1. Describe the structure of chromosome.
Answer: Chromosomes are thread-like structures located inside the nucleus of animal and plant cells. They are composed of DNA tightly coiled many times around proteins called histones.
Structure of a Chromosome:
Chromatid: Each chromosome consists of two identical sister chromatids, which are joined together at a region called the centromere. Each chromatid contains a single DNA molecule.
Centromere: The centromere is the constricted region where the two chromatids are held together. It is the attachment point for spindle fibers during cell division. The position of the centromere determines the shape of the chromosome (metacentric, submetacentric, acrocentric, telocentric). Telomeres: The ends of chromosomes are called telomeres. They are repetitive DNA sequences that protect the chromosome from degradation and fusion with other chromosomes. Chromatin: The material of chromosomes is called chromatin, which consists of DNA and proteins. The fundamental unit of chromatin is the nucleosome, which consists of 147 base pairs of DNA wrapped around a core of eight histone proteins. Euchromatin and Heterochromatin: Chromatin can be divided into euchromatin (loosely packed, transcriptionally active) and heterochromatin (tightly packed, transcriptionally inactive). Kinetochore: A protein structure on the centromere where spindle fibers attach during cell division.
Levels of Organization:
1. DNA double helix
2. Nucleosome (DNA + histones)
3. Chromatin fiber (nucleosomes coiled)
4. Looped domains
5. Chromosome (condensed)
In summary, a chromosome is a highly organized structure of DNA and proteins that carries genetic information. Its structure allows for the compact storage of DNA and its accurate segregation during cell division.
2. Describe in detail the composition and organization of chromosome.
Answer:
Chromosomes are composed of DNA, RNA, and proteins. The organization of chromosomes involves multiple levels of compaction to fit the long DNA molecule into the small nucleus.
Composition:
DNA: The genetic material that carries the hereditary information. Histones: Basic proteins (positively charged) that bind to the negatively charged DNA. The five main types are H1, H2A, H2B, H3, and H4. Non-histone proteins: Involved in DNA replication, transcription, and repair. RNA: A small amount of RNA is associated with chromatin.
Organization:
1. Nucleosome: The basic unit of chromatin. It consists of 147 base pairs of DNA wrapped around an octamer of histones (two each of H2A, H2B, H3, and H4). H1 binds to the linker DNA between nucleosomes.
2. Chromatin Fiber: Nucleosomes are coiled into a 30 nm fiber (solenoid).
3. Looped Domains: The 30 nm fiber is organized into loops attached to a protein scaffold.
4. Condensed Chromosome: During cell division, the loops are further coiled and condensed to form the visible chromosome.
Types of Chromatin:
Euchromatin: Loosely packed, transcriptionally active. Heterochromatin: Tightly packed, transcriptionally inactive.
In summary, the organization of chromosomes involves a hierarchical folding of DNA with histones to achieve the high degree of compaction required for the genome to fit inside the nucleus.
3. Describe the experiment of Griffith to prove that DNA is the hereditary material.
Answer: Frederick Griffith's experiment in 1928 was a pivotal step in identifying DNA as the genetic material. He worked with Streptococcus pneumoniae, which has two strains:
Smooth (S) strain: Virulent, has a polysaccharide capsule, causes pneumonia. Rough (R) strain: Non-virulent, no capsule, does not cause pneumonia.
Experiment:
1. Injection of live S strain: Mice died.
2. Injection of live R strain: Mice survived.
3. Injection of heat-killed S strain: Mice survived.
4. Injection of heat-killed S strain + live R strain: Mice died. Live S strain was recovered from the dead mice.
Conclusion: Griffith concluded that a "transforming principle" from the heat-killed S strain had transformed the live R strain into the virulent S strain. He did not know what the substance was, but he hypothesized it was a chemical component of the cell.
Significance:
Provided the first evidence for transformation. Laid the foundation for the discovery that DNA is the genetic material (later confirmed by Avery, MacLeod, and McCarty). Demonstrated that genetic information can be transferred between bacteria.
4. Describe the experiment of Hershey - Chase to prove that DNA is the hereditary material.
Answer: The Hershey-Chase experiment (1952) used bacteriophages (viruses that infect bacteria) to definitively prove that DNA, not protein, is the genetic material.
Experiment:
Radioactive Labeling: They used two batches of bacteriophages. One batch was labeled with radioactive sulfur (³⁵S), which labels proteins. The other batch was labeled with radioactive phosphorus (³²P), which labels DNA. Infection: The labeled phages were allowed to infect E. coli bacteria. Blending and Centrifugation: After infection, the mixtures were blended to separate the phage coats from the bacteria. The bacteria were then pelleted by centrifugation. Analysis: They measured the radioactivity in the bacterial pellet and the supernatant.
Results:
In the ³⁵S (protein) experiment, radioactivity was found in the supernatant (phage coats), not in the bacteria. In the ³²P (DNA) experiment, radioactivity was found inside the bacteria (pellet).
Conclusion: Only DNA entered the bacterial cells during infection, while proteins remained outside. This proved that DNA is the genetic material that carries hereditary information. The experiment provided definitive evidence for DNA as the molecule of heredity.
5. Describe the three models proposed about the mechanism of DNA replication.
Answer: Three models were proposed to explain the mechanism of DNA replication:
1. Conservative Model:
The original double helix remains completely intact. An entirely new double helix is synthesized. One daughter molecule contains both old strands; the other contains both new strands. This model was disproved by the Meselson-Stahl experiment.
2. Semiconservative Model:
The double helix separates into two single strands. Each strand serves as a template for the synthesis of a new complementary strand. Each daughter molecule contains one old strand and one new strand. This is the correct model, as proven by Meselson and Stahl.
3. Dispersive Model:
The original DNA is broken into fragments. New DNA segments are synthesized and interspersed with the old segments. Each daughter molecule is a mixture of old and new DNA. This model was also disproved by the Meselson-Stahl experiment.
Significance:
Understanding the mechanism of DNA replication is crucial for understanding heredity.
The semiconservative model ensures that genetic information is accurately passed to daughter cells. The Meselson-Stahl experiment (1958) used isotopic labeling to distinguish between these models and confirmed the semiconservative model.
6. Describe the work of Meselson and Stahl to justify the semi-conservative replication as the correct method of replication.
Answer: The Meselson-Stahl experiment (1958) provided definitive evidence for the semiconservative model of DNA replication.
Experiment:
Culturing in Heavy Nitrogen: E. coli bacteria were grown in a medium containing heavy nitrogen (¹⁵N) for many generations. This incorporated ¹⁵N into the DNA, making it denser. Transfer to Light Nitrogen: The bacteria were then transferred to a medium containing light nitrogen (¹⁴N) and allowed to replicate. Sampling: Samples were taken after each generation (0, 20, 40 minutes) and analyzed using cesium chloride density gradient centrifugation.
Results:
Generation 0: All DNA was heavy (¹⁵N/¹⁵N). Generation 1: All DNA was of intermediate density (¹⁵N/¹⁴N), indicating a hybrid molecule. Generation 2: DNA was half intermediate and half light (¹⁴N/¹⁴N).
Interpretation:
The results matched the predictions of the semiconservative model. If replication were conservative, there would be two distinct bands (heavy and light) after generation 1. If replication were dispersive, there would be a single band of intermediate density that gradually shifts. The observation of a single intermediate band after one generation, and two bands after two generations, confirmed that each daughter DNA molecule consists of one old and one new strand.
In summary, the Meselson-Stahl experiment conclusively demonstrated that DNA replication is semiconservative, a fundamental principle of molecular biology.
7. Describe the events of the process of DNA replication and explain DNA stability and variability as two characters of the replicating DNA molecule.
Answer: Events of DNA Replication:
1. Initiation: Replication begins at specific sites called origins of replication. Helicase unwinds the double helix, creating a replication fork. Single-strand binding proteins stabilize the separated strands. Topoisomerase/ gyrase relieves supercoiling.
2. Primer Synthesis: Primase synthesizes a short RNA primer complementary to the DNA template.
3. Elongation: DNA polymerase III adds DNA nucleotides to the 3' end of the primer.
Leading Strand: Synthesized continuously in the 5'[source symbol unavailable]3' direction toward the replication fork. Lagging Strand: Synthesized discontinuously in Okazaki fragments, away from the fork.
4. Primer Removal and Gap Filling: DNA polymerase I removes RNA primers and fills the gaps with DNA.
5. Ligation: DNA ligase seals the nicks between Okazaki fragments, creating a continuous strand.
6. Termination: Replication ends when the forks meet.
DNA Stability and Variability:
Stability: DNA is a stable molecule due to the hydrogen bonds between base pairs and the base stacking interactions. The double helix structure protects the genetic information. Proofreading by DNA polymerase ensures high fidelity of replication. This stability is essential for maintaining genetic information across generations. Variability: Despite the stability, DNA can undergo mutations (changes in the nucleotide sequence). These mutations are the source of genetic variability. Variability is essential for evolution and adaptation. Sexual reproduction (meiosis and fertilization) also generates genetic variability through crossing over and independent assortment.
In summary, DNA replication is a highly coordinated process that ensures accurate duplication of the genetic material. The balance between stability and variability is crucial for the survival and evolution of species.
8. Describe the mechanism of polymerase III activity.
Answer: DNA Polymerase III is the primary enzyme responsible for DNA replication in prokaryotes. It is a large, multi-subunit enzyme complex (holoenzyme).
Mechanism:
1. Binding: DNA polymerase III binds to the DNA template at the primer-template junction.
2. Nucleotide Addition: It adds deoxyribonucleotides (dNTPs) to the 3'-OH end of the growing strand. The incoming nucleotide is selected based on complementary base pairing with the template strand.
3. Catalysis: The enzyme catalyzes the formation of a phosphodiester bond between the 3'-OH of the previous nucleotide and the 5'-phosphate of the incoming nucleotide. This releases pyrophosphate.
4. Proofreading: DNA polymerase III has 3'[source symbol unavailable]5' exonuclease activity, which allows it to remove mismatched nucleotides. This proofreading function ensures high fidelity of replication (error rate ~1 in 10⁹).
5. Processivity: DNA polymerase III is highly processive, meaning it can add many nucleotides without dissociating from the template. This is due to the sliding clamp (β-clamp) that holds the enzyme onto the DNA.
6. Strand Synthesis:
Leading Strand: Synthesized continuously by the same polymerase complex. Lagging Strand: Synthesized discontinuously. The lagging strand template loops through the polymerase complex, allowing synthesis of Okazaki fragments.
In summary, DNA polymerase III is a highly efficient and accurate enzyme that synthesizes new DNA strands during replication. Its proofreading and processivity are essential for maintaining genomic integrity.
9. Explain the mechanism of transcription.
Answer: Transcription is the process of synthesizing RNA from a DNA template. It occurs in three stages: initiation, elongation, and termination.
1. Initiation:
Prokaryotes: RNA polymerase binds to the promoter region (-10 and -35 sequences) with the help of sigma factor.
Eukaryotes: Transcription factors and RNA polymerase II bind to the promoter (TATA box). The DNA double helix unwinds, exposing the template strand.
2. Elongation:
RNA polymerase moves along the template strand in the 3'[source symbol unavailable]5' direction. It adds RNA nucleotides (A, U, G, C) to the 3' end of the growing RNA strand, synthesizing in the 5'[source symbol unavailable]3' direction. The DNA rewinds behind the polymerase.
3. Termination:
Prokaryotes: Termination is signaled by specific sequences (e.g., hairpin structures) that cause the polymerase to detach. Eukaryotes: Termination involves specific sequences (e.g., polyadenylation signal) and the polymerase detaches.
Post-transcriptional Modifications (Eukaryotes):
5' Capping: Addition of a 7-methylguanosine cap. 3' Poly-A Tail: Addition of a poly-A tail. Splicing: Removal of introns and joining of exons.
In summary, transcription is the synthesis of RNA from DNA, involving initiation, elongation, and termination. In eukaryotes, the primary transcript undergoes processing to form mature mRNA.
10. Describe the mechanism of protein synthesis.
Answer: Protein synthesis (translation) is the process of synthesizing a polypeptide chain from an mRNA template. It occurs in the cytoplasm on ribosomes.
1. Activation of Amino Acids:
Each amino acid is attached to its specific tRNA by aminoacyl-tRNA synthetase, using ATP. This forms aminoacyl-tRNA.
2. Initiation:
The small ribosomal subunit binds to the mRNA at the 5' cap (eukaryotes) or Shine-Dalgarno sequence (prokaryotes). The initiator tRNA (carrying methionine) binds to the start codon (AUG). The large ribosomal subunit joins, forming the initiation complex.
3. Elongation:
Codon Recognition: The anticodon of the incoming aminoacyl-tRNA pairs with the complementary codon on the mRNA in the A site. Peptide Bond Formation: The amino acid in the P site is transferred to the amino acid in the A site, forming a peptide bond (catalyzed by rRNA/ribozyme). Translocation: The ribosome moves one codon along the mRNA. The tRNA in the P site moves to the E site and exits. The tRNA with the growing polypeptide moves to the P site.
4. Termination:
When a stop codon (UAA, UAG, UGA) enters the A site, release factors bind. The polypeptide is released from the tRNA.
The ribosome dissociates into its subunits.
5. Post-translational Modifications:
The polypeptide may undergo folding, cleavage, or addition of chemical groups to become functional.
In summary, protein synthesis involves the coordinated action of mRNA, tRNA, ribosomes, and various enzymes to translate the genetic code into a functional protein.
11. Write the difference between protein synthesis in prokaryotes and eukaryotes.
Answer:
| Feature | Prokaryotes | Eukaryotes |
| Location | Cytoplasm | Cytoplasm (ribosomes), nucleus (transcription) |
| Ribosomes | 70S | 80S |
| mRNA | Polycistronic (multiple genes) | Monocistronic (single gene) |
| Transcription and Translation | Coupled (occur
simultaneously) | Separated (transcription in nucleus, translation in
cytoplasm) |
| | |
| Initiation | Shine-Dalgarno sequence | 5' cap and Kozak sequence |
| Initiator tRNA | Formylmethionine (fMet) | Methionine (Met) |
| Splicing | Absent | Present (removal of introns) |
| Post-transcriptional
Modifications | Minimal | Extensive (capping, tailing, splicing) |
| Antibiotics | Sensitive to many antibiotics | Insensitive to antibiotics that target 70S ribosomes |
12. What is the importance of the regulation of gene expression?
Answer: Regulation of gene expression is crucial for the proper functioning of cells and organisms.
Importance:
1. Cell Differentiation: Allows cells to develop into different types (e.g., nerve cells, muscle cells) by expressing specific sets of genes.
2. Development: Controls the timing and location of gene expression during embryonic development.
3. Homeostasis: Maintains internal balance by adjusting gene expression in response to environmental changes.
4. Metabolic Efficiency: Ensures that enzymes are produced only when needed, conserving energy and resources.
5. Response to Stimuli: Allows cells to respond to hormones, growth factors, and stress by altering gene expression.
6. Adaptation: Enables organisms to adapt to changing environments.
7. Prevention of Disease: Dysregulation of gene expression can lead to cancer and other diseases.
In summary, regulation of gene expression is essential for the growth, development, and survival of all organisms. It allows for the efficient use of genetic information and the ability to respond to internal and external signals.
13. Describe the negative control of gene expression by repressor proteins and the positive control of gene expression by activator proteins.
Answer: Gene expression is regulated by proteins that bind to DNA and control transcription.
Negative Control (Repressors):
Repressor proteins bind to the operator region of DNA, blocking RNA polymerase from binding to the promoter. This prevents transcription. Example: The lac operon in E. coli. In the absence of lactose, the repressor binds to the operator, blocking transcription. In the presence of lactose, the inducer (allolactose) binds to the repressor, causing it to detach, allowing transcription.
Positive Control (Activators):
Activator proteins bind to DNA sequences (enhancers or activator binding sites) and stimulate transcription. They help RNA polymerase bind to the promoter. Example: The catabolite activator protein (CAP) in E. coli. When glucose is low, cAMP levels are high. cAMP binds to CAP, which then binds to the promoter of the lac operon, enhancing transcription.
In summary, negative control involves repressors blocking transcription, while positive control involves activators promoting transcription. Both mechanisms are essential for precise regulation of gene expression.
14. Explain post-transcription modification of mRNA.
Answer: Post-transcriptional modifications are changes made to the primary transcript (pre-mRNA) before it becomes mature mRNA in eukaryotes.
Modifications:
1. 5' Capping: The addition of a 7-methylguanosine (m⁷G) cap to the 5' end via a 5'-5' triphosphate linkage. This protects the mRNA from degradation, assists in export from the nucleus, and helps ribosomes recognize the mRNA for translation.
2. 3' Poly-A Tailing: The addition of a poly-A tail (200-300 adenine nucleotides) to the 3' end. This protects the mRNA from degradation and aids in export and translation.
3. Splicing: The removal of introns (non-coding sequences) and joining of exons (coding sequences). This is catalyzed by the spliceosome.
4. RNA Editing: Alteration of nucleotide sequence after transcription (e.g., deamination of cytosine to uracil).
Significance:
Increases stability of mRNA. Enables export to cytoplasm. Ensures translation of correct protein. Allows for alternative splicing, producing multiple proteins from one gene.
15. Define mutation and identify various sources of mutations. (a) Differentiate between natural and induced mutations.
(b) Explain the harmful and useful aspects of mutations. (c) Rationalize that mutations might be a contributing factor towards evolution.
Answer: Mutation: A mutation is a permanent change in the DNA sequence of an organism.
Sources of Mutations:
Spontaneous: Errors in DNA replication, repair, or recombination. Induced: Exposure to mutagens (physical agents like UV and X-rays; chemical agents like base analogs, alkylating agents, and intercalating dyes; biological agents like transposons).
(a) Natural vs. Induced Mutations:
Natural (Spontaneous): Occur without external influence, often due to replication errors. Rate is low (10⁻⁹ to 10⁻¹⁰ per base pair per generation). Induced: Caused by exposure to mutagens. Rate is higher and can be controlled.
(b) Harmful and Useful Aspects:
Harmful: Most mutations are harmful. They can cause genetic disorders (e.g., sickle cell anemia, cystic
fibrosis), cancer, or death. Useful: Some mutations are beneficial. They can provide resistance to diseases (e.g., sickle cell trait confers resistance to malaria), new metabolic capabilities, or improved traits in crops (e.g., drought resistance).
(c) Mutations and Evolution:
Mutations are the ultimate source of all genetic variation. They provide the raw material for natural selection and evolution. Beneficial mutations increase the organism's fitness, allowing it to survive and reproduce. Over time, the accumulation of beneficial mutations can lead to the formation of new species.
In summary, mutations are changes in DNA that can be spontaneous or induced. They can be harmful, neutral, or beneficial. They are essential for genetic diversity and evolution.
16. Describe the symptoms, causes and possible treatment of the following chromosomal mutations: (a) Down syndrome
(b) Klinefelter's syndrome (c) Turner's syndrome
Answer:
(a) Down Syndrome (Trisomy 21):
Cause: Trisomy of chromosome 21 (47, XX+21 or 47, XY+21). Usually due to nondisjunction during meiosis. Symptoms: Intellectual disability, characteristic facial features (flat face, upward-slanting eyes), short stature, heart defects, increased risk of Alzheimer's and leukemia. Treatment: No cure. Management of symptoms (early intervention, speech therapy, physical therapy, special education). Treatment of associated medical conditions (heart defects).
(b) Klinefelter's Syndrome (47, XXY):
Cause: Extra X chromosome in males (nondisjunction). Symptoms: Tall stature, small testes, infertility, reduced body hair, gynecomastia (breast development), learning disabilities. Treatment: Testosterone replacement therapy to treat symptoms and improve secondary sexual characteristics. Fertility treatment (in some cases).
(c) Turner's Syndrome (45, XO):
Cause: Monosomy X (one X chromosome missing in females). Symptoms: Short stature, webbed neck, lack of ovarian development (infertility), heart defects, kidney abnormalities. Treatment: Growth hormone therapy to increase height. Estrogen replacement therapy to induce puberty and maintain secondary sexual characteristics. Fertility treatment (in some cases).
17. Describe the symptoms, causes and possible treatment of the following genetic mutations: (a) Sickle cell anaemia (b) Phenylketonuria
Answer:
(a) Sickle Cell Anaemia:
Cause: Point mutation in the beta-globin gene (GAG [source symbol unavailable] GTG), causing glutamic acid to be replaced by valine. This results in abnormal hemoglobin (HbS). Symptoms: Painful episodes (crises), anemia, fatigue, shortness of breath, delayed growth, frequent infections, organ damage. Treatment:
Pain management (analgesics). Hydroxyurea (increases fetal hemoglobin). Blood transfusions. Bone marrow transplant (curative). Gene therapy (emerging).
(b) Phenylketonuria (PKU):
Cause: Mutation in the PAH gene, which codes for the enzyme phenylalanine hydroxylase. Deficiency of this enzyme leads to accumulation of phenylalanine. Symptoms: Intellectual disability, seizures, behavioral problems, musty odor in urine/breath, light skin and hair. Treatment:
Strict low-phenylalanine diet (avoiding high-protein foods and aspartame). Dietary supplements (tyrosine). Early diagnosis (newborn screening) is crucial for preventing intellectual disability.
