How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Solution and Explanation
Hershey and Chase worked with bacteriophage and E. coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein
coat of the bacteriophage.
They grew some bacteriophages on a medium containing radioactive phosphorus (32P) to identify DNA and some on a medium containing radioactive sulphur (35S) to identify protein. Then these radioactive labelled phages were allowed to infect E. coli bacteria. After infecting the protein coat of the bacteriophage was seperated from the bacterial cell by blending and then subjected to the process of centrifugation. Since the protein coat was lighter it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence it was proved that DNA is the genetic material as it was transferred from virus to bacteria.
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:
Non-Mendelian Genetics
The term - non-mendelian inheritance refers to any pattern of heredity in which features do not separate according to Mendel's laws. These principles describe how features linked with single genes on chromosomes in the nucleus are passed down through generations.
Types of Non-Mendelian Inheritance
Codominance Inheritance
It is a form of incomplete dominance in which both alleles for the same feature are expressed in the heterozygote at the same time. For example, the MN blood types of humans.
Incomplete Dominance
In a heterozygote, the dominant allele does not always completely cover the phenotypic expression of the recessive gene, resulting in an intermediate phenotype which is referred to as "incomplete dominance”.