Last 3 BC module exams

Question 1

Three possibilities to introduce genome wide unbiased mutations.

Answer 1

1. Random mutagenesis screen
   
   1. Create library of mutants of your favorite gene
   2. Unbiased
   3. Low-throughput
2. X-rays
3. UV radiation
4. Alkylating agents (like ENU or EMS)
5. Mutagenic chemicals

Question 2

How is spontaneous protein folding thought to overcome the Levinthal paradox? An exhaustive search through the vast number of possible conformations (~10¹⁰⁰ conformations for a protein with 100 residues) during protein folding would take ages.

Answer 2

1. Local propensity for secondary structure formation encoded by amino acid sequence, especially alpha-helices and beta-hairpins. These secondary structure elements might also act as a nucleus. (1)
2. Hydrophobic collapse by excluding hydrophobic sidechains from aqueous solvent. This quickly bypasses a very large number of extended conformations. (1)
3. Nucleation at key contacts of the future native structure, for example a beta-helix-beta motif to initiate beta-sheet formation/extension. This would essentially be like a folding trajectory, afterwards folding intermediates become progressively more stable during compaction. (1)

Better written answer (in my opinion)

Hydrophobic collapse

• Hydrophobic amino acids collapse towards center to avoid aqueous environment.

Molecular chaperones

• ​proteins in vivo that assist with folding of proteins (but are not part of final structure)

Framework models

• Diffusion-collision​
  
  • nucleation occurs at specific points of contact in future native structure.
  • example would be a beta-helix-beta motif to get ready for beta-sheet creation or extension
  • this would act like a folding trajectory
  • then structures collide and pack tightly together, becoming more stable over time
• Nucleation (propagation)
  
  • The amino acid primary sequence will code for local Secondary structures (particularly alpha-helices and Beta-hairpins) which can act as a nucleus

Question 3

Strains in the yeast deletion library are “barcoded” and can thereby easily be identified by PCR and sequencing. Explain a typical competitive growth assay and how the “barcodes” are utilized in this experiment. (3 points

Answer 3

Strains are pooled in grown in one culture. Barcodes are flank by a common sequence, which is used for PCR amplification. Resulting DNA is hybridized to microarray (harboring the barcode sequences) or analyzed by next-gen sequencing. In this way the abundance of each strain in the culture before and after the experiment (e.g. growth under stressful conditions such as DNA damage) can be measured.

Question 4

1. Which of the following statements is true? (3 points) (BC7 – lecture 2, BC7 – lecture 4/5)
   
   1. Chp1 chromodomain binds nucleosome core
   2. H3K79 methylation is deposited by Dot1
   3. Ccr4-Not complex deadenylates RNA
   4. TFIIS promotes RNA cleavage
   5. Mediator interacts with activators and repressors
   6. H2Bub is found in active chromatin

Answer 4

1. Chp1 chromodomain binds nucleosome core
   
   1. ​True
   2. Chp1 has chromodomain that binds H3K9me nucleosomes
2. H3K79 methylation is deposited by Dot1
   
   1. ​true
3. Ccr4-Not complex deadenylates RNA
   
   1. ​True
4. TFIIS promotes RNA cleavage
   
   1. True
   2. ​TFIIS is a transcription cleavage factor involved in proofreading
   3. is a part of RNA polymerase II
5. Mediator interacts with activators and repressors
   
   1. ​true
6. H2Bub is found in active chromatin
   
   1. ​true
   2. found at promotors and open reading frames

Question 5

Please draw the basic reaction catalyzed by all proteases and name three processes in the mammalian cell where this reaction takes place. (3 points)

Answer 5

1. Many different examples, e.g.
2. To degrade proteins:
   
   1. To switch off the signals that peptides and proteins initiate by degrading either them or the proteins they bind to
   2. To recycle amino acids by degrading the proteins.
   3. To destroy potentially lethal or toxic proteins from parasites and pathogens
   4. To release antigenic peptides from parasites and pathogens.
   5. To obtain amino acids from food proteins.
3. To remove the initiating methionine from the newly synthesized, cytoplasmic proteins
4. To remove signal peptides from proteins targeted to the cell’s secretory pathway.
5. To remove targeting signals from proteins targeted to specific organelles such as the mitochondrion or chloroplast.
6. To remove propeptides from enzymes, hormones and receptors that are synthesized as precursors, so that these are activated.
7. To release individual proteins and peptides from polyproteins
8. To release bioactive peptides from protein precursors.
9. Pathogens and parasites also use proteolytic enzymes to invade their hosts, and to inactivate any host protein that could harm them or interfere with their reproduction.

Question 6

In a genetic screen for C. elegans Ced (Ced, cell-death defective) mutants, you identified the recessive mutation x111. Like animals homozygous for ced-3(n717), animals homozygous for x111 have a general defect in cell death. (3 points)

1. From the phenotype that x111 causes, what can you conclude about the normal function of the gene that x111 defines?
2. Through what genetic test can you determine whether x111 is another allele of the ced-3 gene or an allele of a new ced gene, “ced-14”?
3. The recessive loss-of-function mutation n2812 of the ced-9 gene causes the opposite phenotype i.e. in animals homozygous for n2812 too many cells die. You generate a ced14(x111); ced-9(n2812) double mutant and analyze it for cell death. You find that in the double mutant, too many cells dies. What phenomena are you observing in the double mutant.
4. Draw a regulatory pathway that depicts the relationship between ced-9 and the ced-14 gene and cell death. Explain your answer in one sentence.

Answer 6

1. Required for cell death (killer gene; pro-apoptotic gene)
2. complementation test
3. epistasis
4. Answer:
   
   1. Ced-14 acts upstream of ced-9 (schematic)
   2. Ced-9 is epistatic to ced-14 and therefore acts downstream of ced-14
   3. (the “killing” step is a regulatory pathway or cascade with two alternative states, “DEATH” of “NO DEATH” (“ON” or “OFF”) and “EPISTATIC” therefore indicates “DOWNSTREAM”.)

Question 7

1. Name 2 GTPases which are involved in co-translational targeting of proteins to the bacterial inner (plasma) membrane. Which protein domain(s) do both GTPases have in common? Briefly describe the molecular events which lead to the activation of the 2 GTPases.

Answer 7

1. SRP GTPase activity in sec pathway
   
   1. NG domain (nucleotide binding and GTPase domain)
   2. NG and M domain connected and undergo large scale conformational change upon ribosome binding, increasing affinity for GTP
2. SR (SRP receptor)
   
   1. Has GTPase activity as well
   2. Has NG-domain as well
3. SRP and SR form a NG-NG twin
4. NG twin relocates to distal end of SRP-RNA after complex formed
   
   1. Required for GTPase activation
5. Targeting complex docking to the membrane
6. At this point, the signal sequence can be transferred to the Sec-complex, leading to GTP hydrolysis and dissociation of SRP and SR

Question 8

What are the roles of Izumo1 and Juno during mamalian fertilization? (3 points)

Answer 8

1. During fertilization, a single sperm binds to the egg’s membrane.
2. The protein Izumo1, which is tethered to the membrane of sperm, forms an adhesion complex with its receptor protein, Juno, which spans the egg’s membrane.
3. Fertilization does not take place in the absence of this complex.
4. After fertilization, Juno is lost from the egg’s membrane, thereby preventing the binding and fusion of additional sperm (to block polyspermy).

Question 9

1. What are incretin hormones and what are their main actions on pancreatic beta cells?

Answer 9

1. The hormones are:
   
   1. GIP (glucose-dependent insulinotropic polypeptide)
   2. GLP-1 (glucagon-like peptide 1)
   3. Are secreted by the gut in response to eating
2. Effects on beta cells
   
   1. Stimulate insulin biogenesis
   2. Stimulate β-cell proliferation
   3. Stimulate increase in amount of insulin released from pancreatic β-cells after eating, but before blood glucose levels become elevates

Question 10

1. Which of the following statements is true or false? (3 points)
   
   1. Pol eta is a high fidelity DNA polymerase that is both a replicative polymerase and can read over base crosslinks.
   2. RNA ubiquitination is an important functional switch to attract translesion bypass polymerases.
   3. Mitochondria have a specialized DNA polymerase and do not rely on the nuclear DNA polymerases.

Answer 10

1. False
   
   1. Pol eta is not a replicative polymerase
2. False
   
   1. during translesion synthesis, the sliding clamp is ubiquitylated
3. True
   
   1. mitochondria have DNA pol gamma

Question 11

TDG is a human DNA glycosylase that specifically recognizes T-G mismatches and excises the T base. Why is this enzyme important in human cells along with the canonical mismatch repair machinery? (hint: consider CpG islands) (3 points)

Answer 11

1. CpG islands are often methylated at position 5 of the cytosine 1 (1).
2. Deamination results in T:G mismatches (1).
3. These are properly repaired by TDG, but not necessarily by the mismatch repair machinery (1).

Question 12

1. Sketch the eukaryotic replication fork, drawing the approximate location and geometry of DNA and the approximate location of the DNA polymerases, primase clamp loader, clamp and RPA. (4 points)

Answer 12

Need to find good diagram for this

Question 13

1. Name the attacking nucleophile for each of the five main protease classes and name one member of each class.

Answer 13

1. Serine proteases
   
   1. Attacking nucleophile: Ser(195)
   2. Example: Chymotrypsin, trypsin
2. Cysteine proteases
   
   1. Attacking Nucleophile: Cys(25)
   2. Example: Papein, Cathepsin B, Caspases
3. Metalloproteases
   
   1. Attacking nucleophile: Zn2+
   2. Examples: Carboxypeptidases A and B
4. Aspartic protease
   
   1. Attacking nucleophile: Asp(33), Asp(213)
   2. Examples: Pepsin, Renin, HIV-1 retropepsin
5. Threonine protease
   
   1. Attacking nucleophile: Thr(1)
   2. Proteasome

Question 14

1. What class of proteins mediates apoptosis? What is their cleavage specificity? In which two functional categories can one group them?

Answer 14

1. Caspases
2. They are Asp specific
3. Initiator caspases
   
   1. Initiate apoptotic caspase cascade
4. Effector caspases
   
   1. Are initiated by initiator caspases

Question 15

1. CRISPR/Cas: What are the different components?

Answer 15

1. Induces targeted double strand breaks
2. 2 components:
3. Guide RNA (gRNA)
   
   1. Short synthetic RNA
   2. Composed of scaffold sequence
   3. Is necessary for Cas9 binding
   4. Also contains user defined targeting sequence that targets the genomic locus
4. Non-specific CRISPR associated endonuclease (Cas9)
   
   1. Guided to a specific site by the gRNA, induces a double strand break there
   2. Basically, you change the genomic target of Cas9 by editing the gRNA
5. CRISPR: Stands for Clustered Regularly Interspaced Short Palindromic Repeats

Question 16

Properties of an inbred strain; How is it generated?

Answer 16

1. Properties
   
   1. Except for sex difference, mice of inbred strain are as genetically alike as possible
   2. Homozygous at virtually all their loci
   3. Has unique set of characteristic to set it apart from all other inbred strains
2. Generated by:
   
   1. Sibling (sister x brother) matings for 20 or more consecutive generations

Question 17

1. If you want to test whether H3K9 methylation is epigenetically inherited in vivo, (3 points)
   
   1. How would you design the experiment?
   2. Is H3K9me epigenetically inherited in wild type fission yeast cells?
   3. What do you need at the centromeric repeats in fission yeast to establish heterochromatin?

Answer 17

1. Tether H3K9methyltransferase to euchromatic DNA sequence, check if H3K9me is established, release the enzyme, check if H3K9me is maintained (2)
2. No/to some extent (0.5)
3. siRNAs (0.5)

Question 18

In which compartments of the eukaryotic cell can “Brownian ratchets” be found? How do to the drive protein translocation? Would this system work to secrete proteins through the bacterial inner membrane? Why or why not? (3 points)

Answer 18

1. ER-Membrane/inner Mitomembrane/(Chloroplast) (1)
2. Hsp70 binds translocating protein and prevents backsliding (Hsp40 recruits Hsp70 to translocating protein). (1)
3. No, no ATP in periplasm.

Question 19

Defects of different DNA repair pathways lead to a hypersensitivity to UV light. In the experiment the UV sensitivity of either wild type yeast (white circles) or ubc13Δ single mutant (black circles), mms2 single mutant (black triangles) and mm2 ubc13 double mutants (white triangles) was tested. As shown in the experiment below all three mutant strains show the same UV sensitivity. Would you assign MMS2 and UBC13 to different repair pathways or the same? Give a reason for your decision. (3 points)

Answer 19

1. Both mutants show a similar phenotype therefore both are in principle involved in a pathway that is required in the presence of UV light. Double mutant does show no increase in phenotype (=the mutants are epistatic), therefore have to be assigned to the same pathway.

Question 20

1. What are the genetic functions of meiosis? (3 points)

Answer 20

1. Converts diploid cell into haploid gamete/spore
2. Genetic recombination by crossing over

Question 21

Pol III, Pol δ, RNA Pol II: Function in the cell

Answer 21

1. Poll III
   
   1. If this is DNA Pol III in E. coli (is probably this one)
      
      1. Leading and Lagging strand synthesis in DNA replication
   2. If this is RNA Pol III in eukaryotes
      
      1. tRNA
      2. rRNA 5S
      3. other small RNAs found in nucleus and cytosol
2. Pol δ:
   
   1. In humans
   2. Lagging strand synthesis in replication
3. RNA Pol II
   
   1. Transcribes: mRNA, most snRNA, microRNA

Question 22

1. How can a cell get rid of the large amyloid fibers? Describe 3 potential strategies.

Answer 22

1. Disaggregation via chaperones
2. Unfolded protein response (UPR) pathway
3. ERAD: ER associated decay
4. Heat shock factor1 (HSF1) pathway
5. Inclusion bodies

Question 23

1. Xeroderma pigmentosa: Reason, involved repair pathway; What do you understand under “complementation groups”. Complementation group A contains a mutation in XPC, group B in XPA. Do you expect, that the DNA repair pathway works, if you combine cell lysates from both complementation groups? Why?

Answer 23

1. Group A has non-functioning XPC, but presumably functional XPA
2. Group B has non-functioning XPA, but presumably functional XPC
3. Since the proteins are coded from different genes, presumably combining the cell lysates would show that the genes are in fact complimentary, and the repair pathway would work again, since there would be working copies of both XPA and XPC, providing all factors needed for the repair pathway.

Question 24

1. Which of the following statement/s is/are true? (point deduction for wrong answers!) In nucleosome assembly… (3 points)
   
   1. CAF1 deposits acetylated H3/H4 tetramer onto DNA
   2. CAF1 deposits methylated H3/H4 tetramer onto DNA
   3. NASP deposits H3/H4 tetramer onto DNA
   4. H3.3 variant is found at silent chromatin
   5. H2A.Z is enriched at the termination site
   6. Histone chaperone DAXX binds H3.3

Answer 24

1. CAF1 deposits acetylated H3/H4 tetramer onto DNA (+1.5)
   
   1. True,
   2. Rtt109 acetylates H3K56 to increase binding to Caf1
2. CAF1 deposits methylated H3/H4 tetramer onto DNA (-0.75)
   
   1. False,
   2. H3/H4 tetramer is acetylated
3. NASP deposits H3/H4 tetramer onto DNA (-0.75)
   
   1. False
   2. NASP deposits H1
   3. NASP stabilizes reservoir of newly synthesized H3-H4 dimers
4. H3.3 variant is found at silent chromatin (-0.75)
   
   1. False
   2. Soluble H3.3 has H3K9Ac mark
   3. Is incorporated into active chromatin
5. H2A.Z is enriched at the termination site (-0.75)
   
   1. False
   2. Is rapidly deposited at DSBs
   3. Promotes DNA resection, H4 acetylation and chromatin ubiquitylation
   4. Important for recruiting DNA factors
6. Histone chaperone DAXX binds H3.3 (+1.5)
   
   1. True
   2. HIRA and DAXX deposit H3.3
   3. In heterochromatin, DAXX cooperates with Chromatin remodeler ATRX in accumulating H3.3 at pericentric heterochromatin and telomeres

Question 25

What are glial cells? Which functions do they have during development and in the adult? (3 points)

Answer 25

1. The non-neuronal cell population in the nervous system.
2. Function during development of:
   
   1. Axon guidance
   2. Blood brain barrier
   3. Phagocytosis
   4. Survival signaling to neurons
3. Function during homeostasis:
   
   1. Neurotransmitter and ionic homeostasis
   2. Regulation of energy metabolism
   3. Detoxification

Question 26

1. E.coli OriC: What is the function of 9mer, 13mer and GATC for DNA replication?

Answer 26

1. 9mer: spacing, orientation and sequence, replication initiator DnaA binding
2. 13mer: A/T richness is important, unwinds and allows DnaC helicase to bind
3. GATC: Recognition site for dam methylation system

Question 27

1. Cotranslational protein transport in eukaryotes: Typical signal sequence/structure of the signal sequence; name two interaction partners of the signal sequence; how do they interact?

Answer 27

1. Signal sequence (ER import) = N-terminal hydrophobic sequence
2. Signal sequence recognized by Signal recognition particle (SRP), stopping translation
3. SRP brings ribosome to the ER membrane, where it binds to the SRP receptor (SR) which it bound to the ER membrane.
4. Signal sequence is then transferred to the Sec-complex (the major conducting channel - translocon), leading to GTP hydrolysis and dissociation of the SRP from the SR.

Question 28

You assembled a nucleosomal array in vitro. (3 points)

1. Which enzyme and substrate can you use to methylate H3K36?
2. How can you check if your in vitro reaction worked?
3. What do you expect to see if you bind HP1 protein to your array?

Answer 28

1. H3K36 methyltransferase (Set2)/SAM
2. Western blot with H3K36me antibody or MS
3. It doesn’t bind

Question 29

1. You want to protect an in vitro transcribed mRNA from degradation by Xrn. How do you do this? Enzymes?

Answer 29

1. XRN1 is a 5’->3’ exonuclease
2. Can only degrade decapped 5’ end of mRNA, so mRNA needs to be capped to protect it
3. 5’ cap added in 3 step process, with 3 capping enzymes
   
   1. RNA triphosphatase (RTPase)
      
      1. Cet1 in yeast
      2. Cleaves 5’ terminal γ-β phosphoanhydride bond of nascent mRNA molecules
      3. Enables addition of 5’ cap
   2. Guanyltransferase (GTase)
      
      1. Yeast: eg1
      2. Adds a backwards GMP group from GTP
   3. Guanine-N7-methyltransferase
      
      1. Yeast: Abd1
      2. Methylates guanine to form final 5’ cap structure
      3. Positions RNA cap and AdoMet cofactor

Question 30

Xeroderma pigmentosa variant is caused by a mutation in POLH a homolog of Pol eta: Why do these patients have an increased skin cancer rate?

Answer 30

1. Patients with XP phenotype often have Defects in nucleotide-excision repair genes Defects in translesion synthesis
   
   1. Pol eta is main polymerase dealing with UV crosslinks in translesion synthesis repair
   2. UV crosslinking is something that would be a problem leading to skin cancer due to the exposure of skin to UV radiation
   3. Without the machinery to fix these crosslinks, there would very likely be an increased skin cancer rate amongst these patients

Question 31

1. What type of proteases are caspases and what cellular program do the execute? How many active sites are found in an activated caspase? (3 points)

Answer 31

1. Cysteine protease (1)
2. Apoptosis (1)
3. 2 (1)

Question 32

1. Which are the mendelian ratios? Give an example of a monohybrid cross. In which situation are mendelian ratios modified?

Answer 32

1. A mendelian ratio is the ratio of an organism’s offspring with one phenotype to another phenotype
   
   1. 3:1 for monohybrind
   2. 9:3:3:1 for dihybrid
2. An example of a monohybrid cross would create a 3:1 ratio
   
   1. So if a gene for height had a dominant BB or Bb for tall, and a recessive bb for short
   2. Bb x Bb would give 3 tall (Bb, Bb, BB) and 1 short (bb)
   3. So a 3:1 ratio
3. When are they modified?
   
   1. more than 1 recessive or dominant gene,
   2. maternal inherited genes can also influence that ratio
   3. environmental causes and ofc mutations can lead to a different ratio

Question 33

Name four, nucleosome remodelers; Properties of the +1 Nucleosome

Answer 33

Note: not sure about what is wanted with this one, do I need more specific factors, or is this enough? Also, did I say enough about the +1 nucleosome

1. INO80 family
   
   1. Incorporation and removal of histone variants – H2A.Z or H2A.X
   2. Swr1 is in family – incorporates H2A.Z at DNA damage sites
2. CDH1
   
   1. Assembly and deposition of H3.3
   2. NuRD complex is part of family, is often misregulated in many cancers
3. SWI/SNF
   
   1. Removes ectopically deposited CENP-A
4. ISWI
   
   1. Remodeling activity antagonized by H3K16 ac
5. +1 nucleosome
   
   1. Positioned at downstream boundary of nucleosome free region
   2. Often strongly positioned H2A.Z variants

Question 34

1. What kind of protease is Cathepsin B. Where does it occur in the cell? What kind of proteases are Caspases 3/7/8/9. In what pathway are they involved?

Answer 34

1. Cathepsin B is a cysteine protease found in the lysosome
2. Caspases are cysteine proteases in apoptosis

Question 35

1. How does hydrogen/deuterium exchange reflect the folding state of a protein? What are the principal steps of such an experiment?

Answer 35

1. Covalently bonded hydrogen is replaced by deuterium atom
2. Deuterium is heavier than hydrogen
3. It is therefore possible to distinguish between normal hydrogen and Deuterium with NMR spectroscopy
4. The method gives information about solvent access ability of various parts of the molecule
   
   1. Basically, Deuterium₂O (D₂O) is used as the solvent, and can then use NMR to see how much Deuterium was incorporated
   2. Essentially showing how much covalently bonded hydrogen was exposed in folded form to the solvent.

Question 36

1. What is the key characteristic of a totipotent stem cell? Give examples from the mammalian development.

Answer 36

1. Most versatile of stem cells
2. Has the potential to give rise to any and all human cells
3. Can even give rise to entirely new functional organism
4. Spores and Zygotes are both examples of totipotent cells
5. The first few divisions after egg fertilization in mammals are totipotent, so the precursor cells in the late morula are totipotent for instance

Question 37

1. What are glial cells? What functions do they have during development and in the adult?

Answer 37

1. Non-neuronal Cells in the central and peripheral nervous system that provide support and protection for neurons.
2. Development
   
   1. Axon guidance
   2. Ensheathment and BBB formation
   3. Phagocytosis and apoptotic neurons
   4. Neuronal stem cell proliferation
3. Mature nervous system
   
   1. Neurotransmitter reuptake
   2. Ionic homeostasis
   3. BBB maintenance
   4. Immune response

Question 38

What Transcription factor is divided in a DNA binding domain and an activating domain: How is this used to study protein-protein interactions in Yeast-two hybrid assays? What interaction do you study in an One-hybrid assay?

Answer 38

1. Two-hybrid
   
   1. The Gal4 transcription factor gene produces a BD and an AD
   2. When in close proximity, the AD and BD activate transcription of a reporter gene (often LacZ)
   3. Can study protein protein interactions by binding AD to one protein or a library of proteins. (this is usually the “prey protein)
   4. BD is attached to another protein (usually the known “bait” protein)
   5. If the proteins interact, then the AD and BD of the transcription factor will be brought into close proximity and the reporter gene will be expressed.
2. One-hybrid
   
   1. One hybrid used to study protein-DNA interactions
   2. Activating domain bound to protein of interest
   3. If protein of interest binds to DNA sequence of interest, then the activating domain will be in the correct position
   4. This causes transcription of the reporter gene

Question 39

1. Protein Modifications: Which aa`s are modified in N-linked glycosylation? In which form are the sugars mainly used to build up the precursor? Which enzymes (full name) transfers this precursor onto protein and where in the cell does this happens?

Answer 39

1. Asn modified
2. Branched oligosaccharides mostly used to build up precursors
3. Happens in ER membrane at translocon
4. OST: Oligosaccharyl-transferase

Question 40

1. How does a weak base like Choloquine inhibit lysosomal proteases? (3 points)

Answer 40

1. Lysosomal proteases have an acidic pH optima and are largely inactive at cytosolic pH.
2. Uncharged form of the weak base enters cells and lysosomes.
3. Accumulates in charged form in lysosomes and increases pH, this inhibits the lysosomal protease.

Question 41

1. Draw the peptide G-K-Y with correct charges at pH 7.0

Answer 41

Question 42

How many copies of H2A/H2B/H3/H4 has the cell to synthesize during replication of a diploid cell? Assume that the haploid genome contains 3x10⁹ bp DNA, the linker between the nucleosomes contains 50 bp DNA and that all parental histones are maintained.

Answer 42

1. Haploid genome = 3x10^9 bp DNA
2. 146 bp around each histone octamer, + 50 bp for linker = ~200 bp/octamer
3. # of octamers in haploid cell = 3x10^9/200 = 1.5 x10^7
4. # of octamers in diploid cell = 1.5x10^7 * 2 = 3x10^7
5. octamer contains 2 copies each of H2A/H2B/H3/H4
6. 3x10^7 * 2 = 6x10^7 copies of each are needed.

Question 43

1. Briefly explain why inactivation of uracil-N-glycosylase - an enzyme that cleaves deoxy-uracil in DNA - leads to defects in the maturation of antibodies. (2 points)

Answer 43

1. AID generates uracil by deaminating cytosine in immunoglobuline genes in activated B-cells (1).
2. UNG cleaves the uracils DNA to generate abasic sites, these are used for an error prone repair (1).

Question 44

1. Give three reasons why cells need to degrade proteins and three examples where proteases are involved in other cellular processes.

Answer 44

1. Reasons why:
   
   1. Remodeling of cell
   2. Prevent formation of toxic protein species
   3. Control level of functional species
   4. Damaged/misfolded protein
   5. To protect from bacteria or viruses
   6. Standard apoptotic pathway (programmed cell death)
2. Examples of proteases in other cellular processes
   
   1. Apoptosis:
      
      1. Caspases and cathepsin B
   2. Ubiquitin tagging can be used as a regulation method
      
      1. Example: Mdm2 ubiquitin ligase tagging p53 for degradation by proteasome in p53 pathway
   3. Transcription factor regulation via ubiquitylation
   4. Antigenic peptides generated by the proteasome contribute to the immune response

Question 45

1. dUTPase is an enzyme that converts dUTP to dUMP. You find that deletion of the enzyme leads to a hyperrecombination phenotype in E. Coli, i.e. A highly increased level of homologous recombination, in particular in A:T rich sequence regions. The hyperrecombination phenotype can be allowed by deletin Uracil DNA glycosylase. Please explain both observations.

Answer 45

1. dUTPs now present
2. will be used in place of Ts sometimes in replication
3. U is excised by UDG, generating an abasic site, allowing for DBS, and giving a chance for homologous recombination
4. Delting UDG will allow these sites to be relatively stable, and to continue existing.

Question 46

Which biophysical methods are good for studying protein folding at single molecule level?

Answer 46

• NMR spectroscopy
• Hydrogen deuterium exchange
• Single molecule force miscroscopy
• smFRET - single molecule fluorescence resonance transfer

Question 47

In which transport pathways are they following GTPases involved: Ran/Rab/NSF/SRP/Arf1/SecA

Answer 47

1. Ran: involved in gated transport through the nuclear envelope
   
   1. import: dissociates cargo from import receptor
   2. export: Enables cargo binding to the export receptor
2. Rab: Vesicle transport pathway
   
   1. Coordinators of vesicle transport
   2. Rab-GTPases recruit Rab effector proteins
   3. These attract membrane specific carriers
3. NSF: Vesicle Transport from ER-Golgi
   
   1. NSF dissociates Cis-SNARE complex
4. SRP: Signal recognition particle involved in co-translational translocation through the ER membrane in the Sec Pathway.
5. Arf1: vesicular transport
   
   1. Arf1 localized to golgi and has central role in intra-Golgi transport
   2. Arf family plays role in vesicular trafficking as activators of phospholipase D
6. SecA: Posttranslation translocation in Secretory pathway in bacteria

Question 48

In their zygotic embryonic screen Nusslein-Volhard and Wieschaus discovered three classes of mutations with distinct phenotypes. Please name the three classes and their main phenotypic feature. What does the existence of these three classes of mutants say about the formation of the segmented body pattern? (3 points)

Answer 48

1. GAP,
   
   1. Deletion of chunks
2. pair rule
   
   1. every other segment deleted
3. segment polarity genes.
   
   1. deletion of a portion of each segment.
4. Demonstrates Hierarchical pattern formation.

Question 49

1. Histone variants:

Briefly describe the functions of 4 histone variants

Answer 49

1. Briefly describe the functions of 4 histone variants.
   
   1. H3.3
      
      1. Found at transcribed regions
      2. Important for programming leading to chromatin signature reminiscent of pluripotency
   2. CENP-A
      
      1. Specifically deposited at centromeres
      2. Key determinant of centromere identity
   3. H2A.Z
      
      1. Deposited rapidly at DSBs
         
         1. Promotes DNA resection
         2. Histone H4 acetlylation
         3. Chromatin ubiquitylation, is especially important for recruitment of several DNA repair factors
      2. Is also enriched at transcription start sites, as well as enhancers and insulaters
   4. H2A.X
      
      1. Rapidly phosphorylated at DSB sites
      2. Used as marker for DNA damage response activities
2. Describe differences between replicative histones and histone variants.
   
   1. Histone variants can only differe in a few important amino acids
   2. Often expressed in a tissue specific manner
   3. Incorporation independent of DNA synthesis via histone chaperones

Question 50

Genetic inactivation of RNaseH2, a Rnase that can be found in the nucleus and nicks DNA containing ribonucleotides (at the ribonucleotide containing strand) was found to reduce the proper repair of DNA mismatches after DNA replication in yeast. Explain why! (3 points).

Answer 50

1. DNA polymerases occasionally incorporate ribonucleotides in the nascent strand (1)
2. Mismatch repair uses nicks to discriminate the newly synthesized strand from the templating strand (1).
3. Nicking by Rnase-H2 in the newly synthesized strand generates the strand discrimination signal and proper loading of the helicase/exonuclease for repair (1).

Question 51

Differences between Mitosis and Meiosis.

Answer 51

1. Mitosis
   
   1. creates diploid cells
   2. no genetic recombination
2. Meiosis
   
   1. creates haploid cells (gametes or spores) from diploid cells
   2. also allows for genetic exchange through homologous recombination

Question 52

Antibodies: How is the diversity generated by V(D)J-recombination, activation-induced deaminase (AID) and Uracil?

Answer 52

1. V(D)J-recombination
   
   1. Rag1/2 endonuclease assists in cutting at RSS site flanking V,D, or J coding segments on gene coding for antibody
   2. NHEJ repairs segment, but the targeting region is changed, with the additional help of TdT which incorporates N nucleotides
   3. This means that the antigen binding region is highly variable, allowing it to potentially bind to antigens that the organism has never before encountered.
   4. Helpful for pathogens that frequently change or ones that are completely new.
2. Activation induced deaminase (AID)
   
   1. AID induces class switch DNA recombination (CSR)
   2. Deaminases deoxycytosines in transcribed switch (S) regions
   3. Result is deoxyuracils which are removed by UNG
   4. The abasic sites are excised, resulting in SSBs that result in DSBs
   5. DSBs are repair by c-NHEJ or A-EJ (alternative end joining), leading to formation of S-S junctions and class switching
   6. Rearranges DNA that encodes heavy chain variable region, leading to antibody diversity
3. Uracil
   
   1. The change of deoxycytosine to deoxyuracil during CSR by AID leads to the rearrangements and diversity mentioned above

Question 53

Name two recombination systems used in D. melanogaster. How do they work?

Answer 53

1. Mitotic recombination
   
   1. For conditional activation of genes
   2. Generation of mosaics for lethal mutations
   3. Uses FLP recombinase at HRT sites to create heterozygous mosaic organisms
   4. Have different genotype at a specific locus on chromosome
2. Bacteriophage ΦC31 integrase
   
   1. attP - phage attachment site in phage genome
   2. attB - bacterial attachment site in bacterial host genome
   3. ΦC31 integrase catalyzes recombination between attP and attB site
   4. in drosophila
      
      1. attP site previously integrated with a transposon into fly genome
      2. attB site is present in injected plasmid
      3. ΦC31 integrase mediates recombination between the two sites

Question 54

What is the precursor for N-glycosylation?

Where is it synthesized and which enzyme (full name) transfers the sugar moiety where in the cell onto which amino acid? (3 points)

Answer 54

1. What is the precursor for N-glycosylation?
   
   1. Dolichol-P-P-polysugar
2. Where is it synthesized?
   
   1. ER-membrane (flipase)
3. and which enzyme (full name) transfers the sugar moiety where in the cell onto which amino acid? (3 points)
   
   1. OST (oligosaccharyl-transferase)
   2. Is next to translocon
   3. Transfers sugar moiety onto Asn

Question 55

1. Topic Chaperones: (3 points)
   
   1. Describe the differences between Hsp70 (DnaK) and GroEL in terms of:
      
      1. Substance binding
      2. Consequence of ATP hydrolysis and
      3. Cochaperone action

Answer 55

1. DnaK
   
   1. DnaK binds extended polypeptide stretches with hydrophobic amino acids,
   2. ATP hydrolysis switches DnaK from the low affinity state to the high affinity state,
   3. the cochaperone DnaJ stimulates the ATPase of DnaK
2. GroEL
   
   1. GroEL binds substrates at the apical domain and engulfs them in a cavity, substrate size is limited to 60 kDa,
   2. ATP hydrolysis is the timer for the time the substrate spends inside the cavity,
   3. the cochaperone GroES closes the cavity and induces ADP release from the trans ring.

Question 56

Three different reasons, why C. elegans is a good model organism to study especially apoptosis?

Answer 56

1. Complete cell lineage known.
   
   1. The precise order in which cells divide has been established.
   2. This is helpful in knowing when cells would normally be programmed to die
2. Is transparent throughout life cycle,
   
   1. allowing it to be examined at the cellular level in living preparations.
   2. Means that it can be easily seen if cells die or not. Helpful for studying if cells are engulfed or not after dying.
3. Hermaphrodites have vulva and HSN nerve that regulates vulva opening or not.
   
   1. Easy to detect if HSN has died or not visually by buildup of eggs
   2. Therefore, presence of HSN is helpful in studying cell death because if cell death has been turned off, HSN will not die and egg buildup (Egl phenotype) will not occur.

Question 57

1. Chromatin remodelling:
2. Describe the role of chromatin remodelers
3. List 4 groups of chromatin remodelers
4. Describe the characteristics of the 1+ nucleosome

Answer 57

1. Describe the role of chromatin remodelers
   
   1. Provide access to an otherwise tightly packed genome
   2. Regulation of gene expression
   3. Nucleosome movement by remodelers is important in many biological processes
2. List 4 groups of chromatin remodelers
   
   1. ISWI family
   2. CHD family
   3. INO80 family
   4. SWI/SNF family
3. Describe the characteristics of the 1+ nucleosome
   
   1. Positioned at downstream boundary of nucleosome free region
   2. Often strongly positioned H2A.Z variants
   3. And has H3.3 variants as well
   4. +1 nucleosome placed to transcription start site is stabilized by preinitiation complex

Question 58

1. GroEL: Function of ATP-binding; Which domain of GroEL binds ATP; What happens during ATP-Hydrolysis

Answer 58

1. ATP binding to the apical domain causes conformational change and allows protein access to cage
2. After ATP hydrolysis is GroES dissociation and substrate release from the trans ring
3. ATP hydrolysis serves as a timer for the time the substrate spends inside the cage

Question 59

What is a master regulator in development? What type of molecule and what genetic characteristics? Name two examples. (3 points)

Answer 59

1. Is at the top of a gene hierarchy
2. Type of molecule: Transcription factor
3. Determins development of organ (eye) or tissue type (glia, muscle).
4. Is both necessary and sufficient for development of organ/tissue.
5. Master regulators of imaginal disc development are:
   
   1. Eyeless: late embryo and early eye disc
   2. Scalloped: in late embryo and wing disc
   3. Distal-less: late embryo and leg disc

Question 60

1. Describe the function of the following histone chaperones:
2. CAF1
3. FACT
4. DAXX (XPE)

Answer 60

1. CAF1
   
   1. Deposits H3/H4 tetramers onto new DNA after replication fork
2. FACT
   
   1. H2A/H2B eviction
3. DAXX (XPE)
   
   1. Deposition of histone variant H3.3

Question 61

Amyloids: (3 points)

1) What are the three biophysical hallmarks of protein amyloids?

2) Explain the mechanism for amyloid formation! What is the rate limiting step, and how can this step be by-passed?

Answer 61

1)

1. Answers:
   
   1. Binding of specific dyes like Congo Red or Thioflavin-T (0.5)
   2. Fibrillar structure in electron microscopy. (0.5)
   3. Cross-beta structure in X-ray fiber diffraction (0.5)

2)

1. Nucleation mechanism: Formation of nucleus, followed by linear extension. Possibly, fragmentation of growing fibers. (0.5)
2. Nucleus formation is rate-limiting. (0.5)
3. Seeding with amyloid fiber fragments (Prion hypothesis). (0.5

Question 62

1. Which of the listed chaperons uses ATP to support proteins in their folding process and does not work as a monomer?
2. Hsp70
3. DnaJ
4. Trigger factor
5. NAC
6. GroES
7. GroEL
8. Prefoldin
9. Hsp90
10. Hsp100 (ClpB)

Answer 62

1. Hsp70
   
   • uses ATP, and works with Hsp40, which delivers unfolded peptide
2. DnaJ
   
   • works with DnaK
   • stimulates ATPase activity of DnaK through its J domain
3. Trigger factor
   
   • Works alone, and I don’t believe that there is atp support
4. NAC
   
   • Works with RAC in eukaryotes as a heterodimer and does use ATP
5. GroES
   
   • Works with GroEL and uses ATP
6. GroEL
   
   • Works with GroES and the complex uses ATP
7. Prefoldin
   
   • Works with GimC, but is ATP independent
8. Hsp90
   
   • Is dimeric and uses ATP
9. Hsp100 (ClpB)
   
   • Uses ATP, but as a monomer (I think)

Question 63

1. Draw the active site of a Cysteinprotease; Amino acid residues

Answer 63

Question 64

1. Which of the following statements is true and which false which respect to origins of replication?
2. Human origins are non redundant.
3. DnaA marks origins in yeast.
4. TFII H opens the bubbles of origin DNA to facilitate DNA polymerase loading.

Answer 64

1. Human origins are non redundant.
   
   1. True,
   2. each DNA segment is replicated exactly once
2. DnaA marks origins in yeast.
   
   1. False
   2. DnaA marks origin in E. coli
3. TFII H opens the bubbles of origin DNA to facilitate DNA polymerase loading.
   
   1. False
   2. TFIIH is a transcription factor used in nucleotide excision repair, opens bubble at lesion, but not at origin
   3. The helicase (MCM in eukaryotes) opens up the bubble of origin DNA

Question 65

1. You treat 1) wt yeast, 2) yeast carrying a deletion in the mre11 gene with etoposide, a poison that traps topoisomerase II to form covalent protein-DNA adducts. While cells in 1) have a somewhat reduced growth, cells 2) do not grow. Explain phenotype 2).

Answer 65

1. Nuclease dependent repair of DNA-protein crosslink repair
   
   1. uses MRN complex (MRX in yeast)
      
      1. consists of Rad50
      2. NBS1
      3. Mre11
   2. Without functional Mre11, can’t do nuclease dependent repair

Question 66

1. Describe and explain 2 mechanisms how E. Coli ensures that its origin of replication fires only once per cell cycle. Why are there much more parallel mechanisms required to prevent origin re-firing in eukaryotic cells?

Answer 66

1. The Dam methylation system at the OriC
   
   1. Methylates at the GATC site
   2. Creates a period of replicative incompetence to keep origin of replication from firing again in a cell cycle
   3. SecA binds hemimethylated OriC DNA which occurs during replication, prevents oriC from firing
2. The oriC winding and unwinding allow DnaA to bind, DnaA + OriC accessibility are necessary for replication
3. In eukaryotes, need more mechanisms because there are multiple origins of replication, meaning that you need to decrease the likelihood of firing even more
   
   1. Proper timing is extremely important too, so don’t want them firing too early or late
   2. There is an exact coordination of initiation from multiple origins necessary

Question 67

Amyloid structures: Different phases of the nucleation growth mechanism; What happens during the different phases?

Answer 67

1. Have a solution of monomers for amyloid
2. Trigger amyloid formation via shift in pH
3. Lag phase –
   
   1. very little is happening in this phase as monomers are very slowly coming together and nucleus is formed
   2. is energetically unfavorable
4. Fibril growth:
   
   1. Start growing by adding monomers until pool of monomers depleted
5. At higher concentrations of monomers, the lag time is shorter and maximimum fibril formation peak is higher
6. Nucleation is rate limiting step
7. Can by-pass nucleation phase by seeding with preformed aggregates

Question 68

What are the major categories of diabetes mellitus and what are their characteristics?

Answer 68

1. Type 1 – around 5-10% of cases
   
   1. β-cell destruction, usually leading to absolute insulin deficiency
   2. immune-mediated
2. Type 2 – around 90-95% of cases
   
   1. Can be insulin resistance with relative insulin deficiency
   2. Can also be an insulin secretory defect with insulin resistance
   3. Often associated with obesity

Question 69

How are double strand breaks introduced by Cas9 repaired? (3 points)

Answer 69

1. By NHEJ (non-homologous end joining) or by HDR (homology directed repair)

Question 70

1. Draw the dipeptide KE at pH7.0

Answer 70

Lysine and Glutamic acid

Question 71

Before the metamorphosis of insects is triggered, important metabolic conditions must be met. What is critically important for the larvae before pupation? What is a corresponding parameter that can be experimentally tracked? What happens under adverse environmental conditions? (3 points)

Answer 71

1. The larva must have accumulated enough stored resources (nutrients) to survive during metamorphosis (no feeding occurs during this time. (1)
2. This can be traced as the weight of the larva. (1)
3. Under adverse conditions, larvae above the so-called critical weight will undergo pupation (although they have not yet reached the normal weight for pupation). Larvae below the critical weight will not form pupae. (1)

Question 72

Nucleosome: What does it contain? Properties?

Answer 72

1. Contains:
   
   1. Histone core – octamer of proteins
      
      1. H2A, H2B, H3, and H4 (2 of each)
   2. Linker histones
      
      1. H1 and H5
   3. DNA
2. Properties:
   
   1. Is DNA wrapped around histone core
   2. Chromatin consists of nucleosomes strung together (beads on a string model)
   3. DNA can wrap more or less tightly in response to histone modifications or DNA modifications
   4. Keeps DNA organized and protects from damage
   5. How tightly the DNA is wrapped around histone core in nucleosome will also allow transcription of DNA or not.

Question 73

1. Protein targeting and translocation: Give 2 examples for a protein translocation process where ATP hydrolysis is required as a driving force (1 example prokaryote, 1 example eukaryote). Which factor(s) is (are) the ATPase(s)? What exactly is the effect of the ATPase cycle during these processes?

Answer 73

1. The DnaK/DnaJ cycle in eukaryotes
   
   1. ATP binds: DnaK binds peptide lossly (open state)
   2. ATP hydrolysis: DnaK binds peptide tight (closed state)
   3. ATP binds again: DnaK releases peptide
2. SecA in bacteria
   
   1. Inserts protein into translocation channel (SecYEG complex)
   2. ATP binding:
      
      1. clamp widens
      2. two-helix finger domain to binds polypeptide and move it into channel
   3. ATP hydrolysis:
      
      1. Clamp tightens
      2. Fnger slides back into cytosol.

Question 74

1. Draw the tripeptide Lys-Val-Asp (3 points)

Answer 74

Question 75

Characteristic of a totipotent cell. Name examples.

Answer 75

1. Most versatile stem cell
2. Has the potential to give rise to any and all human cells
3. Can even give rise to entirely new functional organism
4. Spores and Zygotes are both examples of totipotent cells

Question 76

1. Your would like to test if H3K9 methylation is epigenetically inheritated in vivo.
2. Design the experiment.
3. Is H3K9me epigenetically inheritated in wild type fission yeast cells?
4. Define „epigenetic inheritance“

Answer 76

1. Design the experiment.
   
   1. Tether H3K9methyltransferase to euchromatic DNA sequence, check if H3K9me is established, release the enzyme, check if H3K9me is maintained (2)
2. Is H3K9me epigenetically inheritated in wild type fission yeast cells?
   
   1. No/to some extent
3. Define „epigenetic inheritance“
   
   1. offspring inherits parental histone that has specific modifications
   2. these modifications are recognized by complexes
   3. can re-establish same type of modifications on newly deposited adjacent nucleosomes

Question 77

1. Which statements are correct:
   
   1. Black chromatin is deacetylated
   2. Ccr4/Not degrades polyadenine-tail of mRNAs
   3. S2P is involved in transcription termination
   4. H2Bub forms Euchromatin (?)
   5. Transcription is regulated by a mediator binding a repressor/activator

Answer 77

1. Black chromatin is deacetylated
   
   1. true
2. Ccr4/Not degrades polyadenine-tail of mRNAs
   
   1. true
3. S2P is involved in transcription termination
   
   1. True
   2. 3’-mRNA processing and termination
4. H2Bub forms Euchromatin (?)
   
   1. True?
   2. This is a sort of weird question, but H2Bub prevents compaction, so I suppose you could say that it helps form euchromatin –
5. Transcription is regulated by a mediator binding a repressor/activator
   
   1. True (I’m pretty sure)
   2. Repressor or activator bind mediator in regulating transcription

Question 78

1. You would like to determine the RNA Polymerase II localization at your gene of interest. (3 points)
   
   1. Which method and antibody against which RNA Polymerase II can you use to determine RNA Polymerase localization at the transcription initiation site?
   2. What do you expect to see at the promotor regarding the RNA Pol II amount?
   3. Which histone modification and RNA processing factors are recruited at the promotor region?

Answer 78

1. Chip with S5P CTD antibody
   
   1. Elongation initiation is done by phosphorylation of Serine 5 via TFIIH
   2. CTD = c-terminal domain
   3. So chromatin immunoprecipitation with an antibody for the phosphorylated Serine 5 c-terminal domain that marks initiation
2. Pol2 peak
3. H3K4me3, capping factors
   
   1. Methylation of H3K4me3 depends on H2Bub, which itself relies on RNAPol2 S5P
   2. This recruits COMPASS which is a signaling platform for downstream factors

Question 79

Heterozygous double mutant in yeast (Δsabc and Δsxyz): After sporulation on average three haploid cells are obtained, but no haploid double mutant. Explain the kind of experiment and the result.

Answer 79

1. Tetrad analysis
   
   1. Watching the outcome of meiosis
   2. After meiosis 2, there is a separation of chromatids, meaning that each spore essentially represents one chromatid
2. Means that homozygous double mutant is lethal (synthetic lethality)

Question 80

1. Topic: Insertion of membrane proteins (3 points)
   
   1. SNAREs are membrane proteins with a C-terminal transmembrane (TM) domain and a long cytoplasmic N-terminal domain. By which targeting and translocation machinery are these proteins inserted into the ER-membrane? Is SRP involved in the process? Why or why not?

Answer 80

1. The GET/Trc40-system (1)
2. No SRPp, because it recognizes N-terminal signal sequences (2)

Question 81

Properties of proteins bound by Calnexin/Calreticulin; How can this proteins be released?

Answer 81

1. Calnexin and calreticulin are lectins
2. Calnexin/calreticulin bind to incompletely folded proteins containing one terminal glucose on N-linked Oligosaccharrides -> traps them in the ER
3. Glucosidase removes terminal glucose -> protein released from calnexin

Question 82

1. 26S-Proteasome: Which subunit contains an ATPase; What is its function

Answer 82

1. 19S regulatory particle
2. Acts as a gatekeeper, recognizes substrate, deubiquitylates it, unfolds it, and translocation of unfolded protein into the core