Unit 1 Random facts

Question 1

In humans, _______ protects the single-stranded DNA during replication. In prokaryotes it is called SSB (Single Stranded Binding protein).

Answer 1

RPA

Question 2

*Sliding clamps enhance the processivity of DNA polymerase.

_______ is the sliding clamp for eukaryotic DNA Pol D.

B-subunit is the sliding clamp of prokaryotic DNA Pol III

Clamp is loaded to DNA by clamp loader proteins.

Answer 2

PCNA

Question 3

Maturation of okazaki fragments in eukaryotes

Answer 3

Question 4

Cellular levels of some DNA replication proteins such as________ can be sentitive biomarkers for early detection and prognosis of many common cancers.

Answer 4

MCM5 (which is a helicase)

Question 5

Describe the Direct reversal of damaged bases repair

Answer 5

This mechanism of DNA repair takes care of simple issues. For example:

• Reversal of a specific type of single-stranded DNA break by DNA ligase. In other words, ligation of a break in the phosphodiester backbone of the DNA by DNA ligase.
• Reversal of UV-caused base damage (T-T T-C dimers) by photolyase.
• Reversal of base alkylation by O6-meG methyltransferase (MGMT).

Question 6

Explain Nucleotide excision repair (NER):

Answer 6

This repair removes DNA lesions that distort the DNA structure and block RNA or DNA polymerase movement on the DNA. Examples of these types of lesions are thymine dimers resulted from exposure to UV and bulky DNA adducts caused by exposure to carcinogens.

Steps of NER

1. Recognition and binding of the damaged site by a multi-protein complex (two different ways depending on local transcription activity).
2. Local unwinding of the DNA duplex by helicases (parts of the TFIIH protein complex) to form a bubble of ~25 bases.
3. Double incision of the damaged strand by two endonucleases and removal of a ~30 base oligonucleotide containing the

lesion.

4. Filling in the gap by a DNA polymerase.
5. Rejoining the two ends by a DNA ligase.

Over 30 different proteins are needed for NER in humans. The lesions are recognized by two different pathways: Global genome NER recognizes distorting DNA lesions in any region of the genome; Transcription-coupled NER recognizes distorting DNA lesions in regions that are actively transcribed. Following recognition of the distorting DNA lesion by proteins unique to each of the two pathways (they only differ in their 1st step), the remainder of the NER occurs in the same manner.

* Defects in the global genome NER —> Xeroderma pigmentosum (Sun hypersensitivity, Skin cancers, neurologic & cognitive dysfunction).

* Defect in the Transcription-coupled NER—> Cockayne’s Syndrome (Sun hypersensitivity, Premature aging (progeria), Impaired development, Neurological degeneration).

Question 7

Describe Base Excision Repair mechanism (BER):

Answer 7

This repair mechanism removes DNA lesions that are missed by the NER process, but do not necessarily block polymerase function or distort the DNA structure.

• BER requires a family of enzymes called *glycosylases, each recognizing a specific type of altered base. For example, uracil glycosylase recognizes uracils in DNA that result from cytosine deamination, and 5- methylcytosine-DNA glycosylase recognizes 5-methylcytosine to initiate DNA demethylation.
• See photo for steps.

Question 8

Explain Mismatch Repair (MMR):

Answer 8

This repair mechanism fixes errors in nucleotide incorporation made by DNA polymerase during DNA replication. The mismatched base pair is recognized shortly after DNA synthesis by the MutS and MutL proteins in bacteria; their mammalian counterparts are *MSH (MutS Homolog) and MLH (MutL Homolog) proteins.

In bacteria, the newly synthesized strand of DNA is identified by the MMR machinery because it is not yet methylated. An endonuclease cleaves the phosphodiester backbone of the new strand of DNA. An exonuclease chews away the new DNA strand including the mismatch nucleotide while a helicase assists with the unwinding of the double helix. DNA polymerase repairs the resulting single strand gap by incorporating complimentary base pairs, and DNA ligase seals the phosphodiester backbone.

It is critical in mismatch repair to remove the wrongly inserted base (on the newly synthesized daughter DNA strand) rather than its mismatching partner (on the parental DNA strand), and that is why mismatch repair is geared to remove ONLY the portion of the newly replicated DNA that contains the wrongly incorporated base.

How does MMR know which strand is new?

1- Bacteria recognize the newly synthesized strand because, in contrast to the parental strand, it has not yet been methylated.

2- The recognition mechanism used in eukaryotes MMR appears to require DNA nicks that are more abundant on the newly replicated strand.

What disease is caused by mutations in the MMR system?

Hereditary non-polyposis colorectal cancer (HNPCC)

-in most cases (60%) MSH2 is mutated and MLH1 in some cases (30%).

Question 9

Explain Lesion Bypass:

Answer 9

If a cell encounters so much DNA damage of the type that normally blocks DNA replication (such as UV-induced thymidine dimers) that the excision repair systems cannot fix it all, cells resort to a pathway called lesion bypass or translesion synthesis. Lesion bypass allows cells to continue replicating and dividing in the face of immense damage. However, it is highly mutagenic because alternate DNA polymerases that lack 3’ to 5’ proofreading exonuclease activity are used to replicate past the DNA lesion. The result is an error rate 100-10,000 higher than normal DNA replication.

See image for details of the mechanism.

Question 10

What mechanism is activated by a single-strand break and adds poly (ADP-Ribose) chains to proteins?

Answer 10

PARP: Poly(ADP-Ribose) Polymerase

What is the purpose of PARP?

It is a reversible poly(ADP-ribosylation) of proteins near a single-strand break sites, facilitates DNA repair by amplification of the damage signal**, **focal enrichments of repair proteins**, and **change in local structure of chromatin*.

Question 11

The RNA pol II large subunit has a unique C-terminal domain, CTD, composed of heptad repeats (YSPTSPS) which are reversibly phosphorylated. The CTD binds to proteins that regulate __________ and processing of the RNA transcript.

How? The serine residues at 2 and 5 positions are phosphorylated to start the elongation process.

Answer 11

elongation

Question 12

____________ from the death cap mushroom Amanita phalloides is a non-competitive inhibitor of RNA pol II. It binds the bridge helix and blocks RNA chain elongation by preventing translocation

Answer 12

alpha-amanatin

Question 13

Binding of TATA box binding protein to the TATA box in the ___________ helps direct assembly of the pre-initiation complex at the promoter.

Answer 13

minor groove

Question 14

Transcription factor II H (TFIIH) is a important for transcription and DNA repair.

Mutations in the XPB, XPD and p44 subunits of TFIIH cause:

Xeroderma pigmentosum

Cockaynes syndrome

Trichothiodystrophy

-________ helicase opens DNA at the promoter to permit initiation

of transcription.

Answer 14

XPB

Question 15

What are the three steps of 5-cap of mRNA?

Answer 15

Question 16

What are the functions of the 5’cap?

Cap binding complex (CBC) in the nucleus is replaced by eIF4E in the cytoplasmm, which promotes initiation of translation in ribosomes.

Answer 16

Question 17

Introns are recognized through short, conserved sequences. They usually start with ____ residues and end with _____ residues. Alsom their branch point A is located 20-50 bp from the 3’ splice site.

Answer 17

GU; AG

Question 18

The 5’ splice site is 1st reognized by base pairing to the ________.

The branch point is recognized by base pairing to the _________.

The 3’ splice site is recognized by a protein complex _________.

Answer 18

U1 snRNA

U2 snRNA

U2AF

Question 19

What is the spliceosome?

Answer 19

The spliceosome is a large ribonucleoprotein complex composed of the pre-mRNA, over 100 proteins, and 5 small nuclear RNAs (snRNAs)

Question 20

Pre-mRNA Splicing Occurs by Two Transesterification Reactions, describe them:

Answer 20

Question 21

Why does splicing gives HIV virus an advantage?

Answer 21

HIV Makes both Unspliced genomic RNA and over 30 Singly and Doubly Spliced Transcripts from a single primary transcript transcript

Question 22

What are the steps of polyadenylation at the 3’ end of the mRNA?

What is the key sequence for polyadenylation to occur?

*This process is key to promoting translation initiation by providing a specific binding for PABP. It also stabilizes mRNA.

Answer 22

AAUAAA

Question 23

Gene Expression I

What are the two classes of Activators and Repressors?

Answer 23

1. Sequence-specific DNA binding proteins- bind to promoter or enhancer elements (DNA control elements) in their target genes to regulate transcription. The elements they bind to are usually 6-8 base pairs long. Usually bind DNA by inserting their alpha-helices into the major groove of DNA, making contacts between the amino acid side chains of the protein and the bases in the DNA .

2. Co-factors:

Do not bind directly to the DNA elements but rather bind to sequence- specific DNA binding proteins and affect transcription through this contact.

Co-activators- activate transcription

Co-repressors- repress transcription

Question 24

Gene Expression I

Families of Sequence-Specific DNA binding proteins.

Homeodomain Proteins (Helix-turn-helix)

Member include: __1___

Disease associated with this homeodomain protein:

Craniosynostosis

2

Answer 24

1- Hox family, Pits1, Msx.

2-

Craniosynostosis is characterized by the *premature closure of one or more sutures in the skull and affects 1/3000 infants.

One particular variant of craniosynostosis (Boston-type) occurs as a result of a mutation in the homeodomain protein MSX2.

What is MSX2 used for?

It is normally required for proper craniofacial development by affecting the transcription of a number of genes important in this process. When the DNA binding domain (or homeodomain) of this protein has a one amino acid substitution, the protein binds DNA more strongly- giving a “gain of function” or “hypermorphic allele”. This mutated hyperactive protein then affects the transcription of other genes critical for suture closure, leading to craniosynostosis.

Question 25

Gene Expression I

Families of Sequence-Specific DNA binding proteins.

Zinc Finger proteins

Members: ____1____

Disease related to it:

Androgen Insensitivity Syndrome (AIS)

Answer 25

1- Nuclear receptors such as estrogen, androgen, and retinoic acid.

What is Androgen Insensitivity Syndrome?

It is an X-linked recessive disorder.

Androgen insensitivity syndrome includes feminization or undermasculinization of the external genitalia at birth, abnormal secondary sexual development in puberty, and infertility. It occurs in males who are a normal karyotype (46 X,Y), but have mutations in either the DNA binding domain or the ligand binding domain of the androgen receptor (a zinc finger DNA binding protein). This makes the patients less responsive to androgens, leading to the aforementioned characteristics. Depending on the degree to which the mutation disrupts the function of the androgen receptor- varying levels of AIS can be observed (complete, partial, mild).

• In this patients the external genitalia remains female in appearence since there is no DHT effect on the external genitalia. However, vaginal pouch ends in a death.
• If testicles are not removed from the inguinal canal or abdominal cavity, they can develop a gonadoblastoma.

Question 26

Gene Expression I

Families of Sequence-Specific DNA binding proteins:

Basic Leucine zipper proteins (bZIP)

Members include: ___1_____

Disease associated:

Waardenburg Syndrome Type II

Answer 26

1- c-fos and c-jun

What is waanderburg syndrome type II?

Disease characterized by:

-Dafness, pigmentation anomalies of the eyes, and other pigmentation defects (hair, skin).

Mutations in the microphthalmia-associated transcription factor (MITF) gene (which encodes a bHLH DNA
binding protein) are observed in 15-20% of the patients. This gene encodes a transcription factor that plays a major role in the development of melanocytes.

Question 27

Gene Expression II

How does HATS and HDACs work?

Answer 27

Histone acetyltransferases (HATS)- acetylate N-termini of histones. The original thought was that acetylating the N-temini of histones would neutralize the positively charged ends and eliminate electrostatic interactions with DNA phosphates (thus opening up nucleosomal DNA for general transcription factors/Pol II transcriptional apparatus).

Histone deacetylases (HDACs)- In presence of HDACs, histones retain positive charge at N-terminal ends (HDACS remove acetyl groups from histones), The original thought was that this would maintain the interaction with DNA and prevent access of transcription factors to promoter.

Question 28

How do KMTs and KDMs work?

Answer 28

Histone Methyltransferases (KMTs)- Methylation does not alter the charge on histone lysine residues, but can be added to different degrees with 1, 2 ,or 3 methyl groups per lysine (mono-, di- or tri-methylated). The position of the histone in relation to the gene and the number of methyl groups can determine if this is stimulatory or inhibitory to transcription.

Histone Demethylases (KDMs)- Methylation can be removed by two classes of enzymes FAD using enzymes and α-ketoglutarate using enzymes. Depending on the degree of methylation removed and location of the methylation these enzymes can be stimulatory or inhibitory to transcription.

Question 29

What is Rubinstein-Taybi syndrome and how is it related to HATs?

Answer 29

Characterized by growth retardation, mental retardation, craniofacial dysmorphism, abnormally broad thumbs and great toes.

What gene is mutated?

Results from mutations in one copy of the CREB binding protein (CBP) gene. CBP is an essential transcriptional coactivator for many different transcription factors and is a histone acetyltransferase. It is normally recruited to many genes to activate transcription, and thus haploinsufficiency can result in widespread transcriptional changes.

Question 30

Gene expression III

Give an example of regulation of nuclear entry to TF?

Answer 30

NF-AT

High intracellular calcium (triggered by the binding of a ligand to a cell membrane receptor) activates calcineurin’s phosphatase activity- which dephosphorylates cytoplasmic NF-AT. This exposes the nuclear localization sequence, allowing NF-AT to enter the nucleus where it affects transcription of genes involved in the immune response and in heart function.

What are two commonly used immunosuppresants that inhibit calcineurin in order to inhibit NF-AT action?

*Cyclosporin and *FK506

Question 31

Gene Expression III

Give an example of regulating amount of transcription factor in the cell?

Answer 31

Beta catenin

In the absence of Wnt signaling, the cytoplasmic pool of β-catenin is targeted for degradation through the ubiquitin-proteasome pathway via phosphorylation that occurs through the action of glycogen synthase 3 (GSK3) in a complex with Axin and APC. In the presence of Wnt signaling, the Axin-APC-GSK3 complex is destabilized, preventing phosphorylation of β-catenin and leading to an increase in the cytoplasmic pool of the protein. This increase allows some of the B-catenin to move to the nucleus, where it interacts with the TCF family of transcription factors and promotes the expression of Wnt responsive genes. Mutation in APC related to colon cancer.

Question 32

Gene expression III

Give an example of how phosphorylation affects activity of trans-activating factors?

Answer 32

CREB (cyclic AMP response element-binding protein)

A series of events initiated by the binding of a ligand to a guanine nucleotide binding protein coupled receptor induces the phosphorylation of the CREB protein, which, while present on the DNA, is inactive to promote transcription unless phosphorylated. Once phosphorylated, the CREB protein recruits the histone acetyl transferase, CBP (CREB binding protein), which has intrinsic HAT activity and recruits RNA Pol II- leading to transcriptional activation of the gene.

Question 33

Translation

What is the difference between binding of the ribosome to mRNA in bacteria and eukaryotes?

Answer 33

Initiation is the step that differs the most between bacteria and eukaryotes. However, in both the goal is the same: assemble a ribosome with the start codon (AUG) and initiator methionine tRNA in the P-site, ready to receive the next aa-tRNA in the A-site.

In bacteria, the ribosome binds essentially right at the start codon due to the Shine- Dalgarno sequence and three initiation factors work to assembly the full ribosome.

In eukaryotes, initiation factor (eIF) 4E is required to bind to the 7-methyl guanosine cap on the 5’ end of the mRNA. This leads to binding of many other eIFs (4G, 4A, 4B, etc.) and eventually to binding of the small ribosomal subunit, which itself is bound by several factors (eIF3, eIF1A, eIF1, eIF2, etc.).

Question 34

Translation:

Explain the cap-independent process to initiate translation:

Answer 34

It is driven by specific RNA sequences and structures called internal ribosome entry sites. Many viruses use IRESs to initiation translation after they shut down host cell cap-dependent synthesis

Question 35

Translation

What factors aid in elongation?

Answer 35

Elongation can be though of a cycle in which each step results in one more amino acid added to the polypeptide chain.

The cycle consists of an aa-tRNA entering the A-site (delivered by EF1A in eukaryotes, EF-Tu in bacteria), where its anticodon loop base- pairs with the right codon in the mRNA, “reading” the message.

Question 36

Translation:

What is a “sense” mutation?

Answer 36

Opposite of a nonsense mutation – now the stop codon encodes an amino acid and the ribosome keeps going.

Question 37

Describe the start “strenght” associated with a start AUG sequence:

Answer 37

Also, in eukaryotes, different start codons have different “strengths” depending on their Kozak context. AUGs with a weak Kozak context can be bypassed, allowing downstream AUGs to be used. The result would be two different proteins produced at certain amounts, from the same mRNA.

Question 38

Translation:

Explain how translation can be regulated with by the cap binding protein eIF4E:

Answer 38

In eukaryotes, the cap binding protein (eIF4E) can be bound by 4E-binding proteins (4E-BPs) that sequester it and blocks its function.

• When these proteins are **phosphorylated, they do not bind to 4E and this allows cap-dependent translation initiation.
• However, under some conditions (for example, stress), the 4E-BPs are dephosphorylated, they bind to 4E, and they block its function.

Which drug can also induce this effect?

**Rapamycin**

This results in a shut-down of much of translation.

However, note that some messages are less dependent on 4E (e.g. IRES-containing messages), and these could still be translated. The role of eIF4E levels and activity in cancer is a hot topic.

Question 39

Translation:

Explain how eIF2-alpha can interfere with translation:

Answer 39

eIF2-alpha is critical for the steps that lead to binding of the initiator tRNA to the ribosome.

• When eIF2-alpha is phosphorylated, its activity is inhibited and this blocks initiation.
• eIF2-alpha can be phosphorylated by several pathways. One is induced by interferon, which is produced when a cell is infected by a virus. Hence, shutting down translation is a response to viral infection.

Also, many other cellular stresses lead to phosphorylation of eIF2-alpha and so this is an important way for the cell to regulate protein synthesis during certain conditions.

Question 40

Study antibiotics that affect translation:

Answer 40

Question 41

Translation:

What is polycistronic?

Answer 41

In bacteria, there is no cap, no poly-A tail, and generally no UTRs. *Several proteins can be encoded on a single mRNA, which is something named polycistronic.

Question 42

Translation:

Transferrin

Answer 42

Translation:

Ferritin