Molecular Basis of Inheritance - Class 12 Biology - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Molecular Basis of Inheritance | NCERT | Biology | Class 12
Which of the following is the temporal bone?
A) B
B) C
C) D
D) E
The correct answer is A) B.
The human skull consists of two main categories of bones:
Cranial bones
Facial bones
In the image referenced, the bones are specifically cranial bones. The options correspond to:
Frontal bone (1)
Parietal bones (2)
Thus, the temporal bone is indicated by A) B.
ABO blood group is based on:
A) Dominance
B) Incomplete dominance
C) Epistasis
D) Multiple alleles
The correct answer is D) Multiple alleles.
Blood type is a classic representation of multiple allele inheritance. Humans can exhibit one of four possible blood type phenotypes: A, B, AB, and O. The genetic determination of blood type is controlled by three alleles for a single gene.
The idea that a combination of three bases should code for each amino acid was proposed by:
A) Francis Crick
B) James Watson
C) George Gamow
D) Alec Jeffreys
The correct answer is C) George Gamow.
George Gamow proposed that a combination of three nucleotide bases can be used in different combinations to code for the 20 known amino acids, which he termed as triplet code. This foundational idea was a significant development in understanding the genetic code's structure.
The molecular action of UV light is mainly reflected through:
A) Photodynamic action.
B) Formation of pyrimidine.
C) Formation of sticky metaphases.
D) Destruction of hydrogen bonds between DNA strands.
The correct answer is D) Destruction of hydrogen bonds between DNA strands.
UV rays are particularly harmful to living organisms because DNA and proteins absorb UV rays preferentially. The high energy associated with UV rays is capable of breaking the hydrogen bonds within these molecules. This damage to DNA can lead to mutations, contribute to skin aging, damage to skin cells, and various types of skin cancers. Hence, the primary molecular action of UV light is the destruction of hydrogen bonds between DNA strands.
In a mutational event, when adenine is replaced by guanine, it is the case of:
A) Frameshift mutation
B) Transcription
C) Transition
D) Transversion
The correct answer is C) Transition.
Transition is a type of mutation where one purine base is replaced by another purine base, or one pyrimidine base is replaced by another pyrimidine base. In this specific case, adenine (which is a purine) is replaced by guanine (also a purine), thereby categorizing this mutation as a transition.
Assertion: Mitochondria and chloroplast are semi-autonomous organelles. Reason: They have their genetic material which codes for their proteins.
A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion C. Assertion is true but Reason is false D. Both Assertion and Reason are false
The correct answer is A. Both the Assertion and Reason are true, and the Reason is indeed the correct explanation of the Assertion.
Mitochondria and chloroplasts are described as semi-autonomous organelles because they possess their own genetic material in the form of DNA and have the capability to synthesize some of their own proteins. This genetic independence supports their semi-autonomous nature, allowing them to perform certain functions independently from the cell nucleus. However, their existence and functionality are also integrated with and regulated by the nucleus, making them not completely autonomous but rather semi-autonomous. Thus, the presence of their own genetic material, which codes for some of their proteins, correctly explains why they are considered semi-autonomous.
"Antiparallel strands of DNA molecule means:
A. One strand turns clockwise.
B. One strand turns anticlockwise.
C. Phosphate groups of the two strands share the same position at their ends.
D. Phosphate groups at the start of two DNA strands are in opposite positions."
The correct answer is D. Phosphate groups at the start of two DNA strands are in opposite positions.
In DNA, antiparallel orientation refers to the arrangement in which the 5' to 3' direction of one strand runs opposite to the 3' to 5' direction of the other strand. Therefore, one strand begins with a 5' phosphate group at one end, while the complementary strand starts with a 3' hydroxyl group at the same end, effectively placing their terminal phosphate groups in opposite positions.
Hershey and Chase's experiment proving DNA as the genetic material was based on the principle:
A) Transduction
B) Transformation
C) Transcription
D) Translation
The correct answer is A) Transduction.
Hershey and Chase conducted an experiment demonstrating that DNA is the genetic material; they based their research on the principle of Transduction. This process involves the transfer of DNA from one bacterium to another by a virus.
Watson and Crick proposed the helical model of DNA in the year:
A) 1943
B) 1953
C) 1962
D) 1973
The correct answer is B) 1953.
James Watson and Francis Crick proposed the helical model of DNA, famously known as the double helix structure, in the year 1953. This groundbreaking discovery has significantly shaped the understanding of genetic information storage and transmission.
In diseases that show X-linked inheritance, the females are usually the:
A) Promoters
B) Carriers
C) Operators
D) None of the above
The correct answer is B) Carriers.
In diseases that follow X-linked inheritance, females often do not express the disorder despite having one defective allele on the X chromosome, provided this allele is recessive. This is because the other normal allele can mask the effects of the defective one. Conversely, males, possessing only one X chromosome, will show symptoms of the disease if they inherit the defective allele, since they lack a second X chromosome that could carry a potential normal allele to counteract the mutation.
Since males receive their X chromosome from their mothers, the transmission of such diseases often traces from a carrier mother to her son, highlighting the role of females largely as carriers in X-linked conditions.
A man with a genetic disorder marries a normal woman. They have three daughters and five sons. All the daughters inherited the disease, and the sons were normal. The gene of this disease is:
A Sex-linked dominant
B Sex-linked recessive
C Autosomal dominant
D Autosomal recessive
The correct response is A: Sex-linked dominant.
Sex-linked illnesses are genetic diseases linked primarily to the sex chromosomes (the X and Y chromosomes). When considering dominant inheritance patterns, a single copy of a mutated gene (on one of these chromosomes) from one parent is sufficient to manifest the disease, overpowering the normal gene from the other parent.
In scenarios involving an X-linked dominant disorder, if the father carries the mutated gene on his X chromosome:
All his daughters will inherit the gene and thus the disorder, because daughters receive their father’s X chromosome.
His sons will only receive his Y chromosome, not the X chromosome carrying the mutation, and therefore will not inherit the disorder.
This inheritance pattern matches precisely with the family described: all daughters inheriting the disease and none of the sons, pinpointing the genetic transmission as X-linked dominant.
A short length of DNA molecule contains 120 adenine and 120 cytosine bases. The total number of nucleotides in this DNA segment is:
A. 120
B. 60
C. 240
D. 480
In understanding DNA structure, it is pivotal to recognize the base pairing rules which state that adenine (A) always pairs with thymine (T) and cytosine (C) always pairs with guanine (G). Knowing this rule helps in determining the total number of nucleotides in a DNA segment if the number of specific bases are given.
Given in the problem, we have 120 adenine and 120 cytosine bases. According to the base pairing rules:
Each adenine pairs with a thymine. Therefore, if there are 120 adenine bases, there must also be 120 thymine bases.
Each cytosine pairs with a guanine. Therefore, if there are 120 cytosine bases, there must also be 120 guanine bases.
Thus, for the entire segment, we have:
120 adenine bases
120 thymine bases
120 cytosine bases
120 guanine bases
Adding all these together, the total number of nucleotides in the DNA segment is: $$ 120 (\text{A}) + 120 (\text{T}) + 120 (\text{C}) + 120 (\text{G}) = 480 $$
Hence, the correct answer is D. 480.
Which of the following is not strictly considered as a part of a neuron? A. Dendrite B. Myelin sheath C. Axon D. Cell body
When considering the components of a neuron and identifying which one is not strictly considered a part of a neuron, we need to analyze each component listed:
Dendrites are vital for neurons as they are the structures that receive signals from other neurons. This makes them indispensable for neuron functionality.
The axon is another crucial part of a neuron; it transmits signals away from the neuron's cell body to other neurons, muscles, or glands.
The cell body contains the nucleus and cytoplasm of the neuron and is essential for the neuron's general functions, including protein synthesis.
However, the myelin sheath is different. While it is associated with many neurons, primarily in the central and peripheral nervous systems to insulate axons and increase signal transmission speed, it is not present in all neurons. Some neurons are unmyelinated yet still function within the nervous system, particularly in the autonomic nervous system. This distinction reveals that myelin sheath is not a mandatory structural component for a neuron's operation but rather an enhancement that exists in specific types of neurons.
Therefore, the correct choice, which is not strictly considered a part of a neuron, is:
B. Myelin Sheath
In living cells, synthesis of ribonucleic acid (RNA) takes place in: A. Cytoplasm B. Nucleus C. Golgi body D. Nephron
To understand where the synthesis of ribonucleic acid (RNA) occurs in living cells, we must first recognize that RNA is a type of nucleic acid primarily found in the nucleus of the cell. The process responsible for RNA synthesis is called transcription, a critical function happening within the nucleus. During transcription, RNA is synthesized from DNA with the help of various enzymes and molecular machinery which are also located in the nucleus.
Given the options: A. Cytoplasm B. Nucleus C. Golgi body D. Nephron
Cytoplasm is incorrect because it contains cell organelles and the site for many metabolic reactions, but not the primary location for transcription.
Nucleus is correct as it is the central location where transcription – the process of synthesizing RNA – takes place.
Golgi body is involved in modifying, sorting, and packaging proteins and lipids but not in RNA synthesis.
Nephron is the functional unit of the kidney, primarily involved in excretion and filtration, and unrelated to RNA synthesis.
Therefore, the correct answer to where RNA synthesis occurs in living cells is: B. Nucleus
Genetic material is carried out by long chain of molecules made up of:
A) enzymes
B) DNA
C) amino acids
D) proteins
The question you are asking about is what constitutes the long chain of molecules that make up genetic material. Genetic material is primarily found within the nucleus of a cell. The crucial component of genetic material is DNA, which stands for Deoxyribonucleic Acid.
DNA consists of long chains of molecules which include:
Nitrogenous bases
Phosphate groups (these provide a negative charge to DNA)
A sugar component, which is ribose in RNA and deoxyribose in DNA.
Given the options:
A) enzymes
B) DNA
C) amino acids
D) proteins
The correct answer is Option B: DNA. DNA forms the long chains that carry genetic information in organisms.
To clarify further, DNA is not the same as proteins or amino acids. Proteins are synthesized using amino acids as building blocks, which are different from the nucleotides that make up DNA. Enzymes, which are frequently proteins, play roles in facilitating various biochemical reactions but do not form the genetic material itself.
Thus, the fundamental molecular composition of genetic material in any organism is DNA.
Chemical composition of a chromosome is: A) DNA and lipids B) DNA and carbohydrates C) Proteins and lipids D) DNA and proteins
When discussing the chemical composition of chromosomes found within the cell nucleus, they primarily consist of DNA (deoxyribonucleic acid) and proteins. DNA is the genetic material responsible for storing and transferring genetic information in an organism. In the nucleus, DNA does not exist as a floating strand; rather, it is structurally supported and organized through interactions with various proteins.
One of the key groups of proteins associated with DNA are the histones, which are positively charged. Histones help in packaging the DNA into structural units known as nucleosomes, enabling the long molecules of DNA to be compactly folded into the limited space of the nucleus. This complex of DNA and proteins, crucial for the formation of chromosomes, is collectively referred to as chromatin.
Given the answer options of DNA with lipids, carbohydrates, and proteins, the correct choice for the chemical composition of a chromosome is clearly:
D) DNA and proteins
This option accurately represents the fundamental biological makeup of chromosomes, emphasizing the role of both DNA and structural as well as regulatory proteins in their function and organization.
DNA synthesis occurs in which stage of the cell cycle?
A. G1 phase B. G2 phase C. M phase D. S phase
DNA synthesis is a crucial process in the cell cycle, which specifically occurs during the S phase (Synthesis phase). This phase is dedicated to the replication of DNA, preparing the cell for division by ensuring that each new cell will have a complete set of DNA.
The S phase follows the G1 phase (the first gap phase) and precedes the G2 phase (the second gap phase). The purpose of the S phase is not just limited to replicating DNA but also involves checking for any errors in the DNA to ensure accuracy in the genetic material passed on to the daughter cells.
To summarize, the correct answer to "DNA synthesis occurs in which stage of the cell cycle?" is: D. S phase. This stage is marked by DNA replication, essential for cell division and the maintenance of genetic continuity across generations.
The gene for hemophilia is present on the X chromosome. If a hemophilic male marries a normal female, the probability of their son being hemophilic is:
A. nil
B. 0.25
C. 0.5
D. 1
The problem requires understanding the inheritance pattern of hemophilia, which is a X-linked recessive disorder. This means the gene causing hemophilia is located on the X chromosome, and the disorder manifests in a male (who has only one X chromosome) if his X chromosome bears the mutant gene.
Given the scenario, we have:
Hemophilic male (genotype $X^hY$)
Normal female (genotype $XX$)
Here's what happens during reproduction, assuming ideal Mendelian inheritance:
Male Contribution: A male can pass either his X chromosome (affected) or his Y chromosome. Since we're discussing the probability of having a hemophilic son, we focus on the Y chromosome (sons inherit Y from their fathers).
Female Contribution: A normal female will pass one of her normal X chromosomes ($X$).
Combining these contributions, the potential genotypes for children can be:
Daughters: Can inherit the hemophilic X from father and a normal X from mother, resulting in the genotype $X^hX$ (carrier but not affected due to the normal X overshadowing the hemophilic one).
Sons: Will inherit the Y chromosome from their father and a normal X from their mother, resulting in the genotype $XY$ (normal).
Thus, if a son is considered, he will always inherit his father's Y chromosome and his mother’s normal X chromosome. This results in a normal son without hemophilia.
Therefore, the probability of a son being hemophilic in this specific cross is nil. Hence, the correct answer is: A. nil.
Which of the following is known as the 'Currency of Energy'?
A. DNA
B. RNA
C. ATP
D. NAD
Answer: C. ATP
ATP, or Adenosine Triphosphate, is commonly referred to as the "currency of energy" in biological systems. This is because ATP acts as a primary energy carrier within cells. In cellular processes, such as cellular respiration, energy derived from the breakdown of food is used to produce ATP molecules.
The process of cellular respiration primarily takes place in the mitochondria of the cell, often called the powerhouse of the cell. Here, nutrients from food are converted into ATP, thus enabling cellular functions by providing them with necessary energy.
Each ATP molecule is made up of a base (adenosine) and three phosphate groups, hence the name triphosphate. The key role of ATP in energy transfer makes it uniquely crucial in cellular function and metabolism.
Note that the other options given:
DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are genetic materials and not involved directly in energy storage or transfer.
NAD (Nicotinamide Adenine Dinucleotide) functions primarily as a coenzyme in redox reactions and is not considered a direct energy currency like ATP.
Thus, among the given choices, ATP (Option C) is correctly known as the "currency of energy".
Which of the following is the genetic material?
A. Protein
B. Carbohydrate
C. Vitamin
D. Nucleic Acid
Genetic material is defined as the substance that carries and transmits genetic information from one generation to another. This transmission of characteristics from parents to offspring occurs primarily during reproduction.
Each option in the question has a distinct biological function:
Protein, Carbohydrate, and Vitamin serve as essential components for nutrition and are vital for the energy needs and overall survival of organisms.
On the other hand, Nucleic Acid is fundamental in the realm of genetics.
Nucleic acids are the molecules that contain the blueprint of life in forms of DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid). DNA, in particular, is the predominant genetic material found in organisms and it houses the genetic instructions needed for the development and function of living things.
Given the roles of the options provided: A. Protein - nutrition B. Carbohydrate - energy source C. Vitamin - body regulation and health D. Nucleic Acid - transmission of genetic information
The correct answer is: D. Nucleic Acid
Which is called currency of energy? A. D.N.A. B. R.N.A. C. A.T.P. D. N.A.D.
The question is which substance is considered the "currency of energy" in our bodies. The correct answer is C. ATP (Adenosine Triphosphate).
ATP is often referred to as the energy currency of the cell, playing a crucial role in powering various biochemical activities, both internal and external, including bodily movements. This energy is produced in the mitochondria of cells mostly through the oxidation of carbohydrates (sugars). During this process, ATP synthesis occurs, resulting in the creation of ATP.
ATP provides the necessary energy for the cell by hydrolyzing its triphosphate group to release energy. This is why it is essential for numerous bodily functions and biochemical processes.
In the context of the options provided, DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid) are nucleic acids, significant for genetic information storage and transmission, not for energy storage. NAD (Nicotinamide adenine dinucleotide) serves as a cofactor and reducing agent in metabolic reactions and does not function as an energy currency. Thus, the correct option is:
C. ATP
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Ask Chatterbot AINCERT Solutions - Molecular Basis of Inheritance | NCERT | Biology | Class 12
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
The grouping of the given molecules as nitrogenous bases and nucleosides:
Nitrogenous Bases:
Adenine
Thymine
Uracil
Cytosine
Nucleosides:
Cytidine
Guanosine
Definitions:
Nitrogenous Bases are molecules that contain nitrogen and have the chemical properties of a base. They include purines (Adenine and Guanine) and pyrimidines (Thymine, Uracil, and Cytosine).
Nucleosides are nitrogenous bases attached to a sugar molecule (ribose in the case of RNA, deoxyribose in the case of DNA). Examples are Cytidine (Cytosine + Ribose) and Guanosine (Guanine + Ribose).
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
To determine the percentage of adenine (A) in a double-stranded DNA with 20% cytosine (C), we rely on Chargaff's rules, which state:
The amount of adenine (A) is equal to thymine (T).
The amount of cytosine (C) is equal to guanine (G).
Given:
%C = 20%
According to Chargaff's rules:
%G = %C = 20%
Since the total percentage of all bases in DNA must add up to 100%, we can write the equation: [ %A + %T + %C + %G = 100 ]
Substituting the known values: [ %A + %T + 20 + 20 = 100 ]
Since %A = %T: [ 2%A + 40 = 100 ]
Solving for %A: [ 2%A = 60 ] [ %A = 30 ]
So, the percentage of adenine (A) in the DNA is 30%.
If the sequence of one strand of DNA is written as follows:
$5^{\prime}$-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of complementary strand in $5^{\prime} \Omega^{\prime}$ direction.
To write the complementary strand of the given DNA sequence in the $5^{\prime} \rightarrow 3^{\prime}$ direction, we need to pair the complementary bases. Adenine (A) pairs with Thymine (T) and Cytosine (C) pairs with Guanine (G).
Starting with the given sequence: $$5^{\prime}-\text{ATGCATGCATGCATGCATGCATGCATGC}-3^{\prime}$$ The complementary sequence in $3^{\prime} \rightarrow 5^{\prime}$ direction is: $$3^{\prime}-\text{TACGTACGTACGTACGTACGTACGTACG}-5^{\prime}$$
To orient this complementary strand in the $5^{\prime} \rightarrow 3^{\prime}$ direction, we simply reverse it: $$5^{\prime}-\text{GCATGCATGCATGCATGCATGCATGCAT}-3^{\prime}$$
Thus, the sequence of the complementary strand in the $5^{\prime} \rightarrow 3^{\prime}$ direction is:
$$5^{\prime}-\text{GCATGCATGCATGCATGCATGCATGCAT}-3^{\prime}$$
If the sequence of the coding strand in a transcription unit is written as follows:
$5^{\prime}$-ATGCATGCATGCATGCATGCATGCATGC-3
Write down the sequence of mRNA.
To determine the sequence of the mRNA transcribed from the given coding strand, we must remember that the coding strand of DNA is the same as the mRNA sequence, except that the DNA base thymine (T) is replaced by uracil (U) in RNA. The sequence of the coding strand is:
$$5^{\prime} \text{-ATGCATGCATGCATGCATGCATGCATGC-} 3^{\prime}$$
By converting thymine (T) to uracil (U), the mRNA sequence becomes:
$$5^{\prime} \text{-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-} 3^{\prime}$$
So, the sequence of mRNA is:
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Watson and Crick hypothesized the semi-conservative mode of DNA replication based on the property of base pairing in the DNA double helix.
Key Points:
Complementarity:
In the DNA double helix, Adenine (A) forms two hydrogen bonds with Thymine (T), and Guanine (G) forms three hydrogen bonds with Cytosine (C).
Each base on one strand pairs specifically with a complementary base on the opposite strand, making the strands complementary to each other.
Strand Separation and Template Function:
Watson and Crick proposed that the two DNA strands separate during replication.
Each parental strand serves as a template for the synthesis of a new strand.
This would result in two DNA molecules, each containing one original (parental) strand and one newly-synthesized strand.
Detailed Explanation:
By understanding the complementarity of nucleotide bases, Watson and Crick realized that if the strands separated, the sequence of bases in each parental strand could direct the formation of a complementary new strand. Consequently, after replication:
Each daughter DNA molecule would have one strand from the parental DNA and one newly-synthesized strand.
This unique property of base pairing led to the name semi-conservative replication because half of the parental DNA (one of the original strands) is conserved in each daughter molecule.
Visual Representation:
*Figure: Double stranded polynucleotide chain showing base pairing*
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.
Types of Nucleic Acid Polymerases
Polymerases are enzymes that catalyze the synthesis of nucleic acids. Depending on the chemical nature of the template and the resulting nucleic acid, the polymerases can be categorized as follows:
DNA-Dependent DNA Polymerase:
Template: DNA
Product: DNA
Function: Catalyzes the replication of DNA using a DNA template.
DNA-Dependent RNA Polymerase:
Template: DNA
Product: RNA
Function: Synthesizes RNA from a DNA template during the process of transcription.
RNA-Dependent RNA Polymerase:
Template: RNA
Product: RNA
Function: Catalyzes the replication of RNA from an RNA template, usually found in RNA viruses.
RNA-Dependent DNA Polymerase (Reverse Transcriptase):
Template: RNA
Product: DNA
Function: Synthesizes complementary DNA (cDNA) from an RNA template, commonly found in retroviruses.
Each type of polymerase has a specific role and function within biological processes, depending on the type of nucleic acid template used and the nucleic acid product synthesized.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
In the Hershey and Chase experiment, the differentiation between DNA and protein was achieved by using radioactive isotopes to label each component distinctly:
Labeling DNA with Radioactive Phosphorus (${}^{32}$P):
DNA contains phosphorus, but proteins do not.
They grew bacteriophages in a medium containing radioactive phosphorus (${}^{32}$P), which was incorporated into the DNA of the phages.
Labeling Protein with Radioactive Sulfur (${}^{35}$S):
Proteins contain sulfur (due to amino acids like cysteine and methionine), but DNA does not.
They grew another batch of bacteriophages in a medium containing radioactive sulfur (${}^{35}$S), which was incorporated into the proteins of the phages.
The procedure of their experiment involved:
Allowing the radioactive phages to infect the bacteria (Escherichia coli).
Using a blender to separate the phage "ghosts" (empty protein shells) from the bacteria.
Using a centrifuge to separate the bacterial cells (which would contain the genetic material that entered the bacteria) from the phage ghosts.
Results:
Bacteria infected with ${}^{32}$P-labeled phages were radioactive (indicating that DNA entered the bacteria).
Bacteria infected with ${}^{35}$S-labeled phages were not radioactive (indicating that proteins did not enter the bacteria).
Hence, this experiment provided unequivocal evidence that DNA, not protein, is the genetic material that is transferred from viruses to bacteria.
Here is a simplified representation of their experiment:
This experiment was crucial in confirming that DNA is the hereditary material, upending the previous assumption that proteins were responsible for genetic inheritance.
Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand
(a) Repetitive DNA and Satellite DNA
Repetitive DNA:
Definition: Repetitive DNA comprises sequences that recur multiple times in the genome. These sequences can be short or long and are often located in non-coding regions.
Types: Includes multiple forms such as tandem repeats (like microsatellites and minisatellites) and interspersed repeats (like transposable elements).
Function: Usually do not code for proteins, but are important for genome structure, replication, and evolution.
Satellite DNA:
Definition: A subset of repetitive DNA, Satellite DNA refers specifically to tandemly repeated sequences that show differences in their base composition and length.
Types: Can be divided into microsatellites, minisatellites, and macrosatellites depending on the repeat unit size and number.
Function: Often located in centromeres and telomeres, and play roles in chromosomal stability and segregation during cell division.
(b) mRNA and tRNA
mRNA (Messenger RNA):
Role: Serves as the template for protein synthesis during translation. It carries genetic information transcribed from DNA to ribosomes, where proteins are synthesized.
Structure: Linear and single-stranded, with a 5' cap and a poly-A tail in eukaryotes.
Function: Encodes the amino acid sequence of a protein with codons (three-nucleotide sequences) each specifying an amino acid.
tRNA (Transfer RNA):
Role: Functions as an adapter molecule in translation, transferring specific amino acids to the ribosome to be added to the growing polypeptide chain.
Structure: Cloverleaf secondary structure, with a specific anticodon loop and an amino acid attachment site at the 3' end.
Function: Decodes the mRNA sequence into a protein by matching its anticodon with the complementary codon on the mRNA, facilitating the incorporation of the correct amino acid.
(c) Template Strand and Coding Strand
Template Strand:
Role: The strand of DNA that is used as a template during transcription to synthesize a complementary RNA strand.
Polarity: Runs in the 3' to 5' direction relative to the RNA synthesis process.
Function: Directly involved in the process of transcription; the sequence of the RNA polymerized is complementary to this strand.
Coding Strand:
Role: Also known as the non-template strand, it has the same sequence as the RNA being synthesized (except that RNA has uracil (U) instead of thymine (T)).
Polarity: Runs in the 5' to 3' direction.
Function: Not used in the transcription, but it represents the gene sequence that will be translated into a protein. It is useful for predicting the sequence of the RNA and the encoded protein.
List two essential roles of ribosome during translation.
During translation, ribosomes play essential roles in two main functions:
Platform for Polypeptide Synthesis:
Binding Sites: Ribosomes provide binding sites for mRNA and tRNAs. The small subunit binds to the mRNA, while the large subunit has sites (A-site, P-site, and E-site) for tRNA attachment. This spatial organization ensures that amino acids are properly positioned to form peptide bonds.
Peptide Bond Formation: The ribosome facilitates the formation of peptide bonds between adjacent amino acids, thereby elongating the growing polypeptide chain.
Catalytic Role:
Ribozyme Activity: In prokaryotes, the ribosomal RNA (23S rRNA) within the large subunit acts as a catalyst for peptide bond formation, a function attributed to its enzyme-like activity—this RNA enzyme activity is termed a ribozyme.
These roles ensure accurate translation of genetic information from mRNA into functional proteins, aligning amino acids in the precise sequence dictated by the genetic code.
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?
The lac operon in *E. coli* is a classic example of an inducible operon regulated by both an inducer (lactose) and the end product of its metabolism (glucose). Initially, when lactose is added to the medium, it acts as an inducer, binding to the repressor protein and inactivating it. This allows RNA polymerase to access the promoter region and transcribe the genes required for lactose metabolism ($z, y$, and $a$ genes).
However, the lac operon shuts down after some time due to the following reasons:
Depletion of Lactose: As the enzymes encoded by the lac operon (like β-galactosidase from the $z$ gene) metabolize lactose, the level of lactose in the medium decreases. Once the lactose concentration drops below a certain threshold, it can no longer effectively bind to and inactivate the repressor protein. The repressor re-binds to the operator region, blocking transcription of the lac operon.
Presence of Glucose (Catabolite Repression): In the presence of glucose, *E. coli* prefers to metabolize glucose over lactose due to the efficiency of glucose metabolism. This preference is regulated through a mechanism called catabolite repression. High levels of glucose lead to low levels of cyclic AMP (cAMP). The cAMP binds to catabolite activator protein (CAP), which, in the presence of cAMP, binds to the promoter region of the lac operon and enhances the binding of RNA polymerase. When glucose is abundant, cAMP levels are low, CAP cannot bind to its promoter region, and thus, transcription of the lac operon genes diminishes.
Therefore, the shutdown of the lac operon after the addition of lactose is due to a combination of lactose depletion and catabolite repression by glucose.
Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons
(a) Promoter: The promoter is a DNA sequence that acts as a binding site for RNA polymerase, initiating transcription by determining the template strand and starting the gene's transcription.
(b) tRNA: tRNA (transfer RNA) functions as an adaptor molecule in protein synthesis, matching an mRNA codon with the specific amino acid it encodes during translation.
(c) Exons: Exons are the coding sequences of a gene that are transcribed into mRNA and translated into protein, remaining in the mature mRNA after splicing.
Why is the Human Genome project called a mega project?
The Human Genome Project (HGP) is called a mega project due to its vast scale, immense challenges, and substantial resource requirements. Here are the key reasons:
Scope and Scale: The project aimed to identify all the approximately 20,000-25,000 genes in human DNA and determine the sequences of about 3 billion chemical base pairs that make up the human genome.
Enormous Data: Sequencing the entire human genome involves generating a massive amount of data. To put it into perspective, if you were to print the DNA sequence in books with 1000 letters per page and 1000 pages per book, it would take around 3300 such books to store the information from a single human cell.
High Costs: The estimated cost at the beginning was about USD 3 per base pair, which totals to nearly 9 billion US dollars.
Complexity: The project required advanced technologies and coordination between numerous research centers worldwide. This includes isolating DNA, cloning it, sequencing millions of fragments, and then assembling them into the complete genome.
Technological and Computational Demands: The project spurred rapid advancements in automated DNA sequencing techniques and computational methods for data storage, retrieval, and analysis.
Research Collaboration: The project was an international effort, involving contributions from the U.S. Department of Energy, the National Institute of Health, the Wellcome Trust (UK), and scientists from Japan, France, Germany, China, and other countries.
Future Implications: The project's findings are expected to revolutionize our understanding of human biology and lead to new ways of diagnosing, treating, and preventing diseases.
Thus, the Human Genome Project was termed a mega project due to its unprecedented scale, complexity, cost, and the collaborative international effort required to achieve its ambitious goals.
What is DNA fingerprinting? Mention its application.
DNA fingerprinting is a technique used to identify individuals based on the unique patterns in their DNA. This method involves analyzing specific regions in the DNA sequence known as repetitive DNA. These sequences consist of small stretches of DNA that are repeated many times. Differences in these repetitive regions, which show high degrees of polymorphism, form the basis of DNA fingerprinting. The technique typically involves several steps such as DNA extraction, digestion with restriction enzymes, separation by electrophoresis, and hybridization using sequence-specific probes.
Applications of DNA Fingerprinting:
Forensic Science:
Criminal Identification and Forensic Investigations: DNA fingerprinting can be used to match biological samples found at a crime scene (e.g., blood, hair, or saliva) to potential suspects, thus assisting in criminal investigations and legal proceedings.
Paternity Testing:
Establishing Familial Relationships: This technique can determine biological relationships between individuals, such as confirming paternity or maternity in cases of disputed parentage.
Personal Identification:
Identifying Disaster Victims: DNA fingerprinting is used in disaster victim identification when other methods (like visual identification) are not feasible.
Genetics and Medical Research:
Genetic Diversity Studies: Researchers use DNA fingerprinting to study genetic diversity within populations, the evolutionary relationships among species, and the inheritance patterns of certain diseases.
Agriculture:
Breeding Programs: In agriculture, it helps in the identification of plant varieties and in maintaining the genetic purity of breeding lines.
Historical and Archaeological Research:
Tracing Ancestry and Lineages: This method can also be used to study ancient DNA and trace ancestral lineage in historical and archaeological research.
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
(a) Transcription
Transcription is the process of copying genetic information from one strand of DNA into RNA. In this process, the enzyme RNA polymerase reads the DNA template strand and synthesizes a complementary RNA molecule. This occurs in several steps:
Initiation: RNA polymerase binds to the promoter region of the gene.
Elongation: The RNA polymerase moves along the DNA, synthesizing RNA in the 5' to 3' direction.
Termination: The RNA polymerase reaches a terminator sequence and releases the newly formed RNA transcript.
(b) Polymorphism
Polymorphism refers to the occurrence of two or more different sequences among individuals at a specific locus in their DNA. These variations arise due to mutations and can manifest as single nucleotide polymorphisms (SNPs) or repetitive DNA sequences such as microsatellites and minisatellites. Polymorphisms are the basis for genetic diversity within populations and play a role in forensic science, genetic mapping, and evolutionary studies.
(c) Translation
Translation is the process in which the genetic code carried by messenger RNA (mRNA) is decoded to produce a specific sequence of amino acids in a polypeptide chain, or protein. This occurs in several steps:
Initiation: The small subunit of the ribosome binds to the mRNA at the start codon (AUG).
Elongation: Transfer RNA (tRNA) molecules bring amino acids to the ribosome, and peptide bonds form between amino acids.
Termination: The ribosome reaches a stop codon (UAA, UAG, or UGA), prompting the release of the newly synthesized polypeptide and disassembly of the ribosome.
(d) Bioinformatics
Bioinformatics is an interdisciplinary field that develops and applies computational methods and tools to analyze biological data. It encompasses the storage, retrieval, and analysis of biological data such as DNA sequences, protein structures, and gene expression profiles. Bioinformatics is crucial for managing the massive amounts of data generated by projects like the Human Genome Project and for advancing our understanding of biological processes and diseases.
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Notes - Molecular Basis of Inheritance | Class 12 NCERT | Biology
Molecular Basis of Inheritance Class 12 Notes: In-Depth Study Guide and Breakdown
Understanding the molecular basis of inheritance is crucial for comprehending how genetic information is stored, transmitted, and expressed in living organisms. This guide aims to simplify these concepts for Class 12 students, providing a thorough overview of the key topics involved.
Introduction to Molecular Basis of Inheritance
Overview of Genetic Material
The molecular basis of inheritance refers to the structures and processes that facilitate the storage, transmission, and expression of genetic information. The two primary types of genetic material are DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
Discovery of DNA as Genetic Material
Historical Experiments
Griffith's Experiment: The Transforming Principle
In 1928, Frederick Griffith discovered the transforming principle through his experiments with Streptococcus pneumoniae bacteria. He found that non-virulent bacteria could transform into virulent forms when mixed with heat-killed virulent bacteria.
Avery, MacLeod, and McCarty: Biochemical Proof
Oswald Avery, Colin MacLeod, and Maclyn McCarty demonstrated that DNA was the transforming substance by isolating and purifying it from heat-killed bacteria and showing it could induce transformation.
Hershey-Chase Experiment: DNA is the Genetic Material
In 1952, Alfred Hershey and Martha Chase used bacteriophages to conclusively prove that DNA is the genetic material. They showed that DNA, not protein, entered bacterial cells and directed viral reproduction.
Structure of DNA
The Double Helix Model
James Watson and Francis Crick proposed the double helix structure of DNA in 1953, explaining how genetic information is stored and replicated.
Chemical Composition of DNA
DNA is composed of nucleotides, each consisting of a nitrogenous base, a pentose sugar, and a phosphate group. The nitrogenous bases are classified into purines (adenine and guanine) and pyrimidines (cytosine and thymine).
DNA Packaging in Cells
In eukaryotic cells, DNA is packaged into chromosomes within the nucleus. DNA wraps around positively charged histone proteins, forming nucleosomes, which further coil to create chromatin.
DNA Replication
Mechanism of DNA Replication
DNA replication is a semiconservative process where each strand of the original DNA molecule serves as a template for a new strand. The main enzyme involved is DNA-dependent DNA polymerase.
graph LR
A[Original DNA Double Helix]
B[Separation of Strands]
C[Template Strands]
D[Formation of New Strands]
A --> B --> C --> D
Experimental Proof
The Meselson-Stahl experiment provided conclusive evidence for the semiconservative nature of DNA replication by using isotopic labelling and density gradient centrifugation.
Transcription
The Transcription Process
Transcription is the process of synthesising RNA from a DNA template. The RNA polymerase enzyme catalyses this process. Transcription involves three main steps: initiation, elongation, and termination.
graph TD
A[DNA Template]
B[RNA Polymerase Binding]
C[RNA Synthesis]
D[mRNA]
A --> B --> C --> D
Types of RNA and Their Functions
The three main types of RNA are:
- mRNA (messenger RNA): Carries the genetic information from DNA to the ribosomes for protein synthesis.
- tRNA (transfer RNA): Brings amino acids to the ribosomes during translation.
- rRNA (ribosomal RNA): Forms the core of ribosome's structure and catalyses protein synthesis.
The Genetic Code
Deciphering the Genetic Code
The genetic code consists of triplet codons in mRNA that correspond to specific amino acids. There are 64 possible codons, of which 61 code for amino acids and 3 are stop codons.
Role of tRNA
tRNA molecules have anticodon loops that pair with the mRNA codons and an amino acid attachment site for the corresponding amino acid.
Translation and Protein Synthesis
Mechanism of Translation
Translation is the process of assembling amino acids into a polypeptide chain based on the sequence of codons in mRNA. This occurs at the ribosome and involves initiation, elongation, and termination phases.
graph LR
A[mRNA]
B[tRNA with amino acids]
C[Ribosome]
D[Polypeptide Chain]
A --> C
B --> C
C --> D
Regulation of Gene Expression
Levels of Gene Regulation
Gene expression can be regulated at multiple levels:
- Transcriptional level
- Post-transcriptional level (splicing)
- mRNA transport and translation
The Lac Operon Model
The Lac operon in E. coli is a classic model of gene regulation. It consists of genes responsible for the metabolism of lactose, regulated by the presence or absence of lactose.
Human Genome Project
Introduction and Goals
The Human Genome Project aimed to sequence the entire human genome, identifying all the genes and understanding their functions.
Key Findings and Implications
The project revealed that the human genome consists of approximately 3 billion base pairs and around 20,000-25,000 genes. It provided insights into genetic variations and opened up new avenues for medical research.
DNA Fingerprinting
Principles and Techniques
DNA fingerprinting is used to compare DNA sequences between individuals. It involves identifying variations in repetitive DNA sequences, which are highly polymorphic.
Applications
DNA fingerprinting is widely used in forensic science, paternity testing, and studying genetic diversity within populations.
Conclusion
Understanding the molecular basis of inheritance is fundamental to the study of genetics and biology. This guide provides a comprehensive overview of the key concepts, experiments, and applications in the field, helping students grasp the intricacies of genetic material and its functions.
With this article, Class 12 students can gain a detailed understanding of the molecular basis of inheritance, ensuring they are well-prepared for their exams and future studies in biology.
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