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Flashcards in Gatti Review I Deck (54)
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1

UV absorption of Nucleobases

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

2

Describe the composition of a nucleotide

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);

3

Describe the three major forms of DNA

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

4

Describe the three DNA polymerases

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)

5

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

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)

6

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

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)

7

Explain what telomerase is

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

8

Describe the key features of homologous recombination

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)

9

Describe the features of site specific recombination

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)

10

Describe the key features of exon shuffling

Non-homologous, non-site specific;
Can transfer pieces from one gene into another (via transposons!)

11

Describe how DNA repair is modulated by base excision

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

12

How is RNA different/similar to DNA?

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

13

Describe the structure of bascterial RNA polymerase

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.

14

Describe some of the features of a promoter:

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

15

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

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

16

Describe the transcription terminators

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")

17

RNA polymerase in prokaryote vs. eukaryotes

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

18

How is transcription initiated in eukaryotes?

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

19

Describe how the 5' cap of mRNA is formed

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 )

20

Describe Group I intron splicing

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)

21

Describe Group II intron splicing

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)

22

Describe polyadenylation and the mechanisms of processing and variability

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

23

How does the lac operon regulate gene expression?

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

24

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

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

25

How is transcription initiated in eukaryotes?

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)

26

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

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

27

How does DNA methylation reinforce changes in gene expression?

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.

28

Describe the distribution of the maternal gene of the drosphilia egg

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

29

The genetic code is pretty universal, with 3 major exception

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)

30

Describe the wobble position in DNA coding

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)