SURVIVORSHIP CURVES

Curves which represents about surviving individuals in different age groups.

1.r -Strategies(For growth rate=r which is very high ) =Type III survivorship curve(population size varible, type of species found is pioneer,density independent )
2.k-strategies(For carrying capacity=k )=Type I survivorship curve(Population size stable,climax species found ,density dependent)

Types-
1.Type-I Low death rate,High survival in early and middle life  (HUMANS,SALMON)
2.Type-II Moderate death rate,survival is constant(CORALS,HYDRA,SONGBIRDS)
3.Type-III High death rate in early life,survival is low at early stage .(PLANTS,OYSTERS,TURTLES,COCKROACH,FROG,SEA URCHIN)

                                                   Image source credit-http://theamazonriver1.weebly.com/additional-information.html

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DNA QUERIES

1.  7 guanines are present in a DNA segment of 20 base pairs. How many adenines are there in the segment?
Ans. 20 base pairs → 40 bases

C = G → 7 pieces of G and 7 pieces of C are there in the segment

A+T = 40-(C+G) = 40-(7+7) = 26 bases (A and T together)

A = T → 26/2 = 13 adenines.

2. What is the melting temerature of DNA which is 450 bp and has 50%  GC  content ?
Ans.Tm=64.9+0.41(% GC)-500/length of bases

              =64.9+0.41* 50-500/450              =84.3 c
3.Multiple RNA polymerase transcribes a DNA template unwinding about 1.5 turns of DNA template per transcription bubble .From the structural information of classical B-DNA how many transcription bubbles are possible for a 180 base pair DNA molecule?                                                                                                        (GATE)
a.12            b.27                                    c.6                              d.270
Ans. In B-form the helix makes a turn every 3.4 nm and the distance between two neighboring base pairs is =0.34 
so,in 180 base pairs DNA molecules the number of turns will be 18(180/10).
Since,each bubble have 1.5 turns so,the number of  bubbles=18/1.5  

4.Human DNA (3*10^6 kb) is replicated in 5 hrs at a rated of 1kb/min. The number of origins of replication utilized are -
a.1                                c.3                                           (GATE )
b.300                            d.10,000
Ans.  d
Rate of relication -1kb/min
Amount of DNA replicated in 5 hrs=5*60=300kb
No. of origins required =3*10^6/300=10,000 .


5.If the ratio of A+G/T+C= 0.7 in one strand what will be ration in complementary strand.
Given ,A+G/T+C= 0.7
                             =7/10
A+G=7, T+C=10
As, A=T  AND G=C  in complementary strand
T+C /A+G=10/7 =1.43


6.How to calculate the molecular weight of a DNA template?
Ans. Using Avogadro’s number, which is 6.022×10²³ molecules/mole, the number of molecules of the template per gram can be calculated using the following formula:

• ng is the amount of DNA (plasmid, primer etc.) you have in nanograms
• 6.022×10²³ = Avogadro’s number
• length is the length of your DNA fragment in base pairs. Just multiply by 1000 if you are working in kb.
• We multiply by 1×10⁹ to convert our answer to nanograms

7.What is the number of hydrogen bonds in B -DNA of 1000 base pairs with base A occuring 75% of the times in one strand and 10 % in other ?
Ans.  A= 75%
          T=10%
A+T=85%
C+G=15%
To calculate the number of hydrogen bonds =  85%of 1000=850  (A+T)
                                                                   and 15% of 1000=150(G+C)
We know that A=T (2 H-BONDS) and G ≡C (3 H- BONDS)
850*2= 1700
150*3=450
So,total number of hydrogen bonds are 2150.

8.No. of bases in a B-DNA =1000                                       IISC 2009 
C=60%
A=30%
T=10%
calculate the number of hydrogen bonds ?
Ans.  60% of 1000= 600
          30% of 1000=300
          10% of 1000=100
total no. of hydrrogen bonds are =600*3+300*2+100*2 =1800+600+100=2600 .

10.The annealing temperture at which primers attach to template can be calculated by determining the melting temp(Tm) of the primer template hybrid .What will be the tm of the primer ?
    5'--------AGACTCAGAGAGAACCC ----------3'

  Ans .
   Tm = 2(A + T ) + 4 (G + C) 
          =2(7+1)+4(4+5)                           NOTE-(TEMPERATURE AT WHICH 50 % OF THE DNA IS DENATURED)
         =52
11.A virus has a genome consisiting of a single DNA molecule with a base composition of A=24.6% ,T=34.7  ,G=20.2%
and C=20.2 %. How would you describe the DNA?
Ans.A is not equal to T that means it not obeys charguff's rule so the DNA is single strandard it is proved.
And  if it is resistant to double strand and single strand specific exonucleases it is circular .


12. Linking number of DNA is 9 with writhe-1 ,then number of twist is -
a.10                                                                c.9
b.11                                                                 d.12
Ans. lk=Wr+t w
          9 = -1+tw
         tw=  10   

13. The genome of a bacterium is composed of a single DNA molecule which is 10^9 bp long .How many moles of genomic DNA is present in the bacterium?[ Consider Avagadro No. 6* 10^23)
        a.1/6*10^-23                                      c.1/6 *10^-14
        b.  6*10^23                                       d.6*10^14
Ans. a
DNA molecule length =Moles* Avagadros N0.
                        10^9    =  x  *  6x10^23
                         x=  1/6 * 10^-23

14.The diploid genome of a species comprises 6.4* 10^9 bp and fits into a nucleus that is 6μm in diameter. If base pairs occur at intervals of 0.34 nm along the DNA helix, what is the total length of DNA in a resting cell?
a. 3.0 nm              b. 3.5 nm

c. 2.2 nm               d. 4.0 nm
Ans.  (c)  Length of DNA double helix can be calculated by multiplying the total number of base pair with distance between two consecutive base pairs. Here the total number of base pair is =6.4 x 10^bp.
Distance between two consecutive base pair =0.34 nm.
To convert 0.34 nm to metre multiply this with 10^9 i.e; 0.34 *10^9
Therefore Total length of DNA = 6.4 *10^9  x 0.34 *10^9 = 2.2 m

15. There are 2 single stranded DNA of 10 nut each .What is the probability that they form a double stranded DNA with all the 10 base pairs in Watson-Crick pairing ?
Ans. a. (1/4)^10   A dna contains 4  nitrogen base base A,G,C,T . (FORMULA IS 1/4^ n)

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ECOLOGY IMPORTANT NUMERICALS

1. The effective population size for completely monogamous species having 40 males and 10 females would be - (2015 dec)
a.42                                               c.20
b.32                                              d.10                 Given -
Ans.   NE= 4 NmNf/Nm+Nf                                    Nm-Number of males(40)
              =4 40.10/40+10                                        Nf=Number of females(10)
             =1600/50
             =32        (Ans is  b)                                   NE=Effective population size ?                               

2.  D = Σ n(n-1)/N(N-1) (Simpson’s Diversity Index)

Example:i

A lake contains 934 brown trout, 733 smallmouth bass, 34 catfish, 2003 carp, 234 steelheads, and 32 northern pikes.

Fish	Number	n(n-1)
Brown Trout	934	934(934-1) = 871422
Smallmouth Bass	733	733(732)=536556
Catfish	34	34(33)=1122
Carp	2003	2003(2002)=4010006
Steelheads	234	234(233)=54522
Northern Pike	32	32(31)=992
Total	N = 3970 N(N-1) = 15756930	Σ n(n-1) = 5474620

 

D=5474620/15756930

4. A fisheries biologist is maximizing her fishing yield by maintaining a population of lake trout at exactly 500 individuals.  Predict the initial instantaneous population growth rate if the population is stocked with an additional 600 fish.  Assume that r for the trout is 0.005 individuals/(individual*day). 

 Ans.For a populations growing according to the logistic equation, we know that the maximum population growth rate occurs at K/2, so K must be 1000 fish for this population.    

If the population is stocked with an additional 600 fish, the total size will be 1100.  From the logistic equation, the initial instantaneous growth rate will be: DN/dt  =  rN [1- (N/K)]   

 = 0.005(1100)[1-(1100/1000)] = = -0.55 fish / day

5.In a population of 10 million individuals birth rate is 19 per 1000 and death rate is 14 per 1000.Annual rise in population would be-

a.50,000                                                          c.14000

b.5000                                                             d.500000                 (2008)

Ans. a 

Birth rate =19/100*1000=190

Death rate =14/100*1000=140

Rise =b-d= 190-140=50`]

Annual rise=50*1000=50,000

7.The population size of a bird increased from 600-645 in one year.If the per capita birth rate of this population is 0.125 . what is the per capita death rate?

Ans. 0.05  b=B/N  d=D/N

  B=b*N=  12.5/100*600=75

  Death by the number in the population = 75-45=30

    d=D/N =30/600*100=5%=0.05

8.In an experimental population the birth rate is 18 per 1000 and death rate is 14 per 1000.Size of population is 10,000 at time,t,then what will be size of population at timee t+1

a.10,000                                                             c.10,040

b.10,140                                                             d.11,040

Ans.Birth rate is 18/100*1000=180

      Death rate is 14/100*1000=140

rate =b-d=180-140=40

so,the size of population at time t+1= 10000+40=10,040

9.The population density of an insect increases from 40-46 in one month .If the birth rate during that period is 0.4 .What is the death rate ?

a.0.25                                                   c.0.15

b.0.87                                                   d.0.40

Ans.c   Death rate=Final population-Initial population/Birth rate

                          =46-40/0.4

                         =6/0.4=   15% i,e 0.15

10.The population was expected to double within 50 years .Calculate the r for the population.

      tdouble =ln2/r

    r=ln2/t=ln2/50=0.0139

    
11.No. of mice =2000
     Born rate =1000
     Death rate=200
What is the Per capita growth rate over month ?
Ans.  Growth rate =rmax= b-d= 2000-200=800
Per capita growth rate =rmax/N=  800/2000=2/5=0.4 

12.A population is growing logistically with a growth rate =o.5/week ,with a carrying capacity =400.
What is the max.growth rate that this population can achieve?
Ans. K= 400
        r=o.5 
         N=k/2=400/2 =200
dN/dt=rN (1-N/K)
        =100*(200/400)

•         =50

NOTE -1.To  calculate the basic reproductive rate, R₀,  formula is 

R₀ = l_{xm x .      ( LIFE TABLE )} 2.dP/dt=rP-aCP   (prey)    PREY r =INTRINSIC GROWTH RATE OF PRAY

                                                     P=PREY DENSITY

                                                     a=PREDATION EFFICIENCY

                                                     C=PREDATION DENSITY

 dC/dt=B(aCP)-DC  predator         B=ASSIMILATION AFFICIENCY

                                                   D=DEATH RATE OF PREDATOR       (LOTKA VOLTERA MODEL )

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POST TRANSLATIONAL MODIFICATIONS

(Phosphorylation, glycosylation . ubiquitination. S-nitrosylation and Acetylation )

1.Phosphorylation-

 (On serine, threonine or tyrosine residues, is one of the most important post-translational modifications involves in regulatory switches.)

Phosphorylation  roles-  In the regulation of many cellular processes including cell cycle, growth, apoptosis and signal transduction pathways.

2.Glycosylation-

Protein glycosylation is one of the major post-translational modifications, with significant effects on protein folding, conformation, distribution, stability and activity. 

Glycosylation ouucrs in Carbohydrates in the form of aspargine-linked (N-linked) or serine/threonine-linked (O-linked) oligosaccharides are major structural components of many cell surface and secreted proteins. Makes protein hydrophillic and in acylation it makes hydrophobic.

3.Ubiquitination

Ubiquitin is a 8-kDa polypeptide consisting of 76 amino acids that is appended to the ε-NH2 of lysine in target proteins via the C-terminal glycine of ubiquitin.  Polyubiquitinated proteins are recognized by the 26S proteasome that catalyzes the degradation of the ubiquitinated protein and the recycling of ubiquitin.

4.S-Nitrosylation

It  reacts with free cysteine residues to form S-nitrothiols (SNOs). SNOs are often stored in membranes, vesicles, the caspases, which mediate apoptosis, are stored in the mitochondrial intermembrane space as SNOs.  

5.Methylation

The transfer of one-carbon methyl groups to nitrogen or oxygen (N- and O-methylation, respectively) to amino acid side chains (lysine or arginine) increases the hydrophobicity of the protein and can neutralize a negative amino acid charge when bound to carboxylic acids. Methylation is mediated by methyltransferases, and S-adenosyl methionine (SAM) is the primary methyl group donor is acts as a co factor. 

 N-methylation is irreversible, O-methylation is potentially reversible. Methylation is a well-known mechanism of epigenetic regulation, as histone methylation and demethylation influences the availability of DNA for transcription. (Helps in unwinding and in reverse process ). It is common in histone 3(H3).

6.N-Acetylation- (unwrapping of Dna from histones)

N-acetylation, or the transfer of an acetyl group to nitrogen, occurs in almost all eukaryotic proteins. N-terminal acetylation requires the cleavage of the N-terminal methionine by methionine aminopeptidase (MAP).(Prevents degradation)

Acetylation at the ε-NH2 of lysine (termed lysine acetylation) on histone N-termini is a common method of regulating gene transcription. Histone acetylation is a reversible event that reduces chromosomal condensation to promote transcription, and the acetylation of these lysine residues is regulated by transcription factors that contain histone acetyletransferase (HAT) activity. histone deacetylase (HDAC) enzymes are co-repressors that reverse the effects of acetylation by reducing the level of lysine acetylation and increasing chromosomal condensation.

.

                           http://1.bp.blogspot.com/_ITzUKgL3xBE/SiOHw_ny6I/AAAAAAAAACo/ON2YDNj6PPQ/s1600/KIPIJ.jpg

Protein acetylation can be detected -

1.By chromosome immunoprecipitation (ChIP) using acetyllysine-specific antibodies 

2. By mass spectrometry, where an increase in histone by 42 mass units represents a single acetylation.

NOTE- Protein stability is enhanced by hydroxylation of P in collagen and carboxylation of E in prothrombin.

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TYPES OF GRAFTINGS

AUTOGRAFT(Within an individual )

SYNGENETIC GRAFT(FROM identical twins or clone ISOGRAFT)

ALLOGRAFT(homograft     between two human beiengs as they both are from same species  )(From non identical individuals  same species) 

XENOGRAFT from different species or from one species to another species.(heterograft) Between a man and monkey or from pig to humans 

Transfer of cells, tisues or an organ between different parts of the body or between different individuals.(TRANSPLANTATION)

In case of plants graft is  a small shoot or scion of a tree inserted in another tree, the stock of which is to support and nourish it the two unite and become one tree, while transplant is an act of uprooting and moving (something). 

Autografts  and isografts are usually accepted  due to the genetic identify between graft and host . Allograft is genetically dissimilar to the rest it is often recognized as foreign by the immune system and is rejected. 

Q1 Type of graft best suited for renal transplantation-

a.Autograft

bxenograft

cAllograft

d  Isograft

Ans. D Isograft (graft given from a genetically identical twin, dose not require immunosuppressive therapy, therefore it is most suitable.)

Q2. NEOGRAFT -Automated hair transplant system 

Q3.In different types of tissue transplantations, the rate of graft rejection in decreasing order is

 (A) Isograft > Xenograft > Allograft

 (B) Allograft > Isograft > Xenograft 

(C) Xenograft > Autograft > Allograft

 (D) Xenograft > Allograft > Isograft 

Ans.D

4.Gold standard for bone regenerative grafting materials? 
A. Xenogenic bone
B. Allogenic bone
C. Autogenous bone
D. Alloplastic bone

Ans. c

5.Graft of dopamine producing cell is needed to cure

A Parkinson disease
B Haemophillia
c Epilepsy

Ans. A

6.Best Bone Graft?

A. Autograft with Alloplastic
B. Allograft
C. Alloplastic
D. Xenograft

Ans. A
7.Graft from sister to brother-

A. Isograft
B. Allograft
C. Autograft
D. Heterograft

Ans. B

8.Organism causing destruction of skin graft is
A. Streptococcus
B. Staphylococcus
C. Pseudomonas
D. Clostridium

Ans. A

9.A woman with infertility receives an ovary transplant from her sister who is an identical twin. What type of graft is it ?

A) Autograft
B) Allograft
C) Isograft
D) Xenograft

Ans. c

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BRIEF INTRODUCTION ABOUT VDJ RECOMBINATION

V-VARIABLE SEGMENT
D-DIVERSITY SEGMENT
J-JOINING SEGMENT

• 1.Which of the following is only contained in heavy chains and not in light chains?

• A. 
  Leader (L)
• B. 
  Joining (J)
• C. 
  Diversity (D)
• D. 
  Variable (V)                                                 Ans. c 

• 3.During the development of B cells VDJ recombination occur the Ig gene segments are rearranged and brought next to each other to form a contiguous functional gene (somatic recombination).

  
  

• 4.In the developing B cell, the first recombination event  occur in between one D and one J gene segment of the heavy chain locus. This D-J recombination is followed by the joining of one V gene segment, from a region upstream of the newly formed DJ complex, forming a rearranged VDJ gene segment.
  
  
  

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GENE KNOCK IN,KNOCK OUT &KNOCK DOWN INTRODUCTION

KNOCK IN - What it implies something is inserted.
KNOCK OUT - It indicates something is omitted
KNOCK DOWN -  Need to down or decrease some expressions.

GENE KNOCK  IN-- Its the opposite of knock down gene. A gene is inserted from outside  in a region to see the function and interaction .

Techniques used -
1.Site specific recombination
2.By using transposons 

GENE KNOCK OUT - It is a method of creation of transgenic animals .The first knockout mouse was created by Mario R,Martin Evans and Oliver in 1889 for which they were awarded with nobel prize on 2007. It is a long term process .

 What do each of the genes present in a body do? Geneticists studied this by inactivating, or “knocking out”, the corresponding gene in mice or other model organisms. Then they observed whether their is any difference from normal behaviour or not. 
Process involves knocking out (made inoperative) of both the alleles of a gene and replacing it with a artificial piece of DNA which is inactive. The apperance of change in any phenotypic signs implies the function of knockout gene. 
Knocking out 2 genes simultanoeusly is called DKO (Double knockout ).

Functions-  1.To know what exactly happens when a gene is absent or mutated.
              2. To study the role of genes which have been sequenced but have unknown function.
               

Techniques used -
1. In test tube with plasmid .                                      
2.By BAC(bacterial artificial chromosomes)
3.With cell cultures                               
4.Homologous recombination or gene targeting 
5.Introducing short sequences of lox-p sites around the gene.

Goal - To create a transgenic animal that have altered gene .


Image source credit-https://www.genome.gov/12514551/

GENE KNOCK DOWN - Idea is same need to ommit the expression of desired gene but not completely . Genes are present but they are blocked . It is not physically deleted we need to block only the expression . It can achived in few weeks .

Technique used- By RNAi 

NOTE-
A knockin model replaces an endogenous gene with a mutant gene, but leaves the original promoter region intact such that expression is controlled by normal regulatory mechanisms. Therefore, gene expression typically occurs at normal levels. 
Knockout models remove a gene from the genome, and thus do not increase levels of gene expression. 

MCQs.  - 
1.Homologous recombination can be employed to generate 
A.  transgenic animals
B. Gene knockout animals 
C. Site specific mutagenesis 
D. Specific promoter sequences
Ans.  B (GATE XL - 2005) 

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