PART C

You are studying the binding of proteins to the cytoplasmic face of cultured livercells and have found a rnethod that gives a good yield of inside-out vesicles fromthe plasma membrane. Unfortunately, your preparations are contaminated withvariable amounts of right-side-out vesicles. Nothing you have tried avoids thiscontamination. Somebody suggests that you pass the vesicles over an afhnitycolumn made of lectin coupled to Sepharose beads. What is the rational of thissuggestion?

(1) Right-side-out-vesicies will be lysed by lectin coupled to Sepharosebeads.

(2) Right-side-out-vesicles will simply bind to the lectin coupledSepharose beads.

(3) Lectin will bind to the carbohydrate residues present only on the inside-out vesicles.

(4) l,ectin will bind to only gl]'coprcteins and glycolipids present on theinside-or.rt vesicles.

72. The overall length of the celi cycle can be measured from the doubling time of apopulation of exponentially proliferating cells. The doubling tirne of a populationof mcuse L ceils was determined by counting the number of cells in samples cfculture at various times. What is the overall length of tire cell cy'cie in mouse Lcells?

11

71"

I

o

€r"z

40 50 80 '100

Tlme (houF)

(2) 20h

73. Using molecular clock, it was estimated that two species A and B must have

diverged from their common ancestor about 9 x 10o years ago. If the rate ofdivergence per base pair is estimated to be 0.0015 per million years, what is theproportion of base pairs that differ between the two species now?

(i) 30 h (3) 10 h (4) 40 h

GJ 0.0270 (2) 0.013s (3) 0.00017 (4) 0.0035

12

74' A protein in 100 mM KCI soiution r'vas heated and the obsened T* (midpoint ofunfolding) v,zas 60'c. when the same protein solution in 500 mM KCI washeated, the.observed T. was 65oC. What is the most probable reason fbr thisrncrease in Tr?

(1) Hydrophobic interaction is increased ancl electrostatrc repulsion isdecreased.

(2) Hydrophobicinteractionisdecreaseci andelectrostatrcrepulsionisincreased.I{ydrogen-bonding is increased.van der Waals interaction is increased

During receptor-mediatecl endocl'tosis, apolipoprotein B on the surface of a LDLparticle binds to the LDi.-receptor p.*r*nt irrcoated pits containing clathrin. Thereceptor-LDl complex is-internalized by endocltosii, trafflcked to lysosomesand the LDl-receptor is finally recycled. A patient reports with familialhypercholesterolemia. This could be due to

(3)(4)

75.

(1)(2)(3)(1)

mutation in the LDL molecule.defect in LDl-receptor recycling.mutation in the LDL-receptor.defect in cholesterol binding rvith its receptor.

76. Budding yeast cells that are deficient for l\{ad2, a component of the spindle_attachment-check point. are killed by treatment with benomyl, which causesmicrotubles to depolymerise. In the absence of benomyl. hou,ever, the cells areT:fjll

r'iable.which expranation our of the foilou,in! is abre to justif.v thisoosen'at ton' /

(1) In the absence of benomyl, the majority of spindles form normaiiv and thespindle-attachment checkpoint 1l.lad:)- plays no ro le.{2) In the presence of benomyl, the majority oi spinoles form normally.and

Mad2 plays critical role in cell suru.ival.(3) other than'the role in cell survival, microtuble depolyrnerization affects

oxidative phosphorylation in the absence of Madj.(4) Benomyl alscr affects protein slnthesis in the absence of Mad2.

Eukaryotic genornes are organized into chromosomes and can be visualized atmltosls by staining rvith specific ilyes. Heat clenaturation followed by staining

ff:19:::"^r1!:i-d alternate derk and right bands. rhe dark banos obtainJd byInls process are mailr ly

(1) AT-rich and gene rich regions"(2) AT-rich and gene desert resions"(3) GC-rich and gene rich regiins.(4) GC-rich and gene desert regions.

77.

13

78. An amino acid contains no ionizable group.in-its side chain (R)' It is titrated from

pH 0 to t+. Wtrictr of the foffowini ionizable state is noi observed during the

entire titration in the pH range 0 - 14?

R*l

1. H3N - CH

R

2. *.fi - cl',

RI

3. H2N - CH

- cooH

- coo-\ ., lIi

79.

RI

4. HzN - CH - CoOH

Aresearcherhasisolatedarestrictionendonucleasethatcleavesatonlyonespecific 10 base Pair site'

A) Would this enzvrne be useful in protecting::]lii:,T viral infections'

given that a typitat viral genome is 5 x l0{ base pairs long?

B) Restriction;il;i;;;: r" rlo* enzymes with turnover number of

i s-,. suppo-seJhe isolated endonuclease was faster with turnover

numbers similar to those ror curuottlc anhydrase (106 tt), would this

increased;;; benefrcial to host ceils, assuming that the fast enzymes

have similar levels of sPecificitY?

The correct combination of answer is

(B) : Yes Q) (A) : No (B) : No

(B) : No (4) (A) : Yes (B) : Yes

S0.Lacrepressor inhibi tsexpressionofgenesinlac-operonrvhereaspunnebiosynthesis is repressed by ttre.r,$ reprrrir. The two proteins have 3la/o identical

sequences ana tiarre simiiar three-dimensional structures' The gene regulatory

ptop"tti.s of these proteins differ in relation to

A) binding of small molecuies to the repressor'

B) pr.,tt'it of recognition sites on the genome'

Ci bligomeric nature of the repressor'

D) DNA binding ProPertY'

The correct statements are

I

I

(1) (A) : No

(3) (A): Yes

(1) A and B (2) A, B and C (3) A and C (4) B, C and D

'l

81.

14

An *-helix in a peptide or protein is characteri zed,byhydrogen bonds andcharacteristic dihedral angles. choose the right .o*uirrution.

(1) Hydrogen bonding between the amide co of residue i and amide NHof residue z + 4_ Dihedral angles in the region 0_ _S0., V_ _60".(2) Hydrogen bonding between ihe amide Nfr of residue i and amide co ofresidue I * 4. Dihedral angles i",lg region of $_-_5g", V_ _60".(3) Hydrogen bonding betureeithe amide co of residue i and amideNH of residue i + 4. Dihedral angles in the region ofg_ _5i.,1_ *OO"{4) Hydrogen bonding between the amioe co or.Jriaue r ano amide NH ofresidue i + 3. Dihedral angles in the region of $_ 50o, V _60o.

Mouse bone marrow cells were fractionated to derive stem cell antigen-1* (Sca-1*)ceils. These ceils were cultured with interleukin-:, o. gru;urocfe_macrophagec olony stimulating ?:l:t

or.macrophage- colony riirnuTuiing factor, or granulocytecolony stimulating factor' Most numerJus and varied coronies were obtained in theculture stimulated with

82.

(1){2)(3)(4)

Interleukin-3.

$"11ul1fte-macrophage colony stimulating factor.M acrophage-colony stimulating factor.uranutoc)de-colony stimulating factor.

83. Precursors of the atoms in the purine skeleton are

Nl, Asp; C2 and, CB, formate; N3 and N9, guanidine of Arg; C4, C5and N7, Gly; C6, CO2.

Ni, Asp; C2 and'cg, citrate; N3 and N9, amide nitrogen of Gln; c4, c5and N7; Gly; C6, CO2.

Nl, Asp; c2 and c8' formate; N3 and N9, amide nitrogen of Gln, c4, c5and N7, Gly; C6, COz.

Nl, Glu; C and C8, acetate; N3 and N9, amide nitrogen of Asn; C4, C5and N7, Gly; C6, CO2.

nocF7

1N2 \3zN

i ir \u" \ /?-N/

39

(1)

(2)

(3)

(4)

15

84. Two E. coli culfixes A and B are taken. Culture A was earlier grown in thepresence of optimum concentration of gratuitous inducer IPTG. Both the culturesare now used to inoculate fresh mediurn containing sub-optimal concentration ofgratuitous inducer. It was obserr,'ed that culture B was unable to utilize lactose,whereas culture A did so efficiently. The reason behind this is

(1) pretreatment with IPTG has resuited in a mutation as a resuit of which /acoperon is constitutively expressed

(2) IPTG has made the cell rnembrane more porous to small molecules and solactose is taken up more efficiently by A as compared to B.

(3) in culture A, lactose perrnease was induced to a high level, during pre-treatment with IPTG, which allowed the preferential uptake of lactose.

(4) in culture A, IPTG activated a receptor which bound lactose moreefficiently, thereby triggering a signal.

85. Cancer causing genes can be functionally classified into mainly three types: (i)genes that induce cellular proliferation, (ii) tumor suppressor genes, (iii) genesthat regulate apoptotic pathway.

Epstein-Barr virus that causes cancer by modulating apoptotic pathway, contains agene having sequence homology rvith u'hich of tire foilorving genes?

(l) bax (2) bct-z (3) ps3 (4) caspase-3

86. It has been observed that in 5-10% of the eukaryotic mRNAs with multipleAUGs,the first AUG is not the initiation site. In such cases, the ribosorne skipsover one or more AUGs before encountering the favourable one and initiatingtranslation. This is postulated to be due to the presence of the following consensussequence (s):

A) CCA CC AUG G

B) CCG CC AUG G

C) CCG CC AUG C

D) AAC GG AUG A

Which of the following sequence sets related to the above postulations is correct ?

(1) AandB (2) AandC (3) CandD (4) BandD

87.

16

values of T* (midpoint of denaturation), AH* (enthalpy change at T*) and aco(constant-pressure heat capacity change) of a protein are measured in adifferential scanning calorimeter. AGD(T), the Gibbs free energy change at anytemperature T(K) can be estimated using the following form of the Gibbs_Helmholtz equation with the values obtained from these *Jurrr"*"nt, ,

/ r r \

^cD g) =au^

iu | _aco [r.-r+ r ln (rrr, )].

The stability curve for the protein simulated using the observed thermodynamicvalues is given beiow:

n'lnT,K

The shape of the stability curl e is due to

(1) hydrogen-bonding an<lelectrostatic interactionsonll..(2) van der Waals and electrostatic interactions onlv.(3) onlyelectrostaticinteractions.(4) only hydrophobic interaction.

Toll-like receptor 4 is associated rvith responsiveness to LpS. an endotoxrn thatcauses lethal endotoxic shock. The mice deficient in Toll-like recepror .l andBALB/c mice were injected with Escherichia coli.ln addition. some BALB b micewere also injected with the same bacteria alone or with anti-interleukin-10 (iL-lo)antibody. The mice resistant to the lethal effect of the bacteria were:

(1) BALB/b mice receiving the bacteria.(2) BALB/b mice receiving the bacteria and the anti-L-10 antibodv.(3) Mice deficient in Toll-like receptor(4) BALB/c mice receiving the bacteria.

Presence of circular mRNAs for a specific protein in an eukaryotic cell reflects arapid rate of synthesis of that protein.Following mechanisms are suggested:

A) eIF-4G and PABP promote this process through 5'-3'interaction ofmRNA.

B) ribosomes are less active in recognizing circular mRNA.C) PABP and eIF-4A promote this process.D) ribosomes can reinitiate translation without being disassembled.

Which of the following is correct ?

88.

89.

(1) A and D (2) B and D (3) A and C (4) B and C