Birla Institute of Technology (BIT Mesra) 2007 B.E GENERAL BIOLOGY - Question Paper

BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI
FIRST SEMESTER 2007-08
BIO G321 CELL BIOLOGY
COMPREHENSIVE exam – PART A (CLOSED BOOK)
DURATION: 60 MINS MARKS: 40 DATE: 13.12.07
1.(a) Qualitatively compare the resolving power of ordinary light microscope, ultraviolet light microscope and electron microscope. discuss the cause for this difference? (3)
(b) Of the things provided below, list which of the subsequent can be seen (without sectioning) by unaided eye, light microscope or electron microscope based on the size and the best resolution of the microscope.
i) Ribosomes ii) Viruses iii) Mycoplasma iv) human muscle cell
v) Human liver cell vi) Frog egg (3)

( c) Chronologically in the form of a table provide the different scientific observations that led to the current accepted model of plasma membrane? (3)

(d) Studies were carried out in unicellular alga Chara about the various molecular parameters that influence the ability of substance to permeate cell membranes. discuss how the subsequent parameters affect membrane permeability? (4)
i) Distribution coefficient ii) Lipid solubility
iii) Molecular size iv) Valency of ions

2. (a) Nobel prize winner Federick Sanger contends that mammalian mitochondrial genetic systems cannot be classified as either prokaryotic or eukaryotic? Can you discuss why? (3)
(b) Compare the 2 membranes of mitochondria with respect to (3)
i) Thickness ii) Composition iii) Permeability
( c) Schematically provide the Hatch Slack pathway of sugarcane ? (4)
(d) Leukocytes especially granulocytes have a rich source of lysosomes? discuss
why? (3)

3. (a) discuss the subsequent terms in 1 or 2 lines (10)
i. Red drop effect
ii. Tonoplast
iii. Heterophagy
iv. Dictyosomes
v. Constitutive heterochromatin
vi. Facultative heterchromatin
vii. Photosystem II
viii. Totipotency
ix. Plueripotency
x. Hapten
( b) What is the basis of graft rejection in humans? discuss briefly? (4)



BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI
FIRST SEMESTER 2007-08
BIO G321 CELL BIOLOGY
COMPREHENSIVE exam – PART B (OPEN BOOK)
DURATION: 120 MINS MARKS: 80 DATE: 13.12.07

Q1) discuss in 1 line the contribution of the subsequent scientists to cell biology. (10)
i. Wallace Coulter
ii. Novick & Sziland
iii. E. Metchnikoff
iv. Benda
v. M. Calvin
vi. Christian deDuve
vii. E. Heitz
viii. Fontana
ix. J. B. Gordan
x. Hans H. Krebs

Q2) a) You are the Director of Drug Development for a Biotech company called Cyclex. The goal of your company is to develop drugs that halt the cell cycle. You send out an e-mail to your team asking for ideas regarding drug targets. Few ideas for potential targets come back which are provided beneath. Evaluate them and justify your option of good and bad targets. (You may presume that any drug your company develops will completely inactivates its target protein, so that a drug-treated cell will have the identical phenotype as a cell in which both alleles of the gene encoding the target had been deleted). (12)

i) Cdk4, ii) Cdk6, iii)E2F, iv) retinoblastoma (Rb) protein, v) p16, vi) cyclin D

b) E. coli is well known as a lactose fermenting bacteria. However, the sugar lactose does not easily permeate E. coli cells. discuss the phenomena used by E.coli cells to overcome this impediment. (4)

Q3) a) In the yeast S. pombe, the cyclin dependent kinase Cdc2 is phosphorylated on tyrosine 15 (Y15) by the protein kinase Wee1. Cdc25 phosphatase removes this phosphate. Site-directed mutatgenesis is used to create an allele with a tyrosine 15 to phenylalanine (Y15F) substitution (Note: phenylalanine has no hydroxyl group and hence cannot be phosphorylated). This allele is then used to change the wildtype allele. What will be the implications of this mutation? discuss on your understanding of cell cycle regulation. (6)

b) Patient’s with Hunter’s syndrome or Hurler’s syndrome rarely live beyond their teens. Analysis shows that patients accumulate glycosaminoglycans in lysosomes due to the lack of specific lysosomal enzymes necessary for their degradation. When cells from patients with the 2 syndromes are fused, glycosaminoglycans are degraded properly, indicating that the cells are missing various degradative enzymes. Even if the cells are just cultured together, they accurate every others’ defects. Most surprisingly of all, the medium from a culture of Hurler’s cells corrects the defect in Hunter’s cells (and vice versa). The corrective factors in the media are inactivated by treatment with proteases, by treatment with periodate, which destroys carbohydrates, and by treatment with alkaline phosphatase, which removes phosphates. (8)

i) What do you think the corrective factors are, and how do you think they accurate the lysosomal defects?
ii) Why do you think the treatments with protease, periodate and alkaline phosphatase inactivates the corrective factors?

Q4) a) It is known that during viral infections, viral proteins activate CD8+ T-cells via MHC I complexes. However, CD4+ T-cells are also activated. Since CD4+ T-cells are being activated via interactions with MHC II, what might be occurring to allow viral proteins to be expressed on MHC II cells?

b) Patients with DiGeorge’s syndrome have a developmental abnormality that outcomes in partial or total lack of development of the thymus. These patients are capable of handling bacterial infections, but are very vulnerable to viral infections. What kinds of immune cells are missing or defective in these patients?

c) HIV selectively infects and destroys CD4+ T-cells. While it is the CD8+ “Killer” T-cells that destroy intracellular invaders, CD4+ “Helper” T-cells are needed for Killer T-cell activation. What effect will it have on the survival of the individual and why?

d) Red blood cells are not nucleated and do not express any MHC molecules. Why is this property fortuitous for blood transfusions?

e) Why is specific immune response polyclonal?

f) A 15-year-old boy having extreme muscle weakness and paralysis, comes to your clinic. Upon exam in his blood you obtain antibodies against Acetylcholine receptors. What is the most likely diagnosis and reason if this illness?

(6 x 3=18)

Q5) a) What is decorated actin? What is the limitation associated with expression studies of myosin?
b) elaborate the characteristics a probe for membrane integrity should employ? provide examples of endogenous and exogenous markers used.
c) What will be the signal for sorting proteins like LDL receptors and B integrins? How do the mutations in critical residues affect them?
d) Why is intraepithelial lymphocytes a potentially important immunology components in study of cancer?
e) Does all the endosomal – lysosomal sorting signals consist of short, linear array of amino acid residues? Justify with example.
f) Why and in which way are a few membranes within lysosomes degraded whereas other membranes, limiting membranes of endosomes and lysosomes remain unaffected?
(6 X two = 12)

Q6) Regeneration of heart tissue is of great medical interest. 1 idea is that by understanding how an embryo forms a heart, scientists will be able to recapitulate that process to make replacement heart tissue. As a new PhD graduate, you decide to study heart development in the chick embryo, with the ultimate goal of growing replacement heart tissue in the lab.

The 1st thing you need to do is figure out which part of the chick embryo forms the heart. You place dots of green dye at different places in an early chick embryo (stage two or stage 4), and obtain that a region near the front of the embryo (anterior), always goes on to form the heart. Stage two and stage four embryos look similar, and do not have a heart, which only appears and begins to beat, three days later, at stage 18.
The embryo has 3 layers of cells, but when you cut the embryo open, you obtain that just 1 layer of cells, the mesoderm, forms the heart.
You then ask when the future heart cells decide to become such. You do this by isolating a small piece of tissue (called an “explant”), of about 200 cells, from the heart forming anterior mesoderm of both a young stage two embryo, and from a slightly older stage four embryo (about six hours older). You place the tissue in culture medium, and examine it 3 days later. You find the subsequent outcomes.


Stage two anterior mesoderm explant 3 day culture no change from original cells
Stage four anterior mesoderm explant 3 day culture beating heart

6a. At the time of explantation, are the stage two cells determined, differentiated or neither? discuss
6b. At the time of explantation, are the stage four cells determined, differentiated or neither? discuss
6c. Are the beating heart cells determined, differentiated or neither? discuss

Intrigued, you ask why the isolated stage two cells did not become a heart. You remember that the anterior mesdoderm cells truly lie on top of a different cellular layer in the embryo, called “endoderm”, and that you did not isolate (explant) these endoderm cells in the previous experiment. You perform the subsequent experiment, and find outcomes indicated.

Stage two anterior mesoderm explant alone 3 day culture no change
(labeled with green dye) (green)
Stage two anterior mesoderm explant (green) 3 day culture beating heart
plus underlying endoderm (unlabeled) (green)
Endoderm alone (unlabeled) 3 day culture no change

6d. Why does the stage two anterior mesoderm explant plus endoderm make a heart, whereas the stage two anterior mesoderm explant alone does not?

Later, in your studies, you obtain that a purified protein, BMP4, is able to substitute for the endoderm. Thus
Stage two anterior mesoderm explant + BMP4 3 day culture beating heart

Posterior mesoderm does not form the heart. However, you would like to know how powerful an influence BMP4 can have, and so you test whether BMP4 is able to turn posterior mesoderm into beating heart, and get the subsequent outcomes.

Stage two posterior mesoderm explant 3 day culture no change
Stage two posterior mesoderm explant + BMP4 3 day culture no change

6e. Suggest why posterior mesoderm does not respond to BMP - four to make a heart.

ans parts 2a – 2e in sequence. (5 X 2=10)

Earning:   Approval pending.