Southern Blotting Principle, Procedure and Application

Southern blotting

• The technique was developed by E.M. Southern in 1975.
• The Southern blot is used to detect the presence of a particular DNA fragment in a sample.
• The DNA detected can be a single gene, or it can be part of a larger piece of DNA such as a viral genome.

Principle
The key to this method is hybridization.
Hybridization: It is the process of forming a double-stranded DNA molecule between a single-stranded DNA probe and a single-stranded target DNA.
There are 2 important features of hybridization:

• The reactions are specific-the probes will only bind to targets with a complementary sequence.
• The probe can find one molecule of target in a mixture of millions of related but non-complementary molecules.

Summary of procedure (Detailed Procedure)
1. Extract and purify DNA from cells
2. DNA is restricted with enzymes
3. Separated by electrophoresis
4. Denature DNA
5. Transfer to nitrocellulose paper
6. Add labeled probe for hybridization to take place
7. Wash off unbound probe
8. Autoradiograph

Application:

• To identify specific DNA in a DNA sample
• To Isolate desired DNA for construction of rDNA
• Identify mutations, deletions, and gene rearrangements
• Used in prognosis of cancer and in prenatal diagnosis of genetic diseases
• In RFLP
• Used in phylogenetic analysis
• Diagnosis of HIV-1 and infectious disease
• In DNA fingerprinting:
  • Paternity and Maternity Testing
  • Criminal Identification and Forensics
  • Personal Identification

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Southern Blotting Procedure Steps

Southern Blotting Principle and Application

Detailed Procedure

Step 1: DNA purification
Isolate the DNA in question from the rest of the cellular material in the nucleus.
Incubate specimen with detergent to promote cell lysis.
Cell Lysis frees cellular proteins and DNA.

• Proteins are enzymatically degraded by incubation with proteinase.
• DNA is purified from solution by alcohol precipitation.
• Visible DNA fibers are removed and suspended in buffer.

Step 2 : Restriction digestion
Cut the DNA into different sized fragments using restriction endonucleases (RE).

Step 3 : Gel Electrophoresis
Nucleic acids have a net negative charge and will move from the left to the right. The larger molecules are held up while the smaller ones move faster. This results in a separation by size.
Gels are Agarose or polyacrylamide with microscopic pores
Gel is soaked in a buffer which controls the size of the pores
Standards should also be run

Gels can be stained with ethidium bromide (EtBr). This causes DNA to fluoresce under UV light which permits photography of the gel.
This will be help us to know the exact migration of DNA standards and the quality of the RE digestion of the test DNA.

Step 4 & 5 : Denaturation & blotting
DNA is then denatured with an alkaline solution such as NAOH. This causes the double stranded to become single-stranded.
The process of transferring the DNA from the gel to a membrane is called as blotting. The blot is usually done on a sheet of nitrocellulose paper or nylon. DNA is then neutralized with NaCl to prevent re-hybridization before adding the probe. Transferred by either electroblotting or capillary blotting.
The blot is made permanent either by:

• Drying at ~80°C
• Exposing to UV irradiation

Step 6 : Hybridization
The labelled probe is added to the membrane in buffer and incubated for several hours to allow the probe molecules to find their targets.
Preparing the probe
Small piece of labelled DNA used to find another piece of DNA usually prepared by making a radioactive copy of a DNA fragment.

Step 7 & 8 :  Wash and autoradiography
Wash excess probe that are bound non-specifically to the membrane
Blot is incubated with wash buffers containing NaCl and detergent to wash away excess probe and reduce background.
Detection: Radioactive probes enable autoradiographic detection.
If the probe is radioactive, the particles it emits will expose X-ray film.
By pressing the filter and film, the film will become exposed wherever probe is bound to the filter.
After development, there will be dark spots on the film wherever the probe bound.

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Difference between r and K Selection

The r and K classification was originally proposed by the biologists Robert McArthur and E. O. Wilson (1967).

The r selected species live in populations that are highly variable. The fittest individuals in these environments have many offspring and reproduce early.
Examples: Oysters, Insects- Mosquitos, Daphnia, Goldenrod

K selected species live in populations that are at or near equilibrium conditions for long periods of time. Competitive for limited resources is very important in these environments.
Example: Lemurs, Bats, Elephants, Giraffes

Difference between r selection and k selection as a reproductive strategy

Difference Between

 r selection vs K selection

Features	r- Selection	K-Selection
Example	Bacteria, insects	Primates, including humans
Development	Rapid	Slow
Reproduction rate	High	Low
Reproductive age	Early	Late
Body size	Small	Large
Reproductive type	Single reproduction (Semelparous)	Repeated reproduction (Iteroparous)
Length of life	short	Long
Competitive ability	weak	Strong

Life Histories and Risk of Extinction

Life history theory may be useful in helping us to determine which types species are more at risk of extinction.  K selected species sets it at risk of extinction because of the following reasons:

• K selected species tend to be bigger, so they need more habitat to live in.
• K selected species tend to have fewer offspring, so their populations cannot recover as fast from a disturbance such as over hunting or fire.
• K selected species breed at a later age, so their generation time to grow from a small population to larger population is long.
• In K selected species, population size are often small, and therefore, individuals run a high risk of inbreeding.

The reproductive characteristics of an r-selected mammal and K-selected mammal
r-selected mammal(the Norway rat) vs K-selected mammal (African elephant)

Thus K-selected species are more prone to extinction than r-selected species because they mature later in life and have fewer offspring with longer gestation times. Large K selected tree such as the giant sequoia, red wood tree, large terrestrial mammals like elephants, rhinoceros and grizzlies, and large marine mammals such as blue whales and sperm whales are all competitively dominant species that run a substantial risk of extinction.

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Effective Population Size Calculation

If a population consists of breeding males and breeding females, the effective population size is given by
 
Where Nm is the actual number of breeding males and Nf is the actual number of breeding females.
Practice Problem : 1
In a population of 500 with a 50:50 sex ratio and all individuals breeding, however 250  females bred with 10 males. Then what will be the effective population size?

Answer:
  Applying the equation

N_m (Number of Breeding Males) = 10
N_f (Number of Breeding Females) = 250
N_E- Effective Population Size = (4 x 10 x 250) / (10 +250)
                                               = 10000 / 260
                                           N_E = 38.5

Exam Questions:
Effective population size for completely monogamous species having 40 males and 10 females would be (CSIR UGC NET JRF, Dec 2006)

(a) 42 (b) 32 (c) 20 (d) 10

Answer:
Effective population size (NE) = 4NmNf/ (Nm+Nf)
Nm- Number of breeding males                                    
Nf -Number of breeding females
Here
Nm-40
Nf -10
(NE) = 4NmNf/ (Nm+Nf)
Effective population size (NE) = (4 x 40 x10) / (40 +10) = 1600/50 =  32

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Prokaryotic and Eukaryotic Protein Synthesis

Eukaryotic Protein Synthesis vs Prokaryotic Protein Synthesis 1. In eukaryotes protein synthesis occurs in the cytoplasm.    In prokaryotes protein synthesis begins even before the transcription of mRNA molecule is completed. This is called coupled transcription - translation. 2. Eukaryotic mRNA molecules are monocistronic, containing the coding sequence only for one polypeptide. In prokaryotes individual bacterial mRNA molecules are polycistronic having transcripts of several genes of a particular metabolic pathway. (Monocistronic vs polycistronic ) 3. In eukaryotes, most of the gene have introns that separate the actual message for the synthesis of one protein into small coding segment called exons. Prokaryotes do not have introns (Except Archaebacteria). 4. Eukaryotes: The first amino acid methionine entering the ribosome is not formylated.     Prokaryotes: The first amino acid methionine is formylated into N formyl methionine. 5. In eukaryotes, the number of initiating factors (IF) is much more than prokaryotes. About ten initiating factors(IFs ) have been identified in reticulocytes an RBC. These are eIF1, eIF2, eIF3, eIF4 , eIF5, eIF6 ,eIF4B, eIF4C,eIF4D, eIF4F Three initiating factors found in prokaryotes. PIF-1 , PIF-2 , PIF-3.     Learn more: Prokaryotic initiating factors and Eukaryotic initiating factors 6. The primary mRNA transcript in eukaryotes undergoes processing and splicing to change into a functional mRNA. In prokaryotes, splicing of mRNA transcript does not occur. 7. In eukaryotes, mRNA molecules are modified by the addition of a 5’G cap formed of methylated guanosine triphosphate. No such cap is formed at 5’end of bacterial mRNA. 8. In eukaryotes 5’cap initiates translation by binding mRNA to small ribosomal subunit usually at the first codon AUG. In bacteria translation begins at an AUG codon preceded by a special nucleotide sequence. 9. A poly A tail formed of about 200 adenine nucleotides is added at the 3’end of mRNA in Eukaryotes. No poly A tail is added to bacterial mRNA. 10. In eukaryotes small subunit of ribosome(40 S) gets dissociated with the initatior amino acyl tRNA (Met-tRNA Met) without the help of mRNA. The complex joins mRNA later on. In prokaryotes 30 S subunit first complexes with mRNA (30S-mRNA) then joins with f Met tRNA f-Met. Learn more: Difference Between Prokaryotic and Eukaryotic Protein Synthesis

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Genomic Library: Definition and Steps in the Construction of Genomic Library

In the last post, we discussed about the steps in recombinant DNA technology or gene cloning. In that post, we mentioned genomic library as one of the source for getting our desired gene for cloning.

Genomic Library: Definition

Genomic library represents an entire genome of an individual animal, bacteria, plant or virus under study.

or

It is the collection of cloned segments of DNA containing at least one copy of every gene from a particular organism.

Construction of Genomic Library of a Phage

The procedure is same for all other organism.

Step 1: DNA isolation

Isolate complete DNA from the phage (or any cell under study)

Step2: Cutting isolated DNA with restriction fragments of suitable size

Restriction enzymes like Eco R1 are used to cut genome into fragments of suitable size.

Step 3: In cooperation of this fragments into a suitable vector

These fragments are cloned into a suitable vector like plasmid, cosmid etc leading to the formation of rDNA molecule.

Step 4: Introduction to a suitable host like bacterium.

Multiplication, screening, identification and characterization of clones

Introduce this rDNA molecule with DNA fragments into a host, most often E.coli bacterium. Plasmid will multiply inside forming numerous copies. We need to identify the host cells with these fragments from the rest of cells without rDNA molecule.

Step 5: Maintenance of set of clones

The host cell multiplies and forms colonies. Each colony contains cells with a DNA fragment of the phage. Maintenance of such clones or colonies containing all fragments of the phage represents genomic library of the phage.

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CSIR UGC NET JRF Life Sciences Previous Questions

1. Among the following, which of the following sequence cannot be a part of gene?
a) AGCGCTGCATGGGCCCCC
b) ATTGGCCCTTTTUGCCGCA
c) AAAAATTTTTAAAATTTGGG
d) GCGCGCGAAACCCTTTTAAAA


2. In mammalian females, two X chromosomes are present. Expression of genes on both chromosomes may lead to gene dosage imbalance. This problem is solved by a process called dosage compensation. Dosage compensation is achieved by

a) hyperactivation of one X-chromosome
b) elimination of one X chromosome
c) hypoactivation of both X chromosomes
d) methylation of one X chromosome


3. Adaptive immunity is mediated by B cells and T cells. T cells interact with other cells through T-cell receptor. B cells lack a typical receptor of its own. B cell receptor is often
a) an IgG molecule
b) an IgM molecule
c) an IgE molecule
d) all of these

4. If bacterial genome and plasmid are allowed to replicate in same manner then
a) bacterial genome will replicate fast
b) plasmid genome will replicate fast
c) both will replicate at the same time
d) depends on GC content

5. On buoyant density gradient centrifugation, a DNA band was observed as high peak corresponding to low density as compare to other DNA. The inference is
a) DNA is GC rich
b) DNA is AT rich
c) [AT]=[GC]
d) single stranded DNA

6. Vibrio cholera causes diarrhea by
a) opening ion channels
b) closing absorption of water from gut epithelium
c) destroys cells of intestinal lining
d) constitutive expression of adenylate cyclase


7. Choose the correct statement regarding the effect of ozone on biosphere
a) Both atmospheric and stratospheric ozone is harmful
b) Both atmospheric and stratospheric ozone is beneficial
c) Atmospheric ozone is harmful but stratospheric ozone is beneficial
d) Atmospheric ozone is beneficial but stratospheric ozone is harmful

8. Split genes are present in
a) all organisms
b) Most of eukaryotes and some eubacteria
c) all eukaryotes
d) Most of eukaryotes and some archaebacteria


9. Many bimolecular are transported from nucleus to cytoplasm through nuclear pores. Which molecule is continuously transported from nucleus to cytoplasm
a) DNA
b) RNA
c) Protein
d) ribosomes

10. Which statement is correct regarding meiosis
a) There is two round of replication and two round of cell division
b) There is one round of replication and one round of cell division
c) There is one round of replication and two round of cell division
d) There is two round of replication and one round of cell division
Learn more:

• CSIR UGC NET JRF Cell biology practice tests
• CSIR UGC NET JRF Ecology Practice Tests
• CSIR UGC NET Previous Questions

Answers
1. b) ATTGGCCCTTTTUGCCGCA
2. d) methylation of one X chromosome
3. b) an IgM molecule
4. b) plasmid genome will replicate fast
5. b) DNA is AT rich
6. d) constitutive expression of adenylate cyclase
7. c) Atmospheric ozone is harmful but stratospheric ozone is beneficial
8. d) Most of eukaryotes and some archaebacteria
9. b) RNA
10. c) There is one round of replication and two round of cell division

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ICAR ASRB ARS Exam Notification 2013

Agricultural Scientists Recruitment Board (ASRB) will hold a competitive ARS Examination – 2013 as per the following scheme wherein two separate examinations will be held:

(i) Preliminary – ARS Examination - 2013 on 29.12.2013 (SUNDAY) from 10.00 A.M. to 12.00 Noon at 33 centres.

(ii) ARS – Main Examination - 2013 on 16.03.2014 (SUNDAY) from 10.A.M. to 01.00 P.M. at 12 centres.

The preliminary ARS Examination is a qualifying examination and its marks will not be carried forward for determining final merit of the candidates. All those candidates desirous of appearing for ARS will have to take both the examinations and viva-voce. Since a pass
in the Preliminary examination is compulsory for all candidates, only those candidates who qualify the Preliminary ARS Examination will be eligible to appear in the ARS Main Examination. Candidates declared successful in ARS Main Examination and viva-voce will
be recommended for appointment as Scientists in Agricultural Research Service (ARS) of Indian Council of Agricultural Research in the Pay Band-III of Rs 15,600 - 39,100 plus Research Grade Pay of Rs 6,000/- .

Vacancies for ARS

The candidate must see the Rules of examination carefully before filling up the Online
Application form.

Education Qualifications:
A candidate must hold a Master’s Degree or equivalent in the concerned discipline
with specialization as defined in Annexure-V completed on or before 16.03.2014. For
details, please refer to Rule 3 of Rules of the Examination. However, candidates who have appeared at an examination, the passing of which would render them eligible to appear at the ARS Examination – 2013 but have not been informed of the results, may apply for admission to the ARS Examination-2013. Candidates who intend to appear at such a qualifying examination may also apply. Such candidates will be admitted to the ARS Examination-2013, if otherwise eligible but their admission would be deemed to be provisional and subject to cancellation, if they do not produce proof of having passed the requisite qualifying examination which will be required to be submitted by the candidates who qualify on the results of the Preliminary ARS Examination-2013 on or before 16th March, 2014, scheduled date of ARS Main Examination-2013.

Age limit: A candidate must have attained the age of 21 years but not have attained the age of 32 years as on 01-08-2013 (i.e. he/she must have been born not earlier than 2nd August, 1981 and not later than 1st August, 1992). Age relaxation is admissible to the various categories as per rules.

How to Apply :A candidate seeking admission to the Examination must apply online only in the Application Form available on the website: http//www.asrb.org.in OR
http://www.icar.org.in

A candidate must read the provisions contained in this Notification for ARS Examination - 2013 carefully and abide by the same. A candidate must fulfill all the conditions of eligibility regarding age limits, educational qualifications, etc. prescribed for admission to the examination.

Useful Links:

• ARS Notification (Download)
• Books to Refer for ICAR ARS NET
• ICAR ARS NET Exam

Previous Question Papers Free (ICAR ARS NET Preparation Resources)

• Basic Plant Sciences / General Agriculture
• Plant Pathology/ Plant Physiology
• Agricultural Biotechnology / Agricultural Biotechnology / Agricultural Biotechnology /Agricultural Biotechnology
• Agricultural Microbiology
• Fruit Science- Horticulture
• Soil Science /Agronomy
• Facts for ICAR NET exam

Plan of Examination

Two separate competitive written examinations followed by Viva-voce shall be conducted as per the following plan of examinations:-
Examination                                          Max. Marks          Duration 
1. (a) Preliminary-ARS (Objective Type)     150                     2(Two) hours
(b) ARS – Main (Descriptive Type)             240                     3(Three) hours
(c) Viva-voce                                             60

2. (a) Preliminary ARS Examination Paper:-
Preliminary ARS Examination shall have one paper of 150 marks with all multiple type
objective questions to be solved in 02 hours (Two hours), from the respective professional subject(s) for which candidate has opted.
1/3 marks will be deducted for each wrong answer in Preliminary ARS Examination
(Objective Type). The examination will be held in 33 centres as given in Annexure-I. This is a qualifying examination and marks scored will not be counted for final selection.
The candidates who attain such minimum qualifying marks in the Preliminary Examination as may be fixed by the Board in its discretion will be allowed to appear in the Main Examination.
(b). ARS – Main Examination Paper:-
ARS – Main Examination Paper will have only one paper of 240 marks to be completed in

3 hours duration. The paper shall be divided in three parts A, B and C. Part ‘A’ will consists of 40 (forty) questions of 2 (two) marks each. In this part, answers required will be of very short, not exceeding 10 (Ten) words at the most. Part ‘B’ will have 20 (twenty) questions of 5 (five) marks each requiring one or two paragraphs and/or graphic explanation. Part ‘C’ will have 6 (six) essay type or descriptive type questions. Each question will carry 10 (ten) marks and may have two or more parts. Answers are required to be written in the space provided below the question. In no case extra sheets will be provided. All questions in parts ‘A’ ‘B’ and ‘C’ will be compulsory.

First and foremost thing is to begin the preparation now onwards

"Wishing the very Best "

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