CC 1B Transmission of genetic information from the gene to the protein

Question 1

Nucleic Acid Structure & Function

Descreibe the nucleic acids

Answer 1

• Nucleic acids can be DNA or RNA, single-stranded or double-stranded
• Protein coat covers the nucleic acid
• The 2 single-strands are anti-parallel to each other. Going from 5’ to 3’ of one strand means going 3’ to 5’ of other strand.

Question 2

Difference between nucleaotide & nucleosides

Answer 2

• nucleaotide: BASE (Adenine, Guanine, Thymine, Cytosine) + sugar + phosphate
• nucleosides: BASE + SUGAR = Adenosine, Guanosine, Thymidine, Cytidine

Question 3

Sugar phosphate backbone

Answer 3

• Important structural component of DNA which consists of the pentose sugar and phosphate groups
• Sugars linked together by a phosphodiester bond

Question 4

Pyrimidine & Purines

How do base pairing pair? How many hydrogen bonds?

Which pair is stronger?

Answer 4

• Pyrimidine: C, T, & U (2 rings) “pyramids CUT”
• Purines: A + G (1 ring) “pure As Gold”

Base pairing specificity: A with T, G with C

o A forms 2 hydrogen bonds with T
o G forms 3 hydrogen bonds with C
o GC bonds are stronger. DNA with high GC content harder to break apart.

o Complementary strands of DNA hydrogen bond with each other.

Question 5

Function in transmission of genetic information

How does DNA transmit genetic information?

Answer 5

• Because of the complementary nature of base pairing, DNA can transmit genetic information through replication

Question 6

DNA denaturation

DNA reanealing

DNA hybridization

Answer 6

• Disruption of the hydrogen bonds, such as with high temperature, can cause the unwinding of the two strands (denaturation), which can then also be brought back together when proper conditions return (reannealing)
• A single strand of DNA will readily bind another single strand DNA in process of _hybridization_ where there is significant amount of base pair matching between their sequences

Question 7

DNA REPLICATION

mechanism of replication: seperation of strands, specific coupling of free nucleic acids

FIRST STEP OF REPLICATION & ENZYMES:

• DNA GYRASE*
• HELICASE*
• SINGLE-STRANDED BINDING PROTEIN*

Answer 7

Double-stranded DNA must separate or unwind. To do this:

• DNA gyrase (class II topoisomerase) responsible for uncoiling the DNA ahead of the replication fork
• Helicase responsible for unwinding DNA at replication fork
• Single-strand binding protein (SSB) responsible for keeping DNA unwound after helicase. SSBs stabilize ssDNA by binding to it.

Question 8

SECOND STEP OF DNA REPLICATION

Primase & DNA Polymerase

Wchi is a leading/lagging strand?

Which stran contains Okazaki fragments?

Answer 8

You start making DNA that is complementary to the unwound/separated DNA. Note, all biological DNA synthesis occurs from 5’ to 3’ end.

• Primase lays down short RNA primer on unwound DNA.
  
  • Primer made of RNA but is complementary to DNA sequence.
  • Later, this RNA is replaced with DNA.
• DNA polymerase takes over and makes DNA that is complementary to unwound DNA.
• DNA synthesis occurs on both strands of unwound DNA.
  
  • Synthesis that proceeds in direction of replication fork is leading strand.
  • Synthesis that proceeds in opposite direction to replication fork is lagging strand.
    
    • ​Lagging strand contains Okazaki fragments.

Question 9

THIRD STEP OF DNA REPLICATION

Answer 9

3. RNA primers replaced with DNA by a special DNA polymerase. Okazaki fragments in lagging strands are stitched together by DNA ligase.

• DNA synthesis is bidirectional: 2 replication forks form and proceeds in opposite directions.
• Biological DNA synthesis always proceeds from 5’ to 3’ end.
• DNA polymerase has proofreading activity, corrects any mistakes (mutations) it makes
• Replication occurs once every cell generation, during the S phase. (Cell division may occur twice in meiosis, but replication still only occurs once)

Question 10

What does it mean to say that DNA is a Semi-conservative nature of replication?

How was this proved?

What if DNA was completetly conseravative?

What if DNA was dispersive?

Answer 10

• Newly synthesized DNA contains one old strand and one new strand
• Meselson and Stahl proved this by experiment: used heavy (15N) DNA as old (pre-replication) DNA and used light (14N) nucleotides for synthesis of new DNA.
  
  • They can tell difference between heavy and light DNA by centrifugation. They found that when heavy DNA undergoes one round of replication in light nucleotides, the DNA is made of intermediate weight. After second round of replication, DNA is split between intermediate and light weight.
• If DNA replication were completely conservative, only heavy and light DNA would be seen, nothing in between.
• If DNA replication were dispersive, everything would be of intermediate weight. This was not the case because after second round of replication, light DNA was seen.

Question 11

Overview of specific enzymes involved in replication

1. _____________: uses hydrolysis of ATP to “unzip” or unwind DNA helix at replication fork to allow resulting single strands to be copied
2. _____________: polymerizes nucleotide triphosphates in a 5’ to 3’ direction. Synthesizes RNA primers to act as a template for future Okazaki fragments to build on to.
3. _____________: synthesizes nucleotides onto leading end in classic 5’ to 3’ direction.
4. _____________: synthesizes nucleotides onto primers on lagging strand, forming Okazaki fragments. This enzyme cannot completely synthesize all the nucleotides.
5. _________: glues together Okazaki fragments, an area DNA Pol I unable to synthesize
6. _____________: catalyzes lengthening of telomeres; enzyme includes molecule of RNA that serves as template for new telomere segments
7. ______________: excises or cuts out unwanted or defective segments of nucleotides in DNA sequence
8. ______________: introduced single-strand nick in the DNA, enabling it to swivel and thereby relieve the accumulated winding strain generated during unwinding of double helix
9. _______________: holds the replication fork of DNA open while polymerases read the templates and prepare for synthesis

Answer 11

1. Helicase
2. Primase
3. DNA Pol 3
4. DNA Pol 1
5. Ligase
6. Telomerase
7. Nuclease
8. Topoisomerase
9. Single Strand Binding Proteins (SSBP)

Question 12

Origins of replication

__________ ORIGINS in Eukaryotes

___________ ORGINS in Prokaryotes

Answer 12

• Process of DNA replication begins at origin of replication, where molecule’s two strands are separated, producing a replication bubble with two replication forks unzipping the DNA bidirectionally away from the origin.
• Eukaryotes have multiple origins of replication across their numerous linear chromosomes
• Prokaryotes have single origin of replication for their single, circular DNA

Question 13

Replicating the ends of DNA molecules

why are the ends of chroosomes unable to be synthesized? What does this result in?

How is this issue resolved?

Answer 13

• Linear chromosomes arrive at issue with replication at ends of their lagging strands by which a portion of the strand at the very end (located in telomere, a region of repetitive sequences at the end of the chromosome) is unable to by synthesized due to lack of 3’ end of a nucleotide to extend from
• This results in shortening of telomeres in linear chromosomes after numerous rounds of replication
• Issue resolved in presence of telomerase which lengthens telomeres with repetitive sequences, thus protecting them from loss during replication

Question 14

DNA REPAIR DURING REPLICATION

WHAT IS THE ROLE OF 3’–>5’ exonuclease activity?

Answer 14

• DNA polymerase has proofreading activity (also called 3’–>5’ exonuclease activity). If a wrong nucleotide gets incorporated, polymerase will “back-up” and replace it with correct one
• Special polymerase that replaces the RNA primers with DNA also have 5’–>3’ activity. This allows polymerase to clear away short stretches of incorrect nucleotides (RNA or incorrect DNA) and replace it with the right ones (DNA).

Question 15

Repair of Mutations:

Mismatch Repair

How does methylation work and which strand contains meythlation?

Answer 15

• Enzymes recognize incorrectly paired base-pairs and cuts out stretch of DNA containing the mismatch. Then polymerase re-adds the correct nucleotides in.
• During mismatch repair, repair enzyme must decide what strand of DNA to cut since DNA contains 2 strands. To do this, the enzyme cuts DNA strand that does not have methylations.
• The original (old) DNA has methylations but newly synthesized DNA does not have them until shortly after replication. Thus, there is a short period when mismatch repair enzymes can find out what strand to cut if mismatch is encountered.

Question 16

Repair of Mutations

____________________: a damaged base gets cut out. Then the base’s sugar phosphate backbone gets cut out. Several more nucleotides next to base get cut out. Finally, polymerase remakes the cut-out nucleotides.

__________________: damaged nucleotide(s) get cut out then polymerase replaces it. This is like mismatch-repair, but not for mismatch. It’s for damages like thymine dimers, and other damages that changes normal nucleotides into abnormal nucleotides.

Answer 16

• base-excision repair
• nucleaotide-excision repair

Question 17

Repair of Mutations

• _____________basically 5’–>3’ exonuclease activity coupled to polymerase activity. The polymerase chugs along, chews off bad nucleotides and replaces them with new nucleotides. This is what happens when RNA primers are replaced with DNA.
• ______________during replication, when there’s too much DNA damage for normal repair to handle, the __________ repair system comes along. Instead of correcting any DNA damages during replication, polymerase replicates over the damaged DNA as if it were normal. By using damaged DNA as template error rates are high, but still better than not replicated at all.

Answer 17

• Nick translation
• SOS reponse in E Coli, SOS

Question 18

Genetic Code

What is the central Dogma?

Where does DNA reside? What is transcription?

What are the working copies of genes?

What is the process when ribosomes read off mRNA to make proteins?

What is a protein?

Answer 18

• DNA: Resides in nucleus. Codes information in genes.
• Transcription: Inside the nucleus, the DNA genes get transcribed into RNA (mRNAs)
• RNA: The mRNAs get transported out of nucleus into cytoplasm. mRNAs are working copies of the gene.
• Translation: Ribosomes read off mRNAs to make proteins.
• Protein: Synthesized by ribosomes. End product of what’s encoded in the genes and they perform all functions in the cell.

Question 19

The Triplet Code

What is a Codon? Anticodon?

Degenerate code, wobble pairing?

Answer 19

• Codon: The mRNA is a sequence of nucleotides, but it codes for a sequence of amino acids. To do this, every 3 nucleotide codes for an amino acid. These triplets of nucleotides are called codons. A single mRNA contains many codons.
• Codons are continuous, non-overlapping and degenerate.
  
  • Continuous because one codon follows right after another. There are no nucleotides in between.
  • Non-overlapping because the 3 nucleotides that consist one codon never serve as part of nother codon
  • Degenerate (see below)
• Anticodon: the 3 bases on the “tip” of the tRNA. A single tRNA contains a single anticodon at the “tip” and the corresponding amino acid at the “tail.” Anticodons are complementary to their corresponding codon.
• Codon-anticodon relationship: During translation, codons pair with anticodons so that the correct amino acids can be linked to a given codon.
• Degenerate code, wobble pairing
  • Genetic code is degenerate because more than one codon codes for a given amino acid
  • We refer to variable third base in the codon as the wobble position. Wobble is an evolutionary development designed to protect against mutations in the coding regions of our DNA.
  • Mutations in the wobble position tend to be called silent or degenerate, which means no effect on the expression of the amino acid and therefore no adverse effec**ts on the polypeptide sequence

Question 20

Missense Codon:

Nonesense Codon:

Initiation & termination codons:

Answer 20

• Missense codon: mutated codon that results in a different amino acid
• Nonsense codon: mutated codon that results in a stop codon
• Initiation codon (AUG): signals the start of translation. Lies just downstream of the Shine Dalgarno sequence (Kozak sequence for eukaryotes)
• Termination codon (UAG, UGA, UAA): signals the end of translation.
  
  • Unlike other codons, tRNA are not involved. A protein called “release factor” comes along and terminates translation.

Question 21

What is the product of transcription & also a template for translation?

Answer 21

• Messanger RNA (mRNA)
• It’s the product of transcription and the template for translation
• The 5’ cap is a modified nucleotide linked in a special way to mRNA. This protects 5’ end from exonuclease degradation.
• The poly-A tail protects 3’ end of mRNA from exonuclease degradation
  
  • Eukaryotic mRNA: 5’ cap - nucleotides - 3’ poly-A tail
  • Prokaryotic mRNAs don’t have 5’ cap or poly-A tail

Question 22

Other products for transcription BUT DO NOT SERVE AS A TEMPLATE FOR TRANSLATION?

Answer 22

• Both tRNA and rRNA are products of transcription. However, they do not serve as template for translation.
  
  • tRNA responsible for bringing in correct amino acid during translation.
    
    • tRNA made of nucleotides, many of which are modified for structural and functional reasons. At the 3’ end, the amino acid is attached to the 3’OH via an ester linkage.
    • tRNA structure: clover leaf structure with anticodon at tip, and amino acid at 3’ tail.
  • rRNA makes up ribosome, enzyme responsible for translation.
    
    • rRNA made of nucleotides, many modified for structural and functional reasons
    • rRNA highly structured because it contains active site for catalysis. The rRNA of large ribosomal subunit is responsible for catalyzing peptide bond formation and can do this even

without ribosomal proteins.

Question 23

• Mechanism of transcription (RNA polymerase, promoters, primer not required)
• initiation, elongation & termination

Answer 23

1. Chain Initiation: RNA polymerase binds to promoter (TATA box) of dsDNA (closed complex). Double stranded DNA template opens up (open complex)

2. Chain Elongation: nucleoside triphosphates (AUGCs) adds corresponding to the DNA template. No primer is required. RNA elongates as RNA polymerase moves down DNA template. RNA made from 5’ to 3’ direction.

3. Chain Termination: 2 ways that transcription can terminate:

1. Intrinsic termination: specific sequences called a termination site creates a stem-loop structure on RNA that causes RNA to slip off template.

2. Rho (ρ) dependent termination: a protein called the ρ factor travels along the synthesized RNA and bumps off the polymerase

Question 24

Ribozymes

Spliceosomes

Small Nuclear Ribonucleoproteins (snRNPs)

Small Nuclear RNAs (snRNAs)

Answer 24

• Ribozyme = RNA molecule that is capable of catalyzing specific chemical reactions
• Spliceosomes:
• snRNPs = RNA-protein complexes that combine with unmodified pre-mRNA and various other proteins to form a spliceosome, a large RNA-protein molecular complex upon which splicing of pre-mRNA occurs
• snRNAs = complexed with proteins to form snRNPs to splice primary RNA transcripts.

Question 25

Functional & evolutionary importance of introns:

alternative splicing:

Answer 25

• Introns and alternative splicing allow different mRNA sequences and ultimately a greater variety of proteins through translation.
  
  • Without introns or alternative splicing, less protein variability.
• Efficient way to generate wide variety of proteins through use of mRNA compared to modifying preexisting proteins.
• Splicing of introns = using snRNA/snRNP to form a spliceosome complex to excise intron lariat loop

Question 26

Post-transcription modification:

Answer 26

• include addition of the 5’ cap, polyA tail and splicing
• In prokaryotes, mRNAs with better Shine-Dalgarno sequence are translated more
• In eukaryotes, post-transcription regulation can involve adding more polyAs to mRNA (longer mRNA life time), modulation of the translation machinery (phosphorylation of initiation factors), or storing mRNAs to be translated at a later time (mRNA masking)
• The cap and polyA tail are added co-transcriptionally, but still considered post transcriptional
• Splicing gives rise to isoforms. Depending on how you arrange the introns/exons, you get different proteins by alternative splicing.

Question 27

Translation: role of mRNA, tRNA, rRNA

Answer 27

• mRNA: contains codons that code for the peptide sequence
• tRNA: contains the anticodon on the “tip” and the corresponding amino acid on the “tail.”
  
  • Link correct amino acid to its corresponding mRNA codon through codon-anticodon interaction.
• rRNA: forms the ribosome.
  
  • Catalyzes the formation of the peptide bond.

Question 28

Role & structure of ribosome

Answer 28

• Ribosome is the enzyme that catalyzes protein synthesis
• Ribosome has 2 subunits - the large and the small
• Large subunit responsible for peptidyl transfer reaction
• Small subunit responsible for recognizing mRNA and binds to Shine-Dalgarno sequence on mRNA (Kozak sequence for eukaryotes)
• Both subunits needed for translation to occur and they come together in a hamburger fashion that sandwiches the mRNA and tRNAs in between

Question 29

Mechanism of translation

• 1. chain initiation*
• 2. chain elongation*
     a. binding
     b. peptidyl transfer
     c. translocation
• 3. chain termination*

Answer 29

1. Chain Initiation: To begin, you need to form initiation complex, basically an assembly of everything. This includes mRNA, initiator tRNA (fmet), and ribosome (initiation factors, and GTP aids in formation of initiation complex).
   
   1. The initiation complex forms around initiation codon (AUG), which is just downstream of the Shine-Dalgarno sequence. The Shine-Dalgarno sequence is the “promoter” equivalent of translation for prokaryotes (Kozak sequence for eukaryotes).
2. Chain Elongation: protein made from N terminus to C terminus. mRNA codons read from 5’ to 3’ end. Elongation consists of:

• a. Binding:* new tRNA with its amino acid (tRNA + amino acid is called aminoacyl- tRNA) enters the A site. GTP and elongation factor required.
• b. Peptidyl transfer:* attachment of the new amino acid to existing chain in P site. The already existing chain in P site migrates and attaches to the aminoacyl-tRNA in the A site.
• c. Translocation:* the lone tRNA in the P site gets kicked off (E site), and the tRNA in the A site, along with peptide attached to it, moves into the P site. The mRNA gets dragged along also - the codon that was in the A site is now in P site after translocation. A site is now empty and ready for the binding of a new aminoacyl-tRNA to a new codon. Elongation factor and GTP required.

3. Chain Termination: When a stop codon is encountered, proteins called release factors, bound to GTP, come in and block the A site. Peptide chain gets cleaved from tRNA in the P site. Peptide chain falls off and the whole translation complex falls apart.

• Amino acid activation: enzymes called aminoacyl-tRNA synthetases attach the correct amino acids to their corresponding tRNAs.
• ATP required.

Question 30

Post-translational modifications to protein

Answer 30

• Modifications of the protein through addition of groups to the protein through covalent bonds or cleavage of the protein
• Deals with activation/inactivation or enhancing the protein’s function
• Examples include phosphorylation, glycosylation, and ubiquitination (inactivation by tagging protein for proteasome degradation)

Question 31

Eukaryotic Chromosome Organization​

Name the Chromosomal proteins

Answer 31

• Histones: responsible for compact packing and winding of chromosomal DNA. DNA winds itself around histone octamers.
• Non-histone: chromosomal proteins: all the other proteins are lumped together in this group.
  
  • Responsible for roles such as regulatory and enzymatic.

Question 32

Eukaryotic Chromosome Organization

Single copy vs. repetitive DNA

Answer 32

• Single copy = DNA sequence that does not repeat (ex. ATCCGTAG)
  
  • Single copy holds most of the organism’s important genetic information.
  • It is transcribed and translated and has a low mutation rate.
• Repetitive DNA = DNA sequence that does repeat (ex. ATCCATCC)
  
  • Repetitive DNA found near the centromeres. They may contain genes that are transcribed/translated; however, there may be parts that are not transcribed/translated.
  • They have higher mutation rate.
  • Highly repetitive DNA contains no genes so not transcribed/translated and very high mutation rate (ex. telomeres)

Question 33

Eukaryotic Chromosome Organization

Superocoiling

Answer 33

• Supercoiling is a wrapping of DNA on itself as its helical structure is pushed even further toward the telomeres during replication.
• to alleviate the torsional stress and reduce risk of strand breakage, DNA gyrase (DNA topoisomerase II) introduces negative supercoils.

Question 34

Eukaryotic Chromosome Organization

Heterochromatin vs. Euchromatin

Answer 34

• Euchromatin: is structured as loose beads on a string. The beads represent nucleosomes. The majority of DNA is in euchromatin form, as it’s generally under active transcription all the time.
  
  • Note: Prokaryotes only have euchromatin as heterochromatin has a more complex structure.
• Heterochromatin: is densely packed, so like coiled beads on a string. It was thought that the genes here were inaccessible for transcription, but recent research says otherwise.
  
  • Additionally, if some euchromatin is not being transcribed, it may be converted into heterochromatin, and vice-versa.

Question 35

Eukaryotic Chromosome Organization

Telomeres, centromeres

Answer 35

• Telomere: the 2 ends of the chromosome
• Centromere: a region on the chromosome, can be at the center or close to one of the ends.
  
  • After replication, sister chromatids are attached at the centromere.
  • During mitosis, spindle fibers are attached at the centromere and pulls the sister chromatids apart.

Question 36

Control of Gene Expression in Prokaryotes

• Operon*
• Jacob-Monod Model (*structural gene, operator site, promotor site, regulator gene)
• Gene repression in bacteria*
• Positive Control in bacteria*

Answer 36

• Operon = a cluster of genes transcribed as a single mRNA. The numerous genes share a single common promoter region on the DNA sequence and are transcribed as a group.
  
  • Two types of operons: inducible and repressible systems.
• ​​Jacob-Monod Model:
• *1. Structural gene:** codes for protein of interest
• *2. Operator site:** nontranscribable region of DNA capable of binding a repressor protein
• *3. Promoter site:** provides a place for RNA polymerase to bind
• *4. Regulator gene:** codes for the repressor protein

Gene repression in bacteria

• Operons have a binding site for regulatory proteins that turn expression of the operon “up” or “down”
• Some regulatory proteins are repressors that bind to pieces of DNA called operators. When bound to its operator, a repressor reduces transcription (e.g., by blocking RNA polymerase from moving forward on DNA)

Positive control in bacteria

• Some regulatory proteins are activators. When an activator is bound to its DNA binding site, it increases transcription of the operon (e.g., by helping RNA polymerase bind to the promoter)

Question 37

Control of Gene expression in Eukaryotes

Enhancers & Silencers

Answer 37

• Transcription factors (protein) bind to enhancers or silencers (DNA) to affect transcription.
• Enhancers: increase transcription when bound
• Silencers decrease transcritpion
• The main difference in eukaryotes from prokaryotes is that enhancers/silencers can be very far away from actual promoter and can be upstream or downstream.
• The DNA must loop back on itself so that the transcription factor bound to enhancer/silencer can make contact with promoter. Intermediate proteins are involved in the process.
• Eukaryotes lack bacterial transcription regulation mechanisms such as the operon and attenuation

Question 38

Control of Gene expression in Eukaryotes

DNA Binding proteins, transcription factor

Answer 38

• DNA-binding proteins and transcription factors bind to DNA.
• TFs have a DNA-binding domain.
• DNA-binding domains interact with the grooves in the double helix (major and minor grooves)

Question 39

Control of Gene Expression in Eukaryotes

Gene amplification and duplication

Answer 39

• Once the transcription complex is formed, basal (or low-level) transcription can begin and maintain moderate, but adequate, levels of the protein encoded by this gene in the cell. There are times, however, when the expression must be increased, or amplified, in response to specific signals such as hormones, growth factors, and other intracellular conditions. Eukaryotic cells accomplish this through enhancers and gene duplication.
• Gene duplication can also increase expression of a gene product by duplicating the relevant gene. Genes can be duplicated in series on the same chromosome, yielding many copies in a row of the same genetic information. Genes can also be duplicated in parallel by opening the gene with helicases and permitting DNA replication only of that gene; cells can continue replicating the gene until hundreds of copies of the gene exist in parallel on the same chromosome.

Question 40

Control of Gene Expression in Eukaryotes

• Post-transcriptional control, basic concept of splicing (introns, exons)
• mRNA modification (RNA splicing, alternate splicing, & 5’ capping & 3’ poly-Atail

Answer 40

• tRNA and rRNA modifications: some normal nucleotides are modified to control the structure of these RNAs
• mRNA modifications:
  
  • RNA splicing: sequences called introns cut out, sequences called exons are kept and spliced (joined) together
  • Alternate splicing: different ways of cutting up the RNA and rejoining the exons pieces makes different final RNA products
  • 5’ capping and 3’ poly-A tail: these help to protect the RNA from degradation, so they can last longer
• After the correct modifications, RNA is transported out of nucleus where they can function in translation
• After some time, RNA is degraded. The rate and timing of RNA degradation can be controlled by the cell.

Question 41

Control of Gene Expression in Eukaryotes

Cancer as a failure of normal cellular controls, oncogenes, tumor suppressor genes

Answer 41

• Failure of normal cellular controls:
  
  • Cancer cells continue to grow and divide in situations normal cells would not
  • Cancer cells fail to respond to cellular controls and signals that would halt this growth in normal cells
  • Cancer cells avoid apoptosis (self-destruction) that normal cells undergo when extensive DNA damage is present
  • Cancer cells stimulate angiogenesis (cause new blood vessels to grow to nourish the cancer cell)
  • Cancer cells are immortal while normal cells die after a number of divisions
  • Cancer cells can metastasize - break off and grow in another location
• Oncogenes: genes that cause cancer when activated. The product of many oncogenes is involved in speeding up cell division. Before an oncogene is activated, it is a harmless proto-oncogene. Something occurs that changes proto-oncogene to an oncogene. Classic example is src.
• Tumor suppressors: if the oncogene is the “bad” gene, tumor suppressors are the “good” genes. The product of many tumor suppressors is involved in slowing down or controlling cell division. If something happens that cause tumor suppressor to no longer function, then the cell becomes cancerous. Classic example is p53.

Question 42

Control of Gene Expression in Eukaryotes

Regulation of chromatin structure

what is chromatin made up of?

Answer 42

• Chromatin made up of DNA, histone proteins, and non-histone proteins.
• There are repeating units of chromatin, called nucleosomes, which are made up of double helical DNA wrapped around a core of eight histones.
• DNA comes in two flavors: densely packed, and transcriptionally inactive DNA called heterochromatin, and the less dense, transcriptionally active DNA called euchromatin
• Methylated DNA may be bound to methyl cpg-binding domain proteins that recruit additional proteins to the locus certain genes and other chromatin remodeling proteins, and this modifies the histones, forming condensed inactive heterochromatin that is basically transcriptionally silent.
• Acetylation promotes open DNA (aka active chromatin or euchromatin)

Question 43

Control of Gene Expression in Eukaryotes

DNA Methylation

Answer 43

• Involved in chromatin remodeling and regulation of gene expression levels in the cell.
• Methylation often linked with silencing of gene expression. During development, methylation plays an important role in silencing genes that no longer need to be activated.
• Heterochromatin regions of the DNA are more heavily methylated, hindering access of the transcriptional machinery to the DNA.

Question 44

Control of Gene Expression in Eukaryotes (BIO)

Role of non-coding RNAs

Answer 44

• Non-coding RNA is a functional RNA transcribed from DNA but not translated into proteins. They function to regulate gene expression at the transcriptional and post-transcriptional level.
• Three major classes are microRNAs (miRNAs), short interfering RNAs (siRNAs), and piwi- interacting RNAs (piRNAs)

Question 45

Recombinant DNA and Biotechnology (BIO)

Gene Cloning

• Retrieve gene of interest and plasmid that has one area with similar sequence. Cut both with the same __________ Enzyme. Hybridize, then seal with DNA ligase. This produces a ______________. Insert plasmid into bacteria and allow for replication inside bacteria.
• Plasmid must have a ______________ because you need to open it up for the insertion of your gene.
• Plasmid must have an _______________because you want to clone your gene, which is inside your plasmid.
• Plasmid must have ________________ because this lets you kill competing, useless bacteria that don’t have your plasmid. When you add an antibiotic, only the bacteria with _________________will live.
• Plasmids replicate independently of the ____________________.

Answer 45

• Restriction enzyme
• Recombinant Plasmid
• restriction site
• origin of replication
• antibiotic resistant gene
• antibiotic resistant plasmid
• genomic DNA of the bacteria

Question 46

Recombinant DNA and Biotechnology (BIO)

Restriction enzymes

PAlindrome sequence

What do they cut, what are the results?

Sticky ends & blunt ends, which hybridize?

Answer 46

• cut double stranded DNA at palindrome sequences.
  
  • The resulting fragments are called restriction fragments.
• ​If you read from 5’—>3’ of one strand, then read from 5’—->3’ of the other strand, and they are the same, then the section of the double stranded DNA that you read is palindrome sequence.
• Some restriction enzymes cut to make sticky ends, which can hybridize.
• Some restriction enzymes cut to make blunt ends, which cannot hybridize.

Question 47

Recombinant DNA and Biotechnology (BIO)

DNA libraries

name two types

Answer 47

• DNA library is a collection of DNA fragments that have been cloned into vectors so that researchers can identify and isolate the DNA fragments that interest them for further study.
• There are two kinds of libraries: genomic and cDNA libraries.
  
  • ​Genomic libraries contain large fragments of DNA in either bacteriophages or bacterial or P1- derived artificial chromosomes (BACs and PACs).
  • cDNA libraries are made with cloned, reverse-transcribed mRNA, and therefore lack DNA sequences corresponding to genomic regions that are not expressed, such as introns and 5’ and 3’ noncoding regions.
    
    • ​ cDNA libraries generally contain much smaller fragments than genomic DNA libraries and are usually cloned into plasmid vectors.

Question 48

Recombinant DNA and Biotechnology (BIO)

Generation of cDNA

Answer 48

• Once mRNA is purified, oligo-DT (a short sequence of deoxy-thymidine nucleotides) is tagged as a complementary primer which binds to the poly-A tail providing a free 3’-OH end that can be extended by reverse transcriptase to create the complementary DNA strand.
  
  • Now, the mRNA is removed by using RNAse enzyme leaving a single stranded cDNA (sscDNA). This sscDNA is converted into double stranded DNA with the help of DNA polymerase.
• However, for DNA polymerase to synthesize a complementary strand a free 3’-OH end is needed. This is provided by sscDNA itself by generating a hairpin loop at the 3’ end by coiling on itself. The polymerase extends the 3’-OH end and later the loop at 3’ end is opened by the scissoring action of S1 nuclease.
  
  • ​Restriction endonucleases and DNA ligase are then used to clone the sequences into bacterial plasmids.
• The cloned bacteria are then selected, commonly through the use of antibiotic selection. Once selected, stocks of the bacteria are created which can later be grown and sequenced to compile the cDNA library.

Question 49

Recombinant DNA and Biotechnology (BIO)

Hybridization

Southern Blotting

Answer 49

• Hybridization, also called annealing, is where DNA strands base pair with each other.
• In Southern blotting, DNA probes are used to hybridize onto DNA fragments containing a target sequence.
• In gene cloning, hybridization refers to the process where sticky ends from a restriction fragment of a gene base pairs with the same sticky ends on a plasmid.

Question 50

Recombinant DNA and Biotechnology (BIO)

Expressing clonned genes

Answer 50

• cDNA transformed into plasmid, then add antibiotic resistant gene
• Infect bacteria with plasmid and add antibiotics. This allowed only the successfully infected bacteria to survive which contain our gene of interest.
• The bacteria replicates creating many copies of our gene of interest.

Question 51

Recombinant DNA and Biotechnology (BIO)

Polymerase Chain Reaction (PCR)

4 STEPS

Answer 51

1. Denaturation: heat (90 ̊C) to separate double stranded DNA template
2. Annealing: cool reaction in order for primers to anneal to the now single stranded DNA template

• Excess amount of primers, so they outcompete re-annealing of the template strands

3. Elongation: use heat stable polymerase to extend the primers

4. Repeat steps 1 to 3 for n cycles. The resulting amplification of the original DNA template after n cycles is 2n

Question 52

Recombinant DNA and Biotechnology (BIO)

Gel Electrophoresis

What will the gell be for small molecules? large molecules?

bottom ______charged anode

top _________charged cathode

since a voltage source is applied to gel electrophoresis, it follows the same principles as an____________cell.

Answer 52

• separation of proteins, DNA, or RNA based on size and/or charge. For proteins and small molecules of DNA and RNA, the gel will be polyacrylamide. For larger molecules of DNA (> 500bp), the gel will be agarose. An electrical charge is placed across the gel. At the bottom is the positively charged anode and the top is the negatively charged cathode.
• Keep in mind, since a voltage source is applied to gel electrophoresis, it follows the same principles as an electrolytic cell.
  
  • _​_Negatively charged molecules will travel toward the anode. Because of resistance of the gel, larger molecules will have a harder time moving and thus, the molecules will be separated by size with the smallest molecules toward the bottom.
    
    • The gel can be stained for visualization, typically using Coomassie Blue dye. A lane will be loaded with a collection of molecules of a known size, called a ladder, which can be used to determine the size of the molecules being ran.

Question 53

Recombinant DNA and Biotechnology (BIO)

Southern Blotting

Answer 53

1. DNA strand of interest is exposed to restriction enzymes that cut the DNA strand into smaller fragments.
2. The newly cleaved strands of DNA are denatured using a solution of NaOH to create ssDNA strands
3. The single stranded cleaved strands of DNA undergo gel electrophoresis, separating them by size. Smaller fragments will be found at the bottom of the gel. Polyacrylamide is used if the stands are less than 500 base pairs. Agarose is used if the strands are over 500 base pairs.
4. The DNA from the gel is transferred to a sheet of nitrocellulose paper and then exposed to a 32P radiolabeled DNA probe that is complementary to DNA of interest.
5. Nitrocellulose paper is then viewed using autoradiography to identify the strand of interest.

Question 54

DNA Sequencing (Sanger)

determines what?

what is the name of the nucleaotides used?

What are the steps?

Answer 54

• Used to determine the sequence of nucleotides in a strand of DNA
• Modified nucleotides, known as “dideoxynucleotides” (ddNTPs), are used in this method.
• ddNTPs are missing the OH group on the 3’ carbon, thus unable to create a new 5’–>3’ phosphodiester bond. This allows us to control termination of replication.

1. DNA strand of interest is denatured using an NaOH solution to create a ssDNA strand that we can use for replication

2. ssDNA strand of interest is added to a solution containing:

a. A radiolabeled DNA primer that is complementary to the gene of interest

• b. 2. DNA polymerase
  c. 3. All four dNTPs (dATP, dTTP, dCTP, dGTP)
  d. 4. A very small quantity of a single ddNTP (e.g., ddATP)*
• This is done once for each of the four nucleotides in separate solutions

3. Each solution containing a specific dNTP and ddNTP are placed in their own lane of a gel and ran under gel electrophoresis. The gel is transferred to a polymer sheet and autoradiography is used to identify the strands in the gel.

• For each respective nucleotide, the insertion of a ddNTP will terminate replication and create various strands of different length that correspond to that specific nucleotide. Therefore, the gel can be read from bottom-to-top to determine the nucleotide sequence. The smaller the fragment, the further it travels in the gel.

Question 55

Analyzing Gene Expression

Northern Blotting

Western Blotting

RtT-qPCR

Answer 55

• Northern Blotting:
  
  • similar steps to sourthern except Northern uses RNA so steps 1 & 2 are not done
• Western Blotting: Detection of a specific protein in a sample.

1. Proteins from a sample are loaded into an SDS-PAGE gel and separated based on size.

2. Proteins from gel are transferred to a polymer sheet and exposed to a radiolabeled antibody (sometimes using two antibodies; one specific to protein of interest and another radiolabeled antibody that binds to first antibody) that is specific to protein of interest

3. The polymer sheet is viewed using autoradiography. The protein of interest that is bound to the radiolabeled antibody will be visible.

• RT-qPCR: used when starting material is RNA. In this method, RNA is first transcribed into complementary DNA (cDNA) by reverse transcriptase from total RNA or messenger RNA (mRNA).
  
  • cDNA then used as template for qPCR reaction. RT-qPCR is used in a variety of applications including gene expression analysis, RNAi validation, microarray validation, pathogen detection, genetic testing, and disease research
  • RT-qPCR can be performed in a one-step or a two-step assay.
    
    • One-step assays combine reverse transcription and PCR in a single tube and buffer, using a reverse transcriptase along with a DNA polymerase. One-step RT-qPCR only utilizes sequence-specific primers.
    • In two-step assays, the reverse transcription and PCR steps are performed in separate tubes, with different optimized buffers, reaction conditions, and priming strategies.

Question 56

How do you determine gene function?

Answer 56

• Knocking out the gene allows us to observe what functions the gene are responsible for.

Question 57

Stem Cells

• ____________\_can differentiate into any cell of an organism, including the placenta, the amnion and chorion
• ___________\_ can give rise to all cell types, excluding the placenta, the amnion and chorion. They arise from Totipotent cells, and are more specialized
• ______________can develop into more than one cell type, but are more limited than pluripotent cells; adult stem cells and cord blood stem cells are considered multipotent
• ______________within first couple of cell divisions after fertilization are the only cells that are totipotent

Answer 57

1. Totipotent

2. Pluripotent

3. Multipotent

4. Embryonic

Question 58

Practical applications of DNA technology:

• a. medical applications*
• b. human gene therapy*
• c. pharmaceuticals*
• d. forensic evidence*
• e. environmental cleanup*
• g. agriculture*

Answer 58

• Medical applications: some vaccines we use recombinant DNA technology including hep B virus and the herpes virus and malaria.
• Gene therapy: intended for diseases in which a given gene is mutated or inactive, giving rise to pathology. By transferring a normal copy of the gene into the affected tissues, the pathology should be fixed.
  
  • Efficient gene delivery vectors must be used to transfer the cloned gene into the target cells’ DNA. Because viruses naturally infect cells to insert their own genetic material, most gene delivery vectors in use are modified viruses. A portion of the viral genome is replaced with cloned gene such that the virus can infect but not complete its replication cycle.
• Forensics: There are parts of the genome known as non-coding regions of the genome. These regions can help forensic scientists identify specific individuals by looking at things like short tandem repeats, STRs, which are short sequences of DNA, two to six base pairs long. They are found in high amounts and to varying degrees between individuals. Thus, if they sequence these STRs, they could identify specific individuals given a DNA sample.
• Agriculture: scientists can now create crops resistant to insects and resistant to herbicides. Some can delay ripening of crops, so you can transport crop from farm to store.