Explain(in one or two lines)the function of the followings:
(a)Promoter
(b)tRNA
(c)Exons
(a)Promoter
(b)tRNA
(c)Exons
Solution and Explanation
(a) Promoter Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.
(b) tRNA tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.
(c) Exons Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.
Top Questions on Molecular Basis of Inheritance
Study the given below single strand of deoxyribonucleic acid depicted in the form of a “stick” diagram with 5′ – 3′ end directionality, sugars as vertical lines and bases as single letter abbreviations and answer the questions that follow.
Name the covalent bonds depicted as (a) and (b) in the form of slanting lines in the diagram.
How many purines are present in the given “stick” diagram?
Draw the chemical structure of the given polynucleotide chain of DNA.- CBSE CLASS XII - 2025
- Biology
- Molecular Basis of Inheritance
Study the given molecular structure of double-stranded polynucleotide chain of DNA and answer the questions that follow.
(a) How many phosphodiester bonds are present in the given double-stranded polynucleotide chain?
(b) How many base pairs are there in each helical turn of double helix structure of DNA? Also write the distance between a base pair in a helix.
(c) In addition to H-bonds, what confers additional stability to the helical structure of DNA?- CBSE CLASS XII - 2025
- Biology
- Molecular Basis of Inheritance
Use the given information to select the amino acid attached to the 3′ end of tRNA during the process of translation, if the coding strand of the structural gene being transcribed has the nucleotide sequence TAC.
- CBSE CLASS XII - 2025
- Biology
- Molecular Basis of Inheritance
Student to attempt either option-(A) or (B):
(A) Write the features a molecule should have to act as a genetic material. In the light of the above features, evaluate and justify the suitability of the molecule that is preferred as an ideal genetic material.
OR
(B) Differentiate between the following:- [(i)] Polygenic Inheritance and Pleiotropy
- [(ii)] Dominance, Codominance and Incomplete dominance
- CBSE CLASS XII - 2025
- Biology
- Molecular Basis of Inheritance
- Distance between two consecutive base pairs of DNA is \rule{2cm}{0.1mm}
- CBSE Compartment XII - 2025
- Biology
- Molecular Basis of Inheritance
Questions Asked in CBSE CLASS XII exam
- If \( \vec{\alpha} = \hat{i} - 4 \hat{j} + 9 \hat{k} \) and \( \vec{\beta} = 2\hat{i} - \hat{j} + \lambda \hat{k} \) are two mutually parallel vectors, then \( \lambda \) is equal to:
"There is widely spatial variation in different sectors of work participation in India." Evaluate the statement with suitable examples.
- CBSE CLASS XII - 2025
- Population
Alexia Limited invited applications for issuing 1,00,000 equity shares of ₹ 10 each at premium of ₹ 10 per share.
The amount was payable as follows:- On application ₹ 9 per share (Including premium ₹ 6 per share)
- On allotment ₹ 8 per share (Including premium ₹ 4 per share)
- On first and final call ₹ 3 per share.
Applications were received for 1,50,000 equity shares and allotment was made to the applicants as follows:
Category A: Applicants for 90,000 shares were allotted 70,000 shares.
Category B: Applicants for 60,000 shares were allotted 30,000 shares.
Excess money received on application was adjusted towards allotment and first and final call.
Shekhar, who had applied for 1200 shares failed to pay the first and final call. Shekhar belonged to category B.
Pass necessary journal entries for the above transactions in the books of Alexia Limited. Open calls in arrears and calls in advance account, wherever necessary.- Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 20.2 mL.
(i) Why was there a volume change after mixing the liquids?
(ii) Will there be an increase or decrease of temperature after mixing?
(iii) Give one example for this type of solution.- CBSE CLASS XII - 2025
- Solutions
On $31^{\text {st }}$ March, 2024, following is the Balance Sheet of Bhavik Limited :
Bhavik Ltd.
Balance Sheet as at $31^{\text {st }}$ March 2024
I. Equity and Liabilities :
Particulars Note No. $31-3-2024$ (₹) $31-3-2023$ (₹) 1. Shareholders funds (a) Share Capital 12,00,000 10,00,000 (b) Reserves and Surplus 1 4,00,000 3,00,000 2. Non-current liabilities Long-term borrowings 2 6,00,000 10,00,000 3. Current Liabilities 5,00,000 1,00,000 (a) Trade Payables 3 3,00,000 4,00,000 (b) Short-term provisions Total 30,00,000 28,00,000 II. Assets :
1. Non-current Assets (a) Property, Plant and Equipment and Intangible Assets Property plant and equipment 4 19,00,000 15,00,000 (b) Non-current Investments 3,00,000 4,00,000 2. Current Assets (a) Inventories 4,50,000 3,50,000 (b) Trade Receivables 2,50,000 4,50,000 (c) Cash and Cash Equivalents 1,00,000 1,00,000 Total 30,00,000 28,00,000 Notes to Accounts :
Note Particulars $31-3-2024$ (₹) $31-3-2023$ (₹) No. 1. Reserves and Surplus i.e. Balance in Statement of Profit and Loss 4,00,000 3,00,000 2. Long-term borrowings 10% Debentures 6,00,000 10,00,000 3. Short-term provisions Provision for tax 3,00,000 4,00,000 4. Property plant and equipment Plant and Machinery 21,50,000 16,00,000 Less : Accumulated Depreciation 2,50,000 1,00,000 19,00,000 15,00,000 Additional Information :
- During the year, a piece of machinery costing ₹ 8,00,000 with accumulated depreciation of ₹ 50,000 was sold for ₹ 6,50,000.
- Debentures were redeemed on $31^{\text {st }}$ March, 2024.
Calculate :
- Cash flows from Investing Activities
- Cash flows from Financing Activities
- CBSE CLASS XII - 2025
- Balance sheet statement
Concepts Used:
The Molecular Basis of Inheritance
DNA Replication:
DNA synthesis is commenced at particular points within the DNA strand referred to as ‘origins’, which are certain coding regions. There are numerous origin sites, and when replication of DNA starts, these sites are mentioned as replication forks. Within the replication, the complex is the enzyme DNA Helicase, so that they can be utilized as a template for replication. DNA Primase is another enzyme that's essential in DNA replication.
RNA:
Ribonucleic acid (RNA) is an essential biological macromolecule that exists all together in biological cells. It is principally involved in the synthesis of proteins, that carry the messenger instructions from DNA, which itself contains the genetic instructions needed for the event and maintenance of life. In some viruses, RNA, in spite of DNA, carries genetic information.
Genetic Code:
Genetic code is the term we use in the manner that the four bases of DNA--the A, C, G, and Ts--are strung together in a way that the ribosome, the cellular machinery, can read them and switch them into a protein. In the ordering, every three nucleotides during a row count as a triplet and code for one amino alkanoic acid.
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