M W F; 2:30 - 3:20 pm

 Molecular Biology

 Douglas W. Smith

Centre 101

 BIMM 100

 5254 Muir Biology Building

Winter, 2000  

 x42620; dsmith@ucsd.edu

Final Exam - 1995

with Key

 

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                                            Name: _____________________


                                                                  Pages


                        BIMM 100                               2.    50


                                                               3.   160

                   Molecular Biology 
                                                               4.    85

                      D. W. Smith 
                                                               5.    70


              Final Examination                       6.    65


                                                               7.    75
                 645 Points Possible 

                                                               8.    65
       

                                                               9. ___75____


                                               Final Exam Score:    645

                                             Total Course Score:

                      Fall, 1995                          Grade:





     I hereby give permission to have my graded exam placed in a box 
     in the hall outside of 3306 Bonner Hall for me to pick up. 
     I realize this will render my exam available for public 
     examination and scrutiny.


  Name ___________________________________________ Date _________________



      

15 pts 1. In the Gross et al HtpR paper, runoff transcripts were used as an assay system.

(5 pts) a. What is a runoff transcript?

  A runoff transcript is the product of an in vitro RNA polymerase 
  transcription experiment using a linear DNA template such that 
  transcription elongation proceeds to the end of the DNA template 
  and "runs off" the end.

(5 pts) b. What was the purpose of the assay?

  The runnoff transcript assay was used to assess ability of fractions 
  obtained from protein purification procedures to serve as a sigma 
  factor for transcription from a heat shock promoter.

 


(5 pts) c. Two runoff transcripts were used. What was the difference in templates used to generate the two runoff transcripts?

  The difference in DNA templates was the restriction enzyme used to generate
  the template. Both templates used the same heat shock promoter

  Hence the difference in size of RNA product was well defined in relation
  to the restriction sites used as ends of the DNA templates, and hence the
  transcription start position could be precisely determined, and should be
  the same for both templates.

35 pts 2. Draw the structure of the terminal nucleotide (C) of a tRNA molecule and an amino acid, using R for the R group, in their aminoacyl linkage. Then draw SCHEMATICALLY the rest of the tRNA molecule showing the secondary structure features of the tRNA molecule. Label each stem-loop of the tRNA molecule and explain its function. Assuming the codon recognized by this tRNA is CUA, draw the anticodon, with its sequence, on your tRNA drawing. Label 5' and 3' ends of the tRNA molecule.



      

10 pts 3. Write the LETTER for the protein or protein complex that matches each the molecular site in the protein column.
Choose the SINGLE BEST protein or protein complex from the 'Protein choices' list on the right.
Write the name the biological process involving the site and protein in the 'Biological Process' column.

 Molecular Site

 Protein

 Biological Process

 Protein Choices

1. ribosome A site

B

 Translocation elongation: binding 
 of charged tRNA to A site
A. NusB protein
2. oriC 9-mer

H

Initiation of round of DNA replication
B. Met-tRNAm
3. Thymine Dimer

G

Nucleotide Excision repair
C. Releasing Factor
4. Primosome Assembly Site

J

Okazaki fragment primer synthesis
D. Glucocorticoid Receptor
5. TATA box

I

Construction of Basal Transcription 
Apparatus; HoloRNApol
E. fMet-tRNAf
6. GRE

D

Enhanced Transcription in response to 
Steroid-Cell Interaction
F. poly(A) synthetase
7. Nut site

A

AntiTermination in Lambda life cycle
G. UvrABC complex
8. Amber codon

C

Translation termination
H. DnaA protein
9. ribosome P site

E

Translation initiation in prokaryotes
I. TBP
10. pre-mRNA 3'-end

F

poly(A) addition, in maturation of 
pre-mRNA
J. Primase



      

30 pts 4. Suppose you have the following completely overlapping, comma (G is the comma), sequential
(codons are read one after the other), triplet
Genetic Code:

...GACCTGACTTGCTATGTTCAGCAATG...

(20 pts) a. Write down the codons.

  ACC, CCT, ACT, CTT, CTA, TAT, TTC, TCA, CAA, AAT

 

 

(5 pts) b. How many possible codons are there for this Genetic Code?

  3 nucs possible at each of 3 positions (triplet code): 33 = 27

 

 

(5 pts) c. Assuming no redundancy, how many amino acids could a base substitution
mutation change?

  1 or 2 if any base is substituted EXCEPT a G
  If G mutates to A,T,or C: 3 new amino acids are "inserted"
  If non-G base mutates to a G: 1 or 2 amino acids are "deleted"

30 pts 5. Match one or more of the items A-L in the two right columns with the 4 gene expression stages in the left column.
Enter the letters A-L for
your answers in the middle column marked Answers:

 Gene Expression Stage

 Answers

Items

Transcription initiation

 C, E, I, J

A. Alternative Splicing G. Ribosome Slippage
Transcription termination

 F, H

B. Nonsense Suppression H. Lambda Q protein
Pre-mRNA maturation

 A

C. Catabolite Repression I. GATC Methylation State
Translation initiation or termination

 B, G, L

D. Feedback Inhibition J. Lambda CII protein
    E. HtpR gene product K. Missense Suppression
    F. Attenuation L. Ribosome unwinding mRNA structure.

 

 

90 pts 6. The E. coli gal operon has 3 adjacent, coordinately expressed structural genes galE, galT, and galK, with Promoter (galP) and Operator (galO) just upstream of these genes.

  Control of gene expression is via a Repressor from a non-adjacent galR gene
  and via catabolite repression. Galactose is an Inducer of the operon.

 

 

(20 pts) a. Draw the structure of the operon and its regulatory elements.



      

(10 pts) b. On your diagram, show the effects of gal Repressor on Gal operon gene expression when Galactose is absent and show the effects when Galactose is present.



      

(5 pts) c. Define coordinate expression.

  Coordinate expression is the simultaneous or common regulation of a group
  of genes (structural genes). They are all turned on (induced or derepressed)
  or turned off (repressed) together.

(5 pts) d. How are the effects of an Inducer different from those of a Co-Repressor?

  Inducer and Co-Repressor molecules are both small molecules that interact
  with a Repressor of an Operon in prokaryotes. However, an inducer (eg b-galactosides
  for the lac operon) INACTIVATES a Repressor whereas a Co-Repressor (eg tryptophane
  for the trp operon) ACTIVATES the Repressor.

(5 pts each) e. State the expected phenotypes (Gal operon expression in presence and absence of galactose) of the following types of mutations:

1) DEL(galR) [deletion of galR] Constitutive expression; no Repression; always induced whether galactose is present of not.
2) galRs Gal expression always turned on, independent of presence of galactose [super-repressor]
3) galOc Gal expression always turned on, independent of presence of galactose [operator defective]
4) DEL(galO) [deletion of galO] Gal expression always turned on, independent of presence of galactose [operator defective]
5) galP down mutant Reduced Gal expression independent of galactose [defective Promoter; little transcription of gal genes]
6) DEL(galE) [deletion of galE] Gal operon expressed normally, shortened mRNA due to absence of galE sequences: expression with galactose, only basal expression without galactose

 

(5 pts each) f. State the expected dominance relations of the following types of merodiploids:
1) galR+ / D(galR) galR+ dominant to DEL(galR): normal repression in absence of galactose
2) galR+ / galRs galRs dominant to galR+: gal expression always turned off due to Super-Repressor
3) galO+ / galOc Cis-Dominance of galOc: gal genes ON SAME DNA as galOc operator expressed constitutively; gal genes on galO+ DNA show normal induction/repression

 

 

(5 pts) h. Briefly explain Catabolite Repression, using the gal operon as an example.

  Catabolite Repression is the POSITIVE control system whereby cAMP and CAP
  (catabolite activator protein) bind to promoters of catabolic operons to
  ACTIVATE transcription by RNA polymerase. Presence of Glucose reduces cAMP
  concentrations, turning off operon expression, and vice versa: Glucose Effect.

50 pts 7. Bacteriophage Lambda:
(15 pts) a. What is the function of N protein and what lambda DNA sites
are involved in N function?

  N protein functions as an AntiTerminator of transcription, binding with
  NusB to NutR and NutL sites, permitting RNA polymerase to read-through at
  the tR1 and tL1 transcription terminator sites. This Positive regulation
  leads to expression of the CII, CII, O, P, Q genes, and some xis and int
  expression.

(5 pts) b. If Q encoded a new sigma factor, what change in lambda DNA sites would be needed to elicit the normal temporal expression of lambda genes?

  Simplest answer: a new promoter near tR' specific for CoreRNApol + this
  Q sigma factor which would then express Late genes (S, R, Head, Tail genes)

  Alternative: modify the PR' promoter to make it specific for CoreRNApol
  + this Q sigma factor, and delete the tR' transcription terminator.

(25 pts) c. Diagram using arrows the antagonistic behavior of Cro and CI repressors, indicating whether the effect of the arrow is positive or negative. Provide names of the relevant promoters and operators.



      

(5 pts) d. What is the role of cAMP in the lambda life cycle?

  The major role of cAMP is to control the decision lytic vs lysogenic via
  its concentration level. cAMP does this at the PRE promoter: cAMP is required
  for activation of this promoter (together with CII, CIII, CAP proteins)
  which results in initial CI repressor expressor and turn off of Cro, CIII,
  O, P, Q expression via the AntiSense RNA activity of the PRE transcript.

80 pts 8. RNA splicing:
(15 pts) a. What are the three main steps found in all mechanisms of RNA
splicing?

  1. 3'-OH attack at Donor Site [5' end of intron]; source of 3'-OH: G-OH
     in Group I introns, intron A-OH with lariat formation in Spliceosomes, etc

  2. 3'-OH attack of the terminal nuc of Exon 1 (left exon) on the last nuc
     of intron at intron 3' end, with release of Intron and joining of the 2
     exons

  3. Debranching, etc of the Intron

(25 pts) b. What are the 5 main classes of RNA splicing? Indicate which use a Ribozyme and which use a Spliceosome.

  1. Group 1 introns - Ribozyme - [rRNA genes of lower eukaryotes]

  2. Group 2 introns - Ribozyme - [fungal mitochondrial introns]

  3. Hammerhead introns in viroids - Ribozyme - [viroids]

  4. tRNA processing intron type - - [yeast tRNA genes]

  5. Introns using snRNPs - Spliceosome - [nearly all euk mRNA genes]

(5 pts) c. Draw an example of alternative splicing.



      

(10 pts) d. Explain what a "snerp" is and how it functions.

  A Snerp is an snRNP: small nuclear ribonucleoprotein particle composed of
  protein(s) and a snRNA (small nuclear RNA) species.

  Snerps function in Spliceosome-mediated Intron splicing events, each playing
  a specific role in the splicing. When the complete complex of Snerps has
  been constructed, the complex is called a Spliceosome.

(5 pts) e. What is the GT-AG rule?

  GU is always present at 5' end of Intron; AG always present at 3' end of
  Intron

(5 pts) f. Suppose the Internal Guide Sequence of a mutant L-19 RNA were: UUUUUU.
What substrate would be a good choice for an RNA polymerase experiment with
this RNA?

   (A)5 or AAAAA

(10 pts) g. Write down the 2 substrate-product reactions in the overall net reaction with this UUUUUU substrate.



      

(5 pts) h. How did Zaug and Cech show that a deoxyoligonucleotide is a competitive inhibitor of the normal substrate?

  They did enzyme kinetics, measuring amount of product at varying concentrations
  of CCCCC (rCrCrCrCrC) substrate as a function of fixed amount of dCdCdCdCdC
  present. Data are presented as a Lineweaver-Burk plot which shows that the
  dCdCdCdCdC acts as a Competitive Inhibitor of the rCrCrCrCrC reaction

45 pts 9. Control of Eukaryotic gene expression:
(20 pts) a. List the four major properties of Enhancer elements.

  1. DNA region containing a cluster of Transcription Factor binding sites

  2. can be located upstream or downstream of a given gene promoter

  3. can be found very distant from the promoter

  4. can be in either orientation relative to the promoter

  5. must be on same DNA molecule

(15 pts) b. Name three types of general biological processes where Enhancers are used for control of gene expression.

  1. tissue specific gene expression

  2. temporal control of gene expression in development

  3. gene expression in response to external signals)

  4. yeast catabolism, eg GAL4

  Other answers were also possible ...

(10 pts) c. What are Response Elements? How are they similar and how are they different from the TATA box?

  Response Elements are short DNA binding sites for in Enhancers and as Upstream
  Promoter Elements for Transcription Factors that bind in response to changes
  in the cell environment (temperature change in heat shock, presence of small
  molecule, eg heavy metals, steroids, etc)



  Similarities to TATA: short DNA binding site, bind TFs, enhance transcription,...

  Differences from TATA: often part of an Enhancer element, not positional
  element

30 pts 10. In the Metallothionein paper,
(10 pts) a. what was the recombinant DNA methodology used to identify positions
of the upstream regulatory elements?

  Deletion mutagenesis with generation of set of specific 5' deletion derivatives
  or mutants and generation of set of specific 3' deletion derivatives

(10 pts) b. what are 2 of the 3 methods used to identify the binding site for the glucocorticoid receptor?

  1. filter binding assay

  2. DNase I footprinting assay

  3. base modification protection experiments [DiMethylSulfate (DMS) specific
     for G's]

(10 pts) c. what two external signals activate expression of this gene, and what was the major result from this paper concerning activation by these two types of external signals?

  External signals: steroid binding to Glucocorticoid Receptor; presence of
  heavy metal, eg Cadmium

  Major result: presence of either signal sufficient to activate MT II gene
  expression ... the response elements to the 2 external signals act independently
  of each other.

20 pts 11. For lagging strand DNA synthesis in E. coli, diagram and briefly describe the events in synthesis and maturation of two Okazaki fragments adjacent to the mature daughter DNA strand. Name an enzyme that is used to catalyze each event, and label 5' and 3' ends.



      

55 pts 12. Retroviruses:
(25 pts) a. Draw the structure of a typical retroviral genome, labeling
each genetic element present.



      

(5 pts) b. The retrovirus genome is a linear molecule. During its replication, briefly explain how the "end problem" (replication of all nucleotides at the ends of the linear genome) is solved.

  The end problem is solved by duplicating the 3' end of the RNA genome at
  the 5' end of the DNA provirus, and the 5' end of the RNA genome at the
  3' end of the provirus, resulting in the LTR structures (U3-R-U5) at both
  ends of the provirus. Transcription of the provirus to yield progeny RNA
  genomes then can begin at a promoter in U3, to fully duplicate both R sites
  and all sequences in between in the resulting RNA genome molecules.

(10 pts) c. Briefly describe (one sentence each) how this solution to the "end problem" differs

1) from that used by eukaryotic chromosomes.
2) from that used by bacteriophage lambda.

  1. eukaryotic chromosomes use the telomere solution

  2. Lambda circularizes its genome upon infection, permitting replication
     with no end problem

(5 pts) d. When in frame, the pol "gene" is translated via nonsense suppression. Briefly describe nonsense suppression.

  In Nonsense Suppression, a first point mutation results in presence of a
  Stop codon in the middle of a gene, resulting in a truncated, inactive protein
  upon translation. 

  The effects of this mutation are Suppressed by a second mutation, usually
  in a tRNA gene such that the mutant tRNA 1) can still be charged by a cognate
  Aminoacyl Synthetase and 2) has an anticodon that will recognize the Stop
  Codon of the first mutation. The result is that an amino acid is put into
  the growing polypeptide chain during translation when the Stop codon (first
  mutation) is encountered and no truncation of the protein occurs.

(10 pts) e. What are the functions of Reverse Transcriptase and RNaseH in replication of the retrovirus genome?

  Reverse Transcriptase is an RNA-dependent DNA polymerase. It replicates
  the retroviral RNA genome into a double-stranded DNA species which, when
  integrated into the host genome, is called the Provirus.

  RNaseH is an RNase that degrades the RNA part of an RNA:DNA duplex nucleic
  acid molecule. It degrades the retroviral RNA genome after the genome has
  been replicated into DNA by Reverse Transcriptase.

  The two enzymatic activities work in synchrony during this replication process.

65 pts 13. Translation:
(10 pts) a. How do the prokaryotic Tu and Ts factors function in translation,
and what is the comparable eukaryotic protein?

  Elongation factor EF-Tu brings charged tRNA molecules to the A site of the
  ribosome during protein synthesis, using energy from GTP to do this. Upon
  GTP cleavage, EF-Tu is released as a Tu-GDP complex. The GDP is released
  from this complex via displacement by EF-Ts, with formation of a Tu-Ts complex.
  Ts is replaced by GTP when Tu interacts with another charged tRNA molecule.

  The comparable eukaryotic protein is eEF-T or eEF-1. This "T factor"
  does what Tu and Ts do together in prokaryotes.

(35 pts) b. Draw a figure of a ribosome functioning in translation elongation just before peptide bond formation. Include each of the major molecular structures. Label 5' and 3' ends and indicate direction of translation.



      

(20 pts) c. Translation initiation in eukaryotes differs in what major ways from translation initiation in prokaryotes?

  In prokaryotes, 30S ribosome binds to the Shine-Delgarno sequence near the
  Start Codon in an mRNA molecule, and the AUG Start Codon is placed in the
  Psite of the 30S ribosome. fMet-tRNAf then binds to the P site of the 30S-mRNA
  complex, with the AUG Start Codon of the mRNA. Finally 50S ribosome is added.

  In eukaryotes, there is no Shine-Delgarno sequence. There is also no fMet.
  Further, Met-tRNAi binds to 40S ribosome BEFORE 40S ribosome binds to the
  mRNA. The Met-RNAi-40S ribosome complex then binds to the Cap region at
  the 5' end of the mRNA. This complex then translocates down the mRNA to
  the Start Codon, using ATP for energy and placing the Start Codon in the
  P site of the 40S ribosome. finally 60S ribosome is added.






| BIMM100 | Syllabus | Sections / Off Hrs | Grading Policy | DNASYSTEM |
| Lectures | Journal Articles | Study Qs | Lab Techniques | Exams |


 

If you have problems or comments, send email to Doug Smith