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MIMM 466 Final: Solutions - MIMM 466 Final Examination, 2002.doc

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McGill University
Department
Microbiology & Immunology
Course
MIMM 466
Professor
Various

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Solutions mimm 466 final examination, 2002: 1 burkitt"s lymphoma and nasopharyngeal carcinoma a. It also causes a polyclonal b cell lymphocytic leukemia lymphoma: 2 shingles, also called zoster, and is a local recurrence of latent infection, primary infection is called varicella (chickenpox, 1 ebv. Susceptibility of fertilized mouse eggs to minute virus of mice. Involved in contraction/relaxation of the vessels to regulate blood flow: the adventitia provides support for the blood vessels, more prominent in arteries than in veins iv. In order to gauge the effect of autoantibodies, researchers chose to look at the model of experimental autoimmune myocarditis (eam) It"s been found that only mhc-alpha produces cardiac myocarditis. If you put in the beta form, you do not get as much vi: there are also specific immunodominant peptides (i. e. m7a- muscle alpha, basic sequence required for myocarditis formation: Once inside, hpv infects host epithelial cells, tricking them into producing new viruses.

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Related Questions

1. Describe the allosteric regulation and the covalent modification (phosphorylation/dephosphorylation, if available!) of the following enzymes: a) Phosphofructokinase-1. b) Phosphofructokinase 2/Fructose-phosphatase 2 c) Hexokinase. d) Pyruvate kinase. e) Glycogen phosphorylase. f) Glycogen synthase. g) Pyruvate carboxylase and PEP carboxykinase.

1. When a growth factor binds to a receptor tyrosine kinase, the receptor becomes phosphorylated on tyrosine residues.

a. How does this occur?

b. Why is this phosphorylation important in initiating an intracellular signaling cascade?

c. Why does phosphorylation change the activity of proteins? (What happens to the protein physically when it is phosphorylated?)

QUESTION 1:

(a)Examine the Figure below relating to the S.pombe cellcycle:

1) Describe the function of Cdc25 and Wee1and why deficit of Cdc25 or excess of Wee1 can cause G2arrest

2) b) Describe how deficit of Wee1 or excess ofCdc25 may result in abnormally small cells

(b) Theactivities of Wee1 tyrosine kinase and Cdc25 tyrosine phospatasedetermine the state of phosphorylation of tyrosine 15 in the Cdk1component of mitotic cyclin-Cdk complexes (MPF). When tyrosine 15is phosphorylated, M-Cdk is inactive; when tyrosine 15 is notphosphorylated, M-Cdk is active. Just as the activity of M-Cdkitself is controlled by phosphorylation, so too are the activitiesof Wee1 kinase and Cdc25 phosphatase. The regulation of thesevarious activities can be studied in extracts of frog oocytes. Insuch extracts Wee1 tyrosine kinase is active and Cdc25 tyrosinephosphatase is inactive. As a result M-Cdk is inactive because itsCdk1 component is phosphorylated on tyrosine 15. M-Cdk in theseextracts can be rapidly activated by addition of okadaic acid,which is a potent inhibitor of serine/threonine phosphatases. Usingantibodies specific for each component it is possible to examinetheir phosphorylation states by changes in mobility upon gelelectrophoresis.

(Phosphorylated proteins generally run slower than theirnon phosphorylated counterparts).

1) Based on theresults with okadaic acid, decide whether the active forms of Wee1and Cdc25 are phosphorylated or unphosphorylated. In the figureindicate the phosphorylated forms of Wee 1 and Cdc25 and label thearrows connecting their active and inactive forms to show whichtransitions are controlled by protein kinases and which by proteinphosphatases.

2) Are the proteinkinases and phosphatases that control Wee1 and Cdc25 specific forserine/threonine side chains or for tyrosine side chains? How doyou know?

3) How doesaddition of okadaic acid cause an increase in phosphorylation ofWee1 and Cdc25, but a decrease in phosphorylation ofCdk1?

4) If you assumethat Cdc25 and Wee1 are targets for phosphorylation by activeM-Cdk, can you explain how the appearance of a small amount ofactive M-Cdk would lead to its rapid and completeactivation?