Gatti Review I

Question 1

UV absorption of Nucleobases

Answer 1

For mixtures of nucleotides, a wavelength of 260 nm is used for absorption measurements

Question 2

Describe the composition of a nucleotide

Answer 2

1. Sugar (pentose sugar = ribose or deoxyribose, difference at the 2’ OH)
2. Nucleic base (purine or pyrimidine) that is 1’ Beta
3. Phosphate = BACKBONE of nucleic acid (has 2- charge)

The sugar and bases are neutral at neutral pH; the phosphate has 2 negative charges (its acidic);

Question 3

Describe the three major forms of DNA

Answer 3

1.B form (most common)
Right Handed; diameter = 20A, bases per turn = 10.5, helix rise per base = 3.4 Period = 36 A
2. A form
Right Handed, diameter = 26A, bases per turn = 11, helix rise per bp = 2.6 (compressed; fAt) period = 28A
3. Z form
Left Handed, diameter = 18 A, 12 bp/turn, helix rise per bp = 3.7 A
period = 44 A

Question 4

Describe the three DNA polymerases

Answer 4

ALL HAVE 3’–>5’ endonuclease (proofreading), only DNA Pol I also has 5’ –> 3’ endonuclease

DNA pol I: repair slowest; most abudnant but its primary function is in CLEAN UP during replication, repair, and recombination
DNA pol II: Repair
DNA pol III: replication; fastest (fastest polymerization rate and processivity = nucleotides added before polymerase dissociates)

Question 5

Describe the energetics of twists/ writhes and how to produce +/- supercoiling

Answer 5

supercoiling = the change in the sum of twist and writhe; the twist number of helical turns in the DNA and the writhe is the number of times the double helix crosses on itself; twists and writhes are interconvertible (the energy of a twist is converted into a writhe and vice versa)

such that: delta L = S = delta T + delta W

adding helical twists via counterclockwise = positive supercoiling (RH helix) increase the twist #
subtracting twisting = negative supercoiling = left handed helix)

Question 6

Describe the composition of the human genome (what is the percentage breakdown of each component)

Answer 6

1. Largest component = transposons (transposable elements; a DNA sequence that can change its position within the genome) = 45%. includes LINEs, SINEs, retroviruslike;
2. 30% = genes (28.5% of which are introns/noncoding segments, and only 1.5% are exons that are encoding proteins/expressed)
3. Miscellaneous: 5% large segmental duplications (SD) = segments that appear more than once in different locations, 3% = simple sequence repeats (SSR), 17% = unknown (ie: promotors, etc)

Question 7

Explain what telomerase is

Answer 7

maintains the integrity of the chromosome at each replication by adding DNA sequence repeats (GGGTTA in all vertebrates) to the 3’ end of DNA strands in the telomere regions; does so by carrying its own RNA template ;

Has a REVERSE TRANSCRIPTASE which carries its own transcript

After it adds a bunch of those repeat units, the G’s overhang the end of the DNA strand and forms a hairpin with the free 3’ OH that acts as the primer for the complementary strand. Eventually the hairpin is removed by a nuclease

THUS: ITS SOLVES THE PROBLEM OF filling the gap that is left behind by the removal of the primer at the end of the chromosome during replication

Question 8

Describe the key features of homologous recombination

Answer 8

homologous DNA = two strands that are essentially the same sequence but with small difference between two versions;

General Reaction Scheme:

1. Nick
2. Exchange of DNA
3. Ligation and formation of chemical bond between DNA
4. Extension/branching of DNA up until areas where they are not homologous anymore (extent of branching = depends on extend of homology)
5. Holliday structure is formed by rearranging the branches
6. Cut the Holliday structure via resolvase and this introduces more variation

an endonuclease (recBCD) is introduced and causes nicks int each of the strands; Crossing over is then a strand exchange/invasion and since the strands are homologous they easily base pair with each other; the strand exchange requires a protein rec A. the nicks are sealed by DNA ligase;

The cross-over point can undergo a branch migration, by which longer segments of each strand become part of the opposite molecule resulting in the formation of heteroduplex where the two DNA strands are slightly different ;

If the branch point is rotated in 3-D it gives origin to the Holliday structure which is resolved by cutting with an endonuclease (resolvase, the product of the ruvC gene) and ligrating with DNA ligase; two possible ways to resolve the structure which gives rise to more variations because of the resolved products occur with equal frequency

the extent of recombination = proportional to homology (of the site of exchange)

Question 9

Describe the features of site specific recombination

Answer 9

generates antibody diversity;
large number of possible molecules by virtue of the site speciic recombination between variable and junction during embryonic life

Site Specific Recombination only accounts for the variability (Ab variability) at the GENE LEVEL: only account for the removal of certain V/J genes which gives rise to the major variations in the Ab

Alternative Splicing: gives variation to Ab at the level of RNA (doesn’t have the capacity to change the genes that are going to be in the primary transcript (ie: with introns/exons), but does influence which/how they are expressed)

Question 10

Describe the key features of exon shuffling

Answer 10

Non-homologous, non-site specific;

Can transfer pieces from one gene into another (via transposons!)

Question 11

Describe how DNA repair is modulated by base excision

Answer 11

1. DNA glycosylase recognizes damaged base and cleaves between the base and the deoxyribose ** unless there is a depurination, in which the base is already removed, so this step is skipped
2. An endonuclease cleave the phosphodiester bond near the mutation (produces a nick in the backbone)
3. DNA Pol I starts repair from the free 3’ at the nick because it has a 5’ –> 3’endonuclease; it removes and replaces the damaged part simultaneously!!! Simultaneous removal of a small number of bases and addition of new bases 5’—> 3’
4. The remaining nick is sealed by DNA ligase

BASICALLY: Base excision = base removed; DNA Pol I progresses forward and can get rid of bad bases and put in new ones

Question 12

How is RNA different/similar to DNA?

Answer 12

Difference = primarily due to teh 2’ OH:

1. RNA is much less stable in basic condition than DNA (RNA has a lower resistance to basic pH): this is because under basic conditions the 2’ OH in the RNA can be deprotonated and the free O- can act as a nucleophile and attack the phosphate attached to the ribose attaching strand of RNA and cause the formation of an intermediate ring, which is eventually broken and the RNA is broken down into nucleotide.
2. Hydolysis of RNA is catalyzed by RNase:
   RNase P is a ribozyme that breaks up RNA into tRNA “molecules”
   Dicer = cleaves dsRNA into oligonucleotides : to protect from viral genome and used in RNA interference technology

Question 13

Describe the structure of bascterial RNA polymerase

Answer 13

Bacterial RNA polymerase has at least 6 subunits:
2 alpha = function in assembly and binding to the upstream promotor (UP) elements
Beta = main catalytic site
Beta ‘ = responsible for DNA binding
omega = protect the polymerase from denaturation
SIGMA = directs the polymerase to a specific promoter and determines the types of genes expressed; many different types but the most prevalent one is the sigma70 that is directed to transcribe housekeeping * the more common/frequent a sigma unit, the more likely the genes it transcribes for will be expressed. SO housekeeping genes are always transcribed

** sigma subunit presents the first level of regulation of transcription in bacteria because they direct polymerase to certain class of genes.

Question 14

Describe some of the features of a promoter:

Answer 14

the sigma subunit of RNA polymerase binds to the promoter sequences
promoter = 0 to -35
typically promoters has a TATA sequenene ~ 10 bp before transcription starts (downstream, - 10)
-35 region = polymerase binding
opening region starts in -10 since its TATA rich
RNA starts at +1
Also have upstream elements (UP) that are NOT part of the promoter, but can have areas that modulate activity of polymerase **regulates the promoter

Question 15

What happens as the RNA polymerase transcribes DNA? (with respect to coiling)

Answer 15

For synthesis of an RNA strand, complementary to one of two DNA strands in a double helix, the DNA is transiently unwound. Movement of an RNA polymerase along DNA tends to create local SUPERCOILING: positive supercoiling (overwound DNA) is AHEAD of the transcription bubble (downstream) and negative supercoiling (underwound DNA) is BEHIND it (upstream)
But we need to remove the supercoils; done so via topoisomerases in which the positive supercoils are eliminated and the negative supercoils are regulated

Question 16

Describe the transcription terminators

Answer 16

transcription terminator factors
1. Rho independent = no Rho, termination is due to INTRINSIC FACTORS (basically the RNA sorta folds back up on itself because the bases interact and it forms a hairpin which is then causes the DNA to be dissociated from the RNA pol) = allosterically controls it

2. Rho dependent: Rho subunit = upstream (behind) and it picks up a signal sequence for termination and then moves up the RNA pol via helicase like and causes allosteric modulation (like a hairpin does) and then the RNA polymerase comes off at the TERMINATOR SEQUENCE (note: terminator sequence IS NOT the signaling sequence for the Rho dependent transcription factor, it is the sequence at which the RNA pol comes off) ;; binidng for Rho = rich in cytosine, poor in guanine and is UPSTREAM of actual terminator sequence

** Antitermination = causes the enzyme to continue transcription past the terminator sequence (event = “readthrough”)

Question 17

RNA polymerase in prokaryote vs. eukaryotes

Answer 17

prokaryotes: only one type of RNA pol (has a sigma unit that binds to the promoter in order to initiate transcription)
eukaryotes: RNA pol I transcribes ribosomal RNAs (rRNAs), RNA pol II transcribes RNAs that will become messenger RNAs (mRNAs) and also small regulatory RNAs, and RNA pol III transcribes small RNAs such as transfer RNAs (tRNAs).
Transcription initiation doe not involve a sigma unit, instead it involves a TATA binding protein

Question 18

How is transcription initiated in eukaryotes?

Answer 18

1. RNA pol II is assembled via an interaction of the TATA binding protein (TBP) with the promoter **NO CONNECTION BETWEEN TBP AND SIGMA UNIT; TFIID or SAGA are complexed with TBP the cascade then begins and various transcription proteins are added. result = CLOSED COMPLEX
2. Then the complex is activated by TFIIH which has A). Helicase = unwinds DNA WHICH CAUSES A TRANSFORMATION TO THE OPEN COMPLEX B). KInase = phosphorylates the carboxyl end of the polymerase which ACTIVATES THE CATALYTIC SUBUNIT

then transcription begins

Question 19

Describe how the 5’ cap of mRNA is formed

Answer 19

cap forms while the mRNA is being transcribed
BASICALLY:
1. GTP added and GMP get actually added on and forms a unique 5’ 5’ triphosphobridge with mRNA
2. Guanine (that was just added) is methylated by transferase
3. 2’ OH of the next two bases (first two bases of the actual mRNA) get methylated
5’ end is capped with methylguanosine (before synthesis of primary transcript is complete), and it functions in protecting mRNA from 5’ exonuclease degradation

it is catalyzed by enzymes tethered to the C-
terminal domain of polymerase II; after their release, the cap remains bound to the polymerase through the cap binding protein (CBC).

GTP donates guanosine
adoMet (S-adenosyl methionine) donates the METHYL group going to adoHcy (S-adneosylhomocysteine )

Question 20

Describe Group I intron splicing

Answer 20

Self-splicing
The nucleophile in the first step may be guanosine, GMP, GDP, or GTP (intron spliced is eventually degraded);

The nucleophile in these splicings are are EXTERNAL

How it works:
3’OH on guanine nucleophile attacks the phosphate at the 5’ splice site; then the 3’ OH of the 5’ exon becomes the nucleophile and completes the reaction (attacks the 5’ end of the 3’ exon)

Question 21

Describe Group II intron splicing

Answer 21

Self-splicing
Different from Group I by
1. The first nucleophile attack is not external, its an adenosine ON THE iNTRON ITSELF;
2. The first nucleophile attacks the 2’ OH (not the 3’ OH)

How this makes a difference:
The internal nucleophile attacks the 2’ OH which forms a lariat structure from 2’ –> 5’ phosphodiester bond, THEN the 3’ OH of the exon acts a nucleophile to release the lariat (completes the reaction, similar to the group I)

Question 22

Describe polyadenylation and the mechanisms of processing and variability

Answer 22

RNA continues after the DNA transcript and it makes a signal sequence that is recognized by “cleavage and polyadenylation factor” (CPSF) protein;
3 different complexes reconize, cleave, and polyadenylate RNA (poly P polymerase); this allows for the enzymes to be positioned at different points and thus different RNA sequences are produced; The location of polyadenylation is redirected through different splicing

Basically:
How is the poly A tail added? 2 methods:
1. Alternative cleavage and polyadenylation patterns: there are signal sequences that are recognized as polyadentylation sites (multiple per RNA strand) so this causes cleavage and polyadenylation at multiple sites producing multiple strands that have poly A tail and 5’ cap

2. Alternative splicing: two different 3’ splice sites, a 5’ splice site, and a poly A site: this causes cleavage of the introns at the two 3’ splice sites resulting in 2 different mature RNA’s from the same primary transcript

Question 23

How does the lac operon regulate gene expression?

Answer 23

the lac operon has 3 sites for the lac repressor to bind: O1 (main) O2, and O3. The repressor is ALWAYS bound, which inhibits transcription of genes for lactose digestion enzyme unless it is necessary ( it inhibits the RNA polymerase from binding to the promotor upstream at the TATA box)

There are 2 factors that regulate the lac operon-lac repressor binding (both are necessary)

1. the concentration of lactose: if [lactose] is high, the allolactose (a deriv) can bind to the repressor, cause the repressor to conformationally change and thus it will dissociate from the operon
2. the concentration of glucose: if [glucose] is low then that means [cAMP] is high, and thus it is available to bind cAMP receptor hormone (CRP) ; this binding favors RNA polymerase to bind and then transcription (of lactase) is active **CRP DIRECTLY ATTACHES TO THE RNA POL

Question 24

What associates with TBP to assist in transcription initiation? How do they differ? What is their function generally

Answer 24

TBP = TATA box binding protein; it is a single polypeptide that sits along TATA as a molecular saddle and acts as the universal transcription factor that is essential for pol I, II, III;
Associated with 2 complexes: 1. TFIID 2. SAGA which do two essential things for the RNA Pol II initiation: 1. contain a subunit with histone acyltransferease (to acylate histone so RNA Pol can bind DNA) 2. possess TBP binding activity

SAFA: is stress induced, highly regulated and TBP Associate Factor (TAF) independent

TFIID: Houskeeping; TAF depended

Question 25

How is transcription initiated in eukaryotes?

Answer 25

eukaryotes can have activators and repressors which bind upstream, downstream, or anywhere!
Enhancers = promote, silencers = inhibit

In general:

1. Activator binds to enhancer site
2. Histone modification and remodeling complexes are activated ( to expose DNA so RNA pol can bind)
3. Mediator binds: it mediates the activity of the activator to favor TBP complexes (TFIID or SAGA) binding (communicates changes to the promoter)
4. TBP and TFIIH bind: causes an open complex (helicase) and transcription is initiated (kinase)

Question 26

What are the various modules for binding of protein to DNA?

Answer 26

All allow for protein to recognize a certain sequence of DNA
1. helix turn helix: helix interacts with DNA via the major groove
2. helix loop helix : DNA binding helix extends into the helix-loop-helix domain, the second helix of themotif extends into the carboxy-terminal ends of the subunit; interaction of the carboxy-terminal helices of the two subunits describes a coiled coil similar to that of leuricine zipper but with only one pair of leucine zipper
INTERACTS VIA ONE LEU PAIR
3. Leucine Zippers: Leu residues are found at every 7th position in the zipper, and lots of Lys (K) and Arg (R) in the DNA-binding region; The leucine zipper interacts with each other outside of DNA

Question 27

How does DNA methylation reinforce changes in gene expression?

Answer 27

occurs on C5 of cytosines;
in general there is a temporary loss of gene regulatory proteins for the purpose of decreasing its transcription. This leads to naked DNA which is exposed to the action of methylases. When DNA is methylated, the binding of transcription factors to methylated DNA is often lost.
DNA methylation increases with aging.

Question 28

Describe the distribution of the maternal gene of the drosphilia egg

Answer 28

In general: mRNA (materna) is injected into the egg via follicular / nurse cells; the mRNA is transcribed and forms proteins that act as transcription factors that regulates other proteins; so if you dont have the maternal mRNA, the proteins wont form! Bicoid = anterior, nanos = posterior
For the drosophilia, it only starts transcribing during development, no transcription factors are transcribed during embryology, so the DNA of the egg doesnt know how to start replicating; this is provided by the materna nurse cells and follicle cells which deposit maternal mRNAs ( bicoid and nanos gene transcripts) and proteins are in the developing oocyte. After fertilization, the two nuclei of the fertilized egg divide in synchrony within the common cytoplasm, and then migrate to the periphery. The membrane invaginates and creates a monolayer of cells at the periphery and cells in the posterior become germ cells.
The bicoid and nanos mRNAs are localized near the anterior and posterior poles, respectively. The caudal, hunchback, and pumillo mRNA’s are distributed through the egg cytoplasm. The gradients of bcd and nanos proteins lead to accumulation of hunchback protein in the anterior and caudal protein in the posterior of the egg ;
** removal of the bicoid gene in the mother shows that the polarity in the embryo is destroyed which proves that the mRNA for those proteins is provided from maternal mRNA.

The drosophila has one HOX gene which is responsible for the development of structures in a defined part of the body and expressed in defined regions of the embryo

Question 29

The genetic code is pretty universal, with 3 major exception

Answer 29

in mitochondial DNA (which is most similar to nuclear DNA)

1. the UGA sequence codes for TRP instead of STOP (in normal DNA)
2. AUA codes Met instead of Ile (normal)
3. AGA/AGG code STOP instead of Arg (normal)

Question 30

Describe the wobble position in DNA coding

Answer 30

caused by alternative base pairing from the traditional Watson-Crick model,
Involves codon at the 3rd base position (on mRNA) and anticodon at the 1st base position (on tRNA)
Only base pairings that DO NOT decrease the distance between the C1’ glycosidic carbons of the 2 bases are tolerated (G = to U, or I (deaminated adenosine) = C,A, U)

mainly only associated with the 6-codon family amino acids (Leu, Ser, Arg)

Question 31

Describe the two theories of the origin of the genetic code

Answer 31

1. Stereochemical Theory: phsico-chemical interactions between codons and amino acids: suggests that proximity favors catalysis of peptide bonds and that aa were binding directly to the codons when the code was established and that such complementarity by itself imposed the code

** that the original synthesis of proteins occurred directly on the RNA by the stacking of aa onto the codons because those aa had a DIRECT affinity/capacity/proximity to bind (a specific binding affinity associated with codon sequence and aa)

2. Coevolution: The earliest proteins encoded by the code were aa that dont depend on other aa precursors for their formation (prebiotically synthesized amino acids), later some of these precursors were transformed to new product amino acids

Question 32

How is translation initiated?

Answer 32

takes 3 steps to form the initiation complex (in bacteria) each step requires GTP–> GDP

1. initiation factors IF-1 and IF-3 bind to the rRNA
2. the mRNA (ATG initiation codon) is guided to the 30S subunit (at the 16s rRNA area ) by complementarily pairing the ATG mRNA to the rRNA region that is rich in purines (the Shine Dalarno sequence aka the ribosome binding site) that is UPSTREAM of the first codon (and recognized by 16s)

Question 33

How does the first translation elongation happen?

Answer 33

Note: the first amino acid (met) is at the p-site due to initiation with AUG;

1. 2nd aminoacyl-tRNA enters the A site of the ribosome, bound to enlongation fator-Tu (EF-Tu)
2. EF-Tu contains a GTP, and the binding of the 2nd aa to Asite is accompanied by hydrolysis of the GTP to GDP + Pi. and this causes a realease of EF-Tu-GDP complex from the ribosome;

** the bound GDP is released when EF-Tu-GDP binds to EF-Ts and EF-Ts is then released when it binds to GTP making EF-Tu (and the the process repeats)

Question 34

How is the peptide bond formed? what enzymes catalyzes it?

Answer 34

catalyzed by 23S RNA ribozyme (AKA peptidyl transferase)

The N-formylthrionyl on the P site is transferred to the amino group of the second aminoacyl tRNA on the A site, and forms a dipeptidyl tRNA; NOTE: the growing aa chain remains always in the P-site ** aa2 “lifts up” aa1 so that the aa stay in the P-site even though the tRNA move

both tRNAs bound to ribosome shift position to the 50S (large subunit) for a hybrid binding site, the uncharged tRNA shifts so that its 3’ and 5’ ends are in the E site. Similarly the 3’ and 5’ ends of the peptidyl tRNA shift to the P site;

THE ANTICODONS REMAIN IN THE A AND P SITES

Question 35

What is the function of the EF-G protein? How does it work?

Answer 35

Elongation Factor - G: it catalyzes the translocation of the protein;
hydrolyzes GTP for energy
How it works :
EF-G mimics EF-Tu complexed with tRNA (how acylated t-RNAs come into the Asite and then they are hydrolyzed and EF-Tu released) by mimicking the structure of the anticodon loop of tRNA in both shape and charge distribution;

The EF-G pushes the peptidyl tRNA from the P/A state to the P/P state by mimcking the binding of a new aminoacyle-tRNA at the A/T state

GTP Hydrolysis decreases the affinity of EF-G for the A site and the real aminoacyl tRNA can take its place

Question 36

What are some of the factors that affect DNA denaturation?

Answer 36

midpoint of melting (Tm) depends on

1. base composition: high CG increases Tm (because they are stronger 3 Hbonds) (GC >AT)
2. DNA length: longer DNA has a higher Tm
3. pH and ionic strength: high salt increases Tm by masking the negative charges

Question 37

What is unique about Palindromic sequences?

Answer 37

they read the same backwards/forwards

can form alternative structures with intrastrand base pairing. When only a single DNA or RNA strand is involved, the structure is called mirror, if its on two separate (opposite) strands (same sequence though) then its called a pallindrome and it can form a hairpin

restriction enzymes (endonucleases) recognize palindromic sequences (read the same way on opposite strands!)
if it is on the same strand = mirror repeat, not palindromic sequence

Question 38

Explain the basics behind DNA cloning

Answer 38

1. Cloning vector (plasmid) and eukaryotic chromosome w/DNA fragment of interest are each cleaved separately, but with the same restriction endonuclease
2. The fragments are ligated to the cloning vector via DNA ligase
3. Resulting recominbant vector is introduced to a host cell where it can be propagated/cloned
4. propagation (cloning) repoducing many copies of recombinant DNA

Question 39

Describe the key essential features of a cloning vector:

Answer 39

ori = start of replication
site that is unique: restriction site only appears once
selectable marker that have acquired plasmid vs. not acquired (provides antiobiotic resistance etc.)

Question 40

Describe polymerase chain reaction

Answer 40

IN GENERAL: denature, anneal, elongate, repeat; dont need to keep adding enzyme because the thermostable DNA Pol, but the high temp is still needed to denature (and cooling to anneal)

1. DNA strands are separated by heating
2. DNA strands are then annealed to an excess of short synthetic DNA primer that flank the region to be amplified (synthetic oligonucleotide primer)
3. new DNA is synthesized by polymerization via THERMOSTABLE DNA POLYMERASE (TaqI) which is resistant to heating cycles (not denatured)

Question 41

Describe the basics behind the sanger method. How is it opitimized today?

Answer 41

OLD WAY: = SANGER = 4 separate reactions:
radioactive/fluorescently labelled primer is annealed to the DNA to be sequenced (used as a TEMPLATE)
the ddNTP are added and basically cause the DNA to prematurely terminate to break at the ‘N’ as it is being “synthesized”; the result is a solution in which the labelled ‘N’ is found at the LAST BASE ADDED and then you have to electrophoresis to get the whole chain because ddNTP will randomly choose when to bind over dNTP. then repeat for all 4 and you will get THE COMPLEMENT strand to the DNA being tested and you read the result from the electrophorisis bottom to top correlating 5’ –> 3’ because the shortest strands move the farthest which are the ones closes to the 5’ end

where 1 ddNTP 4 dNTP for 1 reaction (repeated 4 times for all 4 ddNTP)

Today:
fluorescently label each base with a different color so all 4 ddNTP can be added at the same time; where you have 4 dideoxyNTP 4deoxyNTP in 1 reaction

Question 42

What is a Southern Blot/ How is it produced? How is it related to forensic medicine

Answer 42

Allows for recognition of Restriction Length Polymorphisms (RFLPs) (unique to each person on fingertips)
How it works: get DNA sample, cleave with a restriction endonuclease, the separate the DNA fragments and run through get electrophoresis; to this for the suspect and the evidence; the denature the DNA and transfer to a nylon membrane, incubate with a radiolabeled DNA probe and wash, the expose the x-ray film to a membrane and see what matches up

PROOF = EXCEED PROBABILITY

Question 43

How do Short Tandem Repeat work with relation to forensic medicine?

Answer 43

class of polymorphisms that occur when a pattern of 2+ nucleotides are repeated directly adjacent to each other; different small repeats in INTRONS.. 2 alleles are separated by various distances and each gene has a unique distance per individual so that when you run the tests you will get two bands per person that are slightly unique, but compared to a different person they are WAY diff (# of repeat unit varies on 2 alleles of the same person’s same gene, and of different people)

Question 44

Describe an expression vector and how it differs from a cloning vector

Answer 44

a special plasmid that allows transcription of inserted gene;
differs by having:
a promoter sequence
operator sequence
shine-dalgarno upstream of incision site (code for ribosome binding site)
transcription termination sequences
also has an ori, selectable marker (for the selection of cells contiating recombinant proteins)

even better when it is carries a gene that encodes repressor for convenient expression of gene

ex: gene to be expressed is inserted on one of the restriction sites in the polylinker, near the promoter and then the end encodes the amino terminus proximal to the promoter

Question 45

How is cDNA constructed

Answer 45

eukaryotic DNA canot be expressed correctly in bacterium because euk genes have 1. exons = coding regions 2. intron and bacteria can’t splice introns. So we need to transfer mRNA (intron free) into expression vector.
cDNA Library = only the genes in that SPECIFIC CELL (very specific for cell type/time)
So we take advantage of the fact that all mRNA have a poly A tail that helps purify it in a mixture and serves as a universal template, once we have mRNA we can make it into cDNA (DNA complement of mRNA = only exons) and then we can use that!

How to make cDNA:

1. get mRNA from a euk cell; thats the template! anneal a synthetic oligonucleotide primer of TTTTT to complementary pair with the poly A tail
2. Reverse transcriptase and dNTP yield a complementary strand
3. mRNA is degraded
4. Oligonucleotid of know sequence is ligated to 3’ end of the new DNA to as as a primer for a 2nd strang
5. DNA pol I and dNTP extend to yield a dsDNA known as the duplex DNA aka the cDNA

Question 46

Explain the basics behind how a microarray works

Answer 46

general function: to measure the expression levels of large numbers of genes simultaneously or to genotype multiple regions of a genome; need to use mRNA though because we only care about the expressed code (without intron); thus you have to make a cDNA in order for the cDNA to complementary base pair with the DNA which is positioned on a solid surface via little drops.

DNA is crosslinked to the surface of the solid chip via UV light. After the DNA is attached to the surface the microarray can be probed with other fluorescently labeled nucleic acids. Then add the cDNA from the different cells (of two diff stages of development) and use a fluorescently labeled dNTP when you make the cDNA via reverse transcriptase from the mRNA; as you add the cDNA it will anneal to complementary sequences and each fluorescent spot represents a gene expressed (use different colors for different mRNA)

Question 47

what is a selectable marker? How are they dependent on homology? Ex?

Answer 47

gene introduced into a cell, especially a bacterium or to cells in culture, that confers a trait suitable for artificial selection. What this means is, the experimenter can tell the right gene is in the cell because the marker can be seen or detected.
Most often, this is used for bacteria or for cells in culture. Selectable markers show the success of a transfection or other procedure meant to introduce foreign DNA into a cell. It is a technique in gene targeting and gene knockout.

Selectable markers are often antibiotic resistance genes; bacteria that have been subjected to a procedure to introduce foreign DNA are grown on a medium containing an antibiotic. The antibiotic knocks out cells which do not have the resistant marker. Those bacterial colonies that can grow have successfully taken up and expressed the introduced genetic material.

Gene knockout = one of an organism’s genes is switched off or replaced by one which does not work. Its a double selection: take out the gene and non-homologous recombination.

two main ways to introduce genes to euk cells:

1. Transfection: used to disrupt chromosomal genes using a positive-negative selection
2. Infection (via a virus)

Examples:
Introduce APH gene (NEO) which inactivate G418 (which inhibits protein synthesis
Intoduce Thymidine Kinase (TK): positive selection: Aminopterin (inhibits de novo purine and thymidylate synthesis)
negative selection: gancyclovir: kills cells which are TK+
the selection mechanism: TK synthesizes thymidylate and activates gancyclovir (which inhibts incorporation of dGTP by DNA polymerase)

Extent of genes present is proportional to the extent of homology and only the cells with original knockout are resistance: So if the genes are inserted at random place (nonhomologous) then the entire fragment is expressed because there is no secquence that it has to stop at. But if the gene is insert at the target, the length of the exchange depends of the length of the homology such that crossing over only occurs until there is no more homology, so if that occurs before the second gene is inserted, the the second recominant gene is not expressed/transfected (in this case you would get APH resistance but would not activate Gancyclovir)

Question 48

How are transgenic mice made?

Answer 48

The genes of interest can be microinjected into either (transfection) 1. pronuclei of fertilzed eggs or 2. embryonic stem cells of the blastocyst and then 1 or 2 are introduced into pseudopregnant female mouse and then you will get it a chimeric mouse and then you can select aguty progeny

In the first way you are directly injecting the DNA in the pronucleus so then after the egg is implanted in a pseudopregnant mouse some of the mice great the new DNA from that was transfected

In the second way (via an ES cell), you get cells from the blastocyst (embyronic) so they can hypothetically grow into anything and you grow them on a plate with a feeder layer then you inject the cells back into a blastocyst which gives rise to a mouse. You do a DNA analysis and see how it has the gene (that is expressed visually) ie: test for homozygous or heterzygous; you can have two products: one that is a knockout (both alleles are out) or just one and you that by measuring the DNA and seeing if you have 2 or 1 lines (2 = heterozygous, 1 = knock out homozygous); then you know you have the correct gene. so you grow it on a Petri dish with the antibiotic you are testing against extract the colonies that survived but that into a blastocyst and inject that into a pseudopregant female. THen you take the progeny of those babies and you will have the genes in mice

BASICALLY TWO METHODS:

1. direct physical contact interact with transgene and pronuclei
2. Indroduce via transfection into ES cells via PCR procedure and a two step process in which the first generation are chimeric and then select the aguty progeny

Question 49

Explain the basics behind the circuit that represents the muscle contraction/cell membranes (ie all the components and how they effect the current)

Answer 49

Battery = equilibrium potential = reverse potential = diffusion potential; it is an intrinsic quantity that relates how readily an ion will diffuse the semipermeable membrane and in what direction (negative = into the cell) ; it is based on the [ions] inside vs [ion] outside. At some point equilibrium is reached because the chemical gradient in one direction is equal to the electrical gradient in the other direction;

Resistors = conductance; it can be voltage gated (as in a voltage gate channel) or not gated (as in a leak channel); the voltage gated ones change according to voltage and the leak channels function in either direction;
Vi - Vo = V = membrane potential = the relative charges on the inside minus those on the outside of a cell membrane; defines the overall electrical difference of all of the ions in/out and it is formed by the diffusions of a [ion];

Capacitor = the membrane; generates a difference in voltage potential by storing charge on either side of the membrane; ions are pumped from one side of the capacity to the other side

You have a net flow of ions (current) running through the circuit; the overall current determines whether the ions are depolarizing (-) or repolarizing (+) the membrane, regardless of ion charge

Question 50

How do you calculate the current flowing through the membrane? Describe it qualitatively also. Also when dies equilibrium occur with respect to the current and conductance? How is the Na/Ca pump’s current effected?

Answer 50

Current (I) = (V- E) *G where V = Vin -Vout
IF current = + then it is repolarizing
IF current = - then it is depolarizing

A + charge produces a + outward current and a - inward current
A - charge produces a + inward current and a - outward current

EX: the equilibrium potential of Na + is 67 mV the membrane potential was found to be 60mV what is the current running through the cell due to the Na leak channels (G = 1)

I = (-60-67)*1 = -137 mV the current is depolarizing and inward

EX 2: Equilibrium potential of Cl- = 123mV the membrane potential of the cell is -60 mV what is the current via the leak channels?

I = (-60 - 123)*1 = -180 which is depolarizing OUTWARD

Equilibrium occurs when the overall conductances (of voltage gated channels) matches the the ratio of the current through the N and K channels by a 1.5 (to equilibrate the 3 Na+ in 2K+ out)

The Na/K exchange carrier at first it positive (becuase the calcium is being brought in at a much faster rate then the Na and that leaves an overall + current aka doesn’t help in the depolarizing) then later it is reversed as Ca2+ needs to be brought back to reserves in the sarcolemma

Question 51

How is the muscle contracted? What are the muscle bands?

Answer 51

1. At rest there is no Ca 2+ and the troponin is bound to tropomyosin which blocks out the myosin active site on the actin molecule; Once Ach binds to the nicotinic receptor it causes Na release which continues to depolarize the membrane and as a result causes the release of Ca2+ via Sodium-gated Calcium channels, this causes the calcium to stimulate the Sarcoplasmic Reticulum to release more caclium and this calcium is what triggers the contraction of the muscle; Calcium binds troponin which then causes it to dissociate and forces tropomyosin off of actin so that the myosin binding sites are exposed; At rest the myosin head is bound to ADP and pi and is extended forward in a high energy conformation;
2. Now that the myosin is site is exposed on actin, myosin -ADP - Pi form a cross bridge with actin
3. ADP and Pi are released CAUSING THE POWER STROKE and movement of ACTIN (I band is shortened) opposite direction of myosin
4. Ca 2+ is released which causes the myosin binding sites to bind tropomyosin (block contraction) and ATP is hydrolyzed to make the high energy conformation of myosin ADP Pi

Question 52

How does the body store ATP for muscle contractions? Describe everything involved with this molecule

Answer 52

ATP is stored via creatine-Phosphate; creatine is formed by glycine and arginine precursors, and then creatine is phosphorylated which requires an ATP (so when it is dephosphorylated it released ATP), creatine-P is constantly being removed from the body creatinine which is excreted via urine.

Creatine is extremely needed for short duration high intensity muscle contraction because it supplies the necessary (quick and fast) ATP via creatine P (** longer exercises require a larger energy supply, and creatine is quickly used up in the first few minutes of working out)

Question 53

What are the energy supplies for the muscle during exercise?

Answer 53

Initially you use creatine-P, aerobic and anaerobic glycolysis, glyogenolysis (secreting more and more glucose from glycogen for glycolysis) eventually your body turns to fatty acid oxidation but once all of the supplies of oaxaloacetate are run out because you went through the glycolysis too many times so now you need a new source: the purine nucleotide cycle (using amino acids for TCA derivatives) turns on **especially aspartate amino acid

Question 54

Describe the major difference and processing of the two different procedures of microarrays

Answer 54

In general: the point is to allow for simultaneous screening of many thousands of genes to allow for genotyping (What genes are present, tissue specific gene expression, mutation analysis etc.)
1. One -color = affymetrix= photolithographic technique
Target = biotin labeled, BOUND to matrix
Probe = single stranded DNA, free in solution;
based on the fact that the probe will complementary base pair and hybridize with the target creating a fluorescently labeled marker
First need to make the chip:
Uses photolithography:
Use and opaque screen to cover the plate where you dont want a base, thus the other regions (With the “blocked groups”) are exposed; shine a light and wash the chip with activated but light-blocked Base 1 (ie: A) then cover all of the spots you dont want to add Base 2, and shine a light, the light unblocks base 1 and then wash with active but blocked base 2 etc. repeat until you get the 22 bp long oligonucleotide of choice (This is all computer/robotic). Each gene is represent by 22 different oligonucleotide. Now the chip is ready, you have tons and tons of oligs (can be different!) representing the different genes.

Now make the probe: the probe is made from mRNA –> cDNA because we only care about the aspects that are expressed; the cDNA makes a duplicate so you have dscDNA; then this DNA is coverted cRNA that can be labelled with biotin-streptavidinphycoerythrin that is then BOUND TO THE MATRIX

Introduce the probe (cRNA) to the microarray of the all the genes and let them “hybridize” then wash the chip and see which spots illuminate; fluorescent intensity is proportional to the concentration of bound nucleic acid duplexes

METHOD 2: Two color = academic method
Target: cDNA aka cRNA with the fluorescent tag; FREE
Probe: BOUND oligonucleotide sequence to the chip
DNA is bound to the chip via cross linkage due to UV; once the DNA is bound, the chip can be probed with any fluorescently labelled nucleic acid;
The TARGET is radiolabelled, NOT the PROBE (like in version I)

in this version, you obtain ALL the genes in the genome (All the cDNA) and then new synthesis of cDNA from specific cell types and then those are collected

vs. the one color (affymetric) you have a single gene with multiple olig that aneal with gene and directly sunthesize on top.