Genetic Information Transfer

Question 1

Genetic info is stored in a class of molecules called _____.

Answer 1

nucleic acids

Question 2

The central dogma of genetic information transfer…

Answer 2

• DNA polymerase replicates DNA.
• RNA polymerase transcribes DNA > RNA.
• Ribosomes translate RNA to proteins.

Question 3

Key structural difference beween RNA and DNA

Answer 3

• RiboNucleic Acid has an OH group on the 2’ carbon of its sugar.
• DeoxyriboNucleic Acid has a H on the 2’ carbon of its sugar (aka deoxidized).

Question 4

Nucleotides are comprised of…

Answer 4

• a base (a purine or pyrimidine)
• a (deoxy) ribose, aka a sugar
• and phosphate PO4 (one or more)

Question 5

Nucleotides are joined together through…

Answer 5

Nucleotides are joined together through phosphodiester bonds between the 3’ oxygen on each nucleotide’s sugar and the phosphate group of the neighboring nucleotide attached to its sugar’s 5’ carbon.

These phosphodiester bonds form the “backbone” of the polynucleotide.

Question 6

Polynucleotides structures of one-strand and 2-strand DNA are stabilized through…

Answer 6

base stacking.

Question 7

purine (A, G) and pyrimidine (T/U, C) pair via…

Answer 7

Hydrogen bonds.

The conformation of the base can affect H-bonding.

Question 8

“sugar pucker” conformation…

Answer 8

The “sugar pucker” conformation affects nucleic acid’s tertiary structure by influencing whether it is tightly or loosely compacted.

It determines the distances between phosphates in the phosphodiester backbone.
The conformation of the base (syn vs. anti) determines how nucleotide bases interact.

DNA and RNA favor DIFFERENT sugar pucker conformations.
Z-form DNA includes both anti and syn conformations.

Question 9

DNA structure basics

Answer 9

Typically 2-stranded in cells.
The two strands run antiparallel (3’ and 5’ on opposite ends).
Turns right-handed.

Question 10

RNA structure basics

Answer 10

Typically single-stranded in cells.
Widely variable secondary structures.
Multi-functional molecule.

Question 11

In forming nucleotide chains (polynucleotides), which end serves as nucleophile?

Answer 11

The 3’ hydroxyl is deprotonoated and acts as nucleophile, attacking the phophate off of the 5’ C on the next nucleotide.
Because of this, nucleotides are always added in a 5’ to 3’ manner.

Question 12

nucleoside

Answer 12

A nucleoSIDE is a base connected to a pentose (5 carbon) sugar.
(no phosphate group)

Question 13

glycosidic bond

Answer 13

The glycosidic bond links the base and the sugar to form a nucleoSIDE.
Occurs between the 1’ position carbon on the sugar and a nitrogen on the purine or pyrimidine base.

Question 14

nucleoside monophosphate

Answer 14

A nucleotide with 1 phosphate group.
(phosphate-sugar-base)

This is what the final nucleic acid polymer is composed of.

Question 15

nucleoside triphosphate

Answer 15

A nucleotide with 3 phosphate groups.
(phosphate-phosphate-phosphate-sugar-base)

These serve as high-energy building blocks used by the cells to synthesize the nucleic acid polymer.

Question 16

Core chemical mechanism of joining nucleotides to form DNA:

Answer 16

1) Oxygen on 3’ carbon activated via deprotonation by a base.
2) 3’ O acts as nucleophile, attacking alpha phosphate of a nucleoside triphosphate.
3) Pyrophosphate (2 of the 3 phosphates) acts as a leaving group to drive the reaction forward.

The nucleotides always add to the 3’ end, so the strands are always extended from 5’ to 3’.

Question 17

most common nucleotide base pairing pattern

Answer 17

H-bonds occur between the amino and carbonyl groups.
In Watson-Crick-Franklin base-pairing, H-bonds occur between the bases of two strands running anti-parallel to each other.
A-T (2-H bonds between adenine and thymine)
C-G (3 H-bonds between guanine and cytosine)

Question 18

B-form double-stranded DNA helix

Answer 18

• Right-handed helix.
• Strands are antiparallel (maximizing WCF base-pairing).
• Sugar-phosphate backbone is oriented to the outside.
• Nucleobases are oriented to the inside.

Question 19

Does B-form DNA need to unwind to interact?

Answer 19

NO! There is a wide, deep MAJOR groove

and a shallow, narrow MINOR groove in the B-form double helix.

Question 20

Hoogsteen base pairs

Answer 20

When anti/syn-conformation nucleotides form base pairs.

(vs. traditional Watson-Crick-Franklin anti/anti base pairs)

Only the purine nucleotides (A, G) can take on syn conformation.

Question 21

Linking Number

Answer 21

Linking number defines the topology of DNA twists.
LK=TW+Wr.
Linking number = sum of its twist and writhe.
Twist and writhe are inversely proportional.
B-form geometry, with equal numbers of Tw and Wr is very energetically favorable!

Question 22

supercoiling

Answer 22

changing twists in DNA molecule.
(recall yoyo string example).

Supercoiled DNA is more compact.

If delta Lk=0, then DNA is relaxed.

If delta Lk<0, then DNA is unwound and negative supercoiling occurs. (most common in cells)

If delta Lk>0, then DNA is overwound and positive supercoiling will occur.

Question 23

Topoisomerases

Answer 23

Topoisomerases are enzymes that change DNA topology by changing the linking number of DNA. This involves cutting, rearranging, and resealing the structure.

Help DNA compact and deal with disruptive structures that might form during replication and transcription.

Question 24

Melting Temperature of DNA

Answer 24

Tmelting = the temperature at which the helix is half double-stranded and half-single-stranded.
(aka 50% denatured)
Stable helix = high Tm
Unstable helix = low Tm

Question 25

Nucleosome

Answer 25

DNA wrapped around histone proteins. The fundamental unit of eucharyothic genome packaging. DNA must be unwrapped from around histone before a gene can be transcribed.

Question 26

Chromatin

Answer 26

A form of DNA packaging. Tight and very compact, but not very accessible.

Question 27

Histone acetylation

Answer 27

Marker that makes DNA MORE ACCESSIBLE. Found in euchromatic regions.

Question 28

Histone methylation

Answer 28

Marker making DNA INACCESSIBLE. Found in heterochromatic regions. Also marked by binding of heterochromatin proteins.

Question 29

What is the preferred sugar-pucker conformation of DNA nucleotides?

Answer 29

C2 endo conformation; dominates B-strand DNA. | C3 endo conformation can be adapted by A-strand DNA.

Question 30

phosphodiester bond

Answer 30

phosphodiester bond occurs between the 3' OH and the phospho group on the 5' C. Links the nucleotides in the polynucleotide strand. A base activates the OH for nucleophilic attack of the dNTP (deoxyribonucleotide 5'-triphosphates). diphosphates act as leaving group.

Question 31

Why is DNA the principal storage, and not RNA?

Answer 31

RNA is more suseptible to degradation than DNA. The 2' OH makes it more reactive and can break down the chain.

Question 32

DNA Polymerase

Answer 32

the enzyme (and protein) that catalyzes nucleotide addition for replication. Requires a primer--a short double-stranded segment--to start replication. Uses metal ions (Mg2+) to interact with triphosphate groups of incoming nucleotides, stabilizing neg charged transition state.

Question 33

Why does DNA polymerase require a primer?

Answer 33

base-pairing provides a portion of the stability; the primer facilitates base STACKING, which is energetically favorable.

Question 34

exonuclease

Answer 34

When DNA polymerase makes mistakes and its own proofreading doesn't catch it, it can cause instability that leads the DNA strand to the exonuclease site on the polymerase. The exonuclease removes the erroneous base, and polymerase has another chance to add the correct base. Cleaves 1 nucleotide at a time from the END of a strand.

Question 35

frequency of DNA mistakes

Answer 35

DNA polymerase error rate is about every 1 in 100,000 to 10,000,000 additions

Question 36

What happens at the DNA replication fork?

Answer 36

leading strand and lagging strand split in opposite directions. The leading strand undergoes continuous synthesis 5' to 3'. The lagging strand requires a series of primers to synthesize 5' to 3'. Therefore, it is made of a series of discontinuous Okazaki fragments, linked together by DNA ligase.

Question 37

Telomerase

Answer 37

prevents shortening at the end of the lagging strand during DNA replication.

Question 38

helicases

Answer 38

provide the energy to unwind double-strand DNA

Question 39

primase

Answer 39

adds DNA primers for replication.

Question 40

Pol III and Pol I

Answer 40

Two types of polymerase used in DNA genome replication. Pol III is the workhorse, built for speed. Pol I is the craftsman, built for accuracy.

Question 41

General steps in DNA replication

Answer 41

1. helicases unwind the double strand 2. primase adds primers 3. Pol III starts replicating 4. unwinding continues 5. lagging strand needs more activity--replace RNA starters with DNA and link fragments with ligase. The product is semi-conservative replication!

Question 42

5 basic steps to DNA repair

Answer 42

1. detection 2. incision 3. excision 4. synthesis 5. ligation (ligase seals)

Question 43

2 main paths to mutation in the genome:

Answer 43

1. polymerase makes a mistake and misses it proofreading. | 2. chemistry takes a toll; modified base will lead to incorrect base pairing during future replication.

Question 44

The lac operon

Answer 44

model from E. coli using transcriptional activators and repressors to regulate gene expression.

Question 45

DNA binding proteins in gene regulation and expression

Answer 45

transcriptional activators and repressors make sequence-specific interactions with DNA. Sequences are read through side-chains that H-bond with DNA bases. Does NOT require melting the DNA! DNA binding activity is often separate from regulatory activity (proteins are modular).

Question 46

regulation of chromatin structure (a type of epigenetic regulation)

Answer 46

relation of chromatin structure by histone proteins; positioning and post-translational modifications

Question 47

chemical modification of nucleosides (a type of epignenetic regulation)

Answer 47

DNA base is modified

Question 48

heterochromatin regions of DNA...

Answer 48

are transcriptionally silent | aka not accessible.

Question 49

euchromatic regions of DNA...

Answer 49

are transcriptionally active | aka accessible

Question 50

heterochromatic regions of chromatin...

Answer 50

are transcriptionally silent | aka not accessible.

Question 51

euchromatic regions of chromatin...

Answer 51

are transcriptionally active | aka accessible

Question 52

acetylation (lysine)

Answer 52

makes DNA more accessible. Tends to be found in euchromatic regions. Active gene expression site.

Question 53

acetylation (lysine)

Answer 53

makes DNA more accessible. Tends to be found in euchromatic regions. Active gene expression site. Provides binding sites for transcriptional activators.

Question 54

5-methylcytosine acts to

Answer 54

silence gene expression

Question 55

5-methylcytosine acts to...

Answer 55

silence gene expression

Question 56

DNA repair pathways--direct repair of alkylation by methyltransferase

Answer 56

- Carbonyl oxygen of base (G) bonds to an alkyl chain of some sort. - AGT protein bonds with alkyl via SN2, base is L.G. - 1 AGT protein sacrificed for each alkylation!

Question 57

DNA repair pathways--mismatch repair of newly-synthesized strand

Answer 57

- newly-synthesized daughter strand has a mismatch, parent strand still methyl-marked. - proteins assembled to recognize and spool around the mismatch; cleave daughter strand at -CH3. - exonuclease "chews" back fragment on daughter strand to the preceding -CH3 so polymerase can try again.

Question 58

DNA repair pathways-base excision repair

Answer 58

- Deamination leaves DNA with wrong base OR depurination leaves DNA with NO base. - glycosylase enzyme cleaves the base - AP nuclease recognizes the empty space, cleaves backbone - DNA polymerase extends chain with new base - ligase "links" with a new phosphodiester bond

Question 59

DNA repair pathways-nucleotide excision repair

Answer 59

- pathway for chemical modifications to DNA that disrupt base-pair interactions--deamination, alkylation, depurination or UV-induced. - NT excision surveillance complex recognizes, attaches, and cleaves 1 strand on either side of the dimers, leaving a knicked substrate. - helicase separates the strands, removing dimer fragment. - DNA polymerase extends chain at the gap. - Ligase seals.

Question 60

UV dimers

Answer 60

UV-induced dimerization are molecular "lesions" formed on stacked pyrimidine bases (T or C); sometimes many nucleotides long

Question 61

deamination

Answer 61

nucleotide in sequence has NO BASE. Fixed via base-excision repair.

Question 62

depurination

Answer 62

removal of a purine base. Fixed via base-excision repair.

Question 63

endonuclease

Answer 63

An enzyme that cleaves the phosphodiester bond WITHIN a chain of nucleotides (vs. exonuclease cleaving from the end).

Question 64

Homologous Recombination

Answer 64

- For big repair jobs. - Repairs DNA double strand breaks via Holliday Junction. - find homologous strand, close match - chew back strands to have 3' overhang to build off of. - strand invasion is when 3' end interacts with the in-tact chromosome; forms 4-way Holliday Junction - resolved by cutting junction either vertically or horizontally--2 diff outcomes! - this is also a tool of the genome to generate variety

Question 65

Sequence-specific recombination --Tyrosine recombinase

Answer 65

- For big repair/ rearrangement. - Begins with Tyrosine OH attacking phosphate in DNA backbone - 2 sequential single-strand breaks - holliday junction intermediate

Question 66

Sequence-specific recombination -- Serine recombinase

Answer 66

- For big repair/rearrangement - Both strands cleaved simultaneously; rotations; ligase seals - NO holliday junction

Question 67

Recombination -- Transposons

Answer 67

- transposons = mobile DNA elements that move around the genome - developed by Barbara McClintock via studying corn pigmentation - transposons give a mosaic blend of different genotypes - 2 types: 1) Retro-transposons, "copy/paste" via RNA intermediate 2) DNA transposons, "cut/paste" (no RNA intermediate)

Question 68

RNA polymerases

Answer 68

- catalyze RNA synthesis from DNA templates (for eventual translation into proteins) - Don't require a primer (like DNA does) - Chemistry is energetically favorable, so can unwind as it goes (without needing helicase to open like DNA does) - only one strand transcribed at a time (results in a copy of the "coding strand" using the "template strand"). - No exonuclease site, but can still back-track to make corrections

Question 69

promoter sequences

Answer 69

- specifies where RNA polymerase will initiate transcription - RNA polymerase binds to promoter - sigma subunit tells RNA polymerase what sequences are promoters - strands melt and only one strand of DNA can be transcribed at a time.

Question 70

Terminating transcription

Answer 70

in prokaryotes, there are 2 primary mechanisms to STOP transcription. 1) Rho dependent 2) Rho independent They both disrupt RNA polymerase from interacting with template strand.

Question 71

Rho-dependent transcription termination

Answer 71

relies on the helicase (unwinding)

Question 72

Hairpin (Rho-independent) transcription termination

Answer 72

- relies on a G/C rich region to make a hairpin structure, with U-rich region downstream - intrinsic to the sequence

Question 73

epigenetics

Answer 73

changes in gene expression not caused by changes in DNA sequence 3 types for this class: 1) regulation of chromatin structure 2) chemical modification of nucleosides 3) microRNA-mediated regulation

Question 74

DNA wraps around ____ proteins to form _____. Then further compacted to _____, then chromosomes.

Answer 74

DNA wraps around histone proteins to form nucleosomes. Then further compacted to chromatin, then chromosomes.

Question 75

3 major classes of RNA in the cell

Answer 75

1) messenger RNA (mRNA)--encodes protein sequences 2) transfer RNA (tRNA)-- 3) ribosomal RNA--

Question 76

introns vs. exons

Answer 76

Introns - gene sequences removed from the RNA transcript | Exons - sequences that remain in the RNA transcript; EXONS are EXPRESSED

Question 77

micro RNAs

Answer 77

another type of RNA molecule that regulates gene expression - small fragments encoded in the genome - use base-pairing interactions to target mRNAs for regulation

Question 78

spliceosome

Answer 78

- in Eukaryotes, the spliceosome splices/removes introns from RNA - happens in the nucleus

Question 79

Processing of mRNA before exiting nucleus (in Eukaryotes)

Answer 79

To signal the mRNA is mature: 1) splicing of introns 2) 5' cap (7-methylguanosine) put on mRNA 3) 3' poly A tail added on by Polyadenylate Polymerase (PAP) Stabilizes and prevents foreign invaders

Question 80

tRNA

Answer 80

tRNAs are adaptor molecules that link a triple codon to its associated amino acid. - Use Watson-Crick-Franklin base-pairing interactions to read triplet codons in the mRNA - "charged" with their amino acid by tRNA synthetases

Question 81

genetic code for translation

Answer 81

A table mapping codons (each made of 3 nucleotides) to specific amino acids. Redundant and universal.

Question 82

Wobble hypothesis

Answer 82

- ~45 tRNA synthetases, but 64 codons - third base of codon is more relaxed and gives coding flexibility; the same synthetases can read multiple codons - Inosine (I) can be put in 3rd spot and can pair with multiple nucleobases

Question 83

start codon sequence, Shine-Dalgarno sequence

Answer 83

start codon sequence: AUG | The Shine-Dalgarno sequence indicates to the ribosome which AUG codon to start at.

Question 84

stop codon sequences

Answer 84

UAG, UAA, UGA "Release factor" protein in ribosome recognizes the stop codon on AA and promotes polypeptide release and translation termination. Nucleotides that STOP synthesis lack a 3' OH!

Question 85

3 consequences of mutations, defined by genetic code and reading frame

Answer 85

1) Silent mutation - no affect on polypeptide 2) Mis-sense mutation- changes mRNA sequence from a codon for one amino acid to a codon for another; may not have major affect 3) Non-sense mutation - change the mRNA sequence form a codon for an amino acid to a STOP codon, terminating translation; can cause BIG problems

Question 86

The ribosome

Answer 86

- The site of protein synthesis and translation. - Has both RNA and protein components (similar to splicisome) - Responsible for: 1. initiating synthesis at start codon AUG 2. synthesis of peptide bonds between amino acids 3. terminating synthesis at stop codon (UAG, UAA, UGA)

Question 87

Polymerase Chain Reaction (PCR) technique for making synthetic DNA

Answer 87

If we know the end (primer) of a DNA segment of interest, we can massively amplify it via repeated cycles of denaturing, primer annealing, and polymerase extension. The primer is incorporated into the product sequence. Overlapping DNA fragments can be assembled into larger genes via Gibson assembly (like a recombination reaction)

Question 88

plasmid

Answer 88

a circular DNA molecule that replicates separately from the host molecule.

Question 89

oligonucleotide (oligos)

Answer 89

short DNA sequence that can be synthesized chemically for use of changing the genetic composition of cells.

Question 90

Sanger Method of reading DNA sequences

Answer 90

Exploits nucleotide structure; remove 3' hydroxyl (di-deoxynucleotide) from one of the nucleotides (ATC or G). Put a small amount of this in, and every so often (1 in 50 times or so); strand will terminate wherever it is incorporated. By looking at the intermediate products, you know the position of that sequence. Repeat for others (ATC or G), and you can run on gel and read the entire sequence! In modern times, fluorescently labeled ddNTPs (one color for each reaction) so you can do this in a single reaction.

Question 91

Nanopore method of DNA sequence reading

Answer 91

DNA fed through a protein-based pore in a membrane. Changes in conductance as the DNA goes through the membrane distinguishes the slightly different voltage between A, T, C, & G. Less accurate, but can read millions of bases in a single read.

Question 92

self-splicing introns

Answer 92

segments of RNA that can splice themselves out of an RNA polymer without the assistance of any proteins.

Question 93

2 types of self-splicing introns

Answer 93

1. Group I Introns --Use a free guanosine nucleotide to initiate cleavage. 3' to 5' nucleophilic attach on phosphodiester backbone joins exon ends. Released intron is a linear fragment. 2. Group II Introns --Use an adenosine within the intron to initiate cleavage.3' to 5' nucleophilic attach on phosphodiester backbone joins exon ends. Released intron is a lariat structure (loop-like).

Question 94

2 types of self-splicing introns

Answer 94

1. Group I Introns --Use a free guanosine nucleotide to initiate cleavage. 3' to 5' nucleophilic attach on phosphodiester backbone joins exon ends. Released intron is a linear fragment. 2. Group II Introns --Use an adenosine within the intron to initiate cleavage.3' to 5' nucleophilic attach on phosphodiester backbone joins exon ends. Released intron is a lariat structure (loop-like). Splicisome might be evolutionarily related to the group II intron.