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Master how genetic information is passed to offspring through reproduction, cell division, DNA synthesis, and inheritance patterns.
Reproduction is the biological process by which organisms produce new individuals. There are two main types: sexual reproduction (involves fusion of gametes) and asexual reproduction (single parent, no gamete fusion).
| Type | Sexual Reproduction | Asexual Reproduction |
|---|---|---|
| Parents | Two parents | One parent |
| Gametes | Fusion of gametes (fertilization) | No gametes |
| Genetic Variation | High - offspring genetically unique | None - offspring are clones |
| Speed | Slower | Faster |
| Adaptation | Better for changing environments | Better for stable environments |
Sexual Reproduction:
Internal Fertilization: Gametes fuse inside female body. Examples: Mammals, reptiles, birds. Advantages: Protected environment, higher survival rate. Requires copulation.
External Fertilization: Gametes fuse outside body in aquatic environment. Examples: Most fish, amphibians. Advantages: Many offspring produced. Disadvantages: Lower survival rate, requires water.
Asexual Reproduction:
Budding: New individual grows from parent body. Example: Hydra.
Fragmentation: Parent breaks into pieces, each becomes new organism. Example: Starfish, planaria.
Parthenogenesis: Development from unfertilized egg. Example: Some insects, reptiles (Komodo dragons).
Sexual Reproduction - Pollination:
Self-pollination: Pollen from same flower/plant. Low genetic variation.
Cross-pollination: Pollen from different plant. Higher genetic variation.
Vectors: Wind, insects (bees), birds, water.
Process: Pollen (male gamete) lands on stigma → pollen tube grows down style → fertilizes ovule → develops into seed.
Asexual Reproduction - Vegetative Propagation:
Runners/Stolons: Horizontal stems above ground. Example: Strawberries.
Rhizomes: Horizontal underground stems. Example: Ginger.
Bulbs: Underground storage organs. Example: Onions, tulips.
Tubers: Swollen underground stems. Example: Potatoes.
Cuttings: Piece of plant grows roots. Used in horticulture.
Q: Compare the advantages of sexual vs asexual reproduction for a plant species living in (a) a stable rainforest environment, (b) a rapidly changing climate.
(a) Stable rainforest environment:
Asexual reproduction advantageous: In stable environments, the parent's genes are already well-adapted. Asexual reproduction allows rapid colonization without need for pollinators. Runners/bulbs spread quickly. No energy wasted finding mates. However, lack of genetic variation means entire population vulnerable to new disease.
(b) Rapidly changing climate:
Sexual reproduction advantageous: Genetic variation from cross-pollination creates offspring with different traits. Some may be better adapted to new temperature/rainfall patterns. Population more resilient to environmental stress and disease. Higher chance of survival through natural selection acting on varied genotypes.
Mammalian reproduction is controlled by a complex system of hormones that regulate gamete production, ovulation, pregnancy, and birth. Understanding these hormonal pathways is critical for Band 6.
Practice drawing hormone graphs: Sketch FSH, LH, Estrogen, and Progesterone levels across the 28-day cycle from memory. Mark ovulation (day 14). Note that estrogen peaks just before LH surge. Progesterone only rises after ovulation. This visual memory is invaluable for exam questions.
Pregnancy is maintained by hormones that prevent menstruation and support fetal development. Birth is initiated by a positive feedback loop involving multiple hormones.
Explain why a woman who takes a pregnancy test 4 days after fertilization might get a negative result, even though she is pregnant.
Answer:
Pregnancy tests detect HCG hormone in blood or urine. HCG is produced by the embryo after implantation.
Timeline issue:
Conclusion: Testing too early (before implantation) gives false negative. Most tests are accurate from approximately 1 week after fertilization, or around the time of missed period (2 weeks post-fertilization).
Q: Compare the feedback mechanisms in the menstrual cycle (Section 1.2) with the feedback mechanism during childbirth. Why is one negative and the other positive?
Menstrual Cycle - NEGATIVE Feedback:
High progesterone from corpus luteum inhibits FSH/LH release from pituitary. This prevents new follicles from developing. Effect: stabilizes the system. When progesterone drops (no pregnancy), FSH/LH rise again to start new cycle. This maintains homeostasis - keeps hormone levels in a narrow range.
Childbirth - POSITIVE Feedback:
Cervical stretch stimulates more oxytocin release → stronger contractions → more stretch → even more oxytocin. Effect: amplifies the response. Loop continues until baby is born (stimulus removed). This achieves a specific outcome rather than maintaining equilibrium.
Why the difference?
Menstrual cycle: Needs to be cyclical and regular - negative feedback maintains this. Childbirth: Needs to rapidly achieve an endpoint (delivery) - positive feedback creates the necessary intensity. Positive feedback is rare in biology precisely because it doesn't maintain homeostasis - it's only used when a rapid, decisive change is needed.
Humans have developed technologies to manipulate reproduction in both animals and plants for agricultural productivity and to overcome infertility. Understanding both the benefits and ethical concerns is essential.
Definition: Sperm is collected from male and artificially introduced into female reproductive tract.
Definition: Fertilization occurs outside the body (in laboratory) and embryo is implanted into uterus.
Q: Evaluate the impact of IVF on society. Consider social, ethical, and economic factors in your response.
Social Impact:
Ethical Impact:
Economic Impact:
Evaluation questions structure: Always organize biotechnology evaluation answers into three categories: Social (who benefits? equity? access?), Ethical (moral concerns? rights? consent?), and Economic (cost? funding? sustainability?). This shows sophisticated analysis and directly addresses marking criteria.
Mitosis and meiosis are two types of cell division with fundamentally different purposes and outcomes. Understanding their differences is critical for Module 5.
| Feature | Mitosis | Meiosis |
|---|---|---|
| Purpose | Growth and repair of body tissues | Production of gametes (sex cells) |
| Where it occurs | Somatic (body) cells | Germ cells (testes/ovaries) |
| Number of divisions | One division | Two divisions (Meiosis I and II) |
| Number of daughter cells | 2 daughter cells | 4 daughter cells |
| Chromosome number | Diploid (2n) → Diploid (2n) - Maintains chromosome number | Diploid (2n) → Haploid (n) - Halves chromosome number |
| Genetic variation | Daughter cells genetically identical to parent | Daughter cells genetically different from parent and each other |
| Crossing over | No crossing over | Crossing over occurs in Prophase I |
| Pairing of homologous chromosomes | No pairing (synapsis) | Homologous pairs form bivalents |
| Example in humans | 46 chromosomes → 46, 46 (skin, muscle, liver cells) | 46 chromosomes → 23, 23, 23, 23 (sperm and egg cells) |
Meiosis I (Reduction Division) - Separates homologous pairs
Meiosis II (Similar to Mitosis) - Separates sister chromatids
Q: A diploid cell with 6 chromosomes undergoes meiosis. (a) How many chromosomes are in each cell after Meiosis I? (b) How many chromosomes are in each cell after Meiosis II? (c) How many total daughter cells are produced?
(a) After Meiosis I: 3 chromosomes per cell. Meiosis I separates homologous pairs, reducing from diploid (2n = 6) to haploid (n = 3). Each chromosome still consists of 2 sister chromatids joined at centromere.
(b) After Meiosis II: 3 chromosomes per cell. Meiosis II separates sister chromatids, but chromosome number stays the same (n = 3). Each chromosome now consists of a single chromatid.
(c) Total daughter cells: 4 haploid cells. Meiosis I produces 2 cells, then each divides in Meiosis II, giving 4 total gametes.
DNA replication is the process by which a cell copies its DNA before cell division. This ensures each daughter cell receives a complete set of genetic information. The process is semi-conservative: each new DNA molecule consists of one original strand and one newly synthesized strand.
During S phase (Synthesis phase) of Interphase, before mitosis or meiosis begins. This ensures each chromosome consists of 2 identical sister chromatids before cell division.
| Feature | Leading Strand | Lagging Strand |
|---|---|---|
| Synthesis direction | Continuous (toward fork) | Discontinuous (away from fork) |
| Template orientation | 3' to 5' | 5' to 3' |
| Number of primers | One | Many (one per Okazaki fragment) |
| Okazaki fragments | None | Yes (1000-2000 nucleotides each) |
| DNA Ligase needed? | No | Yes (to join fragments) |
Given a DNA template strand: 3'-TACGGATCG-5', what will be the sequence of the newly synthesized complementary strand? Show the direction.
Step 1: Identify template direction
Template: 3'-TACGGATCG-5'
Step 2: Remember base pairing rules
Step 3: Build complementary strand
Genetic variation is essential for evolution and adaptation. Meiosis creates genetic diversity through several mechanisms, ensuring that no two gametes (or offspring) are genetically identical.
| Source of Variation | When | What Happens | Impact |
|---|---|---|---|
| Crossing Over | Prophase I | DNA exchanged between non-sister chromatids | New allele combinations |
| Independent Assortment | Metaphase I | Random orientation of bivalents | 2^n combinations |
| Random Segregation | Anaphase I | Homologs separate randomly | Unpredictable distribution |
| Random Fertilization | Conception | Any sperm + any egg | 2^2n combinations |
Q: Explain why siblings (except identical twins) are genetically unique, even though they have the same parents. Refer to at least three sources of genetic variation in your answer.
Siblings are genetically unique due to multiple sources of variation during meiosis and fertilization:
1. Crossing Over:
During Prophase I of meiosis, homologous chromosomes exchange DNA segments. This creates recombinant chromosomes with new combinations of alleles. Each parent can therefore produce gametes with different mixtures of their genetic material.
2. Independent Assortment:
During Metaphase I, homologous pairs orient randomly at the metaphase plate. In humans with 23 chromosome pairs, this creates 2^23 (over 8 million) possible combinations of chromosomes in each gamete. Each parent produces gametes with different chromosome combinations.
3. Random Fertilization:
Any of the millions of possible sperm can fertilize any of the millions of possible eggs. This multiplies the variation: 2^23 × 2^23 = over 70 trillion possible genetic combinations in offspring.
The cell cycle is the series of events from one cell division to the next. Proper regulation ensures cells divide only when appropriate and prevents uncontrolled growth (cancer).
Checkpoints ensure cell division proceeds only when conditions are favorable and DNA is intact.
| Checkpoint | Location | What is Checked? | If Fail? |
|---|---|---|---|
| G1 Checkpoint | End of G1 | Cell size, nutrients, growth signals, DNA damage | Enter G0 or undergo apoptosis |
| G2 Checkpoint | End of G2 | DNA replication complete? Errors in DNA? Cell large enough? | Repair DNA or trigger apoptosis |
| M Checkpoint (Spindle) | Metaphase | All chromosomes attached to spindle? Properly aligned? | Pause until attachment correct |
Q: Explain why mutations in the p53 gene are so commonly associated with cancer. What is the normal role of p53, and what happens when it is mutated?
Normal Role of p53:
p53 is a tumor suppressor protein that acts at the G1 and G2 checkpoints. When DNA damage is detected, p53 either: (1) Halts the cell cycle and activates DNA repair mechanisms, or (2) Triggers apoptosis (programmed cell death) if damage is too severe to repair. This prevents damaged cells from dividing and passing mutations to daughter cells.
When p53 is Mutated:
Loss of functional p53 means cells with DNA damage can bypass checkpoints and continue dividing. Damaged DNA accumulates more mutations over successive divisions. Cells that should die (apoptosis) survive and proliferate. This allows other cancer-promoting mutations to accumulate.
Why p53 Mutations Are So Common in Cancer:
p53 is mutated in over 50% of human cancers because it is the cell's primary defense against uncontrolled division. Losing p53 function removes a critical "brake" on the cell cycle, accelerating cancer development. It is often called "the guardian of the genome" - when this guardian is disabled, genomic instability increases dramatically.
The organization and structure of DNA differs significantly between prokaryotes (bacteria) and eukaryotes (animals, plants, fungi, protists). These differences affect how genes are expressed and regulated.
| Feature | Prokaryotes (Bacteria) | Eukaryotes (Animals, Plants, Fungi) |
|---|---|---|
| DNA Shape | Circular (single chromosome) | Linear (multiple chromosomes) |
| Location | Nucleoid region (cytoplasm, not membrane-bound) | Nucleus (membrane-bound organelle) |
| Histones | No histones (DNA not wrapped around proteins) | Histones present (DNA wrapped around histone octamers → nucleosomes → chromatin) |
| Size | Small (approximately 4 million base pairs in E. coli) | Large (approximately 3 billion base pairs in humans) |
| Introns/Exons | No introns - genes are continuous coding sequences | Introns (non-coding) and Exons (coding) - genes are interrupted |
| mRNA Processing | None - mRNA used directly | Extensive: 5' cap, 3' poly-A tail, splicing (introns removed) |
| Plasmids | Yes - small circular DNA molecules with extra genes (e.g., antibiotic resistance) | Rare (yeast have plasmids, most eukaryotes don't) |
| Gene Density | High (approximately 90% of DNA codes for proteins) | Low (approximately 2% in humans codes for proteins, rest is non-coding) |
| Transcription & Translation | Coupled (happen simultaneously in cytoplasm) | Separated (transcription in nucleus, translation in cytoplasm) |
Q: A gene in a eukaryotic cell is 10,000 base pairs long, but the mature mRNA is only 3,000 base pairs. A similar gene in a bacterial cell is 3,000 base pairs, and the mRNA is also 3,000 base pairs. Explain this difference.
Eukaryotic Gene: The 10,000 base pair gene contains both exons (coding sequences) and introns (non-coding sequences). During transcription, the entire gene is transcribed into pre-mRNA (10,000 bp). However, during RNA processing in the nucleus, introns are removed by splicing, leaving only 3,000 bp of exons in the mature mRNA. The 7,000 bp difference represents removed introns.
Bacterial Gene: Prokaryotic genes lack introns - they are continuous coding sequences. The entire 3,000 bp gene is transcribed into mRNA, which is used directly for translation without processing. Gene length equals mRNA length.
Transcription is the process of copying a gene's DNA sequence into RNA. In eukaryotes, this produces pre-mRNA, which is then processed into mature mRNA that can be translated into protein.
DNA → RNA → Protein
(Transcription) → (Translation)
"DNA makes RNA, RNA makes Protein, Protein makes us"
After transcription, eukaryotic pre-mRNA must be processed before leaving the nucleus:
Given a DNA template strand: 3'-TACGGCATG-5', write the mRNA sequence that would be transcribed. Include directionality.
Step 1: Identify the template strand
Template DNA: 3'-TACGGCATG-5'
Step 2: Apply RNA base pairing rules
Step 3: Write mRNA sequence
Translation is the process by which ribosomes read mRNA and synthesize proteins. The sequence of nucleotides in mRNA determines the sequence of amino acids in the protein.
Multiple ribosomes can translate the same mRNA simultaneously, forming a polyribosome. This allows rapid production of many copies of the same protein from a single mRNA molecule.
Given the mRNA sequence: 5'-AUGCCGUACUGA-3', determine: (a) the amino acid sequence of the polypeptide, and (b) how many amino acids are in the final protein.
Step 1: Identify codons
mRNA: 5'-AUG CCG UAC UGA-3'
Divide into 3-nucleotide codons starting from 5' end
Step 2: Translate each codon
AUG = Methionine (Met) - Start codon
CCG = Proline (Pro)
UAC = Tyrosine (Tyr)
UGA = STOP - Stop codon (no amino acid)
Q: A mutation changes a single nucleotide in the DNA, which changes codon 5 from CAU (Histidine) to CAC (also Histidine). Will this mutation affect the protein? Explain your answer in terms of the genetic code.
Answer: No, this mutation will NOT affect the protein.
Explanation: This is a silent mutation (also called synonymous mutation). Although the DNA nucleotide changed, both the original codon (CAU) and the mutated codon (CAC) code for the same amino acid, Histidine. The protein sequence remains unchanged.
Why this happens: The genetic code is degenerate (redundant), meaning most amino acids are encoded by more than one codon. These alternative codons often differ only in the third position. This redundancy provides some protection against mutations - not all DNA changes result in protein changes.
Mutations are changes in DNA sequence. They can occur spontaneously or be induced by environmental factors. Understanding mutation types and their effects is crucial for genetics and evolution.
| Mutation Type | Change in DNA | Effect on Protein | Severity |
|---|---|---|---|
| Silent | Nucleotide substitution | Same amino acid | None |
| Missense | Nucleotide substitution | Different amino acid | Variable |
| Nonsense | Creates stop codon | Premature termination | Usually severe |
| Frameshift (Insertion/Deletion) | Nucleotides added or removed | All downstream amino acids changed | Usually severe |
Q: Explain why frameshift mutations are generally more severe than point mutations (substitutions). Use an example in your explanation.
Frameshift mutations are more severe because they affect all codons downstream of the mutation, while point mutations only affect one codon.
Point Mutation (Substitution):
Original: AUG-CCG-UAC-GGU-AAA-UGA → Met-Pro-Tyr-Gly-Lys-STOP
Mutated: AUG-UCG-UAC-GGU-AAA-UGA → Met-Ser-Tyr-Gly-Lys-STOP
Effect: Only ONE amino acid changed (Pro → Ser). Rest of protein is normal.
Frameshift Mutation (Deletion of one nucleotide):
Original: AUG-CCG-UAC-GGU-AAA-UGA → Met-Pro-Tyr-Gly-Lys-STOP
Mutated: AUG-CGU-ACG-GUA-AAU-GA... → Met-Arg-Thr-Val-Asn...
Effect: ALL amino acids after mutation site are different. Reading frame is shifted. Protein is completely altered and likely non-functional.
Autosomal inheritance follows Mendel's laws and involves genes located on autosomes (non-sex chromosomes). Understanding dominance relationships and using Punnett squares are fundamental skills for Module 6.
A cross examining inheritance of one trait. Classic example: tall vs short pea plants.
A cross examining inheritance of two traits simultaneously. Both genes must be on different chromosomes or far apart on the same chromosome.
In pea plants, purple flower (P) is dominant to white flower (p). A purple-flowered plant is crossed with a white-flowered plant, and the offspring are 50% purple and 50% white. What is the genotype of the purple-flowered parent?
Step 1: Identify what we know
Step 2: Test both possibilities
If purple parent is PP:
PP × pp → all Pp (100% purple) ✗ Does not match
If purple parent is Pp:
Pp × pp → 50% Pp (purple), 50% pp (white) ✓ Matches!
Sex-linked traits are controlled by genes located on sex chromosomes (X or Y). Most sex-linked traits in humans are X-linked because the Y chromosome carries few genes.
Most common pattern of sex-linkage. Examples: color blindness, hemophilia, Duchenne muscular dystrophy.
Q: A woman who is a carrier for hemophilia (X^H X^h) marries a man with normal blood clotting (X^H Y). What is the probability their first child will be: (a) an affected son, (b) a carrier daughter, (c) an affected daughter?
Step 1: Set up Punnett square
Mother: X^H X^h (carrier) × Father: X^H Y (normal)
Step 2: Complete cross
Possible offspring:
Not all traits follow simple Mendelian patterns. Several non-Mendelian inheritance patterns produce different phenotypic ratios and outcomes.
| Pattern | Heterozygote Phenotype | F2 Ratio (if applicable) | Example |
|---|---|---|---|
| Complete Dominance | Same as dominant homozygote | 3:1 | Pea plant height |
| Incomplete Dominance | Intermediate (blend) | 1:2:1 | Snapdragon color |
| Codominance | Both expressed | 1:2:1 | ABO blood type |
| Multiple Alleles | Varies | Varies | ABO blood type |
| Polygenic | Continuous variation | Bell curve | Human height |
Q: In a species of flower, red (C^R C^R) and white (C^W C^W) show incomplete dominance. If two pink flowers (C^R C^W) are crossed, what percentage of offspring will be pink?
Cross: C^R C^W × C^R C^W
Offspring:
Pedigrees are family trees that show inheritance patterns of traits across generations. Analyzing pedigrees helps determine whether traits are dominant, recessive, autosomal, or sex-linked.
| Pattern | Skip Generations? | Sex Distribution | Key Clue |
|---|---|---|---|
| Autosomal Recessive | Yes | Equal M/F | Unaffected parents → affected child |
| Autosomal Dominant | No | Equal M/F | Appears every generation |
| X-Linked Recessive | Yes | Mostly males | Carrier mother → affected sons |
Q: In a pedigree, you observe: (1) the trait appears in every generation, (2) affected individuals have at least one affected parent, (3) approximately equal numbers of affected males and females. What is the most likely inheritance pattern?
Answer: Autosomal Dominant
Reasoning: