MCQ
Two nucleotides are joined together by a linkage known as :
  •  Phosphodiester linkage
  • B
    Glycosidic linkage
  • C
     Disulphide linkage
  • D
     Peptide linkage

Answer

Correct option: A.
 Phosphodiester linkage
a
 Phosphodiester linkage 

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Number of $sp$ hybridised carbon atom in the given compound is $C{H_2} = CH-C \equiv C-C{H_2}-CN$
Consider separate solutions of $0.500\, M\, C_2H_5OH\,(aq),$ $0.100\,M\,Mg_3(PO_4)_2\, (aq), $ $0.250\, M\,KBr\,(aq)$ and $0.125\, M\,Na_3PO_4\,(aq)$ at $25^o C.$ Which statement is true about these solutions, assuming all salts to be strong electrolytes ?
Which is the most suitable reagent for the following transformation ?

$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\ 
  {C{H_3} - CH = CH - C{H_2} - CH - C{H_3}} 
\end{array}$  $\longrightarrow $    $C{H_3} - CH = CH - C{H_2}C{O_2}H$

The correct order of solubility in water for $He,\,Ne,\,Ar,\,Kr,\,Xe,\,$ is
Calculate the volume of $CO_2$ at $NTP$ that can be obtained by heating $200\,gms$ of $80\%$ pure $CaCO_3(s)$ ............. $\mathrm{L}$
In the following reactions, the total number of oxygen atoms in $X$ and $Y$ is $........$.

$Na _2 O + H _2 O \rightarrow 2 X$

$Cl _2 O _7+ H _2 O \rightarrow 2 Y$

Percentage of silver in German silver is
Give the correct of initials $T$ or $F$ for following statements. Use $T$ if  statement is true and $F$ if it is false.

$(I)\, Co(III)$ is stabilised in presence of weak field ligands, while $Co(II)$ is  stabilised in presence of strong field ligand. 

$(II)$ Four coordinated complexes of $Pd(II)$ and $Pt(II)$ are diamagnetic and  square planar.

$(III)\,[Ni (CN)_4]^{4-}$ ion and $[Ni (CO)_4]$ are diamagnetic tetrahedral and  square planar respectively. 

$(IV)\,Ni^{2+}$ ion does not form inner orbital octahedral complexes.

For the complexes showing the square pyramidal structure, the $d-$ orbital involved  in the hybridisation is
In the following reaction $C{H_3} - C{H_2} - C{H_2} - C{H_3} \xrightarrow[475\, K]{H_2SO_4}$