- Understand the molecular basis of inheritance.
- Explain the structure and function of DNA and RNA.
- Describe the processes of DNA replication, transcription, and translation.
- Analyze the role of ribosomes in protein synthesis.
- Differentiate between various types of nucleic acids and their functions.
- Discuss the significance of the Human Genome Project.
- Explain the concept of DNA fingerprinting and its applications.
- Understand the implications of mutations in genetic material.
Molecular Basis of Inheritance
CBSE Learning Objectives – Key Concepts & Skills You Must Know
CBSE Revision Notes & Quick Summary for Last-Minute Study
Molecular Basis of Inheritance
Exercises
-
Group the following as nitrogenous bases and nucleosides:
- Nitrogenous Bases: Adenine, Thymine, Guanine, Cytosine, Uracil
- Nucleosides: Cytidine, Guanosine
-
If a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
-
If the sequence of one strand of DNA is written as follows:
- 5'-ATGCATGCATGCATGCATGCATGCATGC-3'
- Write down the sequence of complementary strand in 5'-3' direction.
-
If the sequence of the coding strand in a transcription unit is written as follows:
- 5'-ATGCATGCATGCATGCATGCATGCATGC-3'
- Write down the sequence of mRNA.
-
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
-
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
-
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
-
Differentiate between the following:
- (a) Repetitive DNA and Satellite DNA
- (b) mRNA and tRNA
- (c) Template strand and Coding strand
-
List two essential roles of ribosome during translation.
-
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
-
Explain (in one or two lines) the function of the following:
- (a) Promoter
- (b) tRNA
- (c) Exons
-
Why is the Human Genome project called a mega project?
-
What is DNA fingerprinting? Mention its application.
-
Briefly describe the following:
- (a) Transcription
- (b) Polymorphism
- (c) Translation
- (d) Bioinformatics
CBSE Exam Tips, Important Questions & Common Mistakes to Avoid
Common Mistakes and Exam Tips
Common Pitfalls
- Misunderstanding DNA and RNA Functions: Students often confuse the roles of DNA and RNA. Remember, DNA is primarily the genetic material, while RNA serves various roles including messenger functions.
- Confusing Nucleotides and Nucleosides: Be clear on the difference; nucleotides are the building blocks of nucleic acids, while nucleosides consist of a nitrogenous base and a sugar.
- Errors in Base Pairing: Ensure you remember the correct base pairing rules (A-T and G-C for DNA).
- Neglecting the Importance of Mutations: Many overlook how mutations can affect protein synthesis and lead to diseases like sickle cell anemia.
Exam Tips
- Practice Complementary Strand Writing: Regularly practice writing complementary DNA strands to avoid mistakes in directionality (5'-3' vs 3'-5').
- Understand the Central Dogma: Be clear on the flow of genetic information: DNA -> RNA -> Protein. This is fundamental and often tested.
- Familiarize with Genetic Code: Know how to translate mRNA sequences into amino acids, as this is a common question type.
- Review DNA Fingerprinting Techniques: Understand the significance of repetitive DNA and how it is used in forensic science.
- Clarify the Role of Ribosomes: Remember the essential functions of ribosomes during translation to avoid vague answers.
CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions
Multiple Choice Questions
A.
Identifying all the genes expressed as proteins.
B.
Sequencing the entire genome without any technological assistance.
C.
Assigning functions to all coding and non-coding regions after sequencing.
D.
Sequencing only the exons of the genome.
Correct Answer: C
Solution:
One of the major challenges in the Human Genome Project was to assign functions to the sequenced genome, including both coding and non-coding regions, which involved sequence annotation.
A.
Adenine
B.
Guanosine
C.
Ribose
D.
Deoxyribose
Correct Answer: A
Solution:
Adenine is a nitrogenous base, while guanosine is a nucleoside. Ribose and deoxyribose are sugars.
A.
30%
B.
20%
C.
40%
D.
10%
Correct Answer: A
Solution:
Initially, if adenine is 20%, thymine is also 20% due to base pairing rules. This leaves 60% for guanine and cytosine, which must be equal, so each is 30%. The enzyme specifically mutates A-T pairs, so the percentage of guanine remains unchanged at 30%.
A.
To discover new species
B.
To sequence the entire human genome
C.
To develop new antibiotics
D.
To study plant genetics
Correct Answer: B
Solution:
The primary goal of the Human Genome Project was to sequence the entire human genome and identify all the genes it contains.
A.
BAC vectors can replicate in both prokaryotic and eukaryotic cells.
B.
BAC vectors can carry larger DNA fragments compared to other vectors.
C.
BAC vectors have a higher replication rate than other vectors.
D.
BAC vectors are more stable than YAC vectors.
Correct Answer: B
Solution:
BAC vectors are specifically designed to carry large DNA fragments, making them ideal for cloning large segments of DNA in genetic engineering experiments.
A.
5'-TACGTACGTACG-3'
B.
5'-GCATGCATGCAT-3'
C.
5'-CATGCATGCATG-3'
D.
5'-GTACGTACGTAC-3'
Correct Answer: A
Solution:
The complementary strand is formed by pairing adenine with thymine and guanine with cytosine, resulting in 5'-TACGTACGTACG-3'.
A.
tRNA synthesizes mRNA from DNA.
B.
tRNA provides the ribosomal platform for protein synthesis.
C.
tRNA carries amino acids to the ribosome and matches them to the coded mRNA message.
D.
tRNA catalyzes the formation of peptide bonds between amino acids.
Correct Answer: C
Solution:
tRNA acts as an adapter molecule that carries specific amino acids to the ribosome and matches them to the codons on the mRNA through its anticodon, facilitating protein synthesis.
A.
SNPs can help in identifying new species.
B.
SNPs can be used to trace evolutionary history.
C.
SNPs can assist in locating disease-associated sequences.
D.
SNPs can be used to enhance agricultural yield.
Correct Answer: C
Solution:
SNPs are variations at a single nucleotide position in DNA among individuals. They are valuable in medical research for locating chromosomal regions associated with diseases.
A.
RNA polymerase I
B.
RNA polymerase II
C.
RNA polymerase III
D.
DNA polymerase
Correct Answer: A
Solution:
RNA polymerase I is responsible for the transcription of rRNA (28S, 18S, and 5.8S) in eukaryotic cells.
A.
Capping protects the hnRNA from degradation and assists in ribosome binding during translation.
B.
Capping is involved in the removal of introns from hnRNA.
C.
Capping marks the hnRNA for export out of the nucleus.
D.
Capping is necessary for the synthesis of tRNA.
Correct Answer: A
Solution:
Capping involves the addition of a methyl guanosine triphosphate to the 5'-end of hnRNA, which protects it from degradation and aids in ribosomal recognition.
A.
The operon will be continuously expressed, regardless of the presence of the inducer.
B.
The operon will never be expressed, even in the presence of the inducer.
C.
The operon will be expressed only in the presence of the inducer.
D.
The operon will be expressed only in the absence of the inducer.
Correct Answer: A
Solution:
If the repressor cannot bind to the operator, the operon will be expressed continuously because the repressor is unable to block RNA polymerase from transcribing the operon genes.
A.
The ribosome will initiate translation at random codons.
B.
Translation will not initiate, and no protein will be synthesized.
C.
Translation will initiate but produce truncated proteins.
D.
Translation will initiate correctly after a delay.
Correct Answer: B
Solution:
If the ribosome cannot recognize the start codon, translation cannot initiate, resulting in no protein synthesis.
A.
They code for proteins.
B.
They have no known function.
C.
They help in understanding chromosome structure and evolution.
D.
They are involved in mRNA splicing.
Correct Answer: C
Solution:
Repetitive sequences shed light on chromosome structure, dynamics, and evolution.
A.
To terminate transcription
B.
To initiate DNA replication
C.
To initiate transcription
D.
To bind tRNA
Correct Answer: C
Solution:
The promoter is a specific DNA sequence where transcription begins.
A.
5'-UACGUACGUACG-3'
B.
5'-AUGCAUGCAUGC-3'
C.
5'-TACGTACGTACG-3'
D.
5'-ATGCATGCATGC-3'
Correct Answer: B
Solution:
The mRNA sequence is complementary to the template strand and identical to the coding strand, except that thymine (T) is replaced by uracil (U).
A.
It carries amino acids to the ribosome.
B.
It synthesizes mRNA from DNA.
C.
It forms the structure of ribosomes.
D.
It acts as a template for DNA replication.
Correct Answer: A
Solution:
tRNA acts as an adapter molecule that carries specific amino acids to the ribosome during protein synthesis.
A.
It acts as a template for mRNA synthesis.
B.
It carries amino acids to the ribosome.
C.
It forms the structural component of ribosomes.
D.
It catalyzes the formation of peptide bonds.
Correct Answer: B
Solution:
tRNA acts as an adapter molecule that carries specific amino acids to the ribosome during protein synthesis.
A.
The presence of lactose inhibits the repressor protein.
B.
The depletion of lactose removes the inducer, allowing the repressor to bind to the operator.
C.
The operon shuts down due to the accumulation of glucose.
D.
The operon is permanently activated once induced by lactose.
Correct Answer: B
Solution:
Once lactose is metabolized, the inducer (allolactose) is no longer present to inactivate the repressor, allowing it to bind to the operator and shut down the operon.
A.
RNA polymerase I
B.
RNA polymerase II
C.
RNA polymerase III
D.
DNA polymerase
Correct Answer: C
Solution:
RNA polymerase III is responsible for the transcription of tRNA, 5s rRNA, and snRNAs in eukaryotes.
A.
It acts as a regulatory RNA molecule.
B.
It is a pseudogene.
C.
It is an intron.
D.
It is a promoter sequence.
Correct Answer: A
Solution:
Sequences that are transcribed but not translated often function as regulatory RNA molecules, such as microRNAs or long non-coding RNAs, which play roles in gene regulation.
A.
It is always active, regardless of lactose presence.
B.
It is induced by lactose and shuts down when lactose is depleted.
C.
It is a part of the human genome.
D.
It codes for ribosomal RNA.
Correct Answer: B
Solution:
The lac operon is induced by lactose and shuts down when lactose is no longer present in the medium.
A.
The protein will be longer and non-functional.
B.
The protein will be shorter and functional.
C.
The protein synthesis will be unaffected.
D.
The protein will be longer and functional.
Correct Answer: A
Solution:
Retention of introns in mRNA means that non-coding sequences will be translated into the protein, leading to a longer polypeptide chain that is likely non-functional due to incorrect folding or premature stop codons.
A.
Transcription of mRNA
B.
Translation of proteins
C.
Replication of DNA
D.
Splicing of introns
Correct Answer: A
Solution:
RNA polymerase II is responsible for transcribing mRNA from DNA. Inhibition of this enzyme would directly affect the transcription process.
A.
RNA polymerase I
B.
RNA polymerase II
C.
RNA polymerase III
D.
DNA polymerase
Correct Answer: B
Solution:
RNA polymerase II is responsible for transcribing the precursor of mRNA in eukaryotes.
A.
The DNA double helix will become unstable and denature.
B.
The DNA will replicate more efficiently.
C.
The transcription process will become faster.
D.
The DNA will form a triple helix structure.
Correct Answer: A
Solution:
Hydrogen bonds between adenine and thymine are crucial for the stability of the DNA double helix. Without these bonds, the helix would become unstable and likely denature.
A.
20%
B.
30%
C.
40%
D.
60%
Correct Answer: B
Solution:
In a double-stranded DNA, the amount of cytosine equals the amount of guanine, and the amount of adenine equals the amount of thymine. If cytosine is 20%, guanine is also 20%, making a total of 40% for cytosine and guanine. The remaining 60% must be adenine and thymine, so adenine is 30%.
A.
DNA polymerase
B.
RNA polymerase II
C.
Poly(A) polymerase
D.
Ribonuclease
Correct Answer: C
Solution:
Poly(A) polymerase is responsible for adding adenylate residues to the 3'-end of hnRNA in a process known as tailing.
A.
Transcription of rRNA would be halted.
B.
Transcription of tRNA would be halted.
C.
Transcription of mRNA would be halted.
D.
Transcription of all types of RNA would be halted.
Correct Answer: C
Solution:
RNA polymerase II is responsible for the transcription of mRNA in eukaryotic cells. Inhibiting it would specifically halt mRNA synthesis.
A.
Polymerase chain reaction (PCR)
B.
Sanger sequencing
C.
CRISPR-Cas9 technology
D.
Next-generation sequencing
Correct Answer: B
Solution:
The Human Genome Project primarily used the Sanger sequencing method to determine the sequence of human DNA.
A.
They bind to DNA to initiate transcription.
B.
They provide a platform for joining amino acids.
C.
They splice introns from mRNA.
D.
They transport amino acids to mRNA.
Correct Answer: B
Solution:
Ribosomes bind to mRNA and provide a platform for the joining of amino acids during protein synthesis.
A.
They code for proteins.
B.
They have no direct coding function but shed light on chromosome structure.
C.
They are involved in the synthesis of RNA.
D.
They are responsible for DNA replication.
Correct Answer: B
Solution:
Repetitive DNA sequences do not code for proteins but provide insights into chromosome structure, dynamics, and evolution.
A.
The gene will be transcribed at a higher rate.
B.
The gene will not be transcribed.
C.
The gene will be transcribed at a normal rate.
D.
The gene will be transcribed only under stress conditions.
Correct Answer: B
Solution:
If RNA polymerase cannot bind to the promoter, transcription of the gene cannot initiate, leading to no transcription.
A.
Increased transcription of the gene regardless of lactose presence.
B.
Decreased transcription of the gene regardless of lactose presence.
C.
Transcription of the gene only in the presence of lactose.
D.
No effect on the transcription of the gene.
Correct Answer: B
Solution:
Mutations in the promoter region can affect the binding of RNA polymerase, leading to decreased transcription of the gene, irrespective of the presence of lactose.
A.
It acts as a messenger RNA (mRNA) carrying genetic information.
B.
It functions as a ribosomal RNA (rRNA) involved in protein synthesis.
C.
It serves as a transfer RNA (tRNA) bringing amino acids to the ribosome.
D.
It functions as a regulatory RNA, such as a microRNA (miRNA), involved in gene expression regulation.
Correct Answer: D
Solution:
RNA sequences that are transcribed but not translated often have regulatory roles, such as miRNAs, which regulate gene expression post-transcriptionally.
A.
Negative inducible
B.
Positive inducible
C.
Negative repressible
D.
Positive repressible
Correct Answer: A
Solution:
In negative inducible operons, the operon is normally off because a repressor is bound to the operator. The presence of an inducer (in this case, the novel sugar) binds to the repressor, causing it to detach and allowing transcription to proceed.
A.
The presence of large amounts of repetitive DNA sequences.
B.
The extensive presence of non-coding RNA genes.
C.
The presence of introns in the genome.
D.
The high level of genetic polymorphism in humans.
Correct Answer: A
Solution:
Repetitive sequences make up a large portion of the human genome and do not code for proteins, thus contributing to the low percentage of coding DNA.
A.
Transcribes tRNA
B.
Transcribes rRNA
C.
Transcribes mRNA
D.
Transcribes snRNA
Correct Answer: C
Solution:
RNA polymerase II is responsible for transcribing the precursor of mRNA, known as heterogeneous nuclear RNA (hnRNA).
A.
They are non-coding sequences removed during splicing.
B.
They code for proteins and are joined together after splicing.
C.
They bind to ribosomes during translation.
D.
They initiate DNA replication.
Correct Answer: B
Solution:
Exons are coding sequences in eukaryotic genes that are joined together after splicing to form functional mRNA.
A.
The tRNA will now incorporate leucine instead of phenylalanine.
B.
The tRNA will still incorporate phenylalanine as the mutation is silent.
C.
The tRNA will incorporate serine instead of phenylalanine.
D.
The tRNA will incorporate lysine instead of phenylalanine.
Correct Answer: B
Solution:
The anticodon UUU and UUC both correspond to phenylalanine. Therefore, the mutation is silent, and the tRNA will still incorporate phenylalanine during protein synthesis.
A.
They are involved in regulating gene expression.
B.
They are completely non-functional.
C.
They code for alternative proteins.
D.
They are responsible for DNA replication.
Correct Answer: A
Solution:
Non-coding regions of the genome often play crucial roles in regulating gene expression, maintaining chromosome structure, and other regulatory functions.
A.
The operon is regulated by a repressor that is activated by the substrate.
B.
The operon is regulated by a feedback inhibition mechanism.
C.
The operon is constitutively expressed and not regulated.
D.
The operon is regulated by the depletion of substrate over time.
Correct Answer: D
Solution:
In bacterial operons like the lac operon, the presence of substrate (e.g., lactose) induces the operon, but once the substrate is metabolized and depleted, the operon shuts down as there is no longer any substrate to induce its expression.
A.
Binding to mRNA and providing a platform for amino acid assembly
B.
Synthesizing DNA from RNA
C.
Transporting amino acids to the site of protein synthesis
D.
Unwinding the DNA double helix
Correct Answer: A
Solution:
Ribosomes bind to mRNA and provide a platform for the assembly of amino acids into proteins during translation.
A.
The operon is regulated by a feedback inhibition mechanism.
B.
The sugar is degraded by the bacteria.
C.
The operon is permanently deactivated after initial activation.
D.
The sugar changes its structure and becomes inactive.
Correct Answer: A
Solution:
Feedback inhibition is a common regulatory mechanism where the end product of a metabolic pathway inhibits an enzyme involved in its synthesis, thereby shutting down the pathway.
A.
Adenine - Thymine
B.
Cytosine - Thymine
C.
Guanine - Uracil
D.
Adenine - Uracil
Correct Answer: A
Solution:
In DNA, Adenine pairs with Thymine through two hydrogen bonds.
A.
To store genetic information.
B.
To transport amino acids to ribosomes.
C.
To serve as a template for protein synthesis.
D.
To replicate DNA.
Correct Answer: C
Solution:
mRNA serves as a template for protein synthesis, carrying the genetic code from DNA to the ribosome.
A.
The uniform diameter of the double helix.
B.
The antiparallel orientation of the two strands.
C.
The complementary base pairing between adenine and thymine, and guanine and cytosine.
D.
The helical twist of the DNA strands.
Correct Answer: C
Solution:
Watson and Crick's model proposed that each strand of the DNA double helix serves as a template for the creation of a new complementary strand, due to the specific base pairing (A-T and G-C), which is the basis for the semi-conservative model of replication.
A.
Repetitive sequences directly code for proteins.
B.
Repetitive sequences are thought to have no direct coding function but provide insights into chromosome structure.
C.
Repetitive sequences are only found in non-coding regions of the genome.
D.
Repetitive sequences are unique to the human genome and not found in other organisms.
Correct Answer: B
Solution:
Repetitive sequences do not directly code for proteins but help in understanding chromosome structure and dynamics.
A.
Messenger RNA (mRNA)
B.
Transfer RNA (tRNA)
C.
Ribosomal RNA (rRNA)
D.
Small nuclear RNA (snRNA)
Correct Answer: C
Solution:
Ribosomal RNA (rRNA) is known to have catalytic functions, such as forming peptide bonds during translation, acting as a ribozyme.
A.
Each new DNA molecule consists of two newly synthesized strands.
B.
Each new DNA molecule consists of one old and one new strand.
C.
DNA replication results in two identical RNA molecules.
D.
DNA replication occurs without the use of a template.
Correct Answer: B
Solution:
In semi-conservative replication, each new DNA molecule consists of one original strand and one newly synthesized strand.
A.
It aimed to sequence the entire human genome in one year.
B.
It was completed in 1990.
C.
It was a collaborative effort involving multiple countries.
D.
It focused solely on sequencing human DNA without any technological advancements.
Correct Answer: C
Solution:
The Human Genome Project was a collaborative effort involving contributions from the U.S., U.K., Japan, France, Germany, China, and others.
A.
The lac operon will be induced and express lactose-metabolizing enzymes.
B.
The lac operon will remain inactive and not express lactose-metabolizing enzymes.
C.
The lac operon will be partially induced.
D.
The lac operon will express enzymes only in the presence of glucose.
Correct Answer: B
Solution:
A mutation in the promoter region that makes it unresponsive to lactose means that the operon cannot be induced by lactose, and thus lactose-metabolizing enzymes will not be expressed.
True or False
Correct Answer: True
Solution:
The excerpt states that the lac operon is regulated by the amount of lactose present.
Correct Answer: True
Solution:
The human genome contains 3164.7 million base pairs as stated in the excerpt.
Correct Answer: True
Solution:
The excerpt describes how the genetic code uses combinations of nucleotides to code for amino acids.
Correct Answer: False
Solution:
In eukaryotic cells, the primary transcript (hnRNA) undergoes processing such as splicing to remove introns, and capping and tailing, before it becomes mature mRNA that can be translated into proteins.
Correct Answer: True
Solution:
The complementary hydrogen bonding between adenine and thymine, and guanine and cytosine, allows each strand of the DNA double helix to serve as a template for the synthesis of a new complementary strand during replication, supporting the semi-conservative replication model.
Correct Answer: True
Solution:
The primary transcripts contain both the exons and the introns and are non-functional.
Correct Answer: True
Solution:
The Watson and Crick model introduced the double helix structure, which was simple yet comprehensive in explaining how genetic information is stored and replicated.
Correct Answer: False
Solution:
RNA polymerase II is responsible for transcribing precursor mRNA (hnRNA) in eukaryotic cells, not RNA polymerase III.
Correct Answer: False
Solution:
While RNA primarily functions as a messenger, it also serves as an adapter, structural, and catalytic molecule in various organisms.
Correct Answer: False
Solution:
The excerpt explains that in eukaryotes, mRNA is not directly functional after transcription and requires splicing to remove introns.
Correct Answer: False
Solution:
In eukaryotes, the primary transcript (hnRNA) undergoes splicing to remove introns and additional processing such as capping and tailing to become mature mRNA.
Correct Answer: True
Solution:
The genetic code is nearly universal, with the same codons generally specifying the same amino acids across most organisms, which is a key feature of the genetic code.
Correct Answer: False
Solution:
DNA is chemically and structurally more stable than RNA, making it a better genetic material for long-term storage of information.
Correct Answer: False
Solution:
In eukaryotes, the transcribed mRNA is not immediately functional as it contains introns that need to be removed through splicing before it can be translated.
Correct Answer: True
Solution:
It is stated that less than 2% of the human genome codes for proteins.
Correct Answer: True
Solution:
The Human Genome Project revealed that less than 2% of the human genome codes for proteins.
Correct Answer: False
Solution:
The Human Genome Project aimed to sequence the entire human genome, including both coding and non-coding regions.
Correct Answer: False
Solution:
In eukaryotes, the primary transcript (hnRNA) undergoes processing like splicing, capping, and tailing to become functional mRNA.
Correct Answer: True
Solution:
The Human Genome Project was indeed a mega project aimed at sequencing the entire human genome, and it was completed in 2003 as stated in the excerpts.
Correct Answer: False
Solution:
The lac operon shuts down after some time following the addition of lactose.
Correct Answer: False
Solution:
In eukaryotes, introns are removed and exons are joined to produce functional RNA.
Correct Answer: False
Solution:
In eukaryotes, the primary mRNA transcript undergoes splicing where introns are removed and exons are joined, along with capping and tailing, before it becomes a functional mRNA.
Correct Answer: True
Solution:
The excerpt states that DNA is more stable than RNA, which makes it a better genetic material.
Correct Answer: True
Solution:
The Watson and Crick model of DNA proposed the central dogma of molecular biology, which states that genetic information flows from DNA to RNA to protein.
Correct Answer: True
Solution:
The excerpt mentions that the Human Genome Project aimed to sequence the entire human genome.
Correct Answer: True
Solution:
The genetic code consists of triplets of nucleotides, known as codons, which can generate 64 possible combinations, sufficient to code for 20 amino acids.
Correct Answer: True
Solution:
Translation requires transfer of genetic information from a polymer of nucleotides to synthesize a polymer of amino acids.
Correct Answer: False
Solution:
The genetic code is composed of combinations of three nucleotides, known as codons, to code for amino acids.
Correct Answer: True
Solution:
The genetic code is nearly universal, meaning that the same codons specify the same amino acids in almost all organisms.
Correct Answer: False
Solution:
Adenine pairs with thymine, not cytosine, in the DNA double helix.
Correct Answer: True
Solution:
RNA is believed to be the first to evolve, with DNA being derived from RNA.
Correct Answer: False
Solution:
The human genome is estimated to have about 30,000 genes, much lower than previous estimates.
Correct Answer: True
Solution:
The genetic code is a triplet code, where combinations of three nucleotides (codons) specify each amino acid.
Correct Answer: False
Solution:
The Central dogma states that the genetic information flows from DNA to RNA to protein.
Correct Answer: True
Solution:
The Human Genome Project found that repetitive DNA sequences do not directly code for proteins, but they contribute to chromosome structure, dynamics, and evolution.
Correct Answer: True
Solution:
DNA fingerprinting works on the principle of polymorphism in DNA sequences, which allows for the identification of variations among individuals.
Correct Answer: True
Solution:
The average gene consists of 3000 bases, but sizes vary greatly.
Correct Answer: False
Solution:
The total number of genes is estimated at 30,000, much lower than previous estimates of 80,000 to 140,000 genes.
Correct Answer: False
Solution:
Translation involves the synthesis of proteins from mRNA and does not rely on direct complementarity between nucleotides and amino acids.
Correct Answer: True
Solution:
The distance between base pairs in the DNA helix is approximately 0.34 nm, contributing to the stability of the helical structure.
Correct Answer: True
Solution:
RNA is believed to have been the first genetic material, and its versatility suggests it played a crucial role in early life forms.
Correct Answer: False
Solution:
DNA is considered more chemically and structurally stable than RNA, making it a better genetic material.
Correct Answer: False
Solution:
RNA polymerase III is responsible for transcribing tRNA, not RNA polymerase II.