BIOL5420 - Molecular Biology


Tae Hoon Kim

Week 1

Lecture 1

Course Organization

Two components

Analytical approach
Experimental system
Genetic analysis modules 1-5
Mutations & screens
Molecular & atomic genetics
Recombination & transposition
Suppression, synthetic lethality & epistasis
Genomic approaches
Experimental systems 1-5
lac repressor
Phage λ
Signaling & Network

Molecular Biology

Lecture 2

A Brief Overview of Genetics Molecular & Cloning Techniques





How do you study gene control?

Genetic Terms

Week 2 - Mutations

Reading 2.1 - [cite:@De_Lucia_1969]

E. coli strain with a mutation affecting DNA polymerase was a temperature sensitive strain.

False, it was UV sensitive.

What was the likely purpose of N-methyl-N’-nitro-N-nitrosoguanidine (MNNG)?

Induce mutations across the E. coli genome

What might be the best conclusion of the paper?

Fully viable and replication competent E. coli strain was identified and was shown display less than 1% activity of Kornberg’s DNA polymerase activity found in wildtype

TODO Lecture 3

Sperm mutations

sperm reprogramming egg

slide 19

synthetic lethal genetic screen

Discovery of mutagens

Ames test

Genetic-biochemical analysis of DNA Polymerase discovered by Kornberg

Why were several thousand colonies needed to be screened?

What if no clone had been identified?

What if there are no other mutations in the mutant?


Reading 2.2 - [cite:@Pausch_2020]

The article reports a discovery of Casϕ with a significant sequence identity to type V CRISPR-Cas proteins that have been documented thus far.


Processing of pre-crRNA for the CRISPR-Casϕ system is most likely performed by an endogenous RNase III in the host cells.


Elution profile from a size exclusion column was used to provide evidence for which of the following?

RuvC-inactivating mutations did not mis-form Casϕ

TODO Lecture 4

homologous recombination repair

zinc fingers(ZFN)


Week 3 - Lac Repressor

Reading 3.1 - [cite:@Pardee_1959]

Which chemical was used for measuring beta-galactosidase activity in the studies?


What was the purpose of streptomycin and T6 phage in mating D that generated the data for Figure 4?

To inhibit protein synthesis and remating in the ♂ bacteria

The “inducer” model, in contrast to the “repressor” model, described by the authors is most accurately described by the which of the following:

Presence of an inducer in the constitutive and inducible organism that is destroyed by the i+ gene product

Lecture 5

genetic screens - basic structure of genetic screens

genetic approaches

How to do a genetic screen

Model organisms

Thomas Hunt Morgan

Fly Boys
Nobel Prize in Physiology or Medicine 1933


Hermann Müller

When do screens fail

Type I - False Positives

Type II - False Positives

crosses and recombinants – E. coli

TODO Bacterial genetics terms



TODO Blender Technique

gene mapping

bacterial transcription and control

Transcription In E. coli

TODO E. coli RNA polymerase - ~ 55 minutes
Sigma factor - Transcription Regulator

constitutive and inducible systems

adaptation, propagation, evolution

Reading 3.2 - [cite:@gilbert_muller-hill_1966]

The is mutant is used as a positive control for the experiments to show that the i gene encodes a repressor that binds the inducer.


The it mutant was identified in a screen for cells that is able to produce beta-galactosidase with a very low IPTG concentration but grow in TONPG which is toxic to cells that are constitutively expressing beta-galactosidase.


Which of the following does not support the conclusion that the lac repressor is a protein?   ATTACH

Material that binds IPTG sediments at 10S

Lecture 6 - Lac Repressor

lactose metabolism



Jacques Monod

Francois Jacob


z- and i- Means their missing the gene

FIG. 4. Enzyme formation in mating D



FIG. 1

Got the IT alelle

FIG. 3


Nobel Prize in Chemistry 1980

Negative control of lac operon by the respressor

Postive control of lac operon

Week 4 - Lambda Repressor

Reading 4.1 - [cite:@Ptashne_1967]

Which of the following was not an approach employed by the author to label the lambda repressor specifically?

Use temperature sensitive mutant C1ts857 to label the repressor

What might be the reason for C1tsU32 and C1tst2 mutants not showing an H3 labeled peak in DEAE chromatography?

Both are more insoluble than c1ts857

What is the purpose of the amber mutant C1sus10 in the experiments?

It is used as a negative control for repressor synthesis

Lecture 7


Mark Ptashne

Experiment Explanation

λ Phage



System Used to Study

Positive control mutants define activating region whose basic function is to recruit RNA polymerase (RNAP)

cI dimers cooperatively bind to OR1 and OR2

How does cI regulate lysogen

TODO cI Mechanism of action   question

cro Promotes Lytic fate

cII Controls the lytic-lysogen decision stabilization of cII results in the turning off of cro

Reading 4.2 - [cite:@Cui_2013]

What is not true about the enhancers?

Needs CAT protein to function over distance

Which statement is true based on the data presented in Fig. 2(C)?

UP element increased PRM activity in the presence of C1

UP element likely functions by

Interacting with the alpha subunit of RNAP

screenshot   ATTACH

TODO Lecture 8





Phage DNA injection can be slow and heterogeneous

Week 5

Reading 5.1 - [cite:@Keegan_1986]

Based on the results in Table 1, which of the following is not supported by the data?

All GAL4 fusion with Beta-gal exhibited some basal beta-galatosidase activity in addition to the GAL1 reporter activity

What is the purpose of the lane b in Figure 4? Select the best answer.

A negative control to show that lambda repressor does not binds UASG

GAL4(147-881) is not required for binding to UASG but is required for activation of GAL1.   ATTACH


TODO Lecture 9

TODO Nature of Mutations

TODO Basic Types of Mutations

TODO Mutations Induced by Common Mutagens

Molecular genetics

TODO Molecular analysis - domain bashing

TODO [cite:@Keegan_1986]


Reading 5.2 - [cite:@Hecht_1983]

Lambda virulent phages used in the study are different from wildtype lambda phages in that

They harbor mutations in various operator sites for lambda repressor

Which of the following is not true about thermolysin analysis of lambda cI mutants?

Identification of enhanced denaturation of weak solvent exposed cI mutants

Asn77 to Thr77 mutant is a pseudo-revertant, because threonine is similar in chemical property as serine which is the residue in wildtype cI.


TODO Lecture 10

Week 6

Reading 6.1 - [cite:@Gaiano_1996]

TODO Lecture 11


Barbra McClintok

LINE-1 retrotransposons

Transposons as genetic tools

inverse PCR to identify integrated/transposed site


Nancy Hopkins

Conditional degrons

Exam 1

Following section is for Q1-Q5: The diagram below illustrates an experiment in which a lysogenic ‘male' donor strain of E. coli transfers its chromosome to a ‘female' recipient by the process called conjugation. The outcome is that the lambda prophage is induced as soon as it enters the recipient cell (a phenomenon termed zygotic induction, originally called ‘erotic induction' by its French discoverers). The lambda repressor cI is illustrated as a dumbbell.

file: media/image1.jpeg

Q1. What is the molecular explanation for zygotic induction?

When the DNA is transferred, the recipient cell lacks the lambda repressor(unlike the lysogenic donor), so the cro protein is immediately transcribed due to the lack of the repressor.

Q2. What is the outcome if the cross is reversed (that is, a non-lysogenic donor transfers its chromosome to a lysogenic recipient)?

The recipient cell would not undergo lysis because it has only gained a normal chromosome and that won't induce the lambda prophage because it already has sufficent cI built up.

Q3. What is the outcome if both the donor and the recipient are lysogens?

The donor would continue to be lysogenic. The recipient would also be lysogenic but might be more susceptible to undergo lysis because it does not have sufficent cI to repress the new copy of the PR and PL promoters.

Q4. What is the outcome if the recipient expresses lambda repressor cI from a plasmid constitutively(at a constant rate)?

The recipient would not become lytic because the excess cI will be able to repress the new copy.

Q5 What is the outcome if the recipient expresses cro from a plasmid constitutively?

The recipient will become lytic because the cro will ensure the lytic cycle occurs by block the PRM promoter.

Following section is for Q6-Q8: In an activation complex such as the one shown below, it is typically the case that only one monomer of the activator, and only one C-terminal domain of the alpha subunit of RNA polymerase is required for activation.

file: media/image2.png

Q6. Since one activator monomer is sufficient for contact with RNAP, what is a possible reason for the activator being dimeric?

While only one activator monomer is sufficient, the dimeric could create a stronger bond with the RNAP, a positive cooperativity. If one of the monomers binds to RNAP, it could cause the second to have a stronger bond

Q7. In the diagrammed scenario, a DNA sequence near the activator binding site interacts with the alpha subunits of RNAP and is required for an optimal level of expression. From your readings, which element does this sequence most similar in function?

Sigma factor, the bacterial Transcription Factor (TF) that that neables specific binding of RNAP to gene promoters in bacteria.

Q8. Describe a genetic approach that would allow you to identify amino acids in the activator that make contact with the RNAP alpha subunit.

  1. Create mutants of the activator with different deletion mutations that cause changes in the amino acids.

  2. Test the activator mutants ability to activate a reporter gene such as lacZ

  3. The mutants that have a lower β-galactosidase activity are those with the amino acids that are necessary for activator activity

Following section is for Q9-Q12: There is a gene y that is controlled by a gene x in a newly discovered fungal species. The gene x encodes a 100 amino acids long transcriptional regulator. To study x better, you make several deletion mutants that can be expressed constitutively from a plasmid. The plasmid is pX with one of the following deletions:

file: media/image3.wmf

pX-FL is capable of expressing a full length protein x, while pX-A through pX-G express various deletion mutants of x.

You generate a strain of yeast carrying a reporter plasmid with the promoter of the gene y fused to the lacZ gene (pPr-Y-LacZ). To this strain, you introduce various versions of pX deletions and determine the β-galactosidase activity (in Miller units).

x deletion mutantsβ-gal activity
no x200

Q9. What can you infer about the functional domain organization of the transcription factor x? Domain A, and F is very vital(Perhaps they code for specific DNA binding sites) Domains B, C, D are vital(Perhaps they reduce the confirmation of the protein) Domain E is inhibitory to the functioning and the lack of it increases the effectiveness

To investigate the function of each potential functional domains better, you generate x fusion protein to the GAL4 DNA binding domain (GAL4DBD) and use UASG-LacZ reporter to assess the activity GAL4DBD-x fusion proteins. “1-20” indicates amino acid residues included in the fusion for x(1-20), similarly for other x(##-##) fusions.

x fusionβ-gal activity

You also generate x fusion protein to GAL4 activation domain (GAL4AD) and use the reporter plasmid pPr-Y-LacZ to determine the activity of the GAL4AD-x fusion proteins.

x fusionβ-gal activity

Q10. Which region might be an activation domain of x?

61-80 because it has the highest B-gal activity with GAL4AD

Q11. Which region might be a DNA binding domain of x?

81-100 because it has the highest B-gal activity with GAL4BD

Q12. What might be the function of the region corresponding to 21-40 amino acid residues of protein x?

It has no binding activity, as seen by the GAL4BD experiment. It is a weak activation domain because of the GAL4AD

Week 7

Reading 7.1 - [cite:@Young_2019]

Which gene was used to regulate the expression of potentially deleterious proteins from the library?

lacI gene was expressed

How was the library constructed to express potential RNA polymerase subunits?

Yeast genomic DNA was fragmented with EcoRI restriction enzyme and inserted into lambda gt11 DNA

How was genes for RNA polymerase II identified from the lambda gt11 library?

By identifying clones that expressed proteins that bound the available RNA polymerase II antibody

What might be the benefit of using an expression library constructed from genomic DNA rather than cDNA?

wrong? genomic DNA contains introns

each gene is more likely to remain functionally intact in genomic DNA library

TODO Lecture 12

Reading 7.2 - [cite:@Berger_1992]

How was the genetic screen for ADA (alteration/deficiency in activation) genes performed?

Screening for mutants that can grow despite high toxic levels of GAL4-VP16

ADA2 is an essential adaptor protein for all acidic transcription activation domains.


We conclude that ADA2 potentiates the activity of one class of acidic activation domain but not a second class.

ADA1 was not studied further, because ADA1 mutant was found to reduce the level of GAL4-VP16


The ada 1 mutant was considered a poor candidate to encode the adaptor, because it apparently relieved inhibition by reducing the level of synthesis of GAL4-VP16.

Which of the following is not true about the SPT genes?

SPT genes work like ADA2 as an adaptor to link activators to general Transcription Factor (TF)

The properties of the ada mutant are different from spt mutants, including GALll.

TODO Lecture 13

Week 8 - Epistasis

Reading 8.1 - [cite:@Thompson_1993]

Cells lacking both SRB2 and SRB5 have a more severe phenotype than cells lacking only SRB2 or SRB5 alone.


Of the 85 independent suppressor mutants identified, what fraction were dominant mutants?


Both SRB2 and SRB5 are required for activated transcription, but not basal transcription in vitro.


In transcription assays using extracts from cells lacking SRB5, basal and activated transcription required addition of recombinant SRB5 and SRB2 together.


What is not a G-less cassette?

TODO Lecture 14

carboxyl-terminal domain (CTD)


Both SRB2 and SRB5 are required for preinitiation complex formation

Mediator Complex

RNAP-Mediator complex

Transcription activation requires many distinct steps

Genetic Analysis IV - suppression, synthetic lethality, epistasis

Questions to ask about mutants


Reading 8.2 - [cite:@Cooper_2006]

Which of the following is not a known mutation in the alpha-synuclein gene that have been linked to Parkinson’s disease?   question

Deletion of the alpha-synuclein gene

Which of the following was not a rationale for measuring degradation rates of CPY* and Sec61-2p?   question

WRONG Sec61-2p and CYP* degradation are differentially blocked by alpha-synuclein overexpression.

CYP* degradation was faster in cells overexpressing A53T alpha-synuclein mutant than in cell overexpressing wildtype alpha-synuclein.

GYP8 is a suppressor of toxicity due to alpha-synuclein by activating the GTPase activity of Ypt1p.   question

WRONG True False

TODO Lecture 15

Possible double mutant phenotypes

Classes of second site suppressors

Suppression by a complementary mutation in the second gene product at the interaction interface

Bypass Supression

Intragenic supression

Supressors provide insight into gene function

Week 9 - Translation

Reading 9.1 - [cite:@Lee_1993]

lin-4(e912) mutation is epistatic to lin-14 null.   question


True False other way around

C to T transition at nucleotide position 517 found in ma161 mutation was experimentally validated to disrupt basepairing between lin-4 and lin-14.   question


True False

Which of the following is not one of the possible mechanisms listed by authors for how lin-4 may regulate lin-14?   question

If lin-4 RNA inhibits lin-74 translation by an antisense inter- action with the/in-74 3’UTR,

lin-4 RNA may bind to lin-14 mRNA in the cytoplasm and inhibit its translation

lin-4 could inhibit LIN-14 protein synthesis indirectly

lin-4 RNA probably does not control the stability of lin-14 mRNA

How long is lin4S?   question

lin-4S, is approximately 22 nt in length.

WAIT Lecture 16

Nobel Prize in Chemistry 2020

Nobel Prize in Physiology or Medicine 2009


Translation Control   ATTACH

mRNA maturation

Steady State mRNA levels vs transcription rate


split genes: fundamental eukaryotic specialization

Nobel Prize in Physiology or Medicine 1993

Caping and other modifications

co-transcriptional processing of RNA

TODO RNA sequences and factors in regulated splicing

exon definition

variable rate of transcription elongation across the genome

TODO local Global Run-on Sequencing (GRO-seq) pattern around exon

TODO regulated splicing in the context of chromatin


Functional consequences of alternative Polyadenylation

nonsense mediated decay - post nuclear export RNA surveillance

Nobel Prize in Physiology or Medicine 2006


Victor Ambros

Question that Computational Biologist never talk about

Reading 9.2 - [cite:@Gerber_2006]

A reason for not using TAP-tagged full length Pum construct was   question

Which of the following is not true about the tetranucleotide UGUA?   question

The assay presented is unable to uncover all relevant Pum mRNA targets because targets and nontarget mRNAs may associate and disassociate during the affinity purification steps.   question


TODO Lecture 17

Pat Brown


Nobel Prize in Chemistry 2009

Week 10 - Signal Transduction

Reading 10.1 - [cite:@Velazquez_1992]

IGSF3 is dimerizes in response to interferon association with the receptor and sufficient to activate IFN-inducible genes.   question


Which of the following is not true about the 2fTGH cell line used?   question

How was the mutagenized 2fTGH cells screened to isolate 11,1 cells?   question

Back-selection in hypo- xanthine-aminopterin-thymidine (HAT) medium plus IFN allows isolation of 11 ,l derivatives that have reverted to IFN sensitivity.

Screened for their ability to grow in HAT media supplemented with interferon

How was the complementation screen for the defect in interferon response in 11,1 cells carried out?   question

By transfecting human genomic DNA into 11,1 cells and screening for transfected 11,1 cells that grew in HAT media plus interferon

Which of the following is not true about the K6 clone from DDD3-1,5?   question

K6 clone restored interferon response in 11,1 cells

Lecture 18

Signal Transduction

critical for cells to adapt to its environment

critical for organisms to coordinate development and differentiation

Two-component regulatory system

Basic Components of Signaling

Nuclear Receptor Signaling

JAK-STAT Pathways


Interferon Signaling   ATTACH

various mutations in pathways

what would happen to interferon response, if you introduced a mutation in tyk2 that rendered it constitutively active?
would it be recessive or dominant to wildtype?
what would happened to interferon response, if you introduced a mutation in the catalytic residue within the kinase domain?
what would happen if you over expressed (100 fold excess) this mutant in the presence of wild type tyk2? would this over expression mutant be dominant or recessive to the wildtype?
Y701 in STAT1 is phosphorylated by JAK1/TYK2
what would you predict if Y701 is mutated to alanine?
Similarly Y690 in STAT2 is phosphorylated by JAK1/TYK2
what would you predict if Y690 is mutated to alanine?

G-Protein coupled receptor signaling   ATTACH

Nobel Prize in Physiology or Medicine 1994

Nobel Prize in Chemistry 2012

G-proteins-coupled receptors (GPCRs)

Receptor Tyrosine Kinase

Nobel Prize in Physiology or Medicine 1986

Recurrent and Universal RTK Activation in Cancers


Sonic Hedgehog

& Hedgehog (Hh) signaling pathway



Reading 10.2 - [cite:@Tong_2004]

Synthetic Genetic Array analysis identified an equivalent number of genetic interactions as determined for protein-protein interactions   question

Synthetic genetic interactions were also enriched amongst gene pairs encoding homologous proteins (P ⫽ 10 –22 ), but this accounted for relatively few (2%) of the observed interactions.


132 SGA screens involved performed by analyzing double mutants carrying one of 132 mutants in combination with 4700 other mutant genes (or about 620,000 double mutants).   question


Authors were able to generate a complete genetic network representing the global map of functional relationships between genes in the yeast genome.   question

Authors were able to generate a complete genetic network representing the global map of functional relationships between genes in the yeast genome.


Which of the following human diseases was not noted by the authors for relevant models for synthetic genetic interactions?   question

Lecture 19


Biochemical and Molecular approaches for inferring gene-gene interactions

Network Analysis

Networks are composed of nodes and edges
Network can be undirected, directed or weighted
Benefit of network analysis

Systematic Genetic Analysiswith Ordered Arrays of YeastDeletion Mutants

Budding yeast deletion library


notable observations

Gene Ontology (GO)


TODO An Expanded View of Complex Traits: From Polygenic to Omnigenic

Evan A. Boyle

Week 11 - Genomics

Reading 11.1 - [cite:@Findlay_2018]

Lecture 20


Exam 2

WAIT means internet

1. What might be the differences between genomic and cDNA sequence of a gene in yeast?

cDNA would not have all the necessary noncoding regions, the introns and UTR regions. That might make the cDNA not able to create spliced RNAs using alternative splicing.

2. If you want to express a human gene in E. coli to produce a recombinant version of the protein, how would you accomplish this?

Through DNA cloning. The target gene would be inserted into a circular plasmid, and the plasmid introduced through transformation, and selected using an antibiotic screen. The protein could then produced by inducing the expression of the gene, depending on the plasmid.

3. Which region of the eukaryotic RNA polymerase is critical for pre-mRNA processing?

The carboxyl-terminal domain (CTD)

4. You are studying a neuronal gene whose product is localized to the synapse. You delete a part of the last exon of the gene that corresponds to the most of 3' UTR of the mRNA in the neuronal cells. You stain the mutant and wildtype cells for the protein and found that mutant cells show the protein localized to neuronal cell body, while the wildtype cells show the expected distribution with the protein localized to synapses. What can you infer about the mutant that you have generated?

The deleted region is important for the protein transmission. This could be caused by the lack of polyadenylation, and the stability of the mRNA, preventing it from being translated effectively.

Questions 5-9 is based on your reading [cite:@Velazquez_1992].

5. The authors isolated the Cosmid 3D clone that restored the interferon response in 11,1 cells. It contained the gene for TYK2. This clone was recovered from a Cosmid library containing fragmented genomic DNA from DDD3-1,5 cells. A typical Cosmid plasmid has a cargo limit of 40-45 Kb of DNA sequence. What might be the size of the TYK2 gene in the human genome? What might be a result for the screen if the TYK2 gene was 100Kb long?

The gene would be around 40-45 Kb in length because that's the size that could fit in the Cosmid plasmid. If it were 100Kb long the gene wouldn't be able to fit in the Cosmid plasmid so it wouldn't have been able to restore the interferon response in 11,1 cells. So the screen would have been the attempted rescued 11,1 cells would grow in 6TG and 6TG+IFN but not in HAT, HAT+IFN.

6. What would happen to interferon response in 11,1 cells, if you introduced the Cosmid 3D with a mutation in the TYK2 gene that rendered it constitutively active (specifically what would be the grow pattern in HAT and 6TG media with or without interferon)? Fill in the table with “growth” or “no growth”.

Growth HATGrowth HAT+IFN
no growth 6TGno growth 6TG+IFN

The TYK2 would be unregulated by Interferon Receptor so it will cause the formation of ISGF3, which will lead to the expression of interferon stimulated genes(i.e. gpt)

7. Continuing on the previous question 5. You introduce the constitutively active TYK2 mutant into 2fTGH cells that have an intact wildtype TYK2. What would be the grow pattern in HAT and 6TG media with or without interferon for the 2fTGH cells with constitutively active TYK2? Would this mutant TYK2 be recessive or dominant to wildtype TYK2?

HAT No growthNo growth HAT+IFN
6TG GrowthGrowth 6TG+IFN

It wouldn't matter if there's a wild type because it's constitutively active so it would be dominant. The wild type would not render the cells to become normal. It will become unresponsive to interferon and not produce gpt in the presence of interferon

8. What would happen to interferon response in 2fTGH cells, if you introduced the Cosmid 3D with a damaging mutation in the catalytic residue within the kinase domain and expressed this mutant 100 fold excess of the wild type TYK2? Would this over expression mutant be dominant or recessive to the wildtype?

Same phenotype as the 11,1 cells, it won't be able to respond to Interferon Stimulation. It would be recessive though because the wild type copy will be able to compliment it.

9. In their screen to isolate 11,1 cells, they also identified 11,2 cells. However, multiple attempts to rescue the interferon response in the 11,2 cells have failed. Sequencing of the 6-16gpt reporter in the 11,2 cells found a nonsense mutation in the gpt gene. What can you infer about the 11,2 cells?

The interferon pathway would be functioning but the gpt gene would not be producing gpt. So it would grow in 6TG and 6+IFN because the functional gpt could not be produced regardless. The researchers were interested in the IFN pathway however, so the 11,2 cells wouldn’t be useful.

For questions 10 -- 12. Genes A, B, C and D define a metabolic pathway that you are investigating and the corresponding mutants a, b, c and d are null mutants. A series of enzymatic reactions convert the metabolite S to the metabolite P through intermediate alpha, intermediate beta and intermediate gamma.

Analysis of single and double mutants yielded following observations:

mutantintermediate accumulated

10. Complete the table below by identifying which mutation is epistatic for double mutants.

mutantintermediate accumulatedmutation epistatic
abgammab is epistatic to a
acbetac is epistatic to a
adSd is epistatic to a
bcbetac is epistatic to b
bdSd is epistatic to b
cdSd is epistatic to c

11. "Order the genes in the metabolic pathway using the double and single mutant analysis data."

ab gamma means b before a ac beta means c before a bc beta means c before b cd S means d before c

d -> c -> b -> a

12. Order the intermediates in the metabolic pathway.

beta accumulates if C is mutated gamma accumulates if B is mutated alpha accumulates if A is mutated

S -> beta -> Gamma -> Alpha -> P

Week 13 - Group Presentations

Group A


Group B

Effect of Temperature ON Lin4/Lin-14 Interaction In C. elegans

Group C


Cas3 and DinG

Group D



Group E

Group F - Effects of Human HSP70-5 on Cellular Activity and Protein Folding in Pichia Pastoris

Group N

aSyn in Parkinson's Disease