BC6 old exams

Question 1

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

Answer 1

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

Question 2

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 2

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 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

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 4

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 5

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 5

1. GAP, pair rule and segment polarity genes. Deletion of chunks, every other segment, portion of each segment. Hierarchical pattern formation.

Question 6

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

Answer 6

1. Transcription factor determining development of organ (eye) or tissue type (glia, muscle). Both necessary and sufficient for development of organ/tissue.

Question 7

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

Answer 7

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 8

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

Answer 8

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

Question 9

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

Answer 9

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 10

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 10

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 11

What is a Master Regulator?

Answer 11

1. A master regulator is a gene at the top of a gene hierarchy
   
   1. Particularly in regard to cell fate and differentiation
2. 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 12

Properties of an inbred strain; How is it generated?

Answer 12

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 13

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

Answer 13

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 HLN 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 14

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

Answer 14

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. Catalyzes the recombination between phage attachment (attP) site present in its own genome
   2. And bacterial attachment (attB) site present in bacterial host genome
   3. In Drosophila, recombination is mediated via ΦC31 integrase between an attP site previously integrated with a transposon into the fly genome and an attB site present in an injected plasmid

Question 15

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 15

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 16

Characteristic of a totipotent cell. Name examples.

Answer 16

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 17

Differences between Mitosis and Meiosis.

Answer 17

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 18

Three possibilities to introduce genome wide unbiased mutations.

Answer 18

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

Question 19

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

Answer 19

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 20

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 20

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 21

1. Name and describe (in keywords) a „non-classical 2-Hybrid system“ that does NOT involve reconstitution of the GAL4 transcription factor!

Answer 21

1. Split ubiquitin <- this is a good answer
   
   1. Used when studying insoluble proteins, like integral membrane proteins
   2. Two membrane proteins fused to two different ubiquitin molecules
      
      1. A c-terminal ubiquitin moiety (“Cub) – fused to bait protein
      2. N-terminal ubiquitin moiety (“Nub”) – fused to prey protein
   3. In addition to being fused to the protein, the Cub moiety is also fused to a transcription factor that may be cleaved off by specific proteases
   4. Upon the bait and prey proteins interacting, Nub and Cub moieties assemble, reforming the split ubiquitin.
   5. This split ubiquiting is recognized by ubiquitin specific proteases, which cleave off the transcription factor, allowing it to induce transcription of reporter genes
2. Classical two hybrid (Gal4 transcription factor)
   
   1. Gal4 binding to UAL promotes transcription of a gene
   2. Bind Gal4 to one protein, UAL to another, if the proteins interact, they will be nearby each other and the gene will be transcribed
3. One-hybrid
   
   1. Bind an activating domain to a protein, if protein interacts with a DNA sequence, activating domain will be brought into correct position, allowing transcription of the gene

Question 22

1. A yeast mutant screen has resulted in several mutants deficient for growth on medium lacking the amino acid histidine. a) How can you distinguish between dominant or recessive mutations? b) Would you continue your further analysis with recessive or dominant mutants? Justify your decision!

Answer 22

1. ) How can you distinguish between dominant or recessive mutations?
   
   1. Tetrad analysis
   2. Allow to spore
   3. Plate each spore individually on plates that do not contain histidine
   4. if mutation is recessive, none of the spores will grow
2. b) Would you continue your further analysis with recessive or dominant mutants? Justify your decision!
   
   1. Would continue analysis with the recessive mutants, since would always be able to predict outcome, and can screen for recessive or dominant on that gene by whether or not they can grow with or without Histidine
   2. it also depends on what sort of analysis is happening, but can do tests to see if what mutations may then be performed to make it grow on histidine again to investigate pathway
   3. This is an auxotrophic mutation and it is very useful in genetic analysis

Question 23

Please name three general prerequisites for an organism to become a model organism

Answer 23

1. Breeds quickly and cheaply
2. Short life cycle
3. Known genomic code
4. Conserved genomic code
5. Known traits

Question 24

1. What is the basis of “cloning by high-copy number suppression”?

Answer 24

1. Copy number is number of copies of a gene in the genome
2. High scopy number suppression is where A mutant phonotype is suppressed because an interacting gene is overexpressing,
3. essentially the mutant phenotype is suppressed by another mutated gene that is overexpressing, giving it a wild type phenotype
4. in terms of cloning, can screen for high copy number suppressors by using a yeast genomic library in a high copy number vector to transform mutant of interest.
   
   1. Resulting transformed yeast screened for ones that are wild type
   2. indicates that they either actually have the wild type gene or they have been transformed to wild type phenotype by high copy number suppressor
   3. if in fact there has been a high copy number suppression of the mutant gene, will tell researcher that the protein from the library likely interacts with the protein of interest.

Question 25

1. Programmed Cell Death has multiple functions during embryonic development. Please name three of these functions.

Answer 25

1. Sculpting
2. Deleting structures
3. Adjusting cell numbers
4. Eliminating dangerous or injured cells

Question 26

What is a protein microarray and what is it used for?

Answer 26

1. Protein chips
2. Similar to DNA microarrays
3. Various binding partners on a chip in known places
   
   1. Tag protein of interest with fluorescent or different tag, see where it binds. If it binds you know to what it bound to
4. Used to study protein protein interactions, or how other things interact with proteins
5. Is difficult because biochemical properties of proteins more diverse than DNA