Blue white screening

Blue white screening- Foriegn DNA and plasmid Dna are cut with restriction enzyme. Plasmid has gene for lactose hydrolysis lac z gene which codes for enzyme (β-galactosidase). Foriegn dna insert into lacz gene to form the recombinant. The recombinant dna is then transfered into a host cell , which are then grown in the presence of X-gal. 

Result-
colonies with an insert-containing plasmid have a non-functional β-galactosidase, and remain the whitish-cream color of standard E. coli. On the other hand, intact β-galactosidase ( non recombinant plasmid )produces pigment from x-gal (included in the transformation plate medium), turning the bacterial colony blue.

                                                
Imagehttps://en.wikipedia.org/wiki/Blue_white_screen



                                                                                                                                                                                                                                                                                                                                                            Imagesourcehttp://www.slideshare.net/tejondaru/blue-white-selectionpresentation

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Linkers and adapters

Linkers- Short stretches of dsDNA of known  nucloetide sequence (length 8-14 bp )and have a recognition site for 3-8 restriction enzymes.

Ligated to blunt ends of DNA by ligase.Because of the high concentration of these small molecules present in the reaction, the ligation is efficient when compared with blunt-end ligation of large molecules. The cohesive ends are generated by digesting the DNA with appropriate RE that generates cohesive ends by cleaving in the linkers. The problem with linker is that the sites for the enzyme used to generate cohesive ends may be present in the target DNA fragment. This drawback limits the use of linkers for cloning. 

Adapters are linkers with cohesive ends or a linker digested with RE, before ligation.It already has one sticky end.The most widely used definition is cut linkers also called as adapters. 

They are not perfectly double stranded non single stranded. By adding adaptors to the ends of a DNA, sequences that are blunt can be converted into cohesive ends and used for cloning .

Sticky ends are desirable for DNA cloning , but sometimes blunt ends are present .Blunt ends will binds to the blunt ends of target to form sticky ends.

                         Image source-http://www.slideshare.net/drsandeepaiims/gene-cloning-sandeep-jan-2016

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Bioscience formulae

Bioscience formulae-

1.VP=VG+VE+VGE

2. Crossover value by the square root of the number of individuals (.\/N)

3.Chi-Square Formula-

Degrees of freedom (df) = n-1 where n is the number of 

4.No. of different genotypes-n(n+1)/2,3^n

5.No. of different gametes and phenotype - 2ⁿ

^{6.Recessive homozygous in F2 =1/(}2ⁿ⁾²

7.Frequency of recombination-no.of recombinant/Total no. of progeny*100

8.Map distance - No. of recombinant offspring/Total no. of offspring*100

9.Distance betwen 2 genes- No. of single CO+ No. of DCO/Total no. of progeny *100

10.Coincidence- (observed DCO/ expected DCO)

11.Interference=1-(observed DCO/ expected DCO)

12.In tetrad analysis-

a.Map distance= 1/2*No. of SDS asci  / Total no. of asci *100
b.Recombination frequency-NPD+1/2 TT / Total no. of tetrads *100

13.No.of F2 progeny expressing either extreme phenotype- 1/ 4^n

14.Inbredding coefficient= 2pq-H/2pq( population genetics)

15.Ne= 4NmNf /Nm+Nf( Effective population size)

 Nm= no.of male Nf= no. of female

16.dC/dt= -kC^2    dC/dt=rate of renaturation
   
     C0t1/2 = 1/k 

17.p  ( buoyant density )
=1.660 +0.00098 (% GC)g/cm<sup>3 

18.Packaging ratio of DNA= Extended length of DNA/ Packing length</sup>

19.Dna replication time= base pairs/ no. of bases per sec

20. Size of fragments = 4^n  ,where n= no. of base pairs

21.Time since divergence of two species is given by
(100 - X% sequence similarity) / (% change / years).

22.pH = -log[H⁺] or pH = -log[H₃O⁺]

23. K_w = 1.0 x 10^-14 =  [H₃O⁺] [OH^-]  

24.Viability of cryopreserved cells -


25.Determining the no. of genes=1/4 ^n

26.Nt=No+B+I-D-E     I=Immigration  B=Natality rate   D= Mortality rate 

(change in population size at a given time interval t)




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Pleiotrophy

Pleiotrophy - One gene have the ability to produce two or more unrelated phenotypic traits. 
An example is phenylketonuria.
The disease causes mental retardation, reduded hair and pigmentation.
Due to mutation in a single gene phenylalanine to tyrosine.

                            Image source-http://mumtazticloft.com/PigeonGenetics1.asp

Other examples are-
1.Sickle cell anemia -enlarged spleen,muscle pain,low RBC count
2.Albinism- Albinism is also a pleiotropic trait affecting eyesight as well as resulting in a lack of pigmentation in skin, hair, and eyes.
3.Cytokines that have different biological functions.



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Flow cytometry

Flow cytometry- It is used to measure the number of cells from a suspension of heterogenous population. It can analyze and sort particles on a one by one basis at rates of 50000 particles per second. It consist of 3 main components-

1.The flow system
2.The optical system
3.The electronic system

Flow system-  The sample is injected into the centre of a stream of liquid (water or buffer), called the sheath fluid. the particles are randomly distributed.

The optical system-
After hydrodynamic focusing, each particle passes through one or more beams of light(lasers).Light scattering or fluorescence emission (assumed the particle is labeled by a fluorochrome) provides information about the particle’s properties. Fluorescence measurements taken at different wavelengths can provide quantitative and qualitative data about fluorochrome-labeled cell surface receptors or intracellular molecules such as DNA and cytokines. Light is quantified by detectors.The specificity of detection is controlled by optical filters, which block certain wavelengths while transmitting others. Charged droplets are deflected electrostatically by passage through an electrical field.

The electronic system- It analyze statistically by software to report cellular components, size,phenotype .

Applications of Flow Cytometry-

Typical applications of  flow cytometry include-

• Biomolecular studies
• Immunology
• Single cell analysis
• Medical applications (hematology, genetics)
• Autofluorescence characterization. 

1.Flow cytometer is used to measure the number of-

A. Cells                    c. Proteins

B. Dna                      d. Rna

Ans.  A (cells) 

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Promiscuous dna

Promiscuous dna- Endosymbiotic transfer of DNA from cytoplasmic organelles (mitochondia and chloroplast) to the nucleus.  It is supposed that the chloroplasts and mitochondria were once free-living cells that linked up with the embryonic plant cell to form a symbiotic partnership.

                         Image source -http://www.nature.com/scitable/content/organellar-dna-mobility-and-the-genetic-control-14665064

The eukaryotic mitochondria is derived from a proteobacterial endosymbiotic ancestor . Most of the genes that were originally present in this ancestor's genome have been transferred to the nucleus .With only a small number being retained in the organelle (MITOCHONDRIA). Similarly, most of the genes from the cyanobacterial endosymbiont ancestor of the chloroplast were also transferred to the nucleus.

 So, as a result, cytoplasmic organelles are heavily dependent on nuclear genes and import more than 90% of their proteins from the cytoplasm. The dotted arrows indicate how DNA of mitochondrial (blue) and chloroplast (green) origin is still being transferred to the nucleus.Chloroplasts genes have now been found inside plant mitochondria.

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Flippase /Floppase /Scramblase

1.Flippase(P-type ATpase) -Transporters that move lipids to the cytoplasmic face of the membrane are commonly called “flippases.  It uses ATP to move the aminophospholipids  such as PS(phosphatidyl serine) and to a lesser extent, phosphatidylethanolamine (PE), from the outer leaflet to the inner leaflet of the plasma membrane against the concentration gradient.

2. Floppases (ABC transporter)- Transport lipids from the cytofacial surface to the opposite side of the membrane are called “floppases” .it uses ATP to transport substrates such as phosphatidylcholine (PC), sphingolipid (SL) and cholesterol against concentration gradients in the opposite direction. 

Movement of phospholipid molecules between the two leaflets that compose a cell's membrane (transverse diffusion, also known as a "flip-flop" transition). 

3.Scramblase-It moves lipids across the bilayer in either direction towards equilibrium. It briefly disrupt random assymetry.

1.Flippase  2.Floppase 3.Scramblase

                             Image source-https://www.studyblue.com/notes/note/n/micro-2200-study-guide-2011-12-kang-/deck/9733592

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Lipid rafts

Lipid rafts - The plasma membrane is composed of a lipid bilayer  inserted various proteins(transmembrane ion-channel and transporter proteins). Many of these proteins are found associated with surface of the membrane which are rich in free cholesterol and sphingolipids (lipids containing the amino acid sphingosine ).

Lipid rafts are more ordered and tightly packed than the surrounding bilayer, but float freely in the membrane bilayer.

Lipid rafts generally contain 3 to 5-fold the amount of cholesterol found in the surrounding bilayer. Also, lipid rafts are enriched in sphingolipids such as sphingomyelin and play important roles in signal transduction .

https://www.researchgate.net/figure/230685775_fig6_Fig-12-Structure-and-formation-of-lipid-rafts-based-on-a-lipid-ordering-disordering

Types of lipid rafts-
1.Caveolae type- Flask shaped invaginations in plasma membrance which is riched in caveolin.
2.Planar lipid rafts- Found in neurons and enriched in flotilin.
3.Glycospingolipid enriched membranes (GEM)
4.Polyphosphoinositol rafts 

Types of lipid rafts

                                      http://www.getmededu.com/plasma-or-cell-membrane.html
FUNCTIONS -
1.Provide stability or rigidity .
2.Helps in signal transduction.
3.In endocytosis it helps.
4.Cholesterol transport .

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Cloning

 Cloning: (Dolly)

 Definition-Clones are organisms that are exact genetic copies. 

 Every single bit of DNA is identical. It can happen naturally— identical twins are

 just one of many  examples or can be made in the lab.

a) First clone - tadpoles using embryonic cells and nuclear transfer.

b) Why cloning?

  1) cloned animals produce proteins to treat disease. 
  2) for organ transplantation

  3) therapeutic human cloning and stem cells.

c)Types of cloning-

    1.Therapeutic cloning-by the use of stem cells for research.

    2.Reproductive cloning- for cloning human and animals.

                           Image source-http://mrdclassified.weebly.com/cloning-and-gmos.html

 3.Gene cloning-DNA derives from genome of organism and cloned into a cloning

    vector (Plasmid).

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Problems related to hardy weinberg law

1. If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?Figure out the % of homozygous dominant?

q2 = (98/200) = 0.49 (or 49%)
q =  0.7 (70%)
 As p + q = 1.
p = 1-q
p = 1 - 0.7 = 0.3 (30%)

% of homozygous dominant-

p2 = (0.3)(0.3) = 0.09 (or 9%)

What is the frequency of heterozygotes?


Frequency of heterozygotes = 2pq
                                          = 2X0.3X0.7
                                          =0.42
p2 + 2pq + q2 = 1
0.09 + 0.42 + 0.49 = 1.0(PROVED)

2. Red hair is a recessive trait in humans. In a 

randomly mating population in Hardy-
Weinberg equilibrium, approximately 9% of 
individuals are red-haired. What is the 
frequency of heterozygotes? 
1. 81% 2. 49%

3. 42% 4. 18%
Ans.Given - recessive trait (9%)
q2=9/100=0.09
q=0.3
We know that p+q=1
So,p= 1-q=0.7
Now, the question is to find out frequency of heterozygous i,e 2pq
=2*0.3*0.7
=0.42
So, the correct option is 3.

3.In a population which is in Hardy-Weinberg equilibrium, the frequency of a
recessive allele for a certain genetic trait is 0.40. What percentage of individuals
would be expected to show the dominant trait in the next generation?
(A) 16%
(B) 32%
(C) 84%
(D) 96%

Ans.c(0.4=q recessive gene, so p=0.6 dominant. As p+q=1formula is: p2+q2+2pq=1. (0.4)2+ (0.6)2 + 2*0.4*0.6=1. dominant % is 0.36+0.48=0.84 that is  84%

4.In a population at Hardy-Weinberg equilibrium ,the genotype frequencies are 
f(A1A1 )=0.59 ,f(A1A2)= 0.16 ,f(A2A2)=0.25.What are the frequencies of the two alleles at this locus?
1.A1=0.59  A2= 0.41
2.A1=0.75  A2=0.25
3.A1=0.67 A2= 0.33
4.A1=O.55 A2=0.44

Answer-
A1A1=p2= 0.59 
A1A2=2pq=0.16
A2A2=q2=0.25
A1=p=p2+1/2 2pq=0.59+0.08=0.67(Ans)
q=1-p
  =1-0.67
  =0.33 (Ans)

5.In a random sample of 400 individuals from a population with allele of trait in Hardy weinberg equilibrium 36 individuals are homozygous for allele a.How many individuals in the sample are expected to carry atleast one allele A?
A.36                                                 C.364
B.168                                               D.196                  

Ans.364 ( Let us say q is the frequency of allele a and p is the allele frequency of allele A.p+q=1
36 in 400 individuals are aa=q^2=36/400=0.09 q=0.3
p= 0.7 
Individuals carrying at least one AA and Aa.
Their frequency will be p^2+2pq=0.49+ 2.0.7.0.3=0.91
Therefore 400x 0.91 =364  individuals with either AA or Aa.

6.The following genotypes are found in a population AA:Aa:aa::60:20:20. What are the the allele frequencies of A and a?

a.A=0.7 and a=0.3

b.A=0.6 and a=0.2

c.A=0.8 and a=0.2

d.A=0.3 and a=0.7

Ans.The number of individuals = 100 
The number of alleles = 200 (because they are diploid) 
Calculate p(A) from homozygotes AA and heterozygotes Aa, and do for q(a) as well. 
p(A) = ((2x60) + 20)/200 = 0.7 
q(a) = ((2x20) + 20)/200 = 0.3 
so the answer is
 
a) A=0.7 ; a=0.3

7. If 16% of newborns in a population have sickle cell anemia what percentage of the population will be more resistant to malaria-

a.84%       b.36%                  c.24%  d.     48%

Ans. D (48%) 

Homozygous Hb^s Hb^s= sickle cell anemia (die)
HeterozygousHb^N Hb^s=Normal(survive)
if 16%=Hb^s Hb^s q=16%=.4  p+q=1 So, p=.6 2pq=2*.6*.4=.48= 48%

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Hardy weinberg equilibrium

Hardy Weinberg law or Equilibrium by G. H. Hardy (a mathematician) and Wilhelm Weinberg (a physician).

 “The relative frequencies of genes and alleles in a large and randomly mating (panmictic) population tend to remain constant from generation to generation without  mutation, selection, migration and genetic drift.” Acc.to the law the population is not undergoing evolutionary changes.

1.The evolutionary forces are absent (mutation, selection, drift) in the population.

2. The sizeof the population is large.

3. Its individuals are in random mating. The gametes produced by the mates combine at random and the gene frequency remains constant. 

4. There is no evolution in the population.

5. All the genotypes in a population reproduce equally successfully.

6.No emigration or immigration (no gene flow).

 Example-

Suppose a Mendelian population  allele ‘A’ and ‘a’  is on one locus. the frequency of gametes with gene ‘A’ will be same as the frequency of ‘A’ gene. Similarly the frequency of gametes with ‘a’ will be same as the frequency of ‘a’ gene.

Let the numerical proportion of different genes in this population is as follows:

                           AA                     Aa                  aa

                          36%                   48%                 16%
‘AA’ will contribute approximately 36% of all gametes in the population. ‘aa’ individuals will produce 16% of all the gametes. Gametes from ‘Aa’ individuals will contribute 48% with gene ‘A’ and with gene ‘a’ roughly in equal proportion.

If the frequency of ‘A’ is represented as ‘p’ and the frequency of ‘a’ is represented by ‘q’ and if there is random mating of gametes with genes ‘A’ and ‘a’ at the equilibrium state, the population will contain following frequencies of the genes ‘A’ and ‘a’ generation after generation:

AA + 2Aa + aa as genotype frequency

p2 + 2pq + q2 as gene frequency

In a population of large size the probability of receiving the gene ‘A’ from both parents will be p X p = p2

Similarly the probability of receiving the gene ‘a’ from both the parents will be q X q = q2

The probability of being heterozygous will be  pq + pq = 2pq

The relationship between gene frequency and genotype frequency can be expressed as:          

p2 + 2pq + q2 = 1    

or   (p + q)2 = 1

This relation is known as Hardy-Weinberg formula of binomial expression.

From this equation, it is clear that in a large random mating population not only gene frequencies but also genotype frequencies will remain constant.

                                                              Image source-http://bioap.wikispaces.com/Ch23+Collaboration

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