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Unit 19: Chromosome and DNA — Long Questions

12th Class Biology · Unit 19: Chromosome and DNA

LONG15.Annotate the detailed structure of a chromosome.

A chromosome is a thin, thread-like structure made of DNA and histone proteins (about 40% DNA and 60% histone protein by mass). Key structural features:
• Chromatids: After S-phase duplication, each chromosome has two identical sister chromatids joined at a constricted region, the centromere (a specific DNA sequence of about 220 nucleotides).
• Kinetochore: A disc-shaped protein structure attached to the centromere; it is the point of attachment to the mitotic spindle during cell division.
• Secondary constriction/NOR: Some chromosomes have an additional constriction, the Nuclear Organizer Region, which forms nucleoli during interphase.
• Satellite: The knob-like end beside the secondary constriction, containing highly repetitive (junk) DNA.
• Telomeres: The terminal ends of the chromosome, which prevent chromosome ends from fusing with each other.
• Centromere position types: metacentric (centre), sub-metacentric (slightly off-centre), acrocentric (near the end), telocentric (at the end).
Organisation (levels of packing): DNA (2 nm) wraps twice around a histone octamer (H2A, H2B, H3, H4 — two copies each) every ~146 nucleotides to form a nucleosome; nucleosomes are joined by ~80-nucleotide linker DNA with linker histone H1, giving the 11 nm 'beads-on-a-string' nucleosome string; this coils into the 30 nm chromatin fibre; the fibre supercoils into a 300 nm structure, which further condenses into the 700 nm chromatid.

LONG16.Explain the concept of alleles as the alternative forms of a gene.

A gene is a short segment of DNA that encodes the sequence of amino acids of a particular polypeptide; its fixed position on a chromosome is its locus. In diploid organisms, chromosomes occur in homologous pairs, and each homologous chromosome carries a copy of the same gene at the same locus. However, the exact nucleotide sequence of a gene may differ slightly between the two homologous chromosomes — these alternative forms of the same gene, occupying the same locus, are called alleles. For example, if one chromosome carries an allele for a particular eye colour at a given locus, its homologous partner will carry an allele of the same gene (possibly a different version) at the same locus. One allele is inherited from each parent, and the combination of alleles an individual carries determines its traits.

LONG17.Narrate the experimental work of Griffith that proved that DNA is the hereditary material.

In 1928, Frederick Griffith worked with two strains of Streptococcus pneumoniae: the S-type (with a polysaccharide coat, virulent, forms smooth colonies) and the R-type (lacking the coat, non-virulent, forms rough colonies).
• Injecting live S-type into mice caused pneumonia and death.
• Injecting live R-type caused no disease.
• Injecting heat-killed S-type caused no disease.
• Injecting a mixture of heat-killed S-type plus live R-type unexpectedly killed the mice, and live S-type bacteria were recovered from their blood.
Griffith concluded that genetic information for making the polysaccharide coat (and virulence) had passed from the dead S-type bacteria to the living R-type, transforming it into virulent S-type — a process he called transformation (transfer of genetic material between organisms). Griffith's experiment proved that a hereditary 'transforming principle' exists, though he could not identify whether it was DNA or protein. (This was later confirmed as DNA by Avery, MacLeod and McCarty in 1944, who showed that removing DNA — but not protein — from the dead S-type abolished transformation.)

LONG18.Narrate the experimental work of Hershey and Chase that proved that DNA is the hereditary material.

In 1952, Alfred Hershey and Martha Chase worked with bacteriophages (which consist of a DNA core and a protein coat) to determine which component directs the production of new phages inside bacteria. They labelled phage protein with radioactive sulphur (35S, since sulphur occurs in amino acids but not DNA) and phage DNA with radioactive phosphorus (32P, since phosphorus occurs in DNA but not protein).
• Setup 1: Phages with 35S-labelled protein infected bacteria. After centrifugation, radioactivity was found in the surrounding medium, not in the bacterial cells — showing the protein coat stayed outside.
• Setup 2: Phages with 32P-labelled DNA infected bacteria. After centrifugation, radioactivity was found inside the bacterial cells, not in the medium — showing the DNA entered the cells.
Since only the DNA entered the bacteria and directed the synthesis of new phages, Hershey and Chase concluded that DNA, not protein, is the genetic material of the bacteriophage.

LONG19.Describe the three models proposed about the mechanism of DNA replication.

Three models were proposed for the mechanism of DNA replication:
1. Conservative model: The parental double helix remains completely intact, and an entirely new double helix of new material is produced alongside it. After replication, one daughter DNA has both original (parental) strands, and the other has both newly synthesised strands.
2. Semi-conservative model: The two strands of the parental double helix separate, and each strand acts as a template for a new complementary strand. After replication, both daughter DNA molecules consist of one parental strand and one new strand (hybrid molecules).
3. Dispersive model: The parental double helix is broken and dispersed into individual nucleotides, which then combine with new nucleotides to build four new strands. After replication, both daughter DNAs contain strands that are mixtures of old and new nucleotides.
The Meselson-Stahl experiment later confirmed that DNA replicates by the semi-conservative model.

LONG20.Narrate the work of Meselson and Stahl to justify the semi-conservative replication.

In 1958, Meselson and Stahl grew bacteria in a medium containing heavy nitrogen (15N), so the DNA of these bacteria became denser than normal (14N) DNA. They then transferred the bacteria to a medium containing only normal 14N and collected DNA samples at intervals, separating DNA of different densities by ultracentrifugation in a caesium chloride density gradient (denser DNA settles lower).
Observations:
• At 0 minutes (before replication): DNA formed a single heavy sediment (both strands 15N).
• At 20 minutes (after one replication): DNA formed a single sediment of intermediate density — each daughter DNA was a hybrid of one 15N strand and one 14N strand.
• At 40 minutes (after two replications): DNA formed two sediments — one intermediate and one light (fully 14N).
This pattern — hybrid DNA after the first round, and both hybrid and fully light DNA after the second round — is exactly what the semi-conservative model predicts (each new DNA retains one parental strand and gains one new strand), ruling out the conservative model (which would have produced a fully heavy band and a fully light band after one round) and the dispersive model (which would never produce a fully light band). Meselson and Stahl therefore concluded that DNA replicates semi-conservatively.

LONG21.Describe the events of the process of DNA replication.

DNA replication occurs in the following main steps:
1. Opening up the DNA double helix: An initiator protein binds at the replication origin site. Helicase breaks the hydrogen bonds between base pairs, separating the strands and forming Y-shaped replication forks (together forming a replication bubble). Gyrase (topoisomerase) relieves the supercoiling produced ahead of the fork, and single-strand binding proteins keep the separated strands from re-joining. The two forks move apart so replication proceeds in both directions.
2. Building a primer: The enzyme primase lays down a short (5-10 nucleotide) RNA primer on each template strand at the origin, providing a free 3' end for DNA synthesis to begin.
3. Assembling complementary strands (extension): DNA polymerase III adds complementary nucleotides (from dATP, dCTP, dGTP, dTTP) to the 3' end of the growing strand in the 5'→3' direction, matching A with T and C with G. On the strand running toward the fork, synthesis is continuous, forming the leading strand. On the other strand, synthesis must occur away from the fork in short pieces called Okazaki fragments, each needing its own primer — this forms the lagging strand.
4. Termination: At the termination site, DNA polymerase I removes the RNA primers and fills the gaps with DNA nucleotides, and DNA ligase joins the Okazaki fragments (and seals remaining nicks) to produce two continuous, complete daughter DNA molecules.

LONG22.Explain the mechanism of transcription.

Transcription (synthesis of mRNA from a DNA template) occurs in three phases:
1. Initiation: RNA polymerase binds to a promoter sequence on the DNA template strand (e.g., the –35 and –10 sequences in prokaryotes; the TATA box and CAAT box in eukaryotes). The σ subunit of RNA polymerase recognises the promoter and initiates binding, after which it is removed; the remaining core enzyme unwinds the DNA and forms a transcription bubble.
2. Elongation: Inside the transcription bubble, RNA polymerase adds complementary ribonucleotides (from rNTPs) to the growing mRNA chain in the 5'→3' direction, using the template strand. The RNA-DNA hybrid rotates as nucleotides are added, and the transcription bubble moves along the DNA, with the DNA re-winding behind it.
3. Termination: At the end of the gene, a stop signal (GC-rich region followed by AT-rich region) is transcribed into mRNA, forming a GC hairpin loop with a tail of U ribonucleotides. This hairpin causes RNA polymerase to pause; the weak A-U pairing cannot hold the hybrid together, so the RNA strand dissociates from the DNA, ending transcription.

LONG23.Describe the mechanism of protein synthesis.

Protein synthesis (translation) uses mRNA information to assemble a polypeptide and occurs after activation of amino acids and in three phases:
Activation of amino acids: Aminoacyl-tRNA synthetase enzymes (one of 20 kinds, each specific for one amino acid) attach each amino acid to its specific tRNA, forming aminoacyl-tRNA (aa-tRNA).
1. Initiation: The first aa-tRNA (carrying methionine) binds to the smaller ribosomal subunit with the help of an initiation factor. Another initiation factor attaches the 5' end of mRNA to the small subunit, positioning the anticodon of the aa-tRNA over the start codon (AUG). The larger ribosomal subunit then joins, forming the initiation complex with P (peptidyl), A (aminoacyl) and E (exit) sites.
2. Elongation: A new aa-tRNA with an anticodon matching the next mRNA codon binds at the A site (aided by elongation factor). Peptidyl transferase forms a peptide bond between the amino acid at the P site and the one at the A site. The ribosome then moves three nucleotides along the mRNA (translocation): the growing peptide chain shifts to the P site, the empty tRNA moves to the E site and leaves, and a new codon is exposed at the A site. This cycle repeats for each amino acid.
3. Termination: When a stop codon appears at the A site, no aa-tRNA binds (no matching anticodon exists); instead, a release factor recognises the stop codon, releases the completed polypeptide from the tRNA at the P site, and the ribosomal subunits separate from the mRNA.

LONG24.Describe the difference between protein synthesis in prokaryotes and eukaryotes.

Protein synthesis differs between prokaryotes and eukaryotes in several ways:
1. Introns: Prokaryotic genes lack introns, while most eukaryotic genes possess introns.
2. Post-transcriptional processing: In prokaryotes, mRNA is translated immediately after transcription with no modification. In eukaryotes, the mRNA is modified before translation — introns are spliced out, and a 5' cap and 3' poly-A tail are added.
3. Location: In prokaryotes, translation occurs in the cytoplasm. In eukaryotes, translation occurs in the cytoplasm as well as on the rough endoplasmic reticulum.
4. Ribosomes: Prokaryotic ribosomes are smaller (70S), while eukaryotic ribosomes are larger (80S).
5. Initial amino acid: In prokaryotes, the initial amino acid is modified (N-formyl-methionine); in eukaryotes, it is unmodified methionine.

LONG25.Describe the importance of the regulation of gene expression.

Regulation of gene expression is important because not all genes in an organism are expressed simultaneously — genes are switched ON or OFF, and their expression timed, according to need. This allows cells to respond appropriately to changing environmental conditions and to produce the required proteins only when needed, which is energy-efficient. In multicellular organisms, gene regulation is the basis of cellular differentiation and morphogenesis: although every cell type contains the same genome, different genes are switched on in each cell type, producing cells with different structures and functions (e.g., liver cells express alcohol dehydrogenase genes, while neurons keep these genes turned off even though they carry the same genes).

LONG26.Describe the negative and positive control of gene expression.

Gene expression is regulated mainly at the transcriptional level by regulatory proteins binding to regulatory DNA sequences (operators) near the promoter:
• Negative control: Regulatory proteins called repressors bind to the operator and block RNA polymerase from binding to or moving along the promoter, thereby stopping or reducing transcription.
• Positive control: Regulatory proteins called activators (or inducers) bind to the operator and help RNA polymerase bind to the promoter, thereby starting transcription or increasing its rate.
Genes can also be regulated during mRNA splicing in eukaryotes: specific regulatory proteins in different cell types remove introns/exons in different ways, producing mRNAs of different lengths and hence more than one polypeptide from a single gene (alternative splicing).

LONG27.Explain the post-transcriptional modification of mRNA.

In eukaryotes, newly transcribed (immature) mRNA must travel from the nucleus to the cytoplasm, so it undergoes three modifications before translation:
1. Addition of 5' cap: A methylated GTP is attached to the 5' end (first nucleotide, A or G) of the mRNA, forming the 5' cap, which protects the mRNA from nucleases and phosphatases during its journey.
2. Addition of 3' poly-A tail: A chain of about 250 adenine nucleotides is added to the 3' end, protecting the mRNA transcript from degradation by nucleases.
3. Splicing: The immature mRNA contains non-coding introns and coding exons. During splicing, the introns are removed and the remaining exons are joined together, so the mature mRNA consists only of coding sequence (gene) ready for translation.

LONG28.Write a note on point mutations and their sources.

Point mutations are mutations involving a change in only one or a few nucleotides within a gene. They are of three main types:
1. Substitutions — exchange of one nucleotide for another (e.g., A substituted by G).
2. Insertions — addition of one or more extra nucleotides, usually due to replication errors.
3. Deletions — removal of one or more nucleotides.
Sources of point mutations:
• Physical damage to DNA: Ionizing radiation (X-rays, gamma rays) forms free radicals that can break phosphodiester bonds; ultraviolet radiation causes a double covalent bond to form between adjacent pyrimidine nucleotides, which can block replication.
• Mutagenic chemicals: Some chemicals resemble nucleotides and get incorporated into DNA, preventing proper transcription; others remove the amino group from adenine or cytosine, or add hydrocarbon groups to nucleotides, causing mis-pairing (e.g., mustard gas).
• Spontaneous point mutations: Nucleotides occasionally shift to alternative positions, causing DNA polymerase to pair a different nucleotide during replication; misalignment of homologous chromosomes during pairing can also cause looping-out and deletion of a stretch of nucleotides.

LONG29.Describe the symptoms, causes and treatments Down syndrome and Klinefelter syndrome.

Down syndrome • Cause: Non-disjunction of autosome pair 21 during egg formation, producing an egg with 24 chromosomes; fertilisation by a normal sperm gives a child with 47 chromosomes (trisomy 21, 2n+1).
• Symptoms: Mental impairment, stunted growth, increased skin on back and neck, flat facial features, small head, short neck and limbs, upward-slanting eyes, small ears, poor muscle tone, protruding tongue, and heart defects. Affected males are usually sterile; affected females have reduced fertility.
• Treatment: There is no cure, but management includes screening for common associated problems, medical treatment where indicated, a good family environment, and special education and care to improve quality of life.

Klinefelter syndrome • Cause: Non-disjunction of the X chromosome during gamete formation in either parent, resulting in an XXY zygote (47 chromosomes, 2n+1).
• Symptoms: Affected individuals are sterile males with feminised characteristics — less body hair, breast growth, small testicles, abnormal body proportions (long legs, short trunk), and a voice that may not be as deep.
• Treatment: Testosterone therapy (to compensate for testosterone deficiency at puberty), surgical removal of breast tissue, and behavioural therapy for language disorders, school difficulties, and socialisation problems.

LONG30.Describe the symptoms, causes and treatments Turner syndrome.

Turner syndrome • Cause: Non-disjunction of sex chromosomes produces an egg or sperm lacking a sex chromosome (O). Fusion of an O egg with a normal X sperm (or an O sperm with a normal X egg) produces a 45-chromosome (XO, 2n-1) zygote — monosomy of the sex chromosome, resulting in a sterile female. (An OY zygote, by contrast, is inviable, since the genes on the X chromosome are essential for survival.)
• Symptoms: Short stature, webbed neck, immature sex organs and infertility, low mental abilities, small chin and jaw; associated with liver and kidney problems, obesity, diabetes, and hypertension. It occurs in roughly 1 in 5000 female births.
• Treatment: Growth hormone treatment (started early in childhood can help achieve adult height) and estrogen replacement therapy to promote development of secondary sexual characteristics.

LONG31.Describe the symptoms, causes and treatments of Sickle cell anaemia and phenylketonuria.

Sickle cell anaemia • Cause: A recessive point mutation on chromosome 11 in the gene for the β chain of haemoglobin, where DNA nucleotide T is substituted by A, changing codon CTC (glutamic acid) to CAC (valine). A child with both recessive alleles produces abnormal haemoglobin, causing red blood cells to become sickle-shaped and carry less oxygen.
• Symptoms: Pain, fatigue, shortness of breath, jaundice, and increased risk of infections.
• Treatment: Pain management with NSAIDs and opioids, blood transfusions to increase healthy RBCs, bone marrow transplant for a long-term cure, and regular monitoring with preventative measures such as vaccinations and antibiotics.

Phenylketonuria (PKU) • Cause: A point mutation (substitution or deletion) makes the gene for phenylalanine hydroxylase (PAH, on chromosome 12) recessive and non-functional; if a child inherits two recessive alleles, phenylalanine cannot be converted to tyrosine and builds up to toxic levels.
• Symptoms: Intellectual disability, seizures, behavioural problems, mental disorder, microcephaly, and a musty odour in breath, skin, and urine; excess phenylalanine can cause brain damage.
• Treatment: A diet low in phenylalanine (avoiding meat, dairy, eggs, and some grains), regular monitoring of blood phenylalanine levels, medications with synthetic PAH enzyme to lower phenylalanine, and nutritional supplements.