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1. 1
DNA Structure
Document created: 01.27.2022 23:15
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Passing score: 50%
Total number of questions: 12
Number of questions to ask: 12
Answer submission: Submit one question at a time
Number of quiz retries: None
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Send quiz results to instructor: mbbsblues@gmail.com
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3. 3
Imported Questions (12/12 questions)
Question 1. Multiple Choice, 5 points, 1 attempt
If a sample of DNA if adenine is 23% what will be the amount of guanine present
Correct Choices
23 Percent
25 Percent
46 Percent
V 27 Percent
54 Percent
Feedback
Correct:
Based on Chargaff’ rule:
• Purines = Pyrimidine
• No. of Adenine = No. of Thymine
• So Adenine + Thymine = 46%
• Guanine + Cytosine = 54%
• No. of Guanine = No. of Cytosine
• So the amount of Guanine = 54/2 = 27%
Incorrect:
Based on Chargaff’ rule:
• Purines = Pyrimidine
• No. of Adenine = No. of Thymine
• So Adenine + Thymine = 46%
• Guanine + Cytosine = 54%
• No. of Guanine = No. of Cytosine
• So the amount of Guanine = 54/2 = 27%
Question 2. Multiple Choice, 5 points, 1 attempt
Which model of DNA was discovered by Watson and Crick?

4. 4
Correct Choices
A-DNA
V B-DNA
C-DNA
Z-DNA
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Correct: That's right! You chose the correct response.
Incorrect: You did not choose the correct response.
Question 3. Multiple Choice, 5 points, 1 attempt
Northern blot test is used for

5. 5
Correct Choices
DNA Analysis
V RNA Analysis
Analysis of Proteins
Enzyme Analysis
Feedback
Correct:
Northern blot: RNA analysis Southern blot: DNA analysis Western blot: Analysis of
proteins Answer: RNA analysis
Incorrect:
Northern blot: RNA analysis Southern blot: DNA analysis Western blot: Analysis of
proteins Answer: RNA analysis
Question 4. Multiple Choice, 5 points, 1 attempt
Excessive ultraviolet (UV) radiation is harmful to life. The damage caused to the biological systems by
ultraviolet radiation is by:
Correct Choices
Inhibition of DNA synthesis
V Formation of thymidine dimers
Ionization
DNA fragmentation

6. 6
Feedback
Correct:
Exposure of cell to ultraviolet light can result in the covalent joining of two adjacent
pyrimidines usually thymines producing a dimmer.
These thymine dimmers prevent DNA polymerase from the replicating the DNA strand
beyond the site of dimmer formation.
Important point:
First a UV specific endo-nuclease also called UV specific excinuclease recognizes the
dimmer and cleaves the damage strand. Damaged oglionucleotide is released.
Incorrect:
Exposure of cell to ultraviolet light can result in the covalent joining of two adjacent
pyrimidines usually thymines producing a dimmer.
These thymine dimmers prevent DNA polymerase from the replicating the DNA strand
beyond the site of dimmer formation.
Important point:
First a UV specific endo-nuclease also called UV specific excinuclease recognizes the
dimmer and cleaves the damage strand. Damaged oglionucleotide is released.
Question 5. Multiple Choice, 5 points, 1 attempt
To synthesize insulin on a large scale basis, the most suitable starting material obtained from the beta
cells of the pancreas is:

7. 7
Correct Choices
Genomic DNA
Total cellular RNA
c-DNA of insulin
V m-RNA of insulin
Feedback
Correct:
Explanation:
Insulin is synthesized on a large scale basis from complementary DNA (c-DNA), but this is
not the starting material because these c-DNAs are derived from mRNAs with the help of
Reverse transcriptase and DNA polymerase
1.
The c-DNA produced by this method is inserted into an appropriate cloning vector which
produces a large number of c-DNA clones.
Incorrect:
Explanation:
Insulin is synthesized on a large scale basis from complementary DNA (c-DNA), but this is
not the starting material because these c-DNAs are derived from mRNAs with the help of
Reverse transcriptase and DNA polymerase
1.
The c-DNA produced by this method is inserted into an appropriate cloning vector which
produces a large number of c-DNA clones.
Question 6. Multiple Choice, 5 points, 1 attempt
Cation used in PCR

8. 8
Correct Choices
Lithium
Calcium
Sodium
V Magnesium
Feedback
Correct:
PCR (Polymerase Chain Reaction) is a revolutionary
method developed by Kary Mullis in the 1980s.
PCR is based on using the ability of DNA polymerase to
synthesize new strand of DNA complementary to the offered template strand.
RT-PCR (Reverse Transcription PCR) is PCR preceded with conversion of sample RNA
into cDNA with enzyme reverse transcriptase.
Material required for PCR:
DNA template
Primers
DNA polymerase (Taq polymerase) Deoxyribonucleoside tri-phosphates (d-NTPs) Buffer
solution
Divalent cations-Mg++ or Mn++ (Generally Mg++ is used) Monovalent cation-K+
Incorrect:
PCR (Polymerase Chain Reaction) is a revolutionary
method developed by Kary Mullis in the 1980s.
PCR is based on using the ability of DNA polymerase to
synthesize new strand of DNA complementary to the offered template strand.
RT-PCR (Reverse Transcription PCR) is PCR preceded with conversion of sample RNA
into cDNA with enzyme reverse transcriptase.
Material required for PCR:
DNA template
Primers
DNA polymerase (Taq polymerase) Deoxyribonucleoside tri-phosphates (d-NTPs) Buffer
solution
Divalent cations-Mg++ or Mn++ (Generally Mg++ is used) Monovalent cation-K+
Question 7. Multiple Choice, 5 points, 1 attempt
A post graduate medical student working in a molecular biology laboratory is asked by her guide to
determine the base composition of an unlabeled nucleic acid sample left behind by a former research
officer. The results of her analysis show 10% adenine, 40% cytosine, 30% thymine and 20% guanine.
What is the most likely source of the nucleic acid in this sample?

9. 9
Correct Choices
Bacterial chromosome
Bacterial plasmid
Mitochondrial chromosome
Nuclear chromosome
V Viral genome
Feedback
Correct:
A base compositional analysis that deviates from Chargaff ’s rules (%A = %T, %C = %G) is
indicative of single-stranded, not double- stranded, nucleic acid molecule. All options
listed except VIRAL GENOME (A few viruses e.g. parvovirus have single-stranded DNA) are
examples of either circular or linear DNA double helices.
Incorrect:
A base compositional analysis that deviates from Chargaff ’s rules (%A = %T, %C = %G) is
indicative of single-stranded, not double- stranded, nucleic acid molecule. All options
listed except VIRAL GENOME (A few viruses e.g. parvovirus have single-stranded DNA) are
examples of either circular or linear DNA double helices.
Question 8. Multiple Choice, 5 points, 1 attempt
During RNA synthesis, the DNA template sequence TAGC would be transcribed to produce which of the
following sequences?

10. 10
Correct Choices
ATCG
GCTA
CGTA
AUCG
V GCUA
Feedback
Correct:
RNA is antiparallel and complementary to the template strand. Also remember that, by
convention, all base sequences are written in the 5′ to 3′ direction regardless of the
direction in which the sequence may actually be used in the cell.
Approach:
• Cross out any option with a T (RNA has U).
• Look at the 5′ end of DNA (T in this case).
• What is the complement of this base? (A)
Examine the options given. A correct option will have the complement (A in this example)
at the 3′ end. Repeat the procedure for the 3′ end of the DNA. This will usually leave only
one or two options.
Incorrect:
RNA is antiparallel and complementary to the template strand. Also remember that, by
convention, all base sequences are written in the 5′ to 3′ direction regardless of the
direction in which the sequence may actually be used in the cell.
Approach:
• Cross out any option with a T (RNA has U).

11. 11
Feedback
• Look at the 5′ end of DNA (T in this case).
• What is the complement of this base? (A)
Examine the options given. A correct option will have the complement (A in this example)
at the 3′ end. Repeat the procedure for the 3′ end of the DNA. This will usually leave only
one or two options.
Question 9. Multiple Choice, 5 points, 1 attempt
A 10-year-old girl is brought by her parents to the dermatologist. She has many freckles on her face,
neck, arms, and hands, and the parents report that she is unusually sensitive to sunlight. Two basal cell
carcinomas are identified on her face. Based on the clinical picture, which of the following processes is
most likely to be defective in this patient?
Correct Choices
Repair of double-strand breaks by error-prone homologous recombination
Removal of mismatched bases from the 3 -end of Okazaki fragments by a methyl-directed
process
V Removal of pyrimidine dimers from DNA by nucleotide excision repair
Removal of uracil from DNA by base excision repair
Feedback
Correct:
The sensitivity to sunlight, extensive freckling on parts of the body exposed to the sun,
and presence of skin cancer at a young age indicate that the patient most likely suffers
from xeroderma pigmentosum (XP). These patients are deficient in any one of several XP
proteins required for nucleotide excision repair of pyrimidine dimers in ultraviolet light–
damaged DNA. Double-strand breaks are repaired by nonhomologous end-joining (error

12. 12
Feedback
prone) or homologous recombination (error free). Methylation is not used for strand
discrimination in eukaryotic mismatch repair. Uracil is removed from DNA molecules by a
specific glycosylase in base excision repair, but a defect here does not cause XP.
Incorrect:
The sensitivity to sunlight, extensive freckling on parts of the body exposed to the sun,
and presence of skin cancer at a young age indicate that the patient most likely suffers
from xeroderma pigmentosum (XP). These patients are deficient in any one of several XP
proteins required for nucleotide excision repair of pyrimidine dimers in ultraviolet light–
damaged DNA. Double-strand breaks are repaired by nonhomologous end-joining (error
prone) or homologous recombination (error free). Methylation is not used for strand
discrimination in eukaryotic mismatch repair. Uracil is removed from DNA molecules by a
specific glycosylase in base excision repair, but a defect here does not cause XP.
Question 10. Multiple Choice, 5 points, 1 attempt
Which of the following enzymes can be described as a DNA-dependent
RNA polymerase?
Correct Choices
DNA ligase
V Primase
DNA polymerase III
DNA polymerase I
Reverse transcriptase
Feedback
Correct:
Primase is a DNA-dependent RNA polymerase located in the primosome at the replication
fork of DNA. Primase initiates DNA synthesis by synthesizing a 10-base RNA primer. The
DNA-RNA helix formed binds DNA polymerase III, which synthesizes a DNA fragment (the

13. 13
Feedback
Okazaki fragment) in a 5′ to 3′ direction. When the RNA primer of the previous Okazaki
fragment is met, DNA polymerase I replaces III and digests the RNA primer, replacing it
with appropriate DNA bases. When the RNA primer is completely removed, DNA ligase
synthesizes the last phosphodiester bond, thereby sealing the space. What is left is a new
lagging strand extended by the new Okazaki fragment with the 10-base RNA primer at its
5′ end. Reverse transcriptase is a DNA polymerase that uses RNA as a template found in
retroviruses as well as normal eukaryotic cells. Unlike DNA polymerase I and III, which
proofread for errors during normal synthesis, reverse transcriptase has no proofreading
capabilities. Hence, it has an exceedingly high error rate that contributes to the high rate
of mutation in retroviruses like HIV.
Incorrect:
Primase is a DNA-dependent RNA polymerase located in the primosome at the replication
fork of DNA. Primase initiates DNA synthesis by synthesizing a 10-base RNA primer. The
DNA-RNA helix formed binds DNA polymerase III, which synthesizes a DNA fragment (the
Okazaki fragment) in a 5′ to 3′ direction. When the RNA primer of the previous Okazaki
fragment is met, DNA polymerase I replaces III and digests the RNA primer, replacing it
with appropriate DNA bases. When the RNA primer is completely removed, DNA ligase
synthesizes the last phosphodiester bond, thereby sealing the space. What is left is a new
lagging strand extended by the new Okazaki fragment with the 10-base RNA primer at its
5′ end. Reverse transcriptase is a DNA polymerase that uses RNA as a template found in
retroviruses as well as normal eukaryotic cells. Unlike DNA polymerase I and III, which
proofread for errors during normal synthesis, reverse transcriptase has no proofreading
capabilities. Hence, it has an exceedingly high error rate that contributes to the high rate
of mutation in retroviruses like HIV.
Question 11. Multiple Choice, 5 points, 1 attempt
The figure alongside shows part of a nucleosome. The pentagons (identified by black colour) represent
which of the following?
Correct Choices
V Deoxyriboses

14. 14
Correct Choices
Phosphate groups
Proline residues
Pyrimidine bases
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Question 12. Multiple Choice, 5 points, 1 attempt
Template-directed DNA synthesis occurs in all the following except
Correct Choices
The replication fork
Polymerase chain reaction
V Growth of RNA tumor viruses
Expression of oncogenes
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Incorrect: You did not choose the correct response.

15. 15
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