$A$ histone octamer consists of:
- A$H_2A(2), H_2B(2), H_3(2), H_4(2)$
- B$H_1(2), H_2(2), H_3(2), H_4(2)$
- C$H_1A(2), H_1B(2), H_2(2), H_4(2)$
- D$H_1A(3), H_1B(1), H_2(1), H_4(3)$
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Similar Questions
Supercoiled $DNA$ can be traced in
Medium
View SolutionMatch the following columns:
Column-$I$ Column-$II$ $P$. Bacteriophage $\phi \times 174$ $I$. $5386$ nucleotides $Q$. Bacteriophage lambda $II$. $48502 \, bp$ $R$. Escherichia coli $III$. $3.3 \times 10^9 \, bp$ $S$. Human haploid $DNA$ $IV$. $4.6 \times 10^6 \, bp$
| Column-$I$ | Column-$II$ |
| $P$. Bacteriophage $\phi \times 174$ | $I$. $5386$ nucleotides |
| $Q$. Bacteriophage lambda | $II$. $48502 \, bp$ |
| $R$. Escherichia coli | $III$. $3.3 \times 10^9 \, bp$ |
| $S$. Human haploid $DNA$ | $IV$. $4.6 \times 10^6 \, bp$ |
Easy
View SolutionWhich of the following statements is correct?
$A$ $DNA$ polymer has a free phosphate group at one end of the sugar,which is called the $P$ end of the polynucleotide chain. Similarly,at the other end of the polymer,there is a free $OH$ group of the sugar,which is called the $Q$ end of the polynucleotide chain.
$\quad P \quad Q$
$\quad P \quad Q$
Medium
View SolutionThe length of $DNA$ in $E. coli$ is . . . . . . .
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