Botany lecture notes

HSST Botany Question Paper 2012 by Kerala PSC with Answer Key and Explanations Part 4

Acanthaceae, Strobilanths

Strobilanths kunthiana (source wikipedia)

Kerala PSC HSST Botany 2012 Examination
Higher Secondary School Teacher: Junior & Senior
Question Paper Code 40/2012 (Cat. NO. 449/2010)

Original question paper of Kerala PSC HSST Botany Junior / Senior (Higher Secondary School Teacher Botany), Category No. 449/2010) examination conducted by Kerala PSC (Public Service Commission) on 13/04/2002 (Q. Code 40/2012) for the appointment of HSST Botany in Government Higher Secondary Schools of Kerala under the Directorate of Higher Secondary Education, Trivandrum, Govt. of Kerala. Questions are in MCQ (Multiple Choice Questions) format.

Part – 4 (Questions 76 – 100)

76.  Fossil fuels are:

a.       Renewable resources
b.      Non-renewable resource
c.       Inexhaustible resources
d.      Non-renewable and exhaustible resource

Ans. (d). Non-renewable and exhaustible resource

77.  Parthenocarpy is induced by the hormone:

a.      BAP
b.      GA
c.       IAA
d.      ABA

Question removed due to confusions in the options

Parthenocarpy: formation of fruit without fertilization. Parthenocarpic fruits are devoid of seeds, Eg. Banana

Gibberellin can induce the formation of seedless fruits but not through parthenocarpy but through stenospermocarpy

Stenospermocarpy: a biological phenomenon which induces seedless fruit formation in some fruits especially grapes. In stenospermocarpic fruits, normal pollination and fertilization is essential to ensure fruit formation. But during the course of fruit development spontaneous embryo abortion takes places and it leads to near seedless condition.  

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Biotechnology Eligibility Test Preparation

DBT BET JRF Exam 2012-13 Previous Year Question Paper With Answer Key and Explanations Part 2

What are microsatellites?

DNA samples from specimens of Littorina plena amplified using PCR with primers targeting a variable simple sequence repeat (SSR, a.k.a. microsatellite) locus, run on a 5% polyacrylamide gel, visualized by silver staining (source wikipedia)

Biotechnology Eligibility Test (BET) for DBT JRF Award (2012-13)
Previous Year Question Papers with Answer Key, Explanations & References
Government of India, Ministry of Science & Technology
Department of Biotechnology, New Delhi

PART: A Set 2 (Questions 26 – 50)

26.  Which one of the following is the natural host for pseudo-rabies virus?

a.      Dog
b.      Man
c.       Swine
d.      Horse

Ans. (c). Swine

Pseudorabies (Aujeszky’s disease) is a viral disease of swine caused by Suid herpesvirus 1 (SuHV1) belongs to the family Herpesviridae. Other species such as sheep, cattle, and human are also susceptible to SuHV1. This disease is nothing to do with rabies virus. The notorious Rabies virus belongs to the family Rhabdoviridae which cause rabies in human and animals.

27.  Which one of the following is the causative agent of fowl cholera?

a.       V. cholera
b.       P. multocida
c.       E. coli
d.       S. pullorum

Ans. (b). P. multicida

Fowl cholera or avian cholera is caused by Pasteurella multocida. Fowl cholera is considered as Zoonoses

Zoonoses: an infectious disease of animals that can be transferred to human, Example: Rabies

S. pullorum is a poultry adapted species of Salmonella

28.  The wavelengths of light that penetrate the least into the ocean are

a.       Red and violet
b.      Red and yellow
c.       Blue and brown
d.      Green and blue

Ans. (b). Red and Yellow

Yellow, orange and red waves have the least energy in the visible spectrum, thus they cannot penetrated deep into the sea. This is the reason why most of the deep sea animals (fishes) are red in colour. In the deep sea, due to absence of enough light the red colour appears as black so that these animals can be escaped from their predators.

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Botany lecture notes

MG University PhD Botany Course Work Exam: Course II – Biological Techniques Dec 2015 Question Paper

Botany PhD course Work

Ph.D Programme in Botany
Botany Ph.D Course Work Examination December 2015
Paper/Course – II – Biological Techniques
Time: Three Hours                                                                     Maximum: 80 Marks

Part A
Answer any twelve of the following
Each question carries 5 marks

1.    Describe the physiological and biochemical changes during fruit ripening

2.    Give a diagrammatic representation of photosynthetic electron transport chain.

3.     Briefly describe: (a). Isoelectric focusing; and (b). Henderson-Hasselbalch equation.

4.     What is the working principle of GCMS? Specify the applications.

5.     Describe the technique used for the separation and identification of pigments.

6.     Explain the principles of centrifugation. Describe briefly the application of ultracentrifugation.

7.     Explain the role of chemicals used in the isolation of nucleic acids

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Notification of CSIR JRF NET, Life Science Examination

JNCASR Summer Research Fellowship Programme (SRFP) 2016 Notification

JNCASR SRFP 2016 Notification

Jawaharlal Nehru Centre for Advanced Scientific Research (JNCASR), Bangalore offers summer fellowships for two months to bright undergraduate and MSc students (renewable for a second year for selected students). This programme has proved to be popular and competitive; each year, about 5000 students from all over India apply for the 120 fellowships awarded. Fifty fellowships are supported by the Department of Science & Technology, Government of India, fifteen by the Rajiv Gandhi Institute for Contemporary Studies, New Delhi, and the rest by the Centre. Students are placed with research groups at the Centre or with scientists elsewhere in India. They are paid travel expenses and a monthly stipend of Rs. 6000/- Selected students get the opportunity to participate in cutting-edge research, and several summer projects have led to publications in leading journals. Most of the summer students of past years have gone on to pursue graduate studies and a research career, at the JNCASR or at another leading university

The summer research projects will be reviewed by a committee and the best will be chosen for the award of Rajiv Gandhi Science Talent Research Fellows.


Life Sciences: Infectious and non-infectious diseases, bioinformatics, developmental biology, biochemistry, molecular biology, immunology, genetics, life-history evolution, circadian biology, animal behaviour, behavioural ecology, evolutionary genetics, experimental evolution, population dynamics, phylogeography, neurophysiology, behavioural neurobiology and behavioural genetics.

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Biotechnology Eligibility Test Preparation

DBT BET JRF Exam: Previous Year Question Paper With Answer Key and Explanations: BET 2012 – 2013 Part 1

What is immunoglobulin domain

(Image Source Wikipedia)

Biotechnology Eligibility Test (BET) for DBT JRF Award (2012-13)
Previous Year Question Papers with Answer Key, Explanations & References
Government of India, Ministry of Science & Technology
Department of Biotechnology, New Delhi

PART: A Set 1 (Questions 001 – 025)

1. The immunoglobulin fold is made up of:

a.       Seven alpha helical segments
b.      A beta barrel
c.       A sandwich of two parallel beta sheets
d.      A sandwich of two antiparallel beta sheets

Ans. (d). A sandwich of two antiparallel beta sheets

Immunoglobulin fold consists of a pair of β-sheets each built of antiparallel β-strands connected by a disulfide bridge. This β strands surrounds a central hydrophobic core. Immunoglobulin fold is one of the most predominant domains encoded by the human genome.

For more details refer: Biochemistry by Stryer, Ed.6, Chapter: 33, Immune systems, Page 951

2. RNA is analyzed for the location of hairpin folds. Which of the following sequence could form a mini-hairpin?

a.       AGGUUUCCU
c.       AGGUUUGGA


This RNA molecule can loop around itself and can form double stranded structure due to the sequence complementarity. First A is complementary to last U, second and third nucleotide (UU) is complementary to CC. The middle UUU forms the junction of loop and they do not form the complementary hydrogen bonds.

3. Increasing the concentration of which of the following would most effectively anatomized the inhibition of protein synthesis by puromycin?

a.       ATP
b.      eIF2, GTP
c.       Aminoacyl-tRNAs
d.      Peptidyl-tRNAs

Ans. (c). Aminoacyl-tRNA

Puromycin is an antibiotic obtained from a bacterium Streptomyces alboniger, which inhibit protein synthesis by specifically inhibiting the translation process. Puromycin inhibits protein translation by premature chain termination process. A part of the puromycin structurally mimics the 3’ end of Aminoacyl-tRNA molecule and hence they can enter to the ‘A’ site of ribosome and form a nascent peptide-puromycin adduct. This adducts formation results in the release of peptide from the ribosome. Puromycin compete with aminoacyl-tRNA for the A site of ribosome. Thus the inhibitory effect of puromycin can be theoretically overcome by increasing the concentration of the competitor, i.e., aminoacyl-tRNA.

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