1. A. Below is a diagram of a replication fork in a eukaryotic cell. On the diagram, indicate the location of the following:  (2 pts each)

Primase
Topoisomerase
PCNA
Helicase
DNA polymerase alpha
DNA polymerase delta

See diagram in cabinet near 021 McKinly.
 
 

B. Which of the above is responsible for or contains the following? (2 pts each)
 

Synthesizes the leading strand
DNA polymerase delta
 

Has a 3’-5’ exonuclease activity
 DNA polymerase delta
 

Synthesizes Okazaki fragments
DNA polymerase alpha
 

Relieves supercoiling strain as the replication fork progresses
topoisomerase
 

Participates in allowing the template strand to become accessible
helicase
 
 

2. Outline all steps, catalyzed by DNA Ligase, that connect Okazaki fragments together. (8 pts)

1. DNA Ligase catalyzes the hydrolysis of ATP to AMP and PP.
2. PP breaks down to 2 P.
3. The AMP is linked to the 5’ phosphate of one of the fragments to be joined.
4. Then DNA Ligase removes the bound AMP and uses the energy released to make a    phosphodiester bond between the 5’ phosphate and the 3’OH of the other Okazaki  fragment.
 
 

3. During development, eukaryotic cells produce an enzyme called telomerase. Explain why this enzyme is necessary during development and how it works (all steps of its mechanism). (8 pts)

1. The ends of the chromosomes, called telomeres, gradually shorten with each round of DNA replication because the lagging strands cannot be completed.
2. Cells such as those dividing rapidly during development and some other adult cell types (not many) make an enzyme telomerase to extend the ends of the strands.
3. Telomerase has both an RNA component and a protein component. It works as an RNA dependent DNA polymerase.
4. The DNA sequence at the end of the telomeres is complementary and antiparallel to the RNA sequence of that component of telomerase, so they bind.
5. Then the protein part of telomerase catalyzes the attachment of DNA nucleotides to the 3’end of the telomere, using the RNA as the template.
6. This continues until nucleotide number 35 of the telomerase RNA is reached.
7. Then the DNA is displaced and reattaches further up once again to the RNA template.
8. Steps 5, 6, and 7 repeat several times.
9. This results in repeated DNA sequences at the end of the telomeres and added length to the telomere.

See diagram 12-13 on page 466 of the textbook.
 

4. You are walking down a dark street late at night and hear a strange noise. Describe what happens in your cells to give you the extra energy you need to run away if need be. (10 pts)

1. A stress-related hormone is released into the blood and travels to a receptor in the membrane of cells that will need extra energy to help you run.
2. Binding of the hormone causes the receptor to change shape.
3. This shape change causes the activation of a G protein nearby.
4. The G protein releases GDP from its alpha subunit and GTP replaces it.
5. This causes the beta and gamma subunits to relase from the alpha subunit
6. The alpha subunit can now bind to adenyl cyclase (adenylate cyclase also acceptable) and activate it.
7. The activated adenyl cyclase catalyzes the synthesis of cyclic AMP
8. The increased cyclic AMP concentrations allow for cyclic AMP to be able to bind to the cyclic AMP dependent protein kinase
9. This causes the regulatory subunits to release the catalytic subunits of the kinase, allowing it to phosphorylate its substrates
10. This activates enzymes that cause the concentration of glucose in the cells to increase, allowing more to be available for glycolysis and aerobic respiration and the production of ATP. ENERGY TO RUN.
 
 

5. Identify the following : (3 pts each)

A GAP
Protein that causes a G protein to release GDP so GTP can replace it and activate the protein’s activity (our example was ras)

A GEF
Protein that increases the ability of a GTP-bound G protein’s ability to act as a GTPase and hydrolyze the bound GTP to GDP, thus inactivating the G protein (ras example again)

An SH2 domain
A domain on a protein that allows it to bind to phosphorylated tyrosines and their surrounding amino acids, thus transducing a signal from that tyrosine phosphorylated molecule to another molecule down the signaling chain. (Stands for src homology domain 2. This is not needed in the answer to the question)

MAP Kinase
An enzyme that can phosphorylate transcription factors in the nucleus and thus allow them to bind to promoter elements that activate the transcription of many genes involved in either cell division or differentiation.
 
 

6. Describe the role played by the mitochondria in the induction of apoptosis. (8 pts)

In the outer mitochondrial membrane there is a dimer of Bax proteins. When allowed to be active, this forms an ion channel. Ions flow in to the inner-membrane space through the channel. This causes the release of cytochrome c from the mitochondria. It binds to Apaf1 and activates procaspase 9 to become caspase 9. This activates procaspase 3 to become caspase 3. This then sends the cell into apoptosis.

See page 1049, figure 23-50
 
 

7. Using the information you have learned about p53, explain why it is not surprising that the majority of cancer cells in human have either lost all normal p53 or have only mutated forms. (8 pts)

P53 plays two important roles. First, it is a major checkpoint protein that is stabilized in G1 of the cell cycle in response to signals caused by DNA damage. It gets phosphorylated, stabilized, and then acts as a transcription factor to activate its own transcription and that of p21. p21 binds to cdK/cyclin complexes present in G1, inactivating their ability to phosphorylate their substrates. This includes the Rb protein. Therefore, Rb does not get phosphorylated and remains bound to E2F. This prevents E2F, also a transcription factor, from activating the synthesis of genes involved in DNA synthesis and the cell-cycle is stopped.

Second, p53 can induce apoptosis by activating the bax gene and inhibiting bcl-2. Cells that should have damaged DNA that is unrepaired will thus be killed.

Therefore, when p53 is lost or mutated in cells, these protective functions are lost. The cell is free to divide even though its DNA is damaged and this causes further accumulation of mutations in the DNA that can ultimately lead to a full-blown cancer.
 
 

8. The following are all potentially involved in the formation of a cancer cell. Explain how. (4 pts each)

A. The Philadelphia chromosome
This is the result of a reciprocal translocation between chromosomes 9 and 22. The c-abl gene ( a protein tyrosine kinase) moves from 9 to the region of  22 called bcr. When expressed, this makes a fusion protein of bcr-abl. This causes the continuous activation of abl kinase activity leading to inappropriate signaling for cell division.

B. the retroviral src protein

This is a mutated form of the cellular src proto-oncogene. It was transduced into the retroviral genome when the oncogenic retrovirus was formed. During the transduction, the gene sequence mutated such that an important regulatory amino acid was changed. The src protein now could not be inactivated as a tyrosine kinase. Therefore, when this viral src protein is made in the infected cell, the signals are uncontrolled and lead to cancer
 

C. insertional mutagenesis

When a non-oncogenic retrovirus integrates into the host chromosomes, nearby genes can become activated by the powerful retroviral enhancer and promoter sequences. If the nearby gene is a proto-oncogene, it will be expressed when it should not, leading to a cancer.

D. angiogenesis factors

These are proteins such as VEGF, bFGF, etc. that are secreted from primary tumors. They activate the vascular endothelium and cause the formation of new blood vessels to feed the growing tumor.
 

9. Let us say that you found a gene that you suspect might play a role in causing cells to become cancerous. Describe one experiment you might do to learn more about how it is so involved. (You can use any possibility here that relates to something you learned about how cancers may form in humans. Just give the general outline of your study without worrying about experimental details.) (8 pts)
 

Many answers are possible to this question.
 

     PBL 3 Question

This problem studied a drug called Gleevek. Why is this drug promising as a cancer therapeutic?

 Gleevek binds to the ATP binding site of several oncogenic tyrosine kinases that are activated in cancer cells. The one we studied in this problem was the bcr-abl tyrosine kinase that is caused by the chromosome 9, 22 translocation. (See Philadelphia chromosome question above). The structure of Gleevek is such that it only binds to the abnormal kinase found in cancer cells, not other normal kinases. When it is bound, the usual ATP cannot bind and therefore there is no phosphate source for the kinase to use to phosphorylate its substrates. The signaling is stopped. There is also evidence that this causes the cancer to enter apoptosis and die.