krit.club logo

Continuity and change - DNA replication

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

DNA replication is a semi-conservative process, meaning each daughter DNA molecule contains one original parent strand and one newly synthesized strand. This was proved by the Meselson-Stahl experiment using 15N^{15}N and 14N^{14}N isotopes.

•

The enzyme Helicase unwinds the double helix and separates the two strands by breaking the hydrogen bonds between complementary base pairs (A=TA=T and G≡CG \equiv C).

•

DNA Gyrase (Topoisomerase) moves ahead of helicase to reduce torsional strain and prevent supercoiling of the DNA double helix.

•

DNA Polymerase III adds nucleotides to the 3′3' end of the primer, synthesizing the new strand in a 5′→3′5' \to 3' direction. It requires a free 3′3'-OH group to attach the next phosphate.

•

Replication is continuous on the Leading Strand (moving toward the replication fork) and discontinuous on the Lagging Strand (moving away from the fork), creating short segments known as Okazaki fragments.

•

DNA Primase creates a short RNA primer to provide a starting point for DNA Polymerase III. DNA Polymerase I later removes these RNA primers and replaces them with DNA nucleotides.

•

DNA Ligase catalyzes the formation of phosphodiester bonds to join the sugar-phosphate backbones of Okazaki fragments together.

📐Formulae

A=T,G=CA = T, \quad G = C

%A+%G=%T+%C=50%\%A + \%G = \%T + \%C = 50\%

Number of DNA molecules after n generations=2n\text{Number of DNA molecules after } n \text{ generations} = 2^n

Number of polynucleotide strands=2n+1\text{Number of polynucleotide strands} = 2^{n+1}

💡Examples

Problem 1:

A sample of double-stranded DNA is analyzed and found to contain 18%18\% Guanine (GG). Calculate the percentage of Thymine (TT) present in this DNA sample.

Solution:

According to Chargaff's rule, G=CG = C and A=TA = T.

  1. Given G=18%G = 18\%, then C=18%C = 18\%.
  2. Total G+C=18%+18%=36%G + C = 18\% + 18\% = 36\%.
  3. Since the total percentage must be 100%100\%, then A+T=100%−36%=64%A + T = 100\% - 36\% = 64\%.
  4. Because A=TA = T, we divide by 22: T=64%2=32%T = \frac{64\%}{2} = 32\%.

Explanation:

In double-stranded DNA, the concentration of purines equals the concentration of pyrimidines. Since Guanine pairs with Cytosine and Adenine pairs with Thymine, knowing the percentage of one base allows for the calculation of all others.

Problem 2:

A bacterium containing only 15N^{15}N DNA is moved to a medium containing only 14N^{14}N and allowed to replicate for 33 generations. Determine the ratio of hybrid DNA (15N/14N^{15}N/^{14}N) to light DNA (14N/14N^{14}N/^{14}N).

Solution:

  1. After n=3n = 3 generations, the total number of DNA molecules is 2n=23=82^n = 2^3 = 8.
  2. The original two 15N^{15}N strands act as templates and remain in the population, meaning there will always be exactly 22 hybrid molecules (15N/14N^{15}N/^{14}N) regardless of the number of generations.
  3. Number of light molecules (14N/14N^{14}N/^{14}N) = Total molecules - Hybrid molecules = 8−2=68 - 2 = 6.
  4. The ratio of Hybrid : Light is 2:62 : 6, which simplifies to 1:31 : 3.

Explanation:

Because DNA replication is semi-conservative, the two original heavy strands are never destroyed; they are simply partitioned into two different hybrid molecules. All subsequent synthesis uses the 14N^{14}N available in the medium.