RNA FLASH CARD

1.MYC -----------------------------------------------------------Transcription factor helps in activation(CSIR)
2,Retinoids upregulate trascription----------------------------1 and 7 collagens
3.Heavy methylated Dna----------------------------------------Low transcription

4.Types        (CSIR QUESTIONS )  DNA   OF BIOSCIENCE 
a.Heterohenous nuclear Rna ---------------------------------precursor of other RNAs
b.Sc RNA -------------------------------------------------------selection of protein for export
c.Sno-RNA------------------------------------------------------r-RNA processing
d.Sn-RNA -------------------------------------------------------m-RNA processing

5.TRANSCRIPTION FACTORS-
a.TF II A-------------------------------------------------------promoter recognition,nuclear origin,RNApol II                                                                                                forming pre-                                                                                                                                                             inititaion complex
b.TF II B-------------------------------------------------------                   do ------
c.TF II D-------------------------------------------------------TBP (Tata binding protein) is the subunit .TATA                                                                                        Ssequence is recognized by TFII D.
d,TF II E------------------------------------------------------Melting of DNA,  consist of zinc finger motif
e.TF II F-----------------------------------------------------Stabilize the Rna pol II
f.TH II H ---------------------------------------------------nucleotide excision repair,xeroderma                                                                                                                  pigmentosum  (CSIR QUESTION)

DNA   OF BIOSCIENCE 

6.Retroviral genome(contains 3 proteins)(CSIR QUESTION)
a.Gag --------------------------------------------------------It produce capsid proteins(p24 and p17)
b.Pol----------------------------------------------------------Reverse transcriptase,RNase H ,integrase                                                                                                       function & code for proteases.
c.env---------------------------------------------------------Resides in lipd layer, proteins of viral envelope                                                                                             also called transmembrane proteins(gp 41,gp 120)
DNA   OF BIOSCIENCE 
7.Inhibitors-
a.Alpha amanitin---------------------------------------------eukaryotic RNA pol II (CSIR QUESTION)
b.Actinomycin D------------------------------------------Blocks elongation of bacterial RNA pol.
c.Rifampicin----------------------------------------------bind to beta subunit of RNA pol .
d.Azidothymidine(AZT)-------------------------------Reverse transcriptase inhibition


8. Leucine zipper ------------------------------------Transcription factor

9.2'OH group in RNA ---------------------------------Transesterification

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Founder effect a type of genetic drift

Founder effect is a type of genetic drift.
What is genetic drift?
Random change in allele frequency with out any selection.
In a large population some small group of individuals splits and form  a new colony in a new area with less genetic diversity and the allele frequency of founder population is I would say it's totally different from the parent population.
It ruled out by Ernst Mar in 1942.

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GEOLOGICAL TIME SCALE

Geological time has been separated into eons,eras, periods, and epochs.

• Devonian- Age of fish 
• Cambrian -Age of trilobites
• silurian -terrestial plants
• Carboniferous-Age of amphibians,1st seed plants appered (terrestial life )
• Jurassic- Age of reptiles and gymnosperms
• Quaternary-Age of mammals(humans develop) and birds 
• Miocene- Age of flowering plants
• Permian (origin of conifers).
• Ordivician (First green plants and fungi) and cambrian -Age of invertebrates 
• Cretaceous- first flowering plants ,angiosperms in mid cretaceous 
• Mesozoic -(Age of reptiles)
• Paleiozoic era -first vertebrate animals colonized land.

Present day scale is -Phanerozoic Eon, Cenozoic Era, Quaternary Period, Holocene Epoch.

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SCALES OF BIODIVERSITY

The term was introduced by R. H. Whittaker .

Alpha diversity -It is  the diversity within a community,in a particular area or an ecosystem etc. (It represents number of species=species richness) 
Beta diversity -It represents the differences in species composition among sites(between the communities) with respect to change in environment .
β=γα 

The beta diversity between the woodland and the hedgerow habitats is 7 (representing the 5 species found in the woodland but not the hedgerow, plus the 2 species found in the hedgerow but not the woodland). Thus, beta diversity allows us to compare diversity between ecosystems.

High beta-diversity implies low similarity between species composition of different habitats. It is usually expressed in terms of similarity index between communities (or species turnover rate) between different habitats in same geographical area (often expressed as some kind of gradient).

$\frac{\mathrm{<b><span\; style="color:\; red;">Gamma\; diversity</span></b> -Is\; the\; diversity\; of\; the\; entire\; landscape\; or }}{}$Gamma diversity is a measure of the overall diversity for the different ecosystems within a region.   γ$=$βα 

Delta diversity is the change in diversity as you sample large landscapes along major climatic or other physical gradients.

1.Alpha, beta and gamma diversity refer to :

(A) Genetic diversity
(B) Landscape diversity
(C) Species diversity
(D) Population diversity                                                              Ans.  D

2. As we move from one geographical region to next neighbouring region species diversity tends to change .It is termed as -
a. α-diversity b.β-diversity  c.γ-diversity    d.delta diversity             Ans. b

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4 TYPES OF SPECIATION EASY TO REMEMBER

Allopatric speciation -

(sympatric- a subset forms a new species )

• Gene flow interrupted by geographic isolation(environmentals events) so called geographic speciation  or population is separated by a physical barrier .
• ex. Galapagous finches, Grand canyon squirrels 

Sympatric speciation - (reproductive or behavioural separation)

• It happens when individuals of a population are not isolated physically but they reproductively and genetically isolated  due to genetic drift ,mutation and selection which leads to new species.
•  ex.Different polyploidy in plants(wheat grains) ,sexual selection in animals(genetic           polymorphism)          

GENETICS NUMERICALS FULLY SOLVED

1.The genotype of F₁ individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F₂ offspring would have the following genotypes?
1.aabbccdd
2.AaBbCcDd
3AABBCCDD
4.AaBBccDd
5.AaBBCCdd
a. aabbccdd = 1/4*1/4*1/4*1/4=1/256 
b. AaBbCcDd = 1/2*1/2*1/2*1/2= 1/16
c. AABBCCDD = 1/4*1/4*1/4*1/4 = 1/256
d. AaBBccDd = 1/2*1/4*1/4*1/2 = 1/64
e. AaBBCCdd = 1/2*1/4*1/4*1/4 = 1/128
Probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4.

2. The possible genotypes of endosperms borne on a heterozygous (Rr)plant will 

be -RRR,RRr,rrR,rrr

4.A rare x-linked allele is present in the human population at a frequency of 1 in 

10^5 .Among those who carry the allele ,the proportion of males is about -

a.1 in 2                                                 c.1in 4

b.1 in 3                                                d.1 in 5 

Ans.  If the X-linked allele is present at a frequency of 1/10000, then if the frequency of the allele is given by q, then: q = 1/10000 = 0.0001 p + q = 1 ,p = 1 - 0.0001 ,p = 0.9999 So, the heterozygotes=2pq = 2 x 0.9999 x 0.0001 = 0.00019998 The recessive homozygotes =q^2 = 0.0001 x 0.0001 = 0.00000001 

Since the frequency of males with the allele is equal to the frequency of the X-linked allele in the population, then the frequency of males in the population with the allele is 0.0001 
So, the overall frequency of persons with at least one allele is: 
0.0001 (hemizygotes, all males) + 0.00000001 (recessive homozygotes, all female) + 0.00019998 (heterozygotes, all female) = 0.00029999 
So, the proportion of males with the allele out of all person having at least one allele is 0.0001/0.0002999, which is approximately 1/3, so the answer is B.

6.The likelihood of an individual in a population carrrying two specific alleles of a human Dna marker ,each of which has a frequency of 0.2,will be a.0.4            b.0.32          c.   0.16              d.0.08

Ans.use square .2 twice and add them together to get the combined likelihood 

(.2)^2 + (.2)^2 = .08 

8. Red-green color blindness is caused by a sex-linked recessive allele.  A color-blind man marries a woman with normal vision whose father was color-blind.  Man X^bY  Woman X^BX^b.  We know this because she has normal vision (X^B).  Her father was color-blind and gave her her X^b allele.

    A) What is the probability that their daughter will be color-blind?

	XB	Xb
Xb	XBXb	XbXb
Y	XBY	XbY

        The chance that she will be color-blind is 1/2.    B) What is the probability that their son will be color-blind? 
       The chance that a son will be color-blind is also 1/2.
9.The genetic map of three genes in Drosophila melanogaster is given below:
                                 a ----------------------b-----------c
                                               10cm                  5cm
A cross as given below is made between individuals of the genotypes:
a+--------b+----------- c                 a----------b------------c+
a+--------b+-----------c   X          a  ---------b-------------c+ 
The female F1progeny are test-crossed and1000 progeny are obtained. Assuming that there has been no double crossover, what is the expected number of progeny with the genotypes:
A)  a+--------b---------- c+    B)a+--------b+----------- c+     C)a+--------b+----------- c 
      a  ---------b-----------c            a  ---------b-----------c           a  ---------b-----------c     
set (A)     B   C
a. 100   50 850
b. 50     25 425
c. 100  850 50
d. 0      425 75
Ans. F1 test cross: a+b+c/ a b c+ x a b c/ a b c
(A)  The proportion a+ b c+ will be 10/2 % = 5%. In f2 there are 1000 individuals. So the proportion of one having a+ b c+/ a b c will be 5% of 1000=50.
(B)  The proportion a+ b+ c+ will be 5/2 % =2.5%.In f2 there are 1000 individuals. So the proportion of one having a+ b+ c+/ a b c will be 25.
(C) Proportion of total parental variety (a+ b+ c and a b c+) = Total progeny- (Proportion of recombinants produced as a result of SCO between a and b gene + Proportion of recombinants produced as a result of SCO between b and c gene)
Total Parental proportion = 1000 - [(10% of1000) + (5% of 1000)]
= 1000 - 150 = 850.a+ b+ c/ a b c is one of the parental variety,so its number will be 850/2 = 425.So the answer will be option B. A 50, B 25 and C 425.
10. In doing a chi-square test on the results of a trihybrid cross, the number of degrees of freedom would be

A. 8   B. 7  C. 4 D. 3               Ans. B (n-1) =8-1=7 

11.If two people who are both heterozygous for Huntington's disease (an autosomal dominant trait) marry, what is the probability that they will have three children, all of which are normal?

A. 0.140 B. 0.422 C. 0.250 D. 0.016                   Ans. D

12.Given an individual who is heterozygous at 5 loci, how many different gametic genotypes are possible?

A. 25  B. 10 C. 15  D. 32       Ans. 32 (n=5  formula is 2^n=2^5)

13.In a family of 3 children what is the propbability that all 3 are boys?

Ans.1/2*1/2*1/2=1/8    

14. In a cross between AABBCcDdEe* AaBbCcDdEe,what is the propability that the offspring will be AABbccddEE ?
AAXAa=1/2,  BBXBb=1/2 ,DdxDd=1/4 ,EexEe=1/4

Ans. 1/2x1/2x1/4x1/4x1/4=1/512
15.A recessive inherited disease is expressed only in individuals of blood group O and not expressed in blood groups A B and AB. Alleles controlling the disease and blood groups are independently inherited. A normal woman with blood group A and her  normal husband  with blood group B already had one child with the disease . The woman is pregnant for second time . What is the probability that the second child will also have the disease ?
Ans. 1/2*1/2*1/4 =1/16 
Explanation (Recessive allele probability =1/4 normal allele probability 1/2 )
Here given parents are normal that means probability is 1/2*1/2=1/4
The child is recessive means its probability is 1/4 . So, 1/2*1/2*1/4 =1/16 
16.Two varieties of maize averaging 48 and 72 inches in 42 are crossed. The F1 progeny is quite uniform averaging 60 inches in height. Of the 500 F 2 plants, with shortest 2 are 48 inches and the tallest 2 are 72 inches. What is the probable number of polygenes involved in this trait.
Ans .4 genes are inovled because 1/4^n=1/250

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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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