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The correct choice is (c) 3^rd

Best EXPLANATION: In the 1^st QUADRANT and 4^th quadrant one of the two LABELS are ACTIVE. In the 2^nd both are active and in the 3^rd NONE of the labels are present on the molecule.

Right choice is (B) p27

The explanation is: p27 can bind to the CDK-cyclin complex in a way that the catalytic site is no LONGER able to interact with the substrate. This leads to inactivation; THUS, it’s a CDK INHIBITORY protein or CKI.

The correct option is (b) S

To explain: The S phase is MARKED by DNA REPLICATION. Once this phase is done the cell has to undergo DIVISION else its DNA content will remain double that of normal cell.

Correct answer is (c) By electrostatic force

For explanation: As the cells exit the nozzle of the FLOW cytometer a small amount of charge is added to the droplet. Then it passes between the two PLACES which INDUCE the droplet to be properly SEPARATED depending on its label.

Right answer is (a) DNA

Best EXPLANATION: A constant amount of Cln3 must be SENSED for the cell to regulate its SIZE. This is done via its receptors on DNA. As the amount of DNA is constant than when Cln3 exceeds a certain amount it will be FREE. Then it can TRIGGER cell cycle.

Correct option is (c) Pol kappa

To explain I would say: On sensing the DNA damage p53 level rises which stimulate GADD45 and pol K which tries to REPAIR the DNA damage. Only when the damage is too much to repair apoptosis is TRIGGERED.

Correct option is (b) M, M

Easiest EXPLANATION: M Cdk INACTIVATED SIC1 and Hct1 ACTIVITY. It is however inhibited by the same. With DECREASE in M Cdk activity in the early G1 stage sic1 binds and inactivated it.

The correct answer is (d) Negative feedback regulation

The explanation is: If the dependence was on POSITIVE feedback then it would be hard to DIFFERENTIATE between DIFFERENT levels of partially and fully positive CASES. But in case of negative feedback, even one negative signal can be easily detected and its removal will trigger the process.

Right answer is (b) False

To explain: P53 binds to Mdm2 in NORMAL cell which destroys the p53 leading to its REDUCTION in concentration. On the other HAND Mad2 bind to kinetochores that are not ATTACHED to microtubules.

Right choice is (d) Caspases

To EXPLAIN: Akt is an inhibitory src kinase that is activated DOWNSTREAM of anti- apoptotic pathway. This phosphorylates bad and FOXO generating 14-3-3 protein BINDING SITE, and also phosphorylates Gsk-3. But it doesn’t phosphorylate caspases.

Correct option is (c) M

Easiest explanation: The M-cdk on activation LEADS to the M phase. It BREAKS down the NUCLEAR MEMBRANE by specific phosphorylation ACTIVITY allowing the licensing factor to re-enter the nucleus.

Correct CHOICE is (a) Lung cancer

To elaborate: In lung cancer and in colon cancer we see ‘DECOY’ MOLECULE that binds to the Fas ligand. Thus, the ligand can no longer induce APOPTOSIS by interacting with the Fas receptor.

Right option is (B) BFL 1

For explanation I would say: Bcl-Xs is a Bcl-2 family protein which is pro-apototic, also BIM and NOXA are proapoptotic factors. Bfl 1 is an ANTI apoptotic member of Bcl- 2 family.

The correct ANSWER is (a) Hct1

The explanation is: Hct1 that is TRIGGERED by cdc20 at the end of the anaphase REMAINS high in concentration EVEN in the early G1 phase preventing the CDk activity.

The CORRECT CHOICE is (d) It is phosphorylated, ubiquitin is ADDED and destroyed

Explanation: The continued presence of CDC6 and MCM could LEAD to re-initiation of replication at the ORC. Thus, to prevent this, the CDC6 is phosphorylated and marked for destruction.

Correct choice is (B) False

Easiest explanation: It is ENSURED by dilution that in a flow cytometer the particles pass in a SINGLE file. Then they are individually identified and sorted.

Right OPTION is (a) True

The explanation is: MUTANT p53 doesn’t bind to the PROMOTER and HENCE can’t induce apoptosis by transcribing. Pikan083 can bind to and activate mutant p53 in proliferative CELLS leading to their apoptosis.

Correct choice is (d) p53 activating

The explanation is: When tenovin is administered, this SMALL molecule activates p53 which in TURN leads to apoptosis. This is used as a DRUG to TREAT cancer.

Correct CHOICE is (C) Producing ANTI-apoptotic PROTEIN

Easy explanation: EBV produces a protein similar to Bcl-2, i.e. an anti apoptotic protein. It also increases the production of Bcl-2 protein itself. Thus,, b disrupting the pro-apoptotic / anti-apoptotic BALANCE apoptosis is induced.

The correct option is (b) IAP

The explanation is: CASPASES are the EFFECTORS of apoptosis. IAP binds to the caspases thus preventing apoptosis from taking PLACE. Smac and Diablo BIND to IAP thus inhibiting it so they are also pro-apoptotic.

Right option is (a) True

The explanation is: The M-cyclin Cdk when bound to sic1 in turn also CAUSES SLIGHT transcription of M cyclin gene. Increasing CONCENTRATION of M cyclin further INCREASES its activity. This works as a positive FEEDBACK loop which ultimately leads to M cdk activation.

Correct option is (b) INACTIVATED CDC25

The explanation: CDC25 is a phosphatase that removes the inhibitory phosphate on the S Cycin-CDk complex that leads to the activation. WEE1 kinase ADDS this phosphate leading to inactivation. DNA damage requires sop of cell CYCLE thus CDC2 which helps in progression is inactivated.

Right option is (d) Mitochondria

For EXPLANATION I would say: Mitochondria in ANIMALS is an important MEMBER regulating the intrinsic PATHWAY of apoptosis, it releases CYTOCHROME which triggers the process.

Right option is (d) CASPASE 12

Best explanation: In extreme hypoxic CONDITIONS, the TRAF2 in the ER RELEASES the associated PROCASPASE 12; activated caspase 12 and leads to apoptosis.

The correct CHOICE is (b) Bid

To explain: Bid, when acted upon by CASPASE 8 will give the truncated version of Bid- tBid. tBId can move to the MITOCHONDRIA and release CYTOCHROME which will turn on the intrinsic pathway of apoptosis.

Correct choice is (c) E3

To explain: E3 ubiquitin ligase is finally responsible for adding the POLY ubiquitin tail to the SUBSTRATE to MARK it for destruction. Thus regulating the E3 level will help to regulate the concentration of the substrate.

The correct OPTION is (a) Rhodamine

Easiest explanation: The Rhodamine gives red fluorescence while the FLUORESCEIN gives a green fluorescence. CARMINE is a red dye that is used in food but it’s not used for immunological methods. DAPI stains the nucleus BLUE.

The correct OPTION is (b) Bid

Easy explanation: Among the following OPTIONS bid, bad and Smac are pro-apoptotic and Akt is anti apoptotic. But the only Bid among the options directly HELPS in the RELEASE of cytochrme C from the mitochondria as tBid.

Right choice is (d) All can be true

Explanation: In flow CYTOMETRY we can have all the above COMBINATIONS. The cells can EVEN be SORTED according to the ANTIGENS present on them.

Right answer is (a) True

The best I can EXPLAIN: P21 is a CDI that binds to and inactivates the G1-S CYCLIN THUS leading to the stop of cell cycle PROGRESSION from G1 S phase.

Right choice is (c) Both G1 and G1-S

The best I can explain: The G1 Cdk is RESPONSIBLE for the re-initiation of the cell cycle, this is done by the process that G1 Cdk is resistant to Hct1, it TRANSCRIBES G1-S. Thus EVEN in G1 phase when the Hct1 concentration is high it can be ACTIVATED. G1-S CYCLIN phosphorylates and inactivates Hct1.

The correct option is (d) Rad51

To explain: Rad51 is the human HOMOLOGUE of REC that helps in DSB repair. RHG is a GROUP of drosophila pro- apoptotic PROTEINS which consists of Hid, Reaper, Sickel and grim.

The CORRECT choice is (c) DR4

To explain: DR4 is another receptor on the cell surface like Fas which consists of DEATH domains. This accepts TRAIL or TNF receptor related APOPTOSIS INDUCING ligand.

The correct option is (c) CELL membrane exposes phosphatidyl serine on the outer SIDE

For explanation: Phosphatidyl serine is MAINLY present on the inner side of the cell membrane, this is seen in the outer side of the apoptotic cell membrane that encourages PHAGOCYTOSIS.

Right choice is (d) Mad 2

For explanation: Mad 2 bind to the kinetochores that are not BOUND to the SPINDLE fibers. This is a negative signal. Removal of this negative feedback leads to anaphase.

Correct choice is (C) S

The explanation: S CDK phosphorylates the residues on MCM and on ORC THUS triggering ORC firing leading to DNA REPLICATION in the cell. ORC is FIRED once and only once during cell cycle.

Right OPTION is (a) True

The EXPLANATION: In the normal CELL p53 can interact with Mdm2 which acts as specific E3 ubiquitin ligase for p53 and ubiquitination leads to destruction of p53. It thus has a high TURNOVER number and doesn’t accumulate in healthy cells.

The correct answer is (c) Cytochrome

For EXPLANATION I would say: The DISC COMPLEX (death-inducing signalling complex ) consists of the fas ligand and receptor that FORMS the Death domain. This in turn aggregates and activates procaspase 8 to caspase 8. Cytochrome is not directly a part of DISC complex.

The correct choice is (B) Cystine proteases

The BEST explanation: Caspases are cystine proteases which cleave the peptide USING an ACTIVATED cystine amino ACID in the catalytic site.

The CORRECT answer is (C) p21

For explanation I would say: The increased AMOUNT of P53 as a result of DNA damage LEADS to increased transcription of WIP1/ Cip1^p21 gene which leads to increase in p21 levels.

The correct CHOICE is (b) Activates, inactivates

Best explanation: Both Cdc20 and Hct1 are APC ligases which HELP in escape from mitosis by destruction of M CDK, but cdc20 is activated by M CYCLIN – CDK by while Hct1 is INACTIVATED by the same through phsphorylation.

Correct choice is (a) True

To EXPLAIN I would say: It has been reported that caspase-3 mRNA levels are appreciably downregulated in breast, ovarian, and cervical tumours. LOW levels of caspases or IMPAIRMENT in caspase function MAY LEAD to decreased apoptosis.