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Chapter 6 · Class 12 Biology

Molecular Basis of Inheritance

1 exercises7 questions solved
Exercise 6.1DNA Structure, Replication and Gene Expression
Q1

Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil, Cytosine.

Solution

Nitrogenous bases (just the base, no sugar attached): • Adenine (A) — purine • Thymine (T) — pyrimidine (found in DNA) • Uracil (U) — pyrimidine (found in RNA, replaces thymine) • Cytosine (C) — pyrimidine Nucleosides (nitrogenous base + sugar, NO phosphate): • Cytidine = Cytosine + Ribose (RNA nucleoside) • Guanosine = Guanine + Ribose (RNA nucleoside) Note: Nucleoside = Base + Sugar (pentose) Nucleotide = Base + Sugar + Phosphate group Complete list: • Adenosine = Adenine + Ribose • Guanosine = Guanine + Ribose • Cytidine = Cytosine + Ribose • Uridine = Uracil + Ribose • Thymidine = Thymine + Deoxyribose (DNA-specific nucleoside)
Q2

If a double-stranded DNA has 20% of cytosine, what will be the percentage of adenine in it?

Solution

Given: Cytosine (C) = 20% Using Chargaff's Rules: • In double-stranded DNA: A = T and G = C • The percentage of a base equals its complementary base Step 1: Since C = G (Chargaff's rule): G = 20% Step 2: Total % of C + G: C + G = 20% + 20% = 40% Step 3: Total % of A + T: A + T = 100% − 40% = 60% Step 4: Since A = T: A = T = 60%/2 = 30% Answer: Adenine = 30% Verification: A = 30%, T = 30%, G = 20%, C = 20% Total = 30 + 30 + 20 + 20 = 100% ✓ A = T ✓, G = C ✓ (Chargaff's rules satisfied)
Q3

If the sequence of one strand of DNA is written 5'–ATGCATGCATGC–3', what will be the sequence of the complementary strand written in 3'→5' direction? How will this sequence be written in the 5'→3' direction?

Solution

Given strand: 5'–ATGCATGCATGC–3' Complementary base pairing rules in DNA: • A pairs with T (and vice versa) • G pairs with C (and vice versa) • Complementary strand runs antiparallel Step 1: Write the complementary sequence: Original: 5'–A T G C A T G C A T G C–3' Comp: 3'–T A C G T A C G T A C G–5' Complementary strand written 3'→5': 3'–TACGTACGTACG–5' Step 2: Write complementary strand in 5'→3' direction: (Simply reverse the sequence) 5'–GCATGCATGCAT–3' Answer: • Complementary strand (3'→5'): 3'–TACGTACGTACG–5' • Same strand in 5'→3' direction: 5'–GCATGCATGCAT–3'
Q4

If the sequence of the coding strand in a transcription unit is written 5'–ATGCATGCATGC–3', write down the sequence of mRNA.

Solution

Given: Coding strand (non-template strand) = 5'–ATGCATGCATGC–3' In transcription: • The template strand (antisense strand) is read 3'→5' by RNA polymerase • The coding strand (sense strand, non-template) has the same sequence as the mRNA, except T is replaced by U Coding strand: 5'–ATGCATGCATGC–3' To get mRNA sequence: • Replace all T with U in the coding strand sequence mRNA sequence: 5'–AUGCAUGCAUGC–3' Note: • mRNA has the same sequence as the coding strand but with U instead of T • mRNA is read in the 5'→3' direction • The template strand would be: 3'–TACGTACGTACG–5' The sequence 5'–AUGCAUGCAUGC–3' contains the start codon AUG (methionine) at the beginning.
Q5

Explain the process of DNA replication in Prokaryotes.

Solution

DNA replication in prokaryotes (E. coli model — semi-conservative replication): 1. Initiation: • Replication begins at a specific sequence: oriC (origin of replication) • Initiator proteins bind to oriC and unwind a short region • DnaB (helicase) is loaded and unwinds the double helix • Creates two replication forks moving in opposite directions (bidirectional) 2. Unwinding and strand separation: • Helicase (DnaB): Unwinds DNA at replication fork by breaking H-bonds between base pairs • Topoisomerases (gyrase): Relieve supercoiling ahead of the fork • SSB proteins (Single-stranded binding proteins): Stabilise separated single strands 3. Primer synthesis: • DNA polymerase cannot initiate synthesis; needs a primer • Primase (RNA polymerase): Synthesises a short RNA primer (~5–10 nucleotides) on both strands 4. Elongation: • DNA Pol III (main polymerase): Adds deoxyribonucleotides in 5'→3' direction • Leading strand: Synthesised continuously towards the replication fork • Lagging strand: Synthesised discontinuously as Okazaki fragments (opposite direction) • Each Okazaki fragment begins with an RNA primer 5. Primer removal and gap filling: • DNA Pol I: Removes RNA primers and fills gaps with DNA 6. Ligation: • DNA ligase: Joins Okazaki fragments by sealing nicks in the sugar-phosphate backbone 7. Termination: • In circular prokaryotic DNA, replication forks meet at terminus (ter sequences) • Two daughter DNA molecules are separated by topoisomerases Result: Two identical DNA double helices, each with one original + one new strand (semi-conservative).
Q6

How is the lac operon regulated in the presence and absence of lactose?

Solution

Lac operon (Jacob and Monod, 1961) — negative inducible operon in E. coli: Components: • Regulator gene (lacI): codes for Lac repressor protein • Operator (O): binding site for repressor • Promoter (P): RNA polymerase binding site • Structural genes: lacZ (β-galactosidase), lacY (permease), lacA (transacetylase) Regulation in ABSENCE of lactose (operon OFF): 1. lacI gene is always expressed → produces Lac repressor protein 2. Lac repressor binds to operator (O) 3. Operator is occupied → RNA polymerase cannot move past operator 4. Transcription of structural genes is BLOCKED 5. No β-galactosidase produced (not needed — no lactose to digest) Regulation in PRESENCE of lactose (operon ON): 1. Lactose enters the cell; some is converted to allolactose (by residual β-galactosidase) 2. Allolactose is the inducer — it binds to the Lac repressor 3. Inducer-repressor complex undergoes conformational change 4. Cannot bind to operator anymore 5. Operator is FREE → RNA polymerase can transcribe structural genes 6. mRNA is transcribed → β-galactosidase, permease, transacetylase produced 7. β-galactosidase breaks down lactose → glucose + galactose 8. When lactose is exhausted, allolactose disappears → repressor regains activity → binds operator → operon is switched OFF again This is NEGATIVE regulation (repressor normally blocks transcription). Note: Glucose also plays a role (catabolite repression) — when glucose is present, cAMP levels are low, reducing operon efficiency even if lactose is present.
Q7

Explain briefly: (a) Transcription (b) Polymorphism.

Solution

(a) Transcription: • The process of synthesis of RNA from a DNA template • The information in DNA is 'transcribed' into RNA • Occurs in the nucleus (eukaryotes) / cytoplasm (prokaryotes) Steps: 1. Initiation: RNA polymerase binds to the promoter region on the DNA 2. The DNA double helix is unwound at the transcription bubble 3. Template strand (antisense/anti-coding strand) is read 3'→5' 4. Elongation: RNA polymerase adds ribonucleotides (A, U, G, C) in 5'→3' direction, following base-pairing rules (A→U, T→A, G→C, C→G) 5. Termination: When RNA polymerase reaches terminator sequence, it stops 6. The RNA transcript is released In eukaryotes: primary transcript (pre-mRNA) undergoes processing: • 5' capping (7-methylguanosine cap) • 3' polyadenylation (poly-A tail) • Splicing: introns removed, exons joined → Mature mRNA exported to cytoplasm for translation (b) DNA Polymorphism: • Polymorphism means 'many forms' — the occurrence of multiple allelic forms of a gene at a given locus in a population with a frequency greater than 0.01 (1%) Definition: Heritable mutations that appear in population at a frequency high enough to be useful for genetic analysis Types: • SNPs (Single Nucleotide Polymorphisms): Single base changes at a locus • Variable Number Tandem Repeats (VNTR): Different numbers of repeated sequences at a locus — basis of DNA fingerprinting • Microsatellites: Short tandem repeats (STRs) Significance: • Basis of DNA fingerprinting (forensics, paternity testing) • Used in linkage mapping and association studies • Explains individual variation in disease susceptibility and drug response VNTR size (number of repeats) varies between individuals → unique DNA fingerprint for each person.
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