If a double standard DNA has 20 per cent of cytosine calculate the per cent of adenine in the DNA.
Solution and Explanation
According to Chargaff's rule the DNA molecule should have an equal ratio of pyrimidine(cytosine and thymine)and purine(adenine and guanine).
It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
%A=%T and %G=%C
If dsDNA has 20% of cytosine then according to the law it would have 20% of guanine.
Thus, percentage of G+C content=40%
The remainig 60% represents both A+T molecule. Since adenine and guanine are always present in equal numbers the percentage of adenine molecule is 30%.
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:
Structure of DNA and RNA
DNA Structure:
Human DNA is unique and is made up of 3 billion base pairs. However, 99% of them are the same in all humans. The sequence of these bases determines what information needs to be coded that is necessary for the organism's growth and development. During the process of transcription (when DNA is replicated), a product is formed known as RNA (Ribonucleic Acid). This RNA is responsible for translating genetic information from DNA to protein followed by reading through ribosomes.
Read More: Structure of DNA
RNA Structure:
The ribonucleic acid has all the components same to that of the DNA with only 2 main differences within it. RNA has the same nitrogen bases called the adenine, Guanine, Cytosine as that of the DNA except for the Thymine which is replaced by the uracil. Adenine and uracil are considered as the major building blocks of RNA and both of them form base-pair with the help of 2 hydrogen bonds.
RNA resembles a hairpin structure and like the nucleotides in DNA, nucleotides are formed in this ribonucleic material(RNA). Nucleosides are nothing but the phosphate groups which sometimes also helps in the production of nucleotides in the DNA.
Read More: Structure of RNA