CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship
PREVIOUS SOLVED QUESTIONS – 2011 DECEMBER CSIR – NET EXAM
1.The most important property of any microscope is its power of resolution, which is numerically equivalent to D, the minimum distance between two distinguishable objects. D depends on three parameters namely, the angular aperture, a, the refractive index, N, and wavelength, ?, of the incident light. Below are given few possible options to increase the resolution of the microscope.
A. Decrease the value of ? or increase either N or a to improve the resolution.
B. Moving the objective lens closer to the specimen will decrease sin a and improve the resolution.
C. Using a medium with high refraction index between the specimen and the objective lens to improve the resolution.
D. Increase the wavelength of the incident light to improve the resolution.
Which of the following combination of above statements is correct ?
1. A and C 2. B and C 3.A and D 4.C and D
Expln:- Resolving power of a microscope depends upon the numerical aperture of the objective lens system and its wavelength. Resolution of a microscope increases as the wavelength decreases, so ultraviolet light allows one to detect objects not seen with visible light. Similarly, the wavelength of light is much larger than the wavelength of electrons….
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2.Many cancers carry mutant p53 genes, while some cancers have normal p53 genes. P53 activates p21 (Wafl) which inhibits G1/S-Cdks and phosphorylation of the retinoblastoma protein (Rb). Cancers with normal p53 genes could
1. express non-phosphorylatable form of Rb 2. express high levels of p53-deubiquitinases 3.express inactive forms of G1/ S-cdks 4.express inactive forms of G1/S cyclins
Expln:- When damage is sensed, the activity of the p53 protein aids in the decision between repair and the induction of cell death (apoptosis).
As a transcription factor, p53 stimulates the transcription of a group of target genes. Among them, p21 is one of the most important. The product of the p21 gene is a negative regulator of cyclin-dependent kinases, enzymes that are critical in the progression of the cell cycle and ultimately cell division. By stimulating the transcription of the p21 gene, p53 prevents ….
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3. Pre-mRNAs are rapidly bound by snRNPs which carry out dual steps of RNA splicing, that removes the intron and joins the upstream and downstream exons.
The following statements describe some facts related to this event
A. Almost all introns begin with GU and end with AG sequences and hence all the GU or AG sequences are spliced out of RNA.
B. U2 RNA recognizes important sequences at the 3’ acceptor end of the intron.
C. The spliceosome uses ATP to carry out accurate removal of intron.
D. An unusual linkage with 2’ OH group of guanosine within the intron form a ‘Lariat’ structure.
Which of the following combinations is correct ?
1. A and B 2. B and C
3.C and D 4.D and A
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4. Following are some of the statements regarding the effect of CO2 concentration on photosyn thesis in plants
A. With elevated CO2 levels, C3 plants are much more responsive than C4 plants under well watered conditions
B. In C3 plants, increasing intracellular CO2 partial pressure can stimulate photosynthesis only over a narrow range.
C.In C4 plants, CO2 compensation point is nearly zero
Which one of the following combination of above statments is correct ?
1. A and B 2. B and C 3.A and C 4.Only C
Expln:- The C4 plants show saturation at about 360 µ IL-1 while C3 responds to increased CO2 concentration and saturation is seen only beyond 450 µ IL-1. Thus current availability of CO2 levels is limiting to the C3 plants.
In C4 plants, the CO2 compensation point is zero or nearly zero, reflecting their very low levels of photorespiration. In C4 plants, photosynthetic rates saturate internal concentration values of about 15 pascals, reflecting the effective CO2 concentrating mechanisms operating in these plants. In C3…. More Details Refer Simple Instant Notes
5. Following are some statements regarding plant growth hormones.
A. Ethylene regulates abscission
B. Gibberlins do not play any role in flowering
C.Auxin and cytokinin promote cell division
D.Over expression of cytokinin oxidase would promote root growth
E.ABA inhibits root growth and promotes shoot growth at low water potential
F.ABA promotes leaf senescence independent of ethylene
Which one of the following combination of above statements is correct ?
1. A, C and F 2. B, C and D 3.D, E and F 4.B, D and E
Expln:- ABA promotes leaf senescence independently of ethylene. Abscisic acid was originally isolated as an abscission-causing factor. However, it has since become evident that ABA stimulates abscission of ogans in only a few species and that the primary hormone causing abscission is ethylene. On the other hand, ABA is…. More Details Refer Simple Instant Notes
6. A monkey undergoes cerebellectomy. After the post-operative recovery, the monkey was given a task to press a bar. The possible observations are:
A. Its hand would overshoot the target while reaching the bar
B. It would be unable to move forelimbs
C.It would show intention tremor while trying to press the bar
D.It would press the bar with mouth instead of hand
Which one of the following is correct ?
1. A and C 2. B only 3.D only 4.B and D
Expln:- Cerebellum helps to provide smooth, coordinated body movement. The cerebellum processes input from other areas of the brain, spinal cord and sensory receptors to provide precise timing for coordinated, smooth movements of the skeletal muscular system. A stroke…. More Details Refer Simple Instant Notes
7. Maturation-promoting factor (MPF) controls the initiation of mitosis in eukaryotic cells. MPF kinase activity requires cyclin B. Cyclin B is required for chromosome condensation and breakdown of the nuclear envelope into vesicles. Cyclin B degradation is followed by chromosome decondensation, nuclear envelope reformation and exit from mitosis. This requires ubiquitination of a cyclin destruction box motif in cyclin B. RNase-treated Xenopus egg extracts and sperm chromatin were mixed. MPF activity increased with chromosome condensation and nuclear envelope breakdown. However, this was not followed by chromosome decondensation and nuclear envelope reformation because
1. RNase condamination persisted in the system
2. cyclin B was missing from the system
3.ubiquitin ligase had been overexpressed
4.cyclin B lacking the cyclin destruction box had been overexpressed
Expln:- Maturation-promoting factor(MPF) is also known as as mitosis-promoting factor. It is a heterodimeric protein, composed of cyclin B and cyclin-dependent kinase (Cdki). MPF promotes the entrance into mitosis from the G2 phase by phosphorylating multiple proteins needed during mitosis. MPF is activated at the end of G2 by a phosphatase known as Cdc 25, which removes an inhibitory phosphate group added earlier.
The G2 checkpoint triggers the start of the M phase (mitosis). In order for this checkpoint to be passed, the cell has to check a number of factors to ensure the cell is ready for mitosis. If this checkpoint is ……….. More Details Refer Simple Instant Notes
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