amankg4251

amankg4251

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Aman Kumar GuptaVisvesvaraya Technological University - VTU

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Published22

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History2Sociology1Communications1Ethics2Biology13Physics3
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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?

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Answer:Communication.

3. The activity of myogenin must be carefully controlled lest it trigger premature expression of the gene essential for cell differentiation. The myogenin gene is turned ON in before it is actually needed (and myogenin is expressed), but myogenin is prevented from functioning by phosphorylation of its DNA-binding domain AND by its tight binding to Id, a helix-loop-helix protein that lacks the DNA binding domain.

A. What types of interactions would hold the heterodimer together and you expect them to be the same as those that are necessary for Id binding? Where would you find these bonds/interactions (use the figure).

B. Explain how phosphorylation of the DNA binding domain and dimerization with Id might act to keep myogenin nonfunctional.

Extra credit. Why might this method of protein regulation be preferred over transcriptional control in this case?

4. Aminoacyl-tRNA synthetases attach specific amino acids to their appropriate tRNAs. The synthetase that attaches valine to tRNAval must be able to discriminate valine from threonine. Valyl-tRNA synthetase achieves this discrimination in two steps. In the first, it uses a binding pocket whose contours allow only valine and threonine to bind, but the binding of valine is preferred. This site is responsible for coupling the amino acid to the tRNA. In the second step, the enzyme checks the newly made aminoacyl-tRNA using a second binding site that is very specific for threonine and hydrolyses it from the tRNA. How do you suppose it is that the second binding site can be very specific for threonine, whereas the first binding site has only a moderate specificity for valine?

Answer: 3. The types of interactions that would hold the heterodimer together ...

  1. You are studying a steroid hormone-binding protein in mice. Youhave isolated two proteins that have similar but not identicalamino acid sequences from a mixed tissue homogenate (prepared frombrain, muscle, liver, ovary, adrenal gland and kidney of a mixedpopulation of adult mice). There are at least three differentmolecular mechanisms that can give rise to similar butnon-identical proteins. List and explain two of these.

2. The differentiation of muscle cellsin the developing embryo is controlled by myogenin, ahelix-loop-helix (HLH) gene regulatory protein that functions as aheterodimer with another HLH protein. The activity of myogenin mustbe carefully controlled so it does not trigger premature expressionof muscle cell differentiation. The myogenin gene isturned on in advance of the time when it is needed, but myogenin isprevented from functioning by its tight binding to Id, an HLHprotein that lacks a DNA-binding domain, and by phosphorylation ofits DNA-binding domain. Explain how dimerization with Id andphosphorylation of the DNA-binding domain might act to keepmyogenin non-functional.

3. Imprinting occurs only in mammals,and why it should exist at all is a mystery. One gene that isimprinted in most mammalian species is Igf2 gene, whoseproduct � insulin-like growth factor-2 � is required for prenatalgrowth. Most mammalian females can mate with multiple males,generating multiple embryos with different fathers in each litter.If one father had an Igf2 allele that caused more rapidprenatal growth, embryos carrying his genes, it would prosper atthe expense of the other embryos. While this would be good for thefather�s genes (in an evolutionary sense), it would drain theresources of the mother, potentially putting her life at risk (notgood for her genes). Thus, it is in the mother�s interest tocounter these paternal effects with maternal changes that limit thegrowth of the embryo. Based on this scenario, decide whether theIgf2 gene is more likely to be imprinted in the male or inthe female.

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