Maharashtra Board Class 12 Biology Chapter 4 Molecular Basis of Inheritance Solutions

1. Multiple Choice Questions

 

Question 1. Griffith worked on ………………..
(a) Bacteriophage
(b) Drosophila
(c) Frog eggs
(d) Streptococci
Answer: (d) Streptococci
In simple words: Frederick Griffith used the bacterium Streptococci in his famous 1928 experiment to discover that genetic material can be transferred between organisms.

🎯 Exam Tip: Remember that Griffith's experiment involved the S-strain and R-strain of Streptococcus pneumoniae to demonstrate the "transforming principle".

 

Question 2. The molecular knives of DNA are ………………..
(a) Ligases
(b) Polymerases
(c) Endonucleases
(d) Transcriptase
Answer: (c) Endonucleases
In simple words: Restriction endonucleases act like tiny molecular scissors that cut DNA strands at very specific locations.

🎯 Exam Tip: Do not confuse endonucleases (which cut DNA) with ligases (which act as molecular glue to join DNA pieces together).

 

Question 3. Translation occurs in the ………………..
(a) Nucleus
(b) Cytoplasm
(c) Nucleolus
(d) Lysosomes
Answer: (b) Cytoplasm
In simple words: Translation is the process where proteins are built using the instructions on mRNA, and this happens in the cytoplasm where ribosomes are found.

🎯 Exam Tip: Clearly distinguish between transcription (which happens inside the nucleus) and translation (which happens in the cytoplasm).

 

Question 4. The enzyme required for transcription is ................
(a) DNA polymerase
(b) RNA polymerase
(c) Restriction enzyme
(d) RNase
Answer: (b) RNA polymerase
In simple words: Transcription is the process of making an RNA copy from a DNA template, and the main enzyme that builds this RNA chain is RNA polymerase.

🎯 Exam Tip: Remember that "polymerase" means an enzyme that builds a polymer chain, and since transcription makes RNA, it specifically requires RNA polymerase.

 

Question 5. Transcription is the transfer of genetic information from ................
(a) DNA to RNA
(b) t-RNA to m-RNA
(c) DNA to m-RNA
(d) m-RNA to t-RNA
Answer: (a) DNA to RNA
In simple words: Transcription is like copying a recipe from a master cookbook (DNA) onto a single sheet of paper (RNA) so it can be easily used by the cell.

🎯 Exam Tip: Always associate transcription with the synthesis of RNA from a DNA template. Do not confuse it with translation, which is the step from RNA to protein.

 

Question 6. Which of the following is NOT part of protein synthesis?
(a) Replication
(b) Translation
(c) Transcription
(d) All of the options
Answer: (a) Replication
In simple words: Replication is the process of copying DNA to make more DNA for cell division, whereas transcription and translation are the actual steps used to make proteins.

🎯 Exam Tip: Protein synthesis consists of two main stages: transcription and translation. Replication is only for copying genetic material before a cell divides.

 

Question 7. In the RNA molecule, which nitrogen base is found in place of thymine?
(a) Guanine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer: (d) Uracil
In simple words: While DNA uses the base thymine (T) to pair with adenine (A), RNA replaces thymine with uracil (U).

🎯 Exam Tip: This is a highly scoring, classic question. Just remember the key difference: DNA has A, T, C, G, while RNA has A, U, C, G.

 

Question 8. How many codons are needed to specify three amino acids?
(a) 3
(b) 6
(c) 9
(d) 12
Answer: (a) 3
In simple words: Each single codon codes for exactly one amino acid, so you need three codons to make three amino acids.

🎯 Exam Tip: Remember the 1:1 ratio between codons and amino acids during translation to quickly solve such numerical questions.

 

Question 9. Which out of the following is NOT an example of inducible operon?
(a) Lactose operon
(b) Histidine operon
(c) Arabinose operon
(d) Tryptophan operon
Answer: (d) Tryptophan operon
In simple words: An inducible operon is usually turned off and needs to be switched on, but the tryptophan operon is a repressible operon which is normally active and gets turned off when there is enough tryptophan.

🎯 Exam Tip: Clearly distinguish between inducible operons (like lac) and repressible operons (like trp) as this is a highly tested topic.

 

Question 10. Place the following event of translation in the correct sequence ………………..
i. Binding of met-t-RNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second t-RNA.
iv. Joining of small and large ribosome subunits.
(a) iii, iv, i, ii
(b) i, iv, iii, ii
(c) iv, iii, ii, i
(d) ii, iii, iv, i
Answer: (b) i, iv, iii, ii
In simple words: Translation starts with the initiator tRNA binding to the start codon, followed by the ribosome subunits coming together, then the next tRNA binds, and finally a peptide bond forms between the amino acids.

🎯 Exam Tip: Memorize the step-by-step stages of translation (initiation, elongation, termination) to easily sequence these molecular events.

 

Very Short Answer Questions

 

Question 1. What is the function of an RNA primer during protein synthesis?
Answer: During DNA replication, RNA primer provides 3’ OH to which DNA polymerase enzyme can add nucleotides to synthesize new strand using parental strand of DNA as a template. This initiation step is absolutely essential because DNA polymerase cannot start synthesis from scratch.
In simple words: An RNA primer acts as a starting anchor. It provides a free hook (3' OH group) so that the DNA copying enzyme can attach and start building the new strand.

🎯 Exam Tip: Always mention the '3’ OH group' as it is the key chemical requirement that the DNA polymerase enzyme needs to initiate synthesis.

 

Question 2. Why is the genetic code considered as commaless?
Answer: The triplet codon are arranged one after the other on m-RNA molecule without any gap or space and therefore genetic code is considered as commaless. This continuous arrangement ensures that the reading frame is maintained without any interruptions.
In simple words: The genetic code is read continuously from start to finish. There are no punctuation marks, gaps, or spaces between the codons.

🎯 Exam Tip: Remember to use the term "continuous" or "no gaps" when explaining why the genetic code is commaless to secure full marks.

 

Question 3. Genome
Answer: Genome is the total genetic constitution of an organism or a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism. It contains all the necessary information for an organism to build and maintain itself.
In simple words: A genome is the complete set of genetic instructions or DNA found in an organism. It is like a complete instruction manual for a living thing.

🎯 Exam Tip: Define genome clearly as either the total genetic constitution or a complete haploid set of chromosomes to get full marks.

 

Question 4. Which enzyme does remove supercoils from replicating DNA?
Answer: Super-helix relaxing enzyme (Topoisomerase) removes supercoils from replicating DNA. This enzyme works by transiently breaking and rejoining the DNA strands to relieve tension.
In simple words: As DNA unwinds during replication, it gets tightly twisted ahead of the fork. Topoisomerase is the helper enzyme that untwists these tight coils so replication can continue smoothly.

🎯 Exam Tip: Always write both names, "Super-helix relaxing enzyme" and "Topoisomerase", to show a complete understanding of the enzyme's function.

 

Question 5. Why are Okazaki fragments formed on lagging strand only?
Answer: Okazaki fragments are formed only on lagging template as only short stretch of lagging template becomes available for replication at one time. Since DNA polymerase can only synthesize DNA in the 5' to 3' direction, it must produce the lagging strand in short, discontinuous pieces.
In simple words: DNA can only be built in one direction. Because the two strands of DNA run in opposite directions, one strand has to be made in short, backward steps as the DNA unzips.

🎯 Exam Tip: Mention the 5' to 3' directionality of DNA polymerase as the primary reason for discontinuous synthesis on the lagging strand.

 

Question 6. When does DNA replication take place?
Answer: In eukaryotes DNA-replication takes place during S-phase of interphase of cell cycle and in prokaryotes. DNA replicates prior to cell division. This precise timing ensures that each daughter cell receives an exact copy of the genetic material.
In simple words: In complex cells, DNA copies itself during a specific preparation phase called the S-phase. In simpler bacteria, it happens just before the cell splits into two.

🎯 Exam Tip: Clearly distinguish between eukaryotes (S-phase of interphase) and prokaryotes (prior to cell division) for a complete answer.

 

Question 7. Define term Codogen and Codon
Answer: Codogen is a triplet of nucleotides present on the DNA which specifies one amino acid, whereas a codon is the complementary triplet of nucleotides present on the mRNA that codes for that same amino acid. Together, they ensure that the correct amino acids are joined in the right order during protein synthesis.
In simple words: A codogen is the three-letter code on the DNA template, while a codon is the matching three-letter code copied onto the messenger RNA.

🎯 Exam Tip: Clearly state the location of each: codogen is on the DNA, while codon is on the mRNA.

 

Question 8. What is degeneracy of genetic code?
Answer: Genetic code is degenerate as 61 codons code for 20 amino acids, that is two or more codons can specify the same amino acid. E.g. Cysteine has two codons, while isoleucine has three codons. This redundancy acts as a protective mechanism against harmful mutations during protein synthesis.
In simple words: Degeneracy means that multiple different genetic codes can stand for the exact same amino acid. This helps ensure that small mistakes in the DNA code do not always ruin the final protein.

🎯 Exam Tip: Always mention the specific numbers (61 codons and 20 amino acids) and provide a clear example like Cysteine or Isoleucine to score full marks.

 

Question 9. Which are the nucleosomal ‘core’ histones?
Answer: Two molecules each of histone proteins, viz. \( \text{H}_2\text{A} \), \( \text{H}_2\text{B} \), \( \text{H}_3 \) and \( \text{H}_4 \) are the nucleosomal ‘core’ histones. These eight molecules together form an octamer structure that serves as the central spool for DNA wrapping.
In simple words: The core histones are four specific proteins that work in pairs to form a tiny spool. The DNA wraps around this spool to stay organized and compact.

🎯 Exam Tip: Remember that \( \text{H}_1 \) is not a core histone; it is a linker histone. Clearly list only \( \text{H}_2\text{A} \), \( \text{H}_2\text{B} \), \( \text{H}_3 \), and \( \text{H}_4 \) as core histones.

 

Short Answer Questions

 

Question 1. DNA packaging in eukaryotic cell.
Answer:
1. In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (\( 10^{-6} \) m).
2. DNA is associated with histone and non-histone proteins.
3. Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
4. Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. \( \text{H}_2\text{A} \), \( \text{H}_2\text{B} \), \( \text{H}_3 \) and \( \text{H}_4 \) forming histone octamer) and negatively charged DNA (146 bps) that wraps around the histone octamer by \( 1 \frac{3}{4} \) turns.
5. \( \text{H}_1 \) protein binds the DNA thread where it enters and leaves the nucleosome.
6. Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp). This highly organized packaging allows a very long DNA molecule to fit perfectly inside a microscopic cell nucleus.
In simple words: Because DNA is extremely long, it needs to be tightly folded to fit inside a tiny cell. It does this by wrapping around spool-like histone proteins to form structures called nucleosomes, which then fold even further.

🎯 Exam Tip: Memorize the length of DNA (2.2 m), the number of base pairs in a nucleosome (146 bp), and the number of turns (\( 1 \frac{3}{4} \)) as these specific values are highly valued by examiners.

 

Question 2. Enlist the characteristics of genetic code.
Answer: The characteristics of genetic code are:
1. Genetic code is triplet, commaless and non-overlapping.
2. It is degenerate and non-ambiguous.
3. It is universal.
4. It has polarity. This polarity means it is always read in a specific 5' to 3' direction.
In simple words: The genetic code is the set of rules used by living cells to translate information encoded within genetic material. It has specific features like being read in groups of three letters, being shared by almost all organisms, and always reading in one direction.

🎯 Exam Tip: Memorize the four key features of the genetic code (triplet, degenerate, universal, and polarity) as they are frequently asked in short-answer questions.

 

Question 3. Applications of DNA fingerprinting.
Answer: Applications of DNA fingerprinting are as follows:
1. In forensic science to solve rape and murder cases.
2. Finds out the biological father or mother or both, of the child, in case of disputed parentage.
3. Used in pedigree analysis in cats, dogs, horses and humans. This technique has revolutionized modern genetics and legal investigations.
In simple words: DNA fingerprinting is a technique used to identify individuals by their unique DNA patterns. It helps solve crimes, determine biological parents, and study animal family histories.

🎯 Exam Tip: Clearly list forensic, parentage, and pedigree applications to secure full marks in application-based questions.

 

Question 4. Explain the role of lactose in ‘Lac Operon’.
Answer:
1. A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
2. Lactose binds to repressor and inactivates it.
3. Repressor - lactose complex cannot bind with the operator gene, which is then turned on. This process allows the cell to efficiently digest lactose when it is available.
In simple words: Lactose acts as an inducer that turns on the genes needed to digest it. It does this by binding to and disabling the repressor protein that normally keeps the system switched off.

🎯 Exam Tip: Remember that lactose acts as an inducer; explain how its binding to the repressor prevents the repressor from blocking transcription.

Short Answer Questions

 

Question 1. Write a short note on the Human Genome Project.
Answer:
1. Human Genome Project (HGP) was initiated in 1990 under the International administration of the Human Genome Organization (HUGO) and it was completed in 2003. This mega project mapped the entire human genetic blueprint.
2. The main aims of HGP were:
• To sequence 3 billion base pairs of DNA in the human genome and to map an estimated 33,000 genes.
• To store the information collected from the project in databases.
• To develop tools and techniques for analysis of the data.
• Transfer of the related technologies to the private sectors, such as industries.
• Taking care of the legal, ethical and social issues which may arise from the project.
• To sequence the genomes of several other organisms such as bacteria (e.g., E. coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophila, rice, Arabidopsis, Mus musculus, etc.).
3. Significance:
• HGP has a major impact in fields like Medicine, Biotechnology, Bioinformatics, and the Life sciences.
• It provides a better understanding of the functions of genes, proteins, and human evolution.
In simple words: The Human Genome Project was a massive scientific mission to read and map all the DNA in a human body. This helps scientists understand genetic diseases, design better medicines, and learn how humans evolved.

🎯 Exam Tip: Remember the key years of initiation (1990) and completion (2003), along with the focus on ethical, legal, and social issues (ELSI) to score full marks.

 

Question 2. Describe the structure of operon.
Answer: An operon is a unit of genomic DNA containing a cluster of genes under the control of a single promoter. The basic structure of an operon consists of the following components:
1. Regulator Gene: It codes for a repressor protein that can bind to the operator and switch off transcription.
2. Promoter Gene: The sequence where RNA polymerase binds to initiate the transcription of structural genes.
3. Operator Gene: A DNA segment between the promoter and structural genes where the repressor protein binds to regulate transcription.
4. Structural Genes: The actual genes that transcribe mRNA for protein synthesis (e.g., lacZ, lacY, and lacA in the lac operon).
In simple words: An operon is like a pre-programmed switchboard of genes in bacteria. It has a promoter (the start button), an operator (the switch), a regulator (the controller), and structural genes (the appliances that do the work).

🎯 Exam Tip: Draw a simple schematic diagram showing the sequence of Regulator-Promoter-Operator-Structural genes (R-P-O-S) to make your answer highly presentable.

Operon Concept and Regulation
1. An operon is a unit of gene expression and regulation.
2. It includes the structural genes and their control elements. Control elements are promoters and operators.
3. The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
4. Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
5. Operators are present between the promoters and structural genes.
6. There is repressor protein that binds to the operator region of the operon.
7. There are regulatory genes which are responsible for the formation of repressors which interact with operators.

 

Question 3. In the figure below A, B and C are three types of
[Diagram Description: A shows a single strand with Uracil; B shows a Ribosome; C shows a cloverleaf structure with an Amino acid]
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA. These three molecules work together during the process of translation to synthesize proteins.
In simple words: The three main types of RNA are mRNA (messenger), rRNA (ribosomal), and tRNA (transfer), each playing a unique role in making proteins.

🎯 Exam Tip: Remember that mRNA is linear, rRNA is associated with ribosomes, and tRNA has a characteristic cloverleaf shape.

 

Question 4. Identify the labelled structures on the following diagram of translation.
Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the mRNA.
Part C is the ribosome.
These components are essential for translating genetic information into functional proteins.
In simple words: In translation, the ribosome (C) reads the mRNA strand (B), and the tRNA's anticodon (A) matches with the mRNA's codon to add the correct amino acid.

🎯 Exam Tip: Always look at where the pointer line ends; the anticodon is specifically the three-base sequence at the bottom loop of the tRNA molecule.

 

Question 5. Match the entries in Column I with those of Column II and choose the correct answer.

Column I	Column II
A. Alkali treatment	i. Separation of DNA fragments on gel slab
B. Southern blotting	ii. Splits DNA fragments into single strands
C. Electrophoresis	iii. DNA transferred to nitrocellulose sheet
D. PCR	iv. X-ray photography
E. Autoradiography	v. Produce fragments different sizes
F. DNA treated with REN	vi. DNA amplification

Answer:

Column I	Column II
A. Alkali treatment	ii. Splits DNA fragments into single strands
B. Southern blotting	iii. DNA transferred to nitrocellulose sheet
C. Electrophoresis	i. Separation of DNA fragments on gel slab
D. PCR	vi. DNA amplification
E. Autoradiography	iv. X-ray photography
F. DNA treated with REN	v. Produce fragments different sizes

In simple words: This matching pairs different biotechnology techniques with their main functions, such as using PCR to make many copies of DNA (amplification) and electrophoresis to separate DNA pieces by size.

🎯 Exam Tip: Memorize key biotechnology techniques and their primary purposes, as matching questions on these processes are highly scoring and frequently asked.

5. Long Answer Questions

 

Question 1. Explain the process of DNA replication.
Answer: DNA replication is semi-conservative replication. This highly coordinated process ensures that genetic information is accurately passed down to daughter cells during cell division. It involves the following steps:

DNA Replication Diagram Components:

• Parental strands (3' and 5')
• Lagging template
• Leading template
• RNA primer
• Okazaki fragments
• New strands (3' and 5')

Activation of Nucleotides:
1. Nucleotides (dAMP, dGMP, dCMP, and dTMP) present in the nucleoplasm, are activated by ATP in the presence of an enzyme phosphorylase.
2. This phosphorylation results in the formation of deoxyribonucleotide triphosphates i.e. dATP, dGTP, dCTP, and dTTP.

Point of Origin or Initiation point:
1. Replication begins at a specific point ‘O’ - Origin and terminates at point ‘T’.
2. At the point ‘O’, the enzyme endonuclease nicks (breaks the sugar-phosphate backbone or the phosphodiester bond) one of the strands of DNA, temporarily.

Unwinding of DNA molecule:
1. The enzyme DNA helicase breaks weak hydrogen bonds in the vicinity of ‘O’.
2. The strands of DNA separate and unwind. This unwinding is bidirectional.
3. SSBP (Single strand binding proteins) remain attached to both the separated strands and prevent them from recoiling (rejoining).
In simple words: DNA replication is the process where a cell makes an exact copy of its DNA. The double-stranded DNA unzips like a zipper, and each single strand serves as a template to build a new, complementary partner strand.

🎯 Exam Tip: Remember to mention that DNA replication is semi-conservative, meaning each new DNA molecule contains one original strand and one newly synthesized strand.

Replicating Fork

• 1. Y-shape replication fork is formed due to unwinding and separation of two strands.
• 2. The unwinding of strands results in strain which is released by super-helix relaxing enzyme.

 

Synthesis Of New Strands

• 1. Each separated strand acts as a template for the synthesis of new complementary strand.
• 2. A small RNA primer (synthesized by activity of enzyme RNA primase) get attached to the \( 3' \) end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
• 3. These nucleotides bind to the complementary nucleotides on the template strand by hydrogen bonds (i.e. A = T or T = A; G = C or C = G, CEG).
• 4. The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
• 5. The enzyme DNA polymerase catalyses synthesis of new complementary strand always in \( 5' \rightarrow 3' \) direction.

 

Leading And Lagging Strand

• 1. The template strand with free \( 3' \) is called the leading template.
• 2. The template strand with free \( 5' \) end is called the lagging template.
• 3. The replication always starts at \( C\text{-}3 \) end of template strand and proceeds towards \( C\text{-}5 \) end.
• 4. New strands are always formed in \( 5' \rightarrow 3' \) direction.
• 5. The new strand which develops continuously towards replicating fork is called the leading strand.
• 6. The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
• 7. Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
• 8. Later RNA primers are removed by the combined action of RNase H, an enzyme that degrades the RNA strand of RNA-DNA hybrids, and polymerase I.

9. Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
10. Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of Two Daughter DNA Molecules:
1. In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
2. Thus, 50% part (i.e. one strand of the helix) is contributed by mother DNA. Hence, it is described as semiconservative replication.

 

Question 2. Describe the process of transcription in protein synthesis.
Answer: Transcription involves three stages, viz. Initiation, Elongation and Termination. During this process, genetic information from DNA is copied into a complementary RNA strand.

• Transcription Process Diagram Labels:
• DNA strands (5' to 3' and 3' to 5')
• Capping (mGppp Cap at 5' end)
• Exons and Introns (RNA splicing)
• Polyadenylation (Poly A tail at 3' end)
• Messenger RNA (mRNA)

(1) Initiation:
1. RNA polymerase binds to promoter site.
2. It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
3. Only antisense strand functions as template.

(2) Elongation:
The complementary ribonucleoside tri-phosphates get attached to exposed bases of DNA template chain.
In simple words: Transcription is the first step in making proteins, where the cell copies a gene's DNA recipe into a temporary RNA messenger. This messenger then carries the instructions to the protein-making machinery.

🎯 Exam Tip: Clearly list and describe all three stages—Initiation, Elongation, and Termination—and remember to mention that only the antisense strand of DNA acts as the template for transcription.

 

Question 3. Describe the process of translation in protein synthesis.
Answer:
Diagram Labels:
• Initiator tRNA
• mRNA
• Smaller subunit of ribosome
• Larger subunit of ribosome
• E site, P site, A site
• Peptide link
• Polypeptide chain
• Last tRNA
• Released mRNA

Translation involves the following steps, which ensure that the genetic information is accurately translated into functional proteins:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).
2. Initiation of Polypeptide chain:
i. The small subunit of ribosome binds to the mRNA at the 5' end near the start codon (AUG).
ii. The initiator t-RNA charged with methionine binds to the start codon (AUG) of mRNA by its anticodon (UAC) through hydrogen bonds.
iii. Now, the large subunit of ribosome joins the small subunit to form the complete translation initiation complex.
iv. The initiator t-RNA occupies the P-site (peptidyl site) of the ribosome, leaving the A-site (aminoacyl site) vacant.
3. Elongation of Polypeptide chain:
i. Another charged t-RNA carrying the next amino acid enters the ribosome and binds to the vacant A-site.
ii. A peptide bond is formed between the amino acid at the P-site and the amino acid at the A-site, catalyzed by the enzyme peptidyl transferase.
iii. The ribosome then moves (translocates) along the mRNA by one codon in the 5' to 3' direction.
iv. The uncharged t-RNA from the P-site is shifted to the E-site (exit site) and released, while the t-RNA at the A-site carrying the growing polypeptide chain is shifted to the P-site. This process is repeated for subsequent codons.
4. Termination of Polypeptide chain:
i. Elongation continues until a stop codon (UAA, UAG, or UGA) on mRNA reaches the A-site.
ii. Since there are no t-RNAs corresponding to stop codons, a release factor binds to the stop codon, terminating translation.
iii. The polypeptide chain is released, and the ribosomal subunits dissociate from the mRNA.
In simple words: Translation is the process where the cell reads the genetic code on mRNA to build a protein. Ribosomes act as factories, and tRNA molecules bring the correct amino acids one by one to assemble them into a long chain.

🎯 Exam Tip: Clearly label the three sites of the ribosome (A, P, and E sites) and mention the role of the initiator codon (AUG) and stop codons to secure full marks.

ii. ATP is essential for the reaction.

 

2. Initiation of Polypeptide Chain:

• Small subunit of ribosome binds to the m-RNA at 5’ end.
• Start codon is positioned properly at P-site.
• Initiator t-RNA, (carrying amino acid methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA, by its anticodon (UAC) through hydrogen bonds.
• The large subunit of ribosome joins with the smaller subunit in the presence of \( \text{Mg}^{++} \).
• Thus, initiator charged t-RNA occupies the P-site and A – site is vacant.

 

3. Elongations of Polypeptide Chain:

Addition of amino acid occurs in 3 Step cycle-

i. Codon Recognition

Anticodon of second (and subsequent) amino acyl t-RNA molecule recognizes and binds with codon at A-site by hydrogen bonds.

ii. Peptide Bond Formation

1. Ribozyme catalyzes the peptide bond formation between amino acids on the initiator t-RNA at P-site and t-RNA at A-site.
2. It takes less than 0.1 second for formation of peptide bond.
3. Initiator t-RNA at ‘P’ site is then released from E-site.

iii. Translocation

1. Translocation is the process in which sequence of codons on m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
3. The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
4. As ribosome move over the m-RNA, all the codons on m-RNA are exposed one by one for translation.

 

Question 4. Describe Lac ‘Operon’.
Answer:
Lac Operon Diagram Components:
• (a) Inducer absent: Regulatory gene, Promoter, Operator blocked, Structural genes (1, 2, 3) inactive (no synthesis of mRNA), Template DNA strand, Transcription, Repressor mRNA, Translation, Repressor, Movement blocked.
• (b) Inducer present: Regulatory gene, Promoter, Operator, Structural genes (1, 2, 3), Template DNA strand, Transcription, Repressor mRNA, Translation, Repressor, Inducer (allolactose), Inactivated repressor, Transcription proceeds, mRNA, Translation, Proteins (enzymes to metabolize lactose).

Lac operon consists of the following components:
(1) Regulator gene:
• Regulator gene precedes the promoter gene.
• It may not be present immediately adjacent to operator gene.
• Regulator gene codes for a repressor protein which binds with operator gene and represses (stops) its action.

(2) Promoter gene:
• It precedes the operator gene.
• It is present adjacent to operator gene.

This elegant regulatory mechanism allows the cell to conserve energy by only producing lactose-metabolizing enzymes when lactose is actually present.
In simple words: The Lac Operon is like a genetic switchboard in bacteria that turns on the machinery to digest lactose only when lactose is available. This prevents the cell from wasting energy making tools it doesn't need.

🎯 Exam Tip: Clearly label the roles of the regulator, promoter, and operator genes, and remember that the presence of lactose (allolactose) acts as the inducer that turns the operon 'on'.

• RNA Polymerase enzyme binds at promoter site.
• Promoter gene base sequence determines which strand of DNA acts a template.

(3) Operator Gene:

• It precedes the structural genes.
• When operator gene is turned on by an inducer, the structural genes get transcribed to form m-RNA.

(4) Structural Gene:

• There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.
• Enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
• Inducer Allolactose acts as an inducer. It inactivates the repressor by binding with it.

 

Question 5. Justify the statements. If the answer is false, change the underlined word(s) to make the statement true.
(i) The DNA molecule is double stranded and the RNA molecule is single stranded.
(ii) The process of translation occurs at the ribosome.
Answer:
(i) True statement.
1. DNA as the genetic material has to be chemically and structurally stable.
2. It should be able to generate its replica.
3. Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
4. Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
5. Single stranded RNA can be folded to form complex structures and perform specific functions such as synthesis of proteins.

(ii) True statement.
1. Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form proteins.
In simple words: DNA is double-stranded to keep our genetic code safe and stable, while RNA is single-stranded so it can fold into shapes and help build proteins. Translation is the process where ribosomes read this RNA code to assemble proteins.

🎯 Exam Tip: Always begin your justification by clearly stating whether the given statement is True or False to make your answer precise and easy to grade.

Question. (iii) The job of m-RNA is to pick up amino acids and transport them to the ribosomes.
Answer: The job of t-RNA is to pick up amino acids and transport them to ribosomes. t-RNA is an adapter molecule. It reads the codons of m-RNA and also simultaneously transfers specific amino acids to the m-RNA Ribosome complex. It binds with amino acid at its 3′ end. This precise binding is crucial for accurate protein synthesis.
In simple words: The actual job of carrying amino acids to the ribosomes belongs to t-RNA, not m-RNA. It acts like a delivery truck that reads the genetic code and brings the correct building blocks.

🎯 Exam Tip: Remember that 't' in t-RNA stands for 'transfer', which directly relates to its function of transporting amino acids.

 

Question. (iv) Transcription must occur before translation may occur.
Answer: In prokaryotes, translation can start before transcription is complete, as both these processes occur in the same compartment, i.e. cytoplasm. But in eukaryotes, transcription and processing of hnRNA occurs in nucleus. hnRNA then comes out of the nucleus through nuclear pores and then it is translated at ribosomes in the cytoplasm. This spatial separation in eukaryotes allows for additional regulatory steps.
In simple words: In simple cells like bacteria, translation can start while transcription is still happening because there is no nucleus. In complex cells, transcription must finish inside the nucleus first before translation can start in the cytoplasm.

🎯 Exam Tip: Clearly distinguish between prokaryotes and eukaryotes when explaining the timing of transcription and translation to score full marks.

 

Question 6. Guess
(i) the possible locations of DNA on the collected evidence from a crime scene and
(ii) the possible sources of DNA.
Answer: Identifying these sources is a key step in forensic investigations.

Evidence	Possible location of DNA on the evidence	Sources of DNA

In simple words: This table helps investigators identify where DNA might be found on different pieces of evidence collected from a crime scene.

🎯 Exam Tip: In forensic science questions, always link the physical evidence (like a cup or hairbrush) to the biological source (like saliva or hair follicles) for complete answers.

 

Question 1. Complete the following table showing evidence, possible location of DNA on the evidence, and sources of DNA:

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	___________
___________	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	___________
Bite mark	___________	Saliva
___________	Surface area	Hair, semen, sweat, urine

Answer:

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	Saliva
Door	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	Saliva
Bite mark	Teeth impression	Saliva
Clothes	Surface area	Hair, semen, sweat, urine

Understanding where DNA can be recovered from everyday items is crucial for forensic investigations and genetic analysis.
In simple words: This table shows where we can find DNA on different everyday objects. For example, saliva can be found on a used cigarette butt or a bottle, and sweat can be found on a door handle or eyeglasses.

🎯 Exam Tip: Memorize the common sources of DNA like saliva, sweat, and blood, and associate them with everyday objects to easily fill out forensic science tables in exams.