In a 3.2 text{ Kbp} long piece of DNA,820 ade | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Total length of $DNA$ = $3.2 	ext{ Kbp} = 3200 	ext{ bp}$.According to Chargaff's rule,the amount of adenine $(A)$ is equal to thymine $(T)$,and

Vedclass · Accepted Answer

$780$

Vedclass · Answer

(A) Total length of $DNA$ = $3.2 	ext{ Kbp} = 3200 	ext{ bp}$.According to Chargaff's rule,the amount of adenine $(A)$ is equal to thymine $(T)$,and the amount of guanine $(G)$ is equal to cytosine $(C)$.Given,$A = 820$,therefore $T = 820$.Total $A + T$ content = $820 + 820 = 1640$.Total $G + C$ content = Total base pairs - Total $A + T$ content = $3200 - 1640 = 1560$.Since $G = C$,the number of cytosine bases = $\frac{1560}{2} = 780$.

In a $3.2 \text{ Kbp}$ long piece of $DNA$,$820$ adenine bases were found. What would be the number of cytosine bases?

Similar Questions

The nitrogen base found only in $DNA$ is also called

$DNA$ is a double helix and:

What are the length of one full turn of the $DNA$ helix and the distance between the two strands (polynucleotides)?

$DNA$ differs from $RNA$ because:

$A$ human diploid cell contains ............ bp.

Vedclass Test Series

Exam Paper Generator

Online Exam Module