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Master how genetic information is passed to offspring through reproduction, cell division, DNA synthesis, and inheritance patterns.
Module 5 • Pillar 1 of 4
Understand how organisms pass genetic information to offspring through sexual and asexual reproduction across all kingdoms of life.
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.
Sexual: Spores from fusion of hyphae (different mating types).
Asexual: Budding (yeast), Spores (mold).
Asexual: Binary fission (DNA replicates, cell divides into 2 identical cells).
Genetic transfer: Conjugation (plasmid transfer), not true sexual reproduction.
Asexual: Binary fission (Amoeba), Budding (some protists).
Sexual: Conjugation (Paramecium), alternation of generations (some algae).
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.
Androgens (Testosterone)
Source: Leydig cells in testes (stimulated by LH from pituitary)
Function:
Regulation: Negative feedback - high testosterone inhibits GnRH → less LH/FSH released
FSH (Follicle Stimulating Hormone)
Source: Anterior pituitary
Function: Stimulates Sertoli cells → support sperm maturation
FSH (Follicle Stimulating Hormone)
Source: Anterior pituitary
Function: Stimulates follicle growth in ovary. Follicle contains developing egg (oocyte).
Phase: Days 1-14 (Follicular phase)
LH (Luteinizing Hormone)
Source: Anterior pituitary
Function: LH surge triggers ovulation (release of egg from follicle) around day 14.
Post-ovulation: Converts empty follicle → corpus luteum
Estrogen
Source: Developing follicle (days 1-14), then corpus luteum
Function:
Progesterone
Source: Corpus luteum (days 14-28, Luteal phase)
Function:
28-Day Cycle Summary:
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.
HCG (Human Chorionic Gonadotropin)
Source: Embryo (then placenta from week 2)
Timing: Detected in blood/urine approximately 1 week after fertilization (basis of pregnancy tests)
Function:
Note: HCG is structurally similar to LH, which is why it can maintain the corpus luteum.
Progesterone (Pregnancy Hormone)
Source: Corpus luteum (first trimester) → Placenta (second/third trimester)
Function:
Estrogen
Source: Placenta
Function:
Oxytocin
Source: Posterior pituitary (released by hypothalamus)
Function:
Prostaglandins
Source: Uterine lining
Function:
Positive Feedback Loop in Birth:
1. Fetal head stretches cervix
2. Stretch receptors stimulated
3. Signal to hypothalamus
4. Posterior pituitary releases oxytocin
5. Oxytocin causes uterine contractions
6. Stronger contractions → more cervical stretch
7. More oxytocin released (positive feedback)
8. Cycle amplifies until birth
This is POSITIVE feedback (amplifies response), unlike most body systems which use NEGATIVE feedback (dampens response).
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.
Process (Livestock):
Advantages:
Disadvantages / Concerns:
Definition: Fertilization occurs outside the body (in laboratory) and embryo is implanted into uterus.
Process (Human):
Used for:
Advantages:
Disadvantages / Concerns:
Q: Evaluate the impact of IVF on society. Consider social, ethical, and economic factors in your response.
Social Impact:
Ethical Impact:
Economic Impact:
Conclusion:
IVF has had overwhelmingly positive impact in helping individuals overcome infertility, but requires careful regulation to address ethical concerns around embryo use and genetic selection. Access inequity remains a major social issue. Overall, benefits to individuals and society outweigh concerns when properly regulated.
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.
Module 5 • Pillar 2 of 4
Understand how cells divide through mitosis and meiosis, how DNA replicates, and the sources of genetic variation critical for evolution and inheritance.
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) |
Prophase
Metaphase
Anaphase
Telophase
Result
Two genetically identical diploid daughter cells, each with the same number of chromosomes as the parent cell.
Meiosis I (Reduction Division) - Separates homologous pairs
Prophase I (Longest phase)
Metaphase I
Anaphase I
Telophase I and Cytokinesis
Meiosis II (Similar to Mitosis) - Separates sister chromatids
Prophase II
Chromosomes condense (if they decondensed), nuclear envelope breaks down, new spindle forms
Metaphase II
Chromosomes align at metaphase plate (individually, not as pairs)
Anaphase II
Sister chromatids separate and move to opposite poles (like mitosis)
Telophase II and Cytokinesis
Nuclear envelope reforms, chromosomes decondense, cytokinesis completes
Result
Four genetically unique haploid daughter cells (gametes), each with half the chromosome number of the original cell.
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.
Key concept: Meiosis I reduces chromosome number (2n → n), Meiosis II separates sister chromatids (n → n, but single chromatids).
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.
Helicase
Function: "Unzips" the DNA double helix
Mechanism: Breaks hydrogen bonds between complementary base pairs (A-T, G-C)
Result: Creates a replication fork with two template strands
DNA Polymerase
Function: Synthesizes new DNA strand by adding nucleotides
Direction: Can only add nucleotides in the 5' to 3' direction
Requirement: Needs a primer (short RNA sequence) to start
Proofreading: Checks and corrects errors during synthesis
Primase
Function: Synthesizes short RNA primers
Why needed: DNA Polymerase cannot start synthesis from scratch
Note: Primers are later removed and replaced with DNA
DNA Ligase
Function: Joins Okazaki fragments on the lagging strand
Mechanism: Forms phosphodiester bonds between adjacent nucleotides
Result: Creates continuous DNA strand
Step 1: Unwinding
Helicase unwinds the double helix, breaking hydrogen bonds between bases. Two template strands are exposed at the replication fork.
Step 2: Primer Synthesis
Primase synthesizes short RNA primers complementary to template DNA. These provide the 3'-OH group needed for DNA Polymerase to begin.
Step 3: Elongation - Leading Strand
DNA Polymerase adds nucleotides continuously in the 5' to 3' direction
Template is read 3' to 5'
Only one primer needed
Leading strand: Synthesized continuously toward replication fork
Step 4: Elongation - Lagging Strand
Template runs in opposite direction (5' to 3')
DNA Polymerase must work away from replication fork
Synthesized discontinuously in short segments called Okazaki fragments
Each fragment requires its own RNA primer
Lagging strand: Multiple short segments synthesized backward
Step 5: Primer Removal and Replacement
RNA primers are removed and replaced with DNA nucleotides by DNA Polymerase
Step 6: Ligation
DNA Ligase joins Okazaki fragments by forming phosphodiester bonds between the sugar-phosphate backbones, creating one continuous strand
Result
Two identical DNA molecules, each with one original (template) strand and one newly synthesized strand. This is called semi-conservative replication.
| 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
Template: 3'-T A C G G A T C G-5'
New: 5'-A T G C C T A G C-3'
Answer:
The newly synthesized strand is: 5'-ATGCCTAGC-3'
Note: Always specify the direction (5' and 3' ends) - this is critical for Band 6 responses!
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.
What is it?
Exchange of DNA segments between non-sister chromatids of homologous chromosomes
When does it occur?
Prophase I of Meiosis - during synapsis when homologous chromosomes pair up to form bivalents
Mechanism:
Result:
Chromatids now contain a mix of maternal and paternal DNA. Creates recombinant chromosomes with new combinations of alleles that didn't exist in either parent.
Example:
If one chromosome has alleles AB and its homolog has ab, crossing over can produce recombinant chromosomes Ab and aB. This creates gametes with gene combinations neither parent had.
What is it?
Random distribution of homologous chromosome pairs to daughter cells
When does it occur?
Metaphase I of Meiosis - when bivalents align at the metaphase plate
Mechanism:
Mathematical Impact:
Number of possible chromosome combinations = 2^n
Where n = number of homologous pairs (haploid number)
Humans: n = 23, so 2^23 = 8,388,608 possible combinations
This means one human can produce over 8 million genetically different gametes just from independent assortment alone.
Example:
If you have 2 chromosome pairs, there are 2² = 4 possible combinations: AB, Ab, aB, ab (where capital = maternal, lowercase = paternal)
What is it?
Random separation of homologous chromosomes during Anaphase I
Relationship to Independent Assortment:
Random segregation is the actual separation event that follows from independent assortment. The random alignment in Metaphase I leads to random segregation in Anaphase I. These terms are often used interchangeably.
Result:
Each gamete receives one chromosome from each homologous pair, but which one (maternal or paternal) is random.
What is it?
Random fusion of any sperm with any egg during sexual reproduction
Mathematical Impact:
If each parent can produce 2^n different gametes...
Number of possible offspring combinations = 2^n × 2^n = 2^2n
Humans: 2^23 × 2^23 = 2^46 = 70 trillion possible combinations
This doesn't even include crossing over, which adds even more variation.
Why this matters:
The astronomical number of possible combinations means siblings (except identical twins) are genetically unique. This variation provides raw material for natural selection and evolution.
| 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.
Conclusion:
These three mechanisms ensure that each sibling receives a unique combination of parental alleles. Only identical twins are genetically identical because they come from the same fertilized egg that split.
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).
Interphase (approximately 90% of cell cycle)
G1 Phase (Gap 1) - Growth
S Phase (Synthesis) - DNA Replication
G2 Phase (Gap 2) - Preparation for Division
M Phase (Mitotic Phase) - Division (approximately 10% of cycle)
G0 Phase (Gap 0) - Resting/Quiescent State
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 |
p53: "Guardian of the Genome"
The protein p53 is critical at G1 and G2 checkpoints. It detects DNA damage and either: (1) Halts cell cycle and activates DNA repair enzymes, or (2) Triggers apoptosis if damage is irreparable. Mutations in p53 gene are found in greater than 50% of human cancers.
What is Cancer?
Uncontrolled cell division resulting from mutations in genes that regulate the cell cycle. Cells ignore checkpoints and divide continuously.
Proto-oncogenes vs Tumor Suppressor Genes
Proto-oncogenes (accelerator)
Tumor Suppressor Genes (brake)
Multi-Hit Hypothesis
Cancer typically requires multiple mutations (4-6) accumulated over time. One mutation is not enough - need both "accelerator stuck" and "brakes broken".
Characteristics of Cancer Cells
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.
Clinical Relevance:
Li-Fraumeni syndrome is an inherited condition where individuals have one defective p53 allele from birth. They have 90% lifetime risk of developing cancer, highlighting p53's critical protective role.
Module 5 • Pillar 3 of 4
Master how genetic information flows from DNA to RNA to proteins, understanding the central dogma of molecular biology and how genes determine characteristics.
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) |
Exons (Expressed Sequences)
Introns (Intervening Sequences)
Example:
DNA: [Exon 1] - [Intron 1] - [Exon 2] - [Intron 2] - [Exon 3]
↓ Transcription
Pre-mRNA: [Exon 1] - [Intron 1] - [Exon 2] - [Intron 2] - [Exon 3]
↓ Splicing (introns removed)
Mature mRNA: [Exon 1] - [Exon 2] - [Exon 3]
↓ Translation
Protein: Amino acids from Exons 1, 2, 3 only
Mitochondrial DNA (mtDNA)
Chloroplast DNA (cpDNA)
Endosymbiotic Theory Evidence:
Mitochondria and chloroplasts have their own circular DNA (like bacteria), their own ribosomes (70S, like bacteria), and double membranes. This suggests they were once free-living prokaryotes that were engulfed by ancestral eukaryotic cells billions of years ago.
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.
Key Concept: Eukaryotes must process pre-mRNA (splicing out introns), while prokaryotes do not. This is why eukaryotic genes are often much longer than the proteins they encode.
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"
Function
Synthesizes RNA by reading DNA template strand and adding complementary RNA nucleotides
Direction
Base Pairing Rules (DNA → RNA)
1. Initiation
2. Elongation
3. Termination
After transcription, eukaryotic pre-mRNA must be processed before leaving the nucleus:
1. 5' Capping
What: Modified guanine nucleotide added to 5' end
Purpose: Protects mRNA from degradation, helps ribosome binding
2. 3' Poly-A Tail
What: Chain of approximately 200 adenine nucleotides added to 3' end
Purpose: Protects mRNA from degradation, aids export from nucleus
3. Splicing
What: Introns removed, exons joined together
Carried out by: Spliceosome (complex of proteins and small nuclear RNAs)
Result: Mature mRNA contains only exons
Alternative splicing: Different combinations of exons can be joined, producing different proteins from the same gene
Summary: Pre-mRNA to Mature mRNA
Pre-mRNA: [Exon 1] - [Intron] - [Exon 2] - [Intron] - [Exon 3]
↓ Add 5' cap
↓ Add 3' poly-A tail
↓ Splice out introns
Mature mRNA: 5' cap - [Exon 1] - [Exon 2] - [Exon 3] - poly-A tail
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
Template DNA: 3'- T A C G G C A T G -5'
mRNA: 5'- A U G C C G U A C -3'
Answer:
mRNA: 5'-AUGCCGUAC-3'
Note: Remember U instead of T in RNA! Always include 5' and 3' labels.
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.
Codons
Definition: Three-nucleotide sequence on mRNA that codes for one amino acid
Total possible codons: 4³ = 64 different codons
Number of amino acids: 20
Result: The code is degenerate (redundant) - most amino acids are coded by more than one codon
Start and Stop Codons
Start codon: AUG (codes for Methionine) - signals where translation begins
Stop codons: UAA, UAG, UGA - signal where translation ends (do not code for amino acids)
Properties of the Genetic Code
Transfer RNA (tRNA)
Structure: Cloverleaf shape, approximately 80 nucleotides long
Anticodon: Three-nucleotide sequence that pairs with mRNA codon
Amino acid attachment site: 3' end where specific amino acid binds
Function: Brings correct amino acid to ribosome based on mRNA codon
Example: tRNA with anticodon 3'-UAC-5' pairs with mRNA codon 5'-AUG-3' and carries Methionine
Aminoacyl-tRNA Synthetase
Function: Enzyme that attaches correct amino acid to its corresponding tRNA
Specificity: One synthetase for each of the 20 amino acids
Process: Recognizes both the tRNA anticodon and the amino acid, ensuring correct pairing
Ribosomes
Structure: Two subunits (large and small) made of rRNA and proteins
Prokaryotes: 70S ribosomes (50S + 30S subunits)
Eukaryotes: 80S ribosomes (60S + 40S subunits)
Three binding sites:
1. Initiation
2. Elongation (Repeated Cycle)
Step 1 - Codon Recognition:
Step 2 - Peptide Bond Formation:
Step 3 - Translocation:
3. Termination
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.
Advantages:
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)
Answer:
(a) Amino acid sequence: Met-Pro-Tyr
(b) Number of amino acids: 3 amino acids (stop codon does not code for an amino acid)
Note: Translation always starts at AUG and ends at a stop codon (UAA, UAG, or UGA).
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.
Key Concept: Silent mutations demonstrate the degeneracy of the genetic code. Only mutations that change the amino acid (missense) or introduce stop codons (nonsense) will affect the protein.
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.
Silent Mutation (Synonymous)
Change: Nucleotide substitution that does not change the amino acid
Effect on protein: No effect - protein remains unchanged
Example: CAU → CAC (both code for Histidine)
Why: Genetic code is degenerate
Missense Mutation
Change: Nucleotide substitution that changes the amino acid
Effect on protein: Varies from none to severe depending on amino acid properties
Example: GAA → GUA (Glutamic acid → Valine)
Conservative: Similar amino acid (minimal effect)
Non-conservative: Very different amino acid (major effect)
Famous example: Sickle cell disease (Glu → Val in hemoglobin)
Nonsense Mutation
Change: Nucleotide substitution that creates a stop codon
Effect on protein: Premature termination - protein is shortened
Example: UAC → UAA (Tyrosine → STOP)
Severity: Usually severe - truncated proteins are often non-functional
Insertion
Change: One or more nucleotides added to DNA sequence
Effect: If not a multiple of 3, shifts reading frame of all downstream codons
Result: Completely different amino acid sequence after the insertion
Severity: Usually severe - often produces non-functional protein
Deletion
Change: One or more nucleotides removed from DNA sequence
Effect: If not a multiple of 3, shifts reading frame of all downstream codons
Result: Completely different amino acid sequence after the deletion
Severity: Usually severe - often produces non-functional protein
Frameshift Example:
Original mRNA: AUG CCG UAC GGU AAA UGA
Translation: Met-Pro-Tyr-Gly-Lys-STOP
After deletion of 2nd C:
Mutated mRNA: AUG CG UAC GGU AAA UGA
New reading: AUG CGU ACG GUA AAU GA...
Translation: Met-Arg-Thr-Val-Asn...
(Completely different protein after mutation site)
Deletion (Chromosomal)
Loss of a chromosome segment. Multiple genes lost. Usually severe effects.
Duplication
Chromosome segment copied. Results in extra copies of genes. Can lead to evolution of new functions.
Inversion
Chromosome segment reversed end to end. Genes present but in different order. May disrupt gene function.
Translocation
Segment moved from one chromosome to another. Can cause problems in meiosis. Associated with some cancers.
Spontaneous Mutations
Induced Mutations (Mutagens)
| 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.
Conclusion: Frameshift mutations change the entire reading frame, affecting every subsequent codon. Point mutations only change one codon. This is why insertions and deletions (unless in multiples of 3) are generally more damaging than substitutions.
Module 6 • Pillar 1 of 4
Master Mendelian and non-Mendelian inheritance patterns, analyze pedigrees, and understand population genetics to predict offspring traits and genetic diversity.
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.
Gene vs Allele
Genotype vs Phenotype
Homozygous vs Heterozygous
Dominant vs Recessive
Law of Segregation
Statement: The two alleles for each gene separate during gamete formation, so each gamete receives only one allele
Example: Parent with Bb genotype produces gametes with either B or b (not both)
Result: Offspring inherit one allele from each parent
Law of Independent Assortment
Statement: Alleles for different genes assort independently during gamete formation
Example: Alleles for eye color segregate independently from alleles for hair color
Limitation: Only applies to genes on different chromosomes or far apart on same chromosome
Exception: Linked genes (close together on same chromosome) do NOT assort independently
A cross examining inheritance of one trait. Classic example: tall vs short pea plants.
Example: Tt × Tt (both heterozygous tall)
Let T = tall (dominant), t = short (recessive)
| T (from parent 1) | t (from parent 1) | |
|---|---|---|
| T (from parent 2) | TT (tall) | Tt (tall) |
| t (from parent 2) | Tt (tall) | tt (short) |
Results:
Classic Monohybrid Ratios
Heterozygous × Heterozygous (Aa × Aa):
Heterozygous × Homozygous recessive (Aa × aa):
A cross examining inheritance of two traits simultaneously. Both genes must be on different chromosomes or far apart on the same chromosome.
Example: RrYy × RrYy
Let R = round seeds (dominant), r = wrinkled seeds (recessive)
Let Y = yellow seeds (dominant), y = green seeds (recessive)
Step 1: Determine gametes
Each parent (RrYy) can produce: RY, Ry, rY, ry
Step 2: Create 4×4 Punnett square
| RY | Ry | rY | ry | |
|---|---|---|---|---|
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |
Classic Dihybrid Ratio (RrYy × RrYy):
Phenotypic ratio: 9:3:3:1
Purpose
Determine the unknown genotype of an individual showing dominant phenotype. Is it homozygous dominant (AA) or heterozygous (Aa)?
Method
Cross the individual with unknown genotype with a homozygous recessive individual (aa)
Interpreting Results
If ALL offspring show dominant phenotype:
Unknown parent is homozygous dominant (AA × aa → all Aa)
If offspring show 1:1 ratio (dominant:recessive):
Unknown parent is heterozygous (Aa × aa → 50% Aa, 50% aa)
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!
Answer:
The purple-flowered parent has genotype Pp (heterozygous)
Key concept: A 1:1 ratio in offspring always indicates one parent is heterozygous and the other is homozygous recessive.
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.
Females: XX
Males: XY
Most common pattern of sex-linkage. Examples: color blindness, hemophilia, Duchenne muscular dystrophy.
Notation
Write allele as superscript on X chromosome:
Possible Genotypes
Females:
Males:
Key Patterns
Example Cross: Carrier Female × Normal Male
X^N X^n (carrier) × X^N Y (normal)
| X^N (egg) | X^n (egg) | |
|---|---|---|
| X^N (sperm) | X^N X^N (normal female) | X^N X^n (carrier female) |
| Y (sperm) | X^N Y (normal male) | X^n Y (affected male) |
Results:
Characteristics
Examples in Humans
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:
Answers:
(a) Affected son: 25% (or 1/4, or 50% of sons)
(b) Carrier daughter: 25% (or 1/4, or 50% of daughters)
(c) Affected daughter: 0% (impossible - would need X^h from both parents, but father has X^H)
Key concept: Affected fathers cannot have affected daughters if mother is carrier or normal. Daughters would need X^h X^h.
Not all traits follow simple Mendelian patterns. Several non-Mendelian inheritance patterns produce different phenotypic ratios and outcomes.
Definition
Heterozygote shows an intermediate phenotype between the two homozygotes. Neither allele is fully dominant.
Classic Example: Snapdragon Flower Color
C^R C^R = Red flowers
C^R C^W = Pink flowers (intermediate)
C^W C^W = White flowers
Note: Use superscripts, not uppercase/lowercase, because neither allele is dominant
Cross: C^R C^W × C^R C^W (two pink flowers)
| C^R | C^W | |
|---|---|---|
| C^R | C^R C^R (red) | C^R C^W (pink) |
| C^W | C^R C^W (pink) | C^W C^W (white) |
Ratio:
Human Examples
Definition
Heterozygote shows BOTH phenotypes simultaneously. Both alleles are fully expressed.
Classic Example: ABO Blood Type
Three alleles: I^A, I^B, i
Possible Genotypes and Phenotypes:
Example Cross: I^A i × I^B i
| I^A | i | |
|---|---|---|
| I^B | I^A I^B (AB) | I^B i (B) |
| i | I^A i (A) | ii (O) |
Phenotypic ratio: 1 AB : 1 A : 1 B : 1 O
All four blood types possible from this cross!
Incomplete Dominance vs Codominance
Definition
When a gene has more than two alleles in the population (though each individual still has only two alleles)
Example: ABO Blood System
Three alleles exist in population: I^A, I^B, i
Each person has two of these three alleles
Creates 6 possible genotypes and 4 phenotypes
Other Examples
Definition
Single trait controlled by two or more genes. Results in continuous variation (range of phenotypes).
Characteristics
Human Examples
| 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:
Answer: 50% pink
In incomplete dominance, the heterozygote always shows the intermediate phenotype. The F2 generation from two heterozygotes gives a 1:2:1 phenotypic ratio, so 2 out of 4 (50%) are the intermediate phenotype.
Key difference: In complete dominance, F2 is 3:1 (75% dominant). In incomplete dominance, F2 is 1:2:1 (50% intermediate).
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.
Basic Symbols
Generations and Numbering
Key Characteristics
Examples
Cystic fibrosis, sickle cell anemia, Tay-Sachs disease, albinism
Key Characteristics
Examples
Huntington's disease, achondroplasia (dwarfism), polydactyly (extra fingers/toes)
Key Characteristics
Examples
Hemophilia, color blindness, Duchenne muscular dystrophy
Step 1: Determine if trait is dominant or recessive
Step 2: Determine if trait is autosomal or X-linked
Step 3: Assign genotypes
| 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:
Why not the others?