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Explanation of Nobel Prize in Physiology or Medicine 2017- Molecular Mechanisms Controlling the Circadian Rhythm

The Nobel Prize in Physiology or Medicine 2017 was awarded jointly to Jeffrey C. Hall, Michael Rosbash and Michael W. Young for their discoveries of molecular mechanisms controlling the circadian rhythm.
What is Circadian rhythm? (Circa: around; dies: day in Latin)
To adapt to such changes in light and temperature, organisms have evolved an internal biological clock that anticipates day/night cycles and helps them optimize their physiology and behavior. This internally generated daily rhythm is known as circadian rhythm
Circadian Rhythm was first identified in plants (Mimosa). Later this internal biological clock was found to be present in all organisms including unicellular organisms, fungus, plants and animals.
Explanation of Nobel Prize in Physiology or Medicine 2017- Molecular Mechanisms Controlling the Circadian Rhythm

Period gene First gene identified dealing with Circadian Rhythm
Period gene was the first gene identified dealing with circadian rhythm. (Konopka and Benzer, 1971)
This gene was later cloned and sequenced  by Jeffrey Hall and
Michael Rosbash,, and Michael Young


Period (PER)  gene, mRNA & protein:
cycling of period mRNA and period protein
PER protein shuttles btw nucleus and cytoplasm
Accumulation of PER protein resulted in the reduction of period mRNA expression (negative autoregulatory feedback)
The peak of period mRNA levels
occurred early in the night, several hours before the peak in PER protein abundance.
TIM protein : coded by timeless gene can directly bind to PER protein. The interaction is critical for
PER protein nuclear accumulation and repression of the period gene.


How is Period and TIM gene activated?
 CLOCK (CLK) and CYCLE (CYC) protein interact with each other, and bind to specific elements in the period and timeless genes, thereby positively regulating their transcription.
TIM and PER act as negative regulators of CLK activity, and by this, the circadian feedback loop is closed.

How the transcription (Per mRNA) and translation (Per protein) synthesis is delayed or How Per & Tim protein is degraded?
DOUBLETIME protein a kinase : coded by double time (DBT) gene that phosphorylates PER
and increases its degradation. Light can activate the protein product of the cryptochrome cry gene (CRY) and promote its binding to TIM, leading to its degradation in the When morning arrives, TIM is degraded, leaving PER vulnerable to phosphorylation by DBT and subsequent degradation.


Transcription-Translation Feedback Loop (TTFL).
  The molecular mechanism that regulates Circadian rhythm is called Transcription-Translation Feedback Loop (TTFL).
 Accumulation of PER protein resulted in the reduction of period mRNA expression (negative autoregulatory feedback) called as translational feed back and transcription activators of Per gene, CLOCK (CLK) and CYCLE (CYC) is inhibited by TIM-PER protein accumulation in nucleus
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CSIR UGC NET JRF Life Sciences Biochemistry Questions

CSIR Life Sciences Biochemistry MCQ 
Unit-1 Molecules and their Interaction Relevant to Biology 
1. The ionic strength of a 0.2 M Na2HPO4 solution will be (Dec 2015)
a. 0.2 M
b. 0.4 M
c. 0.6 M
d. 0.8 M

Ionic strength Equation


μ is Ionic strength, Ci is the molar concentration of any ion in solution and zi is its valence.
Ans: c. 0.6 M
2. A cell line deficient in salvage pathway for nucleotide biosynthesis was fed with medium containing 15N labelled amino acids. Purines were then extracted. Treatment with which one of the following amino acids is likely to produce 15N labelled purines? (Dec 2015)
a. Aspartic acid
b. Glycine
c. Glutamine
d. Aspartamine

Ans: a. Aspartic acid or Glycine or Glutamine
3.Enzymes accelerate a reaction by which one of the following strategies? (dec 2015)
a. Decreasing energy required to form the transition state.
b. Increasing kinetic energy of the substrate.
c. Increasing the free energy difference between substrate and the product.
d. Increasing the turn over number of enzymes.
Ans: a. Decreasing energy required to form the transition state.

4.Coupling of the reaction centers of oxidative phosphorylation is achieved by which one of the following? (dec 2015)
a. Making a complex of all four reaction centers.
b. Locating all four complexes in the inner membrane.
c. Ubiquinones and cytochrome C.
d. Pumping of protons.
                                                                   Ans: c. Ubiquinones and cytochrome 
CSIR UGC NET JRF Life Sciences Biochemistry Questions
5.The genome of a bacterium is composed of a single DNA molecule which is 109 bp long. How many moles of genomic DNA is present in the bacterium [Consider Avogadro No=  6 x 10 23] (dec 2015)
a. 1/6 x 10 -23
b. 1/6 x 10 -14
c. 6 x 10 14
d. 6 x 10 23

 Ans: a. 1/6 x 10 -23

6. The solubility of gases in water depends on their interaction with water molecules. Four gases i.e. carbon dioxide, .oxygen, sulphur dioxide and ammonia are dissolved in water. In terms of their solubility which of the following statements is correct? (Jun, 2016)
a . Ammonia > Oxygen > Sulphur dioxide  > Carbon dioxide
b. Oxygen > Carbon dioxide > Sulphur dioxide > Ammonia
c. Sulphur dioxide > Oxygen > Ammonia > Carbon dioxide
d. Ammonia > Sulphur dioxide > Carbon dioxide > Oxygen
 Ans: d. Ammonia > Sulphur dioxide > Carbon dioxide > Oxygen 
7. Penicillin acts as a suicide substrate. Which one of the following steps of catalysis does a suicide inhibitor affect?( Jun, 2016)

a. k1
b. k2
c. k3
d. k4
 Ans: c. k3

8.Which of the following is NOT true for cholesterol metabolism? (Jun, 2016)
a. HMG-CoA reductase is the key regulator of cholesterol biosynthesis.
b. Biosynthesis takes place in the cytoplasm.
c. Reduction reactions use NADH as cofactor.
d. Cholesterol is transported by LDL in plasma.
 Ans:c. Reduction reactions use NADH as cofactor.

9. The -COOH group of cellular amino acids can form which of the following bonds inside the cell? ( Jun, 2016)
1. Ether and ester bonds.
2. Ester and amide bonds.
3. Amide and ether bonds.
4. Amide and carboxylic anhydride bonds

 Ans: 2. Ester and amide bonds. 
10. Choose the most appropriate pH at which the net charge is zero for the molecule from the data shown below : (Dec, 2016)



a. 2.02
b. 2.91
c. 5.98
d. 6.87
 Ans: b. 2.91

11. Excess oxygen consumed after a vigorous exercise is (Dec, 2016)
a. to pump out lactic acid from muscle.
b. to increase the concentration of lactic acid in muscle.
c. to reduce dissolved carbon dioxide in blood.
d. to make ATP for gluconeogenesis.
 Ans: d. to make ATP for gluconeogenesis.

12. The gel to liquid crystalline phase transition temperature in phosphatidyl choline (PC) lipids composed of dioleoyl (DO), dipalmitoyl (DP), disteroyl (DS) and palmitoyl oleoyl (PO) fatty acids in increasing order will be (Dec, 2016)
a. DOPC > DPPC > POPC > DSPC
b. DSPC > DPPC > POPC > DOPC
c. DPPC > DSPC > DOPC > POPC
d. POPC > DPPC > DOPC > DSPC

 Ans: b. DSPC > DPPC > POPC > DOPC
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5 Similarities between Plant cell and Animal cell

We have already discussed in detail the differences between plant cell and animal cells. Now let us see how these two cell types are similar.
5 Similarities between Plant cell and Animal cell

1. Both plant cell and animal cell is Eukaryotic (with true nucleus or genetic material surrounded by a membrane).

2. Both plant cell and animal cell possess cell membrane or plasma membrane which regulates the movement of substances between the cell and the surroundings. (In animal cells, cell membrane is the outermost boundary where as in plant cell, cell membrane is present just inner to cell wall)

3. Both plant cell and animal cell possess a well defined nucleus and cytoplasm, where genetic material DNA is surrounded by a nuclear membrane.

4. Most of the organelles like nucleus, mitochondrion, Golgi apparatus, endoplasmic reticulum etc are present in both plant cell and animal cell.

5. In both plants and animals cells, DNA is double stranded and complexed with histone proteins and  proteins are synthesized by ribosomes in the cytoplasm.

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CSIR UGC NET JRF December 2017 Notification Apply Online Now

CSIR will hold   the Joint CSIR-UGC Test on 17th December, 2017 for determining the eligibility of the Indian National  candidates for the award of Junior Research Fellowships (JRF) 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 (JRF) 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.
CSIR UGC NET JRF December 2017 Notification
Exam DateCSIR-UGC JRF NET :17th December, 2017.

Age Limit & Relaxation:
  • For Junior Research Fellowships  (JRF):Maximum 28 years as on 01-07-2017 (upper age limit may be relaxed up to 5 years in case of candidates belonging to SC/ST/OBC, Physically handicapped/Visually handicapped and female applicants).
  • For LS (NET): No upper age limit.
Important Dates:
CSIR Important dates
Educational Qualification:
  • BS-4 years program/BE/BTech/BPharma/MBBS/Integrated BS-MS/MSc or equivalent degree with at least 55% marks for general and OBC (50% for SC/ST candidates, physically and Visually Handicapped candidates).
  • Candidates enrolled for M.Sc or having completed 10+2+3 years of the above qualifying examination are also eligible to apply in the above subject under the Result Awaited (RA) category on the condition that they complete the qualifying degree with requisite percentage of marks within the validity period of two years to
    avail the fellowship from the effective date of award.
    Such candidates will have to submit the attestation format (Given at the reverse of the application form) duly certified by the Head of the Department/Institute from where the candidate is appearing or has appeared.
  • BSc (Hons) or equivalent degree holders or students enrolled in Integrated MS-PhD program with at least 55% marks for general and OBC candidates; 50% marks for SC/ST candidates, physically and visually handicapped candidates are also eligible to apply.Candidates with bachelor’s degree, whether Science, engineering or any other discipline, will be eligible for fellowship only after getting registered/enrolled for PhD/Integrated PhD program within the validity period of two years.
  • The eligibility for lectureship of NET qualified candidates will be subject to fulfilling the criteria laid down by UGC. PhD degree holders who have passed Master’s degree prior to 19th September 1991, with at least 50% marks are eligible to apply for Lectureship only.
Syllabus
The question paper shall be divided into three parts, (A, B & C) as per syllabus & Scheme of Exam.
Part 'A' shall be common to all subjects including Engineering Sciences. This part shall contain questions pertaining to General Aptitude with emphasis on logical reasoning, graphical analysis, analytical and numerical ability, quantitative comparison, series formation, puzzles etc.
Part 'B' shall contain subject-related conventional Multiple Choice questions (MCQs), generally covering the topics given in the syllabus.
Part 'C' shall contain higher value questions that may test the candidate's knowledge of scientific concepts and/or application of the scientific concepts. The questions shall be of analytical nature where a candidate is expected to apply the scientific knowledge to arrive at the solution to the given scientific problem.
Negative marking for wrong answers, wherever required, shall be applicable as per subject wise scheme of Exam.
Examination fee:
Examination Centres: Bangalore, Bhavnagar, Bhopal, Bhubaneshwar, Chandigarh, Chennai, Cochin, Delhi, Guntur, Guwahati, Hyderabad, Imphal, Jammu, Jamshedpur, Karaikudi, Kolkata, Lucknow, Nagpur, Pilani, Pune, Raipur, Roorkee, Srinagar, Thiruvananthapuram, Udaipur and Varanasi.


Exam Result & Validity period of fellowship: The final result of this Single MCQ test may be declared sometime in the month of March/April, 2018 and fellowship to successful candidates will be effective from 1 st July 2018 with the validity period of 2 years for joining the fellowship under CSIR Scheme.
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Worksheet on Nitrogen cycle PDF

Nitrogen cycle is the cyclic movement of nitrogen (N2) between atmosphere, organisms and soil. Nitrogen is an essential element required for the synthesis of bio-molecules like proteins, DNA, vitamins, chlorophyll, alkaloids etc.
 Learn more: Nitrogen Cycle
 Learn more: Nitrogen Cycle (Video) 
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HSST Botany Junior 2017 Kerala PSC Question Paper with Answers

HSST BOTANY Junior Question Paper with Answers
Exam Conducted by Kerala PSC
Exam Date: 05/06/2017
Maximum Marks : 100
Duration: 1 Hour 15 Minutes
Medium of Questions: English
 Mode of Examination :Online
HSST Botany Question Paper with Answers
HIGHER SECONDARY SCHOOL TEACHER (JUNIOR) BOTANY (SR FOR ST ONLY) 
Category code: 355/2016 
HIGHER SECONDARY EDUCATION DEPARTMENT
Syllabus: An Objective Type Test based on the qualification prescribed for the post.
Main Topics:
  • Part I : Core Subject
  • Part II : Research Methodology and Teaching Aptitude 
  • Part III : Salient features of Indian Constitution , Social Welfare Legislations and Programmes 
  • Part IV : General Knowledge, Current Affairs & Renaissance in Kerala


Learn more:  

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SAT Biology E/M Practice Test on Classical Genetics- Quiz on Mendelian Genetics

Questions from Meiosis, Mendelian Genetics-ABO blood group, Pleiotropism

1. Colorblindness in humans is
in equal proportion in both sexes
caused in females by a heterozygous genotype
inherited by males from their male parent
caused in males by a homozygous genotype
caused by a recessive allele
2. A ratio of 3:1 in the phenotype of an organism occurs when
crossing over has occurred in Anaphase II
the alleles are incompletely dominant
only recessive traits are involved
alleles segregate during meiosis
only dominant traits are involved
3.Which of the following is not indicated by Mendel’s experiments?
independent assortment
dominant
recessive
segregation
incomplete dominance
4. In a heterozygous monohybrid cross, the dominant trait can be expressed in the phenotype of the F1 --- of the time.
0 percent
25 percent
50 percent
75 percent
100 percent
5. Homologous chromosomes line up in pairs in
metaphase I
metaphase of mitosis
metaphase II
interphase
prophase of mitosis
6. Trisomy 21 in humans is the result of
polygenic inheritance
pleiotropy
x-inactivation
nondisjunction
epistasis
7. Which of the following blood types are possible if the parents are A and O blood types?
A, B, and O
A and O
O only
B and O
AB only
8. Which of the following would indicate a test cross, where T represents the dominant, tall, trait and t represents the recessive, short, trait?
Tall × Tt
Tall × TT
TT × Tt
Tall × tt
short × tt
9. Which of the following would be the result of a true dihybrid cross?
AABB
AaBb
aabb
AABb
aaBb
10. ---refers to one gene affecting many traits
Linkage
Polygenesis
Nondisjunction
Epistasis
Pleiotropy
Score =
Correct answers:

Practice more Quiz from GeneticsGenetics TerminologyMeiosis
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