DNA Organization, Replication and Repairs

Q-1. The protein rich in basic amino acids which functions in the packaging of DNA in chromosomes is?
a) Histone
b) Collagen
c) Hyaluronic acid binding protein
d) Fibrinogen

Answer: Histone
Explanation:
Histone:
These small proteins are positively charged at physiologic pH as the result of their high content of basic amino acids like lysine and arginine.
Because of their positive charge, they form ionic bond with negatively charged DNA.
Histones, along with positively charged ions such as Mg++, help neutralize the negatively charged DNA phosphate groups.
Important point:
There are five classes of histones, designated H1, H2A, H2B, H3 and H4.

Q-2. Euchromatin is the region of DNA that is relatively
a) Uncondensed
b) Condensed
c) Over-condensed
d) Partially condensed

Answer: Uncondensed
Explanation:
Euchromatin stains weakly and is highly de-condensed chromatin.
Heterochromatin stains more strongly and is more condensed chromatin.

Q-3. A segment of a eukaryotic gene that is not represented in the mature messenger RNA is known as
a) Intron
b) Exon
c) Plasmid
d) TATA box

Answer: Intron
Explanation:
Maturation of eukaryotic m-RNA usually involves the removal of RNA sequence, which do not code for protein (Introns or intervening sequence) from primary transcript.
The remaining coding sequence, the exons are spliced together to form mature m-RNA.
The molecular machine that accomplishes these tasks is known as the spliceosome.

Q-4. Percentage of coding DNA in genome
a) 0.25
b) 0.1
c) 0.4
d) 0.02

Answer: 0.02
Explanation:
Just two percent of our total DNA is coding, while the remaining 98 percent is non-coding.
With the mapping of the human genome (completed in 2003), scientists discovered about 20,500 genes and of these, only about two percent are coding DNA — meaning, they transcribe the code for making proteins and so help create the overall blueprint for making a human being — while non-coding DNA constituted the vast majority (98 percent) of the genome.

Q-5. Microsatellite sequence is
a) Small satellite
b) Extra chromosomal DNA
c) Short sequences (2-5) repeat DNA
d) Looped-DNA

Answer: Short sequence (2-5) repeats DNA
Explanation:
Microsatellites are short segments of DNA that have a repeated sequence such as CACACACA.
The majority of microsatellites occur in gene introns or other non-coding regions of the genome; thus variation in the number of repeats has no consequence on gene function.
Microsatellites are commonly used for mapping, linkage analysis and to trace inheritance patterns.

Q-6. Mitochondrial DNA is
a) Closed circular
b) Nicked circular
c) Linear
d) Open circular

Answer: Closed circular
Explanation:
The mitochondrial genome is built of double-stranded DNA, and it encodes genes. The mitochondrial genome is closed circular.
The mitochondrial genome is built of 16,569 DNA base pairs.
The mitochondrial genome contains 37 genes that encode 13 proteins, 22 t-RNAs, and 2 r-RNAs.
Mitochondrial DNA in humans is always inherited from a person’s mother.

Q-7. The gaps between segments of DNA on the lagging strand produced by restriction enzymes are rejoined/ sealed by
a) DNA ligase
b) DNA Helicase
c) DNA topoisomerase
d) DNA phosphorylase

Answer: DNA ligase
Explanation:
DNA ligases close nicks in the phosphodiester backbone of DNA. DNA ligases are essential for the joining of Okazaki fragments on lagging strand during replication, and for completing short-patch DNA synthesis occurring in DNA repair process.

Q-8. After digestion by restriction endo-nucleases, DNA strands can be joined again by: (AIIMS May 2010)
a) DNA polymerase
b) DNA ligase
c) DNA topoisomerase
d) DNA helicase

Answer: DNA ligase
Explanation:
Restriction enzymes, found naturally in bacteria, can be used to cut DNA fragment at specific sequences, while another enzyme, DNA ligase, can attach or rejoin DNA fragments with complementary ends.

Q-9. Okazaki fragments are formed during the synthesis of
a) ds-DNA
b) ss-DNA
c) m-RNA
d) t-RNA

Answer: ds-DNA
Explanation:
Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging template strand during process of replication of a ds-DNA.
They are complementary to the lagging template strand, together forming short double-stranded DNA sections.

Q-10. During replication of DNA which one of the following enzymes polymerizes the Okazaki fragments?
a) DNA Polymerase I
b) DNA Polymerase II
c) DNA Polymerase III
d) RAN Polymerase I

Answer: DNA polymerase III
Explanation:
Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging template strand during process of replication of a ds-DNA.
They are complementary to the lagging template strand, together forming short double-stranded DNA sections.
Important points:
Okazaki fragments require RNA priming, which serves as the beginning of an Okazaki fragment polymerization.
This polymerization function is accompanied by DNA polymerase III.
These RNA primers are subsequently removed by DNA polymerase-I at the end of Okazaki fragment polymerization.
The complete but still separated Okazaki fragments are then joined into a single strand of DNA via the enzyme, DNA ligase.

Crack PGMEE: On a Mission to Re-imagine Medical Education
Q-11. Intron is not found in which DNA? (AIIMS Nov 2008)
a) B DNA
b) Z DNA
c) Nuclear DNA
d) Mitochondrial DNA

Answer: Mitochondrial DNA
Explanation:
The human mitochondrial genome is a circular DNA molecule. It encodes 37 genes: 13 for subunits of respiratory complexes, 22 for mitochondrial tRNA and 2 for rRNA.
In general, mitochondrial DNA lacks introns, as is the case in the human mitochondrial genome; however, introns have been observed in some eukaryotic mitochondrial DNA, such as that of yeast and protists.

Q-12. Xeroderma pigmentosa is caused due to a group of closely related abnormalities in: (AIIMS NOV 2005)
a) Mismatch repair
b) Base excision repair
c) Nucleotide excision repair
d) SOS repair

Answer: Nucleotide excision repair
Explanation:
Xeroderma pigmentosum is a rare disorder transmitted in an autosomal recessive manner.
The basic defect in xeroderma pigmentosum is in nucleotide excision repair (NER), leading to deficient repair of DNA damaged by UV radiation.
The most common form of this disease is caused by the absence of the UV-specific excinuclease.

Q-13. Xeroderma pigmentosum is produced as a result of a defect in:
a) DNA polymerase III
b) DNA polymerase I
c) DNA exo-nuclease
d) DNA ligase

Ans: b and d
Explanation:
Xeroderma pigmentosum, or XP, is an autosomal recessive genetic disorder of DNA repair in which the ability to repair damage caused by ultraviolet (UV) light is deficient.
The most common defect in xeroderma pigmentosum is an autosomal recessive genetic defect in which nucleotide excision repair (NER) enzymes are mutated, leading to a reduction in or elimination of NER.
In a healthy, normal human being, the damage is first excised by endo-nucleases. DNA polymerase then repairs the missing sequence, and ligase “seals” the transaction. This process is known as nucleotide excision repair.
The enzymes affected in this condition are:

  1. UV specific endo-nuclease (most common)
  2. DNA polymerase I
  3. DNA ligase

Q-14. Excessive ultraviolet (UV) radiation is harmful to life. The damage caused to the biological systems by ultraviolet radiation is by: (AIIMS May 2004)
a) Inhibition of DNA synthesis
b) Formation of thymidine dimers
c) Ionization
d) DNA fragmentation

Answer: Formation of thymidine dimers
Explanation:
Exposure of cell to ultra-violet 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.
First a UV specific endo-nuclease also called UV specific excinuclease recognizes the dimmer and cleaves the damage strand. Damaged oglio-nucleotide is released.

Q-15. All of the following cell types contain the enzyme telomerase which protects the length of telomeres at the end of chromosomes except?
a) Germinal
b) Somatic
c) Hemopoietic
d) Tumor

Answer: Somatic
Explanation:
The telomeres are special structures on the chromosome ends that are essential for providing protection from enzymatic end-degradation and maintaining chromosomal and genomic stability.
Telomeres are composed of a DNA component characterized by non-coding repetitive sequences rich in guanine (G) and multiple protein components.
Important points:
Most human somatic cells do not produce active telomerase and do not maintain stable telomere length with proliferation.

Q-16. Y chromosome is: (AIIMS May 2007)
a) Acro-centric
b) Meta-centric
c) Sub-meta-centric
d) Telo-centric

Answer: Acro-centric
Explanation:
X chromosome: Sub-Meta-centric chromosome
Y chromosome: Acro-centric
Humans do not posses any telo-centric chromosome.

Q-17. The long and short arms of chromosomes are designated respectively as
a) p and q arms
b) m and q arms
c) q and p arms
d) I and s arms

Answer: q and p arms
Explanation:
The short arm of chromosome is designated as p and the long arm is as q.

Q-18. Which of the following would form basis for karyotyping studies in female?
a) Phenotypic abnormality
b) Testosterone quantity
c) Barr body
d) Abnormal X

Answer: Abnormal X
Explanation:

Q-19. Karyotyping, under light microscope is done by (AIIMS Nov 2009)
a) R Banding
b) Q Banding
c) G Banding
d) C Banding

Answer: G Banding
Explanation:
Karyotype analysis is a technique where chromosomes are visualized under a microscope.
Cells are collected from an individual, induced to divide, and then arrested at metaphase.
The chromosomes are stained with certain dyes that show a pattern of light and dark bands (called the banding pattern). These bands reflect regional differences in the amounts of A and T versus G and C.
The banding pattern for each chromosome is specific and consistent allowing identification of each of the 24 chromosomes.
Karyotype analysis can detect large chromosomal abnormalities such as loss or gain of an entire chromosome or portions of a chromosome.
The chromosomes are stained with a dye (Geimsa), resulting in a banding pattern of light and dark stripes, known as G-banding. The patterns are specific, allowing us to identify each chromosome.

Q-20. In genomic imprinting, DNA undergoes
a) Methylation
b) Acetylation
c) Deamination
d) Phosphorylation

Answer: Methylation
Explanation:
Genomic imprinting is the epigenetic phenomenon by which certain genes are expressed in a parent of origin-specific manner.
Genomic imprinting is an inheritance process independent of the classical Mendelian inheritance.
It is an epigenetic process that involves DNA methylation and histone modifications (Methylation, acetylation and phosphorylation) without altering the genetic sequence.

Q-21. In the mammalian genome, maximum number of genes code for the receptors of:
a) Growth factors
b) Immunoglobulin receptors
c) Interleukins
d) Odorants

Answer: Odorants
Explanation:
Olfactory receptor molecules are homologous to a large family of other G-protein-linked receptors.
Odorant receptor proteins have seven membrane-spanning hydrophobic domains, potential odorant binding sites in the extracellular domain of the protein, and the ability to interact with G-proteins at the carboxyl terminal region of their cytoplasmic domain.
The Olfactory receptor family is one of the largest known mammalian gene families, with around 900 genes in human.