NOTE:  The sequence of topics was not the same as this year. You will need to choose those questions from the three sample exams that apply to the topics we have covered. I will not do this for you. Consider it another way to study. Minimally, you should be able to recognize the topics. The answers to the question follows the question.

                                                                            One Example

1.  Illustrate a deoxytrinucleotide (3 DNA nucleotides linked by phosphodiester bonds). Do not draw the base but
show where it is attached. All other structures must be included. (10 pts)

Consult the text for the appropriate diagram.

2.  To add radioactivity to a DNA strand as it is being made in a laboratory, researchers use
deoxynucleotidetriphosphates (dNTPs) that are radioactively labeled at the alpha phosphate. Why are radioactive
alpha forms used instead of radioactive beta or gamma forms? Base your answer on what you know about nucleic acid
synthesis and explain things thoroughly. (7 pts)

As DNA strands grow, the gamma and beta phosphates are cleaved off as pyrophosphate. Only the alpha phosphate
remains attached to the nucleotide that is then added to the growing DNA strand. Therefore, the radioactivity must be
associated with the alpha phosphate to end up in the DNA strand.
 

3.  The following experiment was done. DNA was obtained from liver cells and from neurons (brain cells). Following
digestion of this DNA with a restriction enzyme, increasing amounts of DNAse was added to the two samples. They
were then run out on an agarose gel and subjected to Southern blotting. The blots were probed with sequences
complementary to a sequence in the alcohol dehydrogenase gene, a gene that codes for an enzyme that detoxifies
ethanol in the liver. Unfortunately, the lab assistant forgot to label the samples and does not know which blot came from
which cell type. Examine the results below and identify the blots. (6 pts)

Blot on the left is from the liver cells where the DNAse can access the DNA to degrade it in this cell type that
expresses the gene. The blot on the right is from neurons (brain cells) where the DNAse cannot degrade the DNA due
to the more compact chromatin structure of this gene region that is not expressed in these cells.
 

4.  Answer the following about B-DNA (Watson and Crick’s) (5 pts each)

A.  Where do we find the nitrogenous bases?
        Inside the double-helix, perpendicular to the axis.

B.  Why does the thickness (viscosity) of DNA disappear when it is heated?
        Hydrogen bonds between the bases on opposite strands are broken making the
        DNA single-stranded instead of double-stranded.
 

C.  What is meant by the term antiparallel?
        The two strands of DNA run in opposite directions in the double-stranded molecules. One
        strand runs 5'-3' and the other strand runs 3'-5'.
 

D.  Illustrate a major and a minor groove.

        Consult the text for this diagram.
 

5.  Illustrate one replication fork and indicate the template strand that is copied continuously and that which is copied
discontinuously. (6 pts)

    Consult the text for this diagram.

6.  Demonstrate one way that the following RNA sequence could undergo a frameshift mutation. (5 pts)

  AUUACCGCUAAUGACCUUGGGAU

    Several possible answers here. Insertions of one or two bases or deletions of one or two bases. Also, reading four
bases instead of three accidentally or accidentally skipping over a base.

7.  Outline the steps you would use to prove that the amino acid leucine is coded for by codon CCU. (Use the
Nirenberg method) (10 pts)

1.    Synthesize CCU
2.    Set up 20 tubes containing all tRNAs and translation components. In one of each tube
        include 19 amino acids not radioactively labeled and one radioactively labeled. A different
        amino acid is labeled in each tube, so all 20 possibilities are there.
3.    Add CCU to each tube and then filter each onto filter paper through which ribosomes cannot
        move.
4.    Locate the radioactivity for each sample. If the radioactivity is in the flow-through, that tube
        does not contain the radioactive amino acid coded by CCU. If the filter paper is radioactive
        that tube contains the radioactive amino acid coded by CCU.
5.    The radioactive filter-paper should be found in the sample that contained the radioactive leucine,
        because the tRNA with leucine attached has associated with the CCU codon on the ribosome
        and therefore cannot flow through the filter.
 

8.  If DNA of the sequence 5’CAGTTGACCAGTTACGAATTCCGG3’ needed to be copied into a daughter strand,
which of the following primers would you choose to use? (5 pts)

A.  5’GUCAAC3’
B.  5’AAGGCC3’
C.  5’CCGGAA3’ This is the correct answer
D.  5’CAACUG3’

9.  Choose from the list below to answer A through G (3 pts each)

 3’-5’ exonuclease
 5’-3’ exonuclease
 PCNA
 primase
 DNA polymerase alpha
 DNA polymerase delta
 helicase
 topoisomerase
 telomerase
 ligase

A.  removes the RNA primer _____________ 5'-3' exonuclease (will accept DNA polymerase alpha also)
B.  prevents ends of lagging strands from shortening ________ telomerase
C.  makes phosphodiester bonds without a DNA template __________ligase
D.  synthesizes the Okazaki fragments ________ DNA polymease alpha
E.  removes incorrect nucleotides from daughter strands during DNA replication ___________ 3'-5' exonuclease (will
accept DNA polymerase delta also)
F.  melts hydrogen bonds in the double helix ___________ helicase
G.  monitors the amount of supercoiling in DNA __________ topoisomerase

10.
A.  What type of mutation is caused by UV irradiation? (3 pts)
        thymine dimers (will accept bulky-adducts)

B.  What type of repair system is used to correct the error? (3 pts)
        excision repair

C.  In mismatch repair, how is the newly made DNA strand recognized by the repair enzymes? (4 pts)

    In our example using the prokaryotic system, discussed in class, GATC sequences are methylated on the C residue
of the parental strand but newly made daughter DNA strands are not methylated for a while following replication.
Therefore, the MutH replication enzyme binds to the hemi-methylated region of these sequences (meaning it binds when
one methyl group is on the sequence of one strand but not on the other strand). This binding will cause the enzyme to
cleave the unmethylated strand when it is activated by the other repair enzymes, initiating removal of the incorrect base
from that strand but not the template strand, containing the correct base.

                                                              A Second Example

1.  What role is played by the following in the initiation of transcription in eukaryotic cells? (4 pts each)

A.  TATA box

General Transcription Factors bind here and help load on RNA Polymerase II.
 

B.  CTD of RNA polymerase II

Contains multiple amino acid side chains that can become phosphorylated allowing the RNA Polymerase II to clear the
transcription factor complex and continue laying down the RNA transcript.
 

C.  histone acetylase

Attaches acetyl groups to the amino terminal end of histones, loosening the chromatin structure and allowing for more
efficient initiation of transcription.
 

D.  enhancer binding proteins

Proteins which bind to enhancer sequence as either activators or silencers, loop the DNA over to the vicinity of the
TATA box and wither upregulate or downregulate the initiation of transcription.
 

2.  Scientists often study transcription regulation by probing cell extracts with radioactive cDNAs derived from certain
genes.

A.  What is a cDNA? (4 pts)

A double-stranded DNA copy of a particular mRNA obtained by using reverse transcriptase.

B.  How does using such a cDNA allow you to tell if the gene is transcribed in the cell extract? (5 pts)

If the gene is transcribed the mRNA should be present. Then, when a tagged copy of the cDNA is used in
single-stranded form, it should hybridize to that mRNA. This hybrid can then be detected by a variety of means,
including dot blots on filters. Alterntively, the mRNA population can be tagged and passaged over the cDNA attached
to a filter. If the mRNA is there, it will hybridize to the cDNA and the filter will be tagged.

3.  The following figure shows the outline of a eukaryotic gene. This gene is transcribed in both liver and brain cells.
However, the protein produced from this gene in liver cells is very different from that produced in brain cells. By
examining the structure of this gene, explain how this could happen. (10 pts)

The two cell types could use different poly-A signals, creating different 3' ends following cleavage and polyadenylation.
If the first is used, the intron 2 will not be spliced out. The mRNA will have exons 1 and 2 and intron 2 up to the
poly-A site. The protein will include all sequences in exon 1, exon 2, and intron 2 up to the stop codon. If the second
poly-A site is used, intron 2 will be removed by splicing. The mRNA will contain exons 1, 2, and 3. The protein will
contain sequences from exons 1 and 2 and from exon 3 up to the stop codon.
 

4.  Draw the result that you would expect if you subjected the following gene to an R-looping analysis using mRNA
from cells that express the gene. Be sure to label the diagram showing what is double-stranded, single-stranded, DNA,
and RNA. (8 pts)

See the figure in the cabinet near 019 McKinly. There will be two looped out regions representing single-stranded
DNA which are the introns. There will be a double-stranded hybrid region consisting of mRNA and DNA strands that
are complementary. It will include a single stranded poly-A tail at the 3' end of the mRNA. It will include
double-stranded DNA at the 5' end of the DNA upstream from the hybrid representing the rest of the DNA in the
DNA fragment used in the R-looping.
 

5.  Control at the level of mRNA stability is illustrated by the cell’s response to low concentrations of  iron. Explain this.
(10 pts)

Iron is brought into the cell by the transferin receptor. This protein has a mRNA that codes for it and contains a 3'
untranslated region that contains multiple stem-loops that are A/U rich. The loops containg iron-response-element
sequences. These stem-loops signal for destruction of the mRNA by nucleases in the cell. The iron-response-element
binding protein is also present in the cell and can exist in an active or inactive form depending upon iron concentrations.
When iron concentration is high, it cannot bind to the IREs but when iron concentration is low it does and this protects
the mRNA from degredation, allowing more transferin receptor to be made in order to bring more iron into the cell.
See page 441, Fig. 11-44.
 

6.  Outline the CHEMICAL reactions that remove an intron from a primary RNA transcript. (8 pts)

A. The 2'OH of the branch point A attacks the 5' Phosphate of the first nucleotide of the intron.
B. This breaks the bond holding the last nucleotide of exon 1 to the intron.
C. A phosphdiester bond forms between the 2'O of the branch-point A and the 5' Phosphate of the first nucleotide of
the intron.
D. The 3'OH of the last nucleotide of exon 1 attacks the 5' Phosphate of the first nucleotide of exon 2.
E. This breaks the bond between the last nucleotide of the intron and the first nucleotide of exon 2.
F. Then the 3'O of the last nucleotide of exon 1 binds to the 5' Phosphate of the first nucleotide of exon 2 creating a
phosphodiester bond and completing the splicing.
G. The intron is displaced in its lariat form and degrades.
 
 

7.  What is the role of the following proteins in determining the sex of a developing fruit-fly? (3 pts each)

A.  early sex-lethal protein

Transcribed from early promoter only in females. It binds to an acceptor sequence before an exon in the late sex-lethal
transcript, causing alternative splicing that removes this exon and the stop codon within it. This allows a functional late
sex-lethal protein to be made. In males, without the early sex-lethal protein, the transcript of the late sex-lethal gene is
spliced normally. The stop codon that remains causes a premature termination of translation and a non-functional late
sex-lethal protein.

B.  late sex-lethal protein

Transcribed from the late promoter in both males and females but made functionally only in females (see above
explanation). It autoregulates its own splicing, taking the place of the early sex-lethal protein to allow continued
functional late sex-lethal protein to be produced. It also regulates the splicing of the transformer gene transcript by
binding to an acceptor site in front of an exon and causing it to be removed along with a stop codon within that
exon.The allows a functional transformer protein to be made in females. In males, without the late sex-lethal protein
function, the transformer transcript is spliced normally, creating a mRNA containing a stop codon that causes a
non-functional transformer to be made.

C.  transformer protein

Transformer protein can bind to transformer two protein in females only. This complex binds near an acceptor before
an exon in the double-sex gene transcript, activating the acceptor that would otherwise not be used. This allows for that
exon to remain in the mRNA being made and the protein from it will be the female form of the double-sex protein. In
males, who make no complex, the weak acceptor is ignored and therefore the exon is spliced out.This creates a
mRNA that gives rise to the male form of the double-sex protein.

D.  double-sex protein

Double-sex female form represses genes for male-specific phenotypes in the fruit-fly, so the fly becomes female.
Double-sex male form represses genes for female-specific phenotypes so the fly becomes male.
 

8.  Into what category would you place the following amino acid side chains? (3 pts each)

A.  A side chain that pulls away from water and is most frequently found in the interior of globular proteins?

hydrophobic, non-polar

B.  A side chain that can ionize by removing hydrogen ions from its surrounding solution.

hydrophilic basic.

C.  A side chain that cannot ionize and that interacts well with a water environment.

hydrophilic, polar uncharged

9.  Illustrate the  chemical reaction that allows two amino acids to be joined by a peptide bond. The side chains can
simply be designated R1 and R2. (8 pts)

See text.
 
 

10.  Answer the following questions about protein structure.

A.  Describe the secondary structure of a protein called the alpha-helix. Be sure to explain what stabilizes this structure.
(4 pts)
Right-handed helix with 3.6 amino acids per turn of the helix. Stabilized by hydrogen bonds between the peptide
linkages. Specifically, between the oxygen of the carbonyl of one linkage and the amide hydrogen of the linkage four
residues away towards the carboxy terminal. The side chains of the amino acids project to the outside of the helix.

B.  What is a motif?  (3 pts)

A combination of secondary structures that take on a certain topology that is frequently seen in proteins performing a
similar activity. Examples: zinc-finger, helix-loop-helix.
 

C.  What feature of the amino acid cysteine allows it to have a significant influence on the structure of a protein? (3 pts)

Cysteine has a sulfhydryl group in its side chain (-SH). If two cysteines find themselves near one another in the folding
of a polypeptide, they can potentially react and form a disulfide bond between their two sulfhydryls. -SH and -SH
become S-S.

                                                                        A Third Example

  1.  Describe the two step reaction that attaches an amino acid to a tRNA molecule. Be sure to properly illustrate the
linkage site and to indicate where on the tRNA the attachment occurs. (8pts)

1.  ATP + amino acid = PP + AMP-amino acid
2.  AMP-amino acid + tRNA = AMP + amino-acid-tRNA

The COOH of the amino acid attaches to either the 2’ or 3’ OH of the 3’ end nucleotide of the tRNA (an A).
This is catalyzed by amino-acyl tRNA synthetase.
 

2.  Starting with unassembled ribosomal subunits, outline in stepwise fashion how a functional ribosome is ultimately
formed at the correct AUG start codon during translation initiation . (9 pts)

1.  Ternary complex forms between eIF2,GTP, and initiator tRNA-Met.
2.  40S ribosomal subunit, bound to other eIFs joins with the ternary complex forming the preinitiation complex.
3.  Preinitiation complex finds eIF4 bound to the Cap on the mRNA.
4.  ATP is hydrolyzed as complex binds and melts the "stem-loop" in the 5’UTR.
5.  Complex scans for AUG after the Cap. This requires ATP hydrolysis also.
6.  Complex assembles at the AUG and the 60S subunit binds.
7.  The GTP is hydrolyzed by eIF2 to GDP and eIF2 leaves. The functional ribosome is formed.

3.

A.  Describe the role of eEF1 (eukaryotic elongation factor 1) in translation. (5 pts)

eEF1-GTP binds to the "charged" tRNAs. This complex moves into the A site of the ribosome. The eEF1 hydrolyzes
GTP to GDP, releasing energy for the stable attachment of the tRNA to the A site. EEF1-GDP leaves the A site.
Peptide bond formation can now occur.
 

B.  Why does this help insure that incorrect amino acids are not put into the growing polypeptide? (5 pts)
 

If an incorrect tRNA enters, the codon-anticodon interaction is not stable. The complex is knocked out of the A site
before the GTP can be hydrolyzed. Therefore, the need to complete the GTP hydrolysis before getting stable
attachment and peptide bond formation provides a kinetic time-lag that prevents the wrong tRNA  from participating in
peptide bond formation.

4.

A.  What do we call the class of proteins that helps other proteins to fold correctly? (3 pts)

   chaperones (chaperonins acceptable)

B.  What modification is found on lysine residues in short-lived proteins? (3 pts)

   polymer of ubiquitinsis attached

C.  What cellular structure is responsible for the degredation of the proteins modified as described in part B? (3 pts)

    proteosome

D.  What role is played by the glycine lid at the active site of  cyclic AMP dependent  protein kinase? (3 pts)

 It closes over the nucleotide binding pocket when ATP binds at the active site of the enzyme, preventing  the
 ATP from leaving the active site.

5.
A.  What is a GEF? (3 pts)

Guanine-nucleotide exchange factor that binds to a GDP-bound G protein and causes it to release its bound GDP.
 

B.  What is a GAP? (3 pts)

A GTPase activating protein that binds to the GTP-bound form of a G protein and stimulates its ability to hydrolyze
GTP to GDP.
 

C.  Match the term to the statement that most closely fits to it. (2 pts each)

 Gs    receptor tyrosine kinase
 phospholipase C   IP3
 Map Kinase   protein kinase C
 adenylyl cyclase   Gp
 diacylglycerol (DAG)  MEK
 calmodulin   SH3 domains
 ras    PIP2
 SH2 domain   cyclic AMP
 

a.  Phosphorylates transcription factors in the nucleus. _Map Kinase_________________
b.  Can phosphorylate serines, threonines, and tyrosines. MEK___________________________
c.  Binds to and opens up calcium channels in the endoplasmic reticulum. _IP3_________________
d.  Cleavage of its substrate releases two molecules that serve as second messengers. _phospholipase
C______________
e.  Binds to phosphorylated tyrosines.__SH2 domain__________________
f.  Mediates signals by binding to calcium. _calmodulin______________
g.  Activates phospholipase C. _Gp____________
h.  Catalyzes formation of cAMP. __adenylyl cyclase_________________

6.
A.  Molecule X has been identified. When overexpressed in cells, the cells shrivel up and die. Molecule X is not a cell
membrane protein but does have characteristics that suggest it can associate with membranes. What possible role in cell
death could molecule X have? Defend your answer using information on known mechanisms that cause cell death. (5
pts)
 

Protein that binds to the mitochondrial outer membrane and helps form a pore allowing influx of ions across the
membrane, releasing cytochrome c and causing apoptosis. Similar to how the bax protein functions.
 

B.  How could you test your hypothesis? (5 pts)

Many possible answers. Some examples include
a.  Look for homology in the amino acid sequence of X and bax.
b.  Make antibodies against X and see if the X protein is located on the outer mitochondrial membrane.
c.  Measure the amounts of cytochrome c in the cytoplasm before and after X overexpression.
d.    Other possible answers.
 

7.  You are working as a technician in a large cancer research laboratory. At the monthly lab meeting, the following
results are reported. Using what you have learned in class about oncogenes, give one possible explanation for each
result. (3 pts each)

A.  Chromosome spreads of cancer cells show several double-minutes when probed with a cDNA for the myc gene.

c-myc gene is amplified in the cancer cells. (Many copies where there should only be one)
 

B.  Chromosome spreads of cancer cells show a Philadelphia chromosome.

A reciprocal translocation has occurred between chromosomes  9 and 22 causing the formation of the bcr-abl fusion
protein
 

C.  Cancerous mouse cells show several more bands on  protein gels that are probed with antibodies directed against
phosphotyrosines than do their normal counterpart cells.

Receptor tyrosine kinases and other tyrosine kinases have been activated in the cancer cells in an unregulated way.
 
 

8.  p53 is known to induce a cell-cycle arrest in response to UV irradiation of cells. This arrest is known to happen in
G1 of the cell-cycle. Outline a pathway that would allow the cell-cycle to be stopped by p53 in response to the
irradiation. (8 pts)
 

1.  p53 is stabilized by phosphorylation. (mdm-2 cannot bind to target it for ubiquitination).
2.  p53 activates its own transcription and the transcription of p21.
3.  p21 inactivates the cdk/cyclin complexes usually active in G1.
4.  cdk/cyclin now cannot phosphorylate Rb.
5.  Rb therefore remains bound to E2F.
6.  E2F will remain inactive and not activate the transcription of DNA synthesis related genes.
 
 

9.  The following are three approaches being studied in cancer labs to inhibit or prevent the growth and further
progression of cancers. Describe why these studies could eventually be useful in the treatment of malignancies. (4 pts
each)

A.  structure-function studies of the telomerase enzyme
 

Telomerase extends telomeres of replicating cells. It should be silent in adult cells but is active in many cancer cells.
Understanding its structure and function could allow for production of inhibitors of its activity, hopefully causing the
cancer cells to senesce (stop dividing) or to die.
 

B.  identification of proteins involved in the control of angiogenesis

Angiogenesis is the growth of new blood vessels. Tumor cells secrete factors that stimulate angiogenesis in order to
feed themselves. Finding proteins to inhibit this could starve tumors and prevent their growth.
 

C.  human gene therapy using replication impaired Adenoviruses expressing p53.

Several possible answers. Examples:

 Introduction of normal genes for p53 into cancer cells could induce the apoptosis of those cells. Also, introduction of
p53 into p53 negative cells that are not yet cancerous could possible prevent them from becoming cancerous by
allowing the p53 DNA damage response checkpoint to be functional again, preventing the accumulation of additional
mutations in the cell.