to prevent aggregation
a. EX: HSP100  can tag and unfold mis – folded proteins
b. HSP 90  essential eukaryotic chaperone for signaling
molecules
g. Identify and distinguish between the nine common posttranslational modifications of amino
acid residues. Identify examples.
a. Phosphorylation  adds a phosphate to serine, threonine or tyrosine
b. Lipidation  attaches a lipid, such as a fatty acid to a protein chain
c. Ubiquination  adds ubiquitin to a lysine residue of a target protein marking it for
destruction
d. Disulfide bond  covalently links the “S” atoms of two different cysteine residues
e. Acetylation  Adds an acetyl group to the N – terminus of a protein to increase stability
f. Glycosylation  attaches a sugar, usually to an “N” or “O” atom in an amino acid side
chain
g. Hydroxylation  proline and lysine residues of the alpha chains of collagen are
hydroxylated by vitamin C – dependent hydroxylases in the ER
h. Methylation
i. Carboxylation
h. Define the term “signal peptide” and explain its function.
a. Signal peptide  cellular addresses to get newly synthesized protein to correct cell
organelles
b. Function  transport of proteins synthesized on RER
i. Distinguish differences in ribosomal protein synthesis for cytosolic proteins versus proteins that
will be exported.
a. Cytosolic protein are not synthesized in the ER, they are synthesized through free
ribosome in the cytosol. Proteins that are going to be exported are synthesized in the
ER.
j. Identify the steps and processes involved in the processing of proteins that are synthesized and
translocated into the ER
a. SRP binds to SRP receptor  signal peptidase cleaves the signal peptide  protein is
synthesized  nascent protein is made and completed  it goes into vesicle and then
to cis Golgi
k. Identify the steps and processes involved in the secretory pathway in cells and the role of
different organelles.
a. Many proteins destined for secretion from the cell are initially made as large, precursor
molecules that are functionally inactive
b. Some precursor proteins are cleaved in the ER or Golgi apparatus
c. Collagen is cleaved after secretion
d. Zymogens are inactive precursors of secreted proteins and becomes activated through
cleavage after reaching a site of action.
l. Define the function of mannose-6-phosphate attached to newly synthesized proteins.
a. Man – 6 – P tag allows transport to lysosome
b. Used to target enzyme replacement therapy for lysosomal storage diseases
m. Distinguish the secretory pathway and targeting of proteins to subcellular organelles (golgi, ER,
cell membrane).
a.
Lecture 8

Compare and contract Prokaryotic and eukaryotic regulation/gene expression
o Prokaryotic  positive and negative lac operons
 Induction of expression of lac operon (lactose)
 Lactose will bind to repressor protein which inactivates it allowing the
lac operon to be transcribed which produces the proteins for the given
genes in the operon.
 Gene regulation (tryptophan)
 trpR (trp repressor) is located upstream from the tryptophan structural
genes. It’s main job is to repress the attenuator. If the attenuator is on it
prevents the transcription of domain 2 (tryptophan). When you have
high levels of tryptophan it will bind to trpR which inactivates trpR
which allows the attenuator to be turned on, so it stops the
transcription of tryptophan.
 Sigma factor  Co – factor in prokaryotic RNA polymerase complex
o Eukaryotic
 Consensus sequence
 GC box + CAAT box  proximal control elements  it’s the mile marker
for the exit (TATA box)
 TATA Box  core promoter, starting point, highway exit (exit now or
run out of gas, don’t miss it)
 Located only upstream
 Terminator
 Red thunderbolt  cleave site following terminator sequence. After you
cleave you add your poly – A – tail.
 Enhancer
 Distal  father away from promoter, can be located 5’ or 3’ gene
(downstream or upstream)
 IT CAN BE LOCATED ON INTRON
 Insulator
 Blocks action between enhancer and promoter (enhancer activated
promoter)
 Cis – trans
 Cis  affects one gene on same DNA strand
 Trans  can affect multiple genes on many DNA strands/different
chromosomes
 Gene expression is regulated by hormonal expression
o Types of gene expression
 Constitutive gene expression
 Housekeeping genes  GAPDH
 Gene essential for cell functions  actin
 Inducible Gene Expression
 Cytochrome P450  family member responsible for drug metaboliosm





 Enzyme in catabolic pathways  hormone
 Repressible gene expression
Compare and contrast induction and repressions.
o Induction  lactose lac operon (objective 1), Sigma factor
o Repression  trp Operon (objective 1), insulator
o Hormone can be both, induction and repression
Identify a promoter complex and how it works.
o TATA box where RNA pol binds and initiates translation
Identify the elements involved in transcriptional regulation (IRE, HRE, TRE).
o IRE  Iron response element (negative response)
 If you don’t have iron, IRE – BP cannot bind to iron, therefore it will bind to IRE
which is located on DNA, and it will prevent initiation of transcription of
transferrin.
o HRE  Hormone response element (positive response)
 Hormone + BP will bind to HRE which initiates transcription of said hormone
o TRE  Thyroid response element (both response)
 T3 (thyroid hormone) will bind to response element on DNA and it has many
activities and tissue targets, many gene targets
 It can be both inducer and repressor (positive and negative)
Identify the steps and processes involved in the regulation of globin production by heme.
o You need iron to make heme.
o FCH adds iron to the heme  heme is synthesized  heme + Erythropoietin (which
causes the production of globin syntheses) will make hemoglobin.
Delineate the hormonal regulation of gene expression using a steroids as an example.
o Look at thyroid hormone and compare it to estrogen
Lecture 9 and 10

Identify techniques that are used by molecular biologists to investigate and diagnose cellular
disease processes, give examples and explain limitations.
o Polymerase Chain Reaction (PCR)
 Amplify region of gDNA to sequence for point mutation  example Gaucher
disease
 Ex: loss of cardiac sympathetic neurotransmitters in heart failure and NE
infusion is associate with reduced NGF.
 Limitations  you need a specific primer which has to be made
o DNA sequencing (whole genome and exome)
 Exome sequencing is a technique for sequencing all the protein – coding genes
in a genome (known as the exome).
 It is used for comparison of normal vs mutated gene in DNA
 Limitations  Very expensive
 Whole – exome sequencing – based discovery of STIM1 deficiency in a
child with fatal classic Kaposi sarcoma.
o Array Comparative Genomic Hybridization
 Take mRNA and make cDNA of both cancer and normal cell
 Red  more expression of cancer mRNA not found in normal cell
 Green  more normal cell gene expression
 Yellow  both cancer and normal cell have same amount of gene
expression
 It shows what gene the cell has lost, or gained
 Limitations  you need a compliment DNA for every mRNA, as well as a
compliment DNA stand for the cDNA.
o Fluorescence In Situ Hybridization (FISH)
 Used to detect and localize the presence or absence of specific DNA sequences
on chromosomes. Can detect translocation as well
 Limitations  fluorescent probe + microscope are needed
 Diseases: cancer, anything unregular about the correct number of
chromosomes, chromosomal translocation
o Genomic and Expression Arrays
 Tells you whether our target sequence of Gene of Interest is detected
 Good for checking deletions
 Limitations  you need DNA/cDNA
o Genome Wide Association Studies (GWAS)
 Determines a specific disease in the population
 Takes a group of normal people vs disease people, runs their genome
and compares the differences between the two groups.
 Hypertension, CVD, T2D.
 Limitation  need a large population (very larger population), usually
not 1 gene involved in disease
o Transgenic Technologies

 Knockout  remove gene/protein from body
 Transgenic  over express a gene
 This is used for gene targeted therapies
 Limitations  Takes years
 AGUTI and EGFP is used to make sure that the mice took your gene.
Identify the mechanisms involved in gel electrophoresis for DNA and proteins
o
o
o


Southern blot  gDNA

Take a particular gene of interest  cut the DNA with restriction enzyme  run
the DNA on a gel  transfer it to a nitrocellulose membrane  probe the DNA
with the labeled probe  expose to UV/film

This is used to detect a gene of interest

Example -> sickle cell anemia (mutation from GAG (glutamic acid) to GTG
(valine), in the mutation form you get 3 cut site, and 2 pieces with the probe. In
the normal you see 2 cut sites, and 1 piece with the probe.)
Northern blot  mRNA

RNA single stranded so to keep strands from internal bonding gel contains
formaldehyde of other denaturing chemical. Probe with known sequence,
capture expressed genes.

Housekeeping gene to show equal load or RNA.
Western blot  Protein

Polyacrylimide gel

protein is separated based on size

you have PVDF membrane

you need primary antibody and secondary antibody.

You can detect any protein
Delineate how a restriction endonuclease can be used in the analysis of patient samples to
determine the presence or absence of a particular genetic locus.
o gDNA cut with restriction enzyme gives you many sites common in genome. The gene
can be run on southern blot, and a specific band is detected based on the probe
Describe FISH (Fluorescent In Situ Hybridization) and its uses in the diagnosis of chromosomal
disorders
o
micro-deletions

Pre – natal detection

Example  Positive Williams syndrome (chromosome #7)
o
o
copy number variants (CNV)

o
Deletion or duplication from normal

HeLa has 2 Her2 genes which is normal

SKBR3 has 14 Her2 genes which is abnormal
aneuploidy

Chromosome is lost

o
The elastin gene is found on only one chromosome. The other
copy carries on elastin gene deletions
Example  chromosome 17 is lost, abnormal number of chromosome
 acute promyelocytic leukemia
Polyploidy

Cancer cell karyotype  carries multiple chromosome sets, including 2Y
chromosomes and 22 translocations.