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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