CSIR UGC NET Examination December 2015: Life Sciences: Notification for Online Application Submission

Free Net Life Science Study Materials

CSIR will hold the Joint CSIR-UGC Test Tentatively on Sunday 20th December, 2015 for determining the eligibility of the Indian National candidates for the award of Junior Research Fellowships (JRF) NET and for determining eligibility for appointment of Lecturers (NET) in certain subject areas falling under the faculty of Science & Technology. The award of Junior Research Fellowship (NET) to the successful eligible candidates will depend on their finding admission/placement in a university/ national laboratory/ institution of higher learning and research, as applicable. A candidate may apply either for ‘JRF’ OR for ‘Lectureship (LS) only’ He/she may indicate his/her preference in the Online Application, as the case may be. CSIR may consider candidates for ‘JRF’, or ‘Lectureship only” depending on number of fellowships available & performance in the test subject to the condition that they fulfill the laid down eligibility criterion. If a candidate is found to be over-age for JRF (NET) he/she will automatically be considered for Lectureship (NET) only. Candidates with M.Sc. qualification OR under M.Sc. Result Awaited (RA) category shall be eligible for LS subject to fulfilling the eligibility criteria as laid down by the UGC. Two separate merit lists, one comprising the candidates qualifying for the award of Junior Research Fellowship (JRF – NET) and the second, of those candidates qualifying the Eligibility Test for Lectureship (NET), will be made on the basis of their performance in the above Test. Candidates qualifying for JRF (NET), will also be eligible for Lectureship (NET) subject to fulfilling the eligibility criteria laid down by UGC. The candidates qualifying for Lectureship will be eligible for recruitment as Lecturers as well as for JRF-ship in a Scheme/Project, if otherwise suitable.

Read more

Model Question Paper for ICMR JRF Examination (Part 3)

mcq biology

*(1). Pick out the correct taxonomic hierarchy in zoological systematics. a.       Phylum – Class – Order – Family – Genus b.      Class – Phylum – Order – Family – Genus c.       Phylum – Class – Family – Order – Genus d.      Phylum – Class – Order – Genus – Species (2). In molecular systematics of … Read more

ICMR JRF Entrance Examination Model Questions with Answers (Part-2)

mcq biology

(1). Which of the following soil will be most porous: a.       Sandy loam b.      Clay loam c.       Clay (2). Photosynthetic formation of glucose can be sited as an example for: a.       Oxidation b.      Reduction c.       Condensation d.      Fixation (3). Members of which family has been extensively employed as phyo-remediation agents for the removal of contaminates … Read more

Kerala Public Service Commission (PSC) Botany Lecturer / Assistant Professor Exam 2009 Question Paper Part 4 (MCQ 013)

mcq biology

Part – 4 (Questions 76 – 100) (MCQ 013)

(76).  A man who is affected with phenylketonuria marries a woman who is heterozygous at that locus. What is the probability that their first child will have phenylketonuria?

a.       1/8
b.      ¼
c.       ½
d.      ¾

(77).  Which of the following statements about heritability are true?

a.       Is a measure of level of gene linkage
b.      Is a measure of inbreeding
c.       Is a measure of proportion of repeated DNA in an organism
d.      Is a measure of the proportion of variation that is due to genetic causes

(78).  Heat shock proteins were originally described as protein produced in response to heat stress. Some are known to act as:

a.       Molecular chaperones that regulate protein folding
b.      Protein-tyrosine kinase protein
c.       Protease that degrade ubiquitone-tagged protein
d.      GTPase activating protein

(79).  The synthesis of glucose from lactate, glycerol, or amino acid is called:

a.       Glycogenolysis
b.      Glycolysis
c.       Lipolysis
d.      Gluconeogenesis

Read more

Botany Lecturer/Assistant Professor Selection by Kerala PSC 2009 Part 3: Cat. NO. 464/2007 (MCQ 012)

mcq biology

Part – 3 (Questions 51 – 75) (MCQ 012)

(51).  Buffers has the ability to

a.       Change in pH
b.      Resist change in pH
c.       No resist to change in pH
d.      All of the above

(52).  Why is the filamentous body of the slime and water molds considered an example of convergent evolution with the hyphae of true fungi?

a.       Fungi are closely related to the slime and water molds
b.      Body shape reflects phylogenetic relationships among organisms
c.       Filamentous shape is an adaptation for a nutritional mode as a decomposer with absorptive nutrition
d.      Filamentous body shape is evolutionary primitive for all eukaryotes

(53).  The amount of light absorbed by a material is proportional to the concentration of the absorbing solution is referred as:

a.       Beer’s law
b.      Boger-Lambert law
c.       Poiseuille’s law
d.      All of the above

(54).  Lateral roots in higher plants arise from:

a.       Cortex
b.      Pericycle
c.       Epidermis
d.      Endodermis

Read more

error: Content is protected !!