DNA Replication | Anatomy2Medicine
DNA Replication

DNA Replication

DNA replication

    • In both prokaryotes and eukaryotes, DNA replication is semiconservative and involves both continuous and discontinuous (Okazaki fragment) synthesis.


  • Origin of replication


      • Particular consensus sequence of base pairs in genome where DNA replication begins.
      • May be single (prokaryotes) or multiple (eukaryotes).


  • Replication fork


      • Y-shaped region along DNA template where leading and lagging strands are synthesized.


  • Helicase


      • Unwinds DNA template at replication fork.


  • Single-stranded binding proteins


      • Prevent strands from reannealing.


  • DNA topoisomerases


      • Create a single- or double-stranded break in the helix to add or remove supercoils.
      • Fluoroquinolonesinhibit DNA gyrase (prokaryotic topoisomerase II).


  • DNA polymerase III


      • Seen in Prokaryotic only
      • Elongates leading strand by adding deoxynucleotides to the 3 end.
      • Elongates lagging strand until it reaches primer of preceding fragment.
      • 3 5exonuclease activity “proofreads” each added nucleotide.
      • DNA polymerase III has 5 to 3 synthesis and proofreads with 3 to 5 exonuclease.


  • DNA polymerase I


      • Seen in Prokaryotic only.
      • Degrades RNA primer; replaces it with DNA.


  • Has same functions as DNA polymerase III but also excises RNA primer with 5′ to 3′ exonuclease.
  • DNA ligase


      • Catalyzes the formation of a phosphodiester bond within a strand of double-stranded DNA (i.e., joins Okazaki fragments).
      • Ligase Seals.


  • Telomerase


    • An RNA-dependent DNA polymerase
    • it adds DNA to 3 ends of chromosomes to avoid loss of genetic material with every duplication.

Exam MCQ

Which of the following reactions is required for proofreading (i.e. correcting replication errors) during DNA replication by DNA polymerase III?

a ) 3′ – 5′ exonuclease activity

  1. b) 5′ – 3′ exonuclease activity
  2. c) 3′ – 5′ endonuclease activity
  3. d) 5′ – 3′ endonuclease activity

Answer c) 3′ – 5′ endonuclease activity

Explanation The proofreading activity possessed by many DNA polymerases is an exonuclease activity that degrades mismatched bases that have been wrongly incorporated into the growing chain. This is, therefore, an exonuclease activity ( exonucleases digest from the end of a DNA chain) and it operates ‘backwards’ from the 3′ growing end, i.e. 3′ – 5′

Exam MCQ

How does the mismatch repair system distinguish between the parental (i.e. correct) DNA strand and the newly synthesised strand containing the mismatched base?

  1. a) Thymine in the parental strand of the helix is methylated at GATC.
  2. b) Thymine in the new strand of the helix is methylated at GATC.
  3. c) Guanine in the parental strand of the helix is methylated at GATC.
  4. d) Guanine in the new strand of the helix is methylated at GATC.

Answer d)

Guanine in the new strand of the helix is methylated at GATC.


Mismatch repair is a system that repairs mismatches that have slipped evaded proofreading during DNA replication. DNA becomes methylated at the G of GATC sequences after replication; however, this does not occur immediately so the new strand, containing the error, can be distinguished from the methylated parental strand

Exam MCQ

What is the name of the DNA repair system in E. coli in which dual incisions are made in the damaged part of the double helix, and a 12-13 base segment is removed and replaced with new DNA ?

  1. a) Mismatch repair
  2. b) Base excision repair
  3. c) Nucleotide excision repair
  4. d) AP site repair

Answer c) Nucleotide excision repair


    • Nucleotide excision repair is an almost universal repair mechanism in which a section of damaged DNA is removed and replaced with new DNA by a DNA polymerase.


  • It is used to repair photoproducts caused by UV damage and bulky DNA lesions caused by a variety of mutagens


Exam MCQ

Which of the following is the name of the human genetic disorder resulting from defects in nucleotide excision repair ?

  1. a) Hereditary nonpolyposis colorectal cancer (HNPCC)
  2. b) Xeroderma pigmentosum (XP)
  3. c) Lynch syndrome
  4. d) Diabetes

Answer b) Xeroderma pigmentosum (XP)

Explanation People born with the disorder, xeroderma pigmentosum , have a mutation in one of the genes coding for nucleotide excision repair enzymes. Therefore they are unable to carry out efficient repair on sunlight damage and they are hypersensitive to sunlight. They have to protect their skin from daylight or risk getting skin cancer

Exam MCQ

In which of the following would you find telomeres?

  1. a) Human mitochondrial DNA
  2. b) Human chromosomes
  3. c) Bacterial chromosomes
  4. d) All of above

Answer b) Human chromosomes


    • Telomeres are found at the ends of the linear double-stranded DNA molecules in human chromosomes.
    • They protect chromosome ends from nucleases and they also provide a special mechanism for replication of chromosome ends, using the enzyme telomerase.


  • Circular molecules such as bacterial chromosomes and most mitochondrial genomes do not need these specialised DNA ends


Exam MCQ

Which of the following enzymes are used to join bits of DNA

  1. a) DNA ligase
  2. b) DNA polymerase
  3. c) primase
  4. d) Endonuclease

Answer DNA ligase

Explanation DNA ligase enzyme joins bits of DNA on lagging strand

Exam MCQ

Why is an RNA primer considered essential during DNA synthesis by DNA polymerase III ?

  1. a) The enzyme requires a free 3′-PO 4 group.
  2. b) The enzyme requires a free 5′-PO 4 group.
  3. c) The enzyme requires a free 5′-OH group.
  4. d) The enzyme requires a free 3′-OH group.
  5. e) There is no particular reason, that is simply the observation .

Answer d)

The enzyme requires a free 3′-OH group


    • Polymerase III requires a free 3′-OH group to begin synthesis of DNA.


  • An RNA primer provides a 3′ hydroxyl group, from which the DNA polymerase can start synthesis


Exam MCQ

Which molecule serves to destabilize the DNA helix in order to open it up, creating a replication fork ?

a)DNA helicase

  1. b) DNA ligase
  2. c) DNA polymerase
  3. d) SSBPs
  4. e) DNA gyrase

Answer a) DNA Helicase

Explanation The DNA helicase enzyme destabilizes the DNA helix by breaking hydrogen bonds

Exam MCQ

For DNA Replicaion , unwinding of DNA is done by

  1. a) Helicase
  2. b) ligase
  3. c) Hexonuclease
  4. d) Topoisomerase

Answer a) Helicase

Exam MCQ

The elongation of the leading strand during DNA synthesis

  1. a) Progresses away from the replication fork
  2. b) Occur in 3’-5’ direction
  3. c) Produces Okazaki fragment
  4. d) Depend on the action of DNA polymerase

Answer d) Depend on the action of DNA polymerase


    • Takes place with advancement of DNA polymerase on the template ,


  • On leading strand it is towards the neck of Replication forlk in 5’—-3’ direction



Exam MCQ

Eukaryotes differ from prokaryote in mechanism of DNA replication due to:

  1. a) Different enzyme for synthesis of lagging and leading strand
  2. b) Use of DNA primer rather than RNA primer
  3. c) Unidirectional rather than bidirectional replication
  4. d) Discontinuous rather than semidiscontinuous replication

Answer d) Discontinuous rather than semidiscontinuous replication

Explanation Okazai fragments are smaller and are formed discontinuousl

Exam MCQ

True replication of DNA is possible due to

  1. a) Hydrogen bonding
  2. b) Phosphate backbone
  3. c) Complementary base pairing rule
  4. d) None of the above

Answer c) Complementary base pairing rule

Explanation Adenine pairs with Thymine only( A = T ) and Guanine pairs with Cytosine ( G C ) only ,because of strict base pairing pattern the new daughter strand is assembled according to bases present on the template

Exam MCQ

All of the following are differences between eukaryotic and prokaryotic DNA replication except

  1. a) the type and number of polymerases involved in DNA synthesis
  2. b) multiple vs. single replication origins
  3. c) the rate of DNA synthesis the ability to form a replication fork
  4. d) the ability to form a replication fork

Answer D

Explanation Both prokaryotes and eukaryotes form replication forks during DNA replication

Exam MCQ

Telomeres have all of the following properties or characteristics except

  1. a) the telomerase enzyme
  2. b) forming “hairpin” loops
  3. c) a link to the aging process in eukaryotic cells being found in both eukaryotic and prokaryotic chromosomes
  4. d) being found in both eukaryotic and prokaryotic chromosome

Answer d

Explanation The telomere sequence occurs at the ends of linear eukaryotic chromosomes . Prokaryotes have circular chromosomes so donot have the terminities or telomere

Exam MCQ

The discovery of Okazaki fragments suggested that DNA synthesis is __________.

  1. a) discontinuous
  2. b) continuous
  3. c) 3 ‘ to 5’
  4. d) semiconservative

Answer a) discontinuous

Explanation DNA must be synthesized on both strands of the double helix at the same time. However, one strand runs in a 5′ to 3′ direction and one runs in the 3′ to 5′ direction.

In order for replication to occur on both strands in the same direction simultaneously, one strand must be made in a discontinuous fashion (in short pieces—Okazaki fragments) and reannealed later

Exam MCQ

In vivo synthesis of DNA is __________

  1. a) 3′ to 5′
  2. b) 5′ to 3′
  3. c) both 3′ to 5′ and 5′ to 3′
  4. d) neither 3′ to 5′ nor 5′ to 3′

Answer b) 5′ to 3‘

Explanation DNA polymerase must add nucleotides in a 5′ to 3′ direction. Therefore, DNA synthesis must also occur in this direction

Exam MCQ


Which of the following forms of DNA can serve as a template for DNA polymerase

  1. Partially double stranded DNA
  2. Circular double stranded DNA
  3. Intact double stranded DNA
  4. Circular single stranded DNA

Answer a) Partially double stranded DNA

Explanation DNA double helix gradually opens by DNA helicse by dissolving H bonds between base pairs giving single strands which behave as template for DNA polymerase

Exam MCQ

All input of energy is not required for which of the following steps of replication ?

  1. Separation of two strand
  2. Unwinding of DNA duplex
  3. Linking together of the individual deoxynucleotides
  4. All of the above require energy

Answer d) All of the above require energy

Explanation Most of the steps of DNA replication require energy which is generally provided by hydrolysis of energy rich comounds like ATP


Exam MCQ

Why should patients with Xeroderma pigmentosa avoid sunlight

  1. a) The UV wavelengths do irrepairable damage to DNA
  2. b) Sunlight inhibits any residual DNA repair activity of the cell
  3. c) These patients lack pigmentation to protect them from burning
  4. d) Sunlight inhibits DNA polymerases

Answer a) The UV wavelengths do irrepairable damage to DNA

Exam MCQ

The exonuclease activity of the DNA polymerases functions to

  1. Remove the RNA primer sequences
  2. Proofread the new DNA strand and remove inappropriate nucleotides
  3. Maximize the fidelity of DNA replication
  4. All of the above

Anwer d) All of the above

Explanation By having exonuclease activity it can remove the nucleotides from the ends of Okazaki fragments and also by moving backwards for proof reading

Exam MCQ

The last DNA to be replicated in the eukaryotic chromosome is

  1. Telomeres at the end of the chromosomes
  2. Heterochromatin
  3. Euchromatin in the arms of the chromosome
  4. Facultative heterochromatin

Answer b) heterochromatin

Explanation As heterochromatin is densely packed it is always late replicating