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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