Class 12 Chemistry Molecular Basis of Inheritance MCQs with Answer
At Horizon Competition School, our Class 12 Chemistry MCQs on the Molecular Basis of Inheritance are meticulously designed to aid students in mastering this intricate topic. Covering essential concepts such as DNA replication, genetic code, mutations, and gene expression, these multiple-choice questions offer comprehensive preparation for Class 12 Chemistry exams.
Our MCQs challenge students to apply their understanding of molecular biology principles to solve problems effectively. With each question, students are encouraged to think critically and analyze genetic scenarios, enhancing their problem-solving skills.
Accompanied by detailed explanations, our MCQs provide invaluable insights into the underlying concepts, ensuring thorough comprehension. Students receive immediate feedback on their answers, enabling them to identify areas for improvement and track their progress effectively.
By practicing Class 12 Chemistry MCQs on the Molecular Basis of Inheritance at Horizon Competition School, students can build confidence, sharpen their analytical skills, and excel in their examinations.
Q1. If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be:
- 5′ – UAC – 3′.
- 5′ – CAU – 3′.
- 5′ – AUG – 3′.
- 5′ – GUA – 3′.
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Ans: 2. 5′ – CAU – 3′.
Base sequence of a codon in mRNA is 5′ – AUG – 3′, the sequence of tRNA pairing with it must be complementary i.e., 3′ – UAC – 5′.
Q2. Who amongst the following scientists had no contribution in the development of the double helix model for the
structure of DNA?
- Rosalind Franklin.
- Maurice Wilkins.
- Erwin Chargaff.
- Meselson and Stahl.
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Ans: 4. Meselson and Stahl.
Meselson and Stahl gives the experimental proof of semiconservative DNA replication. He had no contribution the development of
the double helix model for the structure of DNA.
Q3. With regard to mature mRNA in eukaryotes:
- Exons and introns do not appear in the mature RNA.
- Exons appear but introns do not appear in the mature RNA.
- Introns appear but exons do not appear in the mature RNA.
- Both exons and introns appear in the mature RNA.
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Ans: 2. Exons appear but introns do not appear in the mature RNA.
With regard to mature mRNA in eukaryotes exons appear but introns do not appear in the mature RNA.
Q4. The amino acid attaches to the tRNA at its:
- 5′ – end.
- 3′ – end.
- Anti codon site.
- DHU loop.
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Ans: 2. 3′ – end.
The amino acid attaches to the tRNA at its 3′ – end.
Q5. Which of the following are the functions of RNA?
- It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
- It carries amino acids to ribosomes.
- It is a constituent component of ribosomes.
- All of the above.
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Ans: 4. All of the above.
Functions of mRNA:
Carrier of genetic information from DNA to ribosomes synthesising polypeptides.
Carries amino acids to ribosomes.
Constituent component of ribosomes
Q6. Which was the last human chromosome to be completely sequenced:
- Chromosome 1.
- Chromosome 11.
- Chromosome 21.
- Chromosome x.
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Ans: 1. Chromosome 1.
The Human genome project was completed in 2003. The sequence of chromosome 1 was completed only in May 2006.
Q7. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called:
- A-DNA.
- B-DNA.
- cDNA.
- rDNA.
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Ans: 3. cDNA.
In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called c DNA.
Q8. Control of gene expression takes place at the level of:
- DNA-replication.
- Transcription.
- Translation.
- None of the above.
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Ans: 2. Transcription.
Transcription is the first step of gene expression. In this process, a particular segment of DNA is copied into mRNA. Thus, it controls
gene expression.
Q9. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the
proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering
the Chargaff’s rule it can be concluded that:
- It is a double stranded circular DNA.
- It is single stranded DNA.
- It is a double stranded linear DNA.
- No conclusion can be drawn.
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Ans: 2. It is single stranded DNA.
Chargaff’s rules states that DNA from any cell of all organisms should have a 1 : 1 ratio (base Pair Rule) of pyrimidine and purine
bases. This means that the amount of guanine is equal to cytosine and the amount of adenine is equal to thymine. This pattern is
found in both strands of the DNA. In this case, percentage of adenine is not equal to that of guanine and same holds true for cytosine
and thymine. Hence, it is a single stranded DNA.
Q10. The human chromosome with the highest and least number of genes in them are respectively:
- Chromosome 21 and Y.
- Chromosome 1 and X.
- Chromosome 1 and Y.
- Chromosome X and Y.
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Ans: 3. Chromosome 1 and Y.
Chromosome 1 has 2968 genes, while chromosome Y has 231 genes.
Q11. Both deoxyribose and ribose belong to a class of sugars called:
- Trioses.
- Hexoses.
- Pentoses.
- Polysaccharides.
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Ans: 3. Pentoses
Q12. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is:
5′ – A T G A A T G – 3′,
The sequence of bases in its RNA transcript would be;
- 5′ – A U G A A U G – 3′.
- 5′ – U A C U U A C – 3′.
- 5′ – C A U U C A U – 3′.
- 5′ – G U A A G U A – 3′.
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Ans: 1. 5′ – A U G A A U G – 3′.
Coding strand: 5′ – A T G A A T G – 3′
Template strand: 3′ – T A C T T A C – 5″
RNA transcript: 5′ – A U G A A U G – 3′.
Q13. The promoter site and the terminator site for transcription are located at:
- 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit.
- 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit.
- The 5′ (upstream) end.
- The 3′ (downstream) end.
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Ans: 2. 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit.
The promoter site and the terminator site for transcription are located at 5′ (upstream) end and 3′ (downstream) end, respectively of
the transcription unit.
Q14. The fact that a purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA
double helix:
- The antiparallel nature.
- The semiconservative nature.
- Uniform width throughout DNA.
- Uniform length in all DNA
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Ans: 3. Uniform width throughout DNA.
A purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix, uniform width
throughout DNA.
Q15. The DNA site where DNA-dependent RNA polymerase binds for transcription, is called.
- Operator.
- Promotor.
- Regulator.
- Receptor.
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Ans: 2. Promotor.
FAQs
How do practicing MCQs on the Molecular Basis of Inheritance benefit students preparing for Class 12 Chemistry exams?
Engaging with MCQs on the Molecular Basis of Inheritance is highly beneficial for Class 12 Chemistry students as it reinforces their understanding of essential genetic concepts. These MCQs cover topics such as DNA structure, replication, transcription, translation, and genetic disorders. By practicing MCQs, students can assess their comprehension of these concepts, identify areas of weakness, and improve their problem-solving skills. Additionally, MCQs help students familiarize themselves with the exam pattern, enhancing their confidence and performance in the actual examination.
What strategies can students employ to effectively utilize MCQs on the Molecular Basis of Inheritance for exam preparation?
To effectively utilize MCQs on the Molecular Basis of Inheritance for exam preparation, students should first ensure a solid understanding of the fundamental genetic principles covered in their curriculum. They can then practice solving a variety of MCQs from reputable sources, focusing on topics such as DNA replication, genetic code, gene expression, and genetic disorders. It’s essential to analyze both correct and incorrect answers to deepen understanding and identify areas for improvement. Regular practice with MCQs, coupled with thorough revision of concepts, can significantly enhance students’ performance in Class 12 Chemistry examinations.