Diploma Prep - Cell Division, Genetics & Molecular Biology PDF

Cell Division, Genetics and Molecular Biology Common Diploma Questions (approx 40% of exam - 24 questions) Cell Division Mendelian Genetics Molecular Genetics Mitosis~2 Crosses~4-5 DNA Replication~2 Meiosis~ 3 Pedigree Protein Synthesis~3 Nond...

Cell Division, Genetics and Molecular Biology Common Diploma Questions (approx 40% of exam - 24 questions) Cell Division Mendelian Genetics Molecular Genetics Mitosis~2 Crosses~4-5 DNA Replication~2 Meiosis~ 3 Pedigree Protein Synthesis~3 Nondisjunction ~1 Chrom. Mapping Recombination (what if)~1 Life Cycles~1 What if → mode of inheritance Things to Study for Cell Division Mitosis Meiosis Define and explain the significance of chromosome Describe the process of meiosis (spermatogenesis & number in somatic and sex cells (diploidy vs. haploidy). oogenesis) and the necessity for the reduction of Explain the events of the cell cycle (interphase, mitosis chromosome number. and cytokinesis). Compare the processes of mitosis and meiosis. Demonstrate the behaviour of chromosomes during the four stages of mitosis (prophase, metaphase, anaphase, Describe the processes of crossing over and telophase). nondisjunction and evaluate their significance to organism inheritance and development. Identify the stages of mitosis based on a diagram or images of prepared microscope slides. Compare the formation of fraternal and identical offspring in a single birthing event. Calculate the duration of each stage of mitosis in a sample of cells. Alternation of Generations Interpret models of human karyotypes. Describe the diversity of reproductive strategies by comparing the alternation of generations in a range of organisms. Cell Division  Notes: pg. 2 - 5 Textbook: pgs. 548 - 562 The Cell Cycle & Mitosis Functions of Cell Division: Mitosis: occurs in somatic cells, Meiosis: occurs in sex cells, produces diploid daughter cells produces haploid daughter cells genetically identical to parent cell genetically unique from parent cell 1. Growth 1. Production of gametes 2. Maintenance (sperm and egg cells) 3. Repair The Cell Cycle in 24 hrs 1. Interphase (23 hours) G1 (11 hrs) , S (8 hrs) & G2 (4 hrs) phases 2. Mitosis (1 hour) Prophase, metaphase, anaphase & telophase 3. Cytokinesis Interphase G1 phase = cell growth S phase = DNA is decondensed into chromatin to be replicated G2 phase = cell produces proteins necessary for division - Unpacked DNA + histones = chromatin - DNA typically only condenses into chromosomes for cell division Humans = 46 chromosomes (23 pairs) - 22 pairs = autosomes - 1 pair = sex chromosomes - Homologous chromosomes = same gene sequence as one another; (one homologue from maternal origin, and the other is of paternal origin) - carry the same set of genes but may carry different forms of each gene (alleles) - Human autosomal cells carry two pairs of each chromosome = diploid (2n) - gametes carry one copy of each chromosome = haploid (n)  The Cell Cycle in 24 hrs 1. Interphase (23 hours) G1 (11 hrs) , S (8 hrs) & G2 (4 hrs) phases 2. Mitosis (1 hour) Prophase, metaphase, anaphase & telophase 3. Cytokinesis Mitosis (“PMAT”) 1. Prophase - Chromatin condenses into chromosomes - Nuclear envelope dissolves - Centrioles move to opposite poles of the cell - Spindle fibers start to form 2. Metaphase - Chromosomes line up single file at the “metaphase plate” - Spindle fibers from the centrioles attach to the centromere of each chromosome 3. Anaphase - Spindle fibers shorten, pulling sister chromatids apart (centrosomes break) - Each pole of the cell now has diploid set of unreplicated chromosomes 4. Telophase - Chromosomes reach opposite ends of the cell - Spindle fiber dissolves - Nuclear membrane begins to reform around each set of chromosomes - Chromosomes begin to unwind into chromatin  The Cell Cycle in 24 hrs 1. Interphase (23 hours) G1 (11 hrs) , S (8 hrs) & G2 (4 hrs) phases 2. Mitosis (1 hour) Prophase, metaphase, anaphase & telophase 3. Cytokinesis Cytokinesis  Cytokinesis Animal cells: - Cell pinches in around the equator, forming a cleavage furrow and continues - 2 cells result Plant cells: - No centrioles - Cell plate forms instead of cleavage/cytokinesis Unregulated Cell Cycle: Cancer 1) Oncogenes stimulate the cell cycle and promote cell growth 2) Tumour suppressor genes repress cell division and promote apoptosis (cell death) Mutations which lead to either the over-expression of oncogenes or under-expression of tumour suppressor genes can lead to the development of tumours.  Textbook: pgs. 563 - 572 Notes: pg. 6 - 9 Meiosis & Nondisjunction Key Outcomes of Meiosis 1. Reduction division: daughter cells have half the number of chromosomes of the parent cell (diploid parent cell → 4 haploid daughter cells) 2. Genetic recombination: hkjfs daughter cells are genetically unique from parent cell (due to crossing over and independent assortment) Meiosis contains the same steps as mitosis, but each step occurs twice. Meiosis I: chromosomes are reduced from diploid to haploid (2n → n) Meiosis II: Each haploid cell from meiosis I undergoes “mitosis” (n → n) Interphase - Chromosomes replicated in S phase - Occurs before meiosis Prophase I - Chromatin condenses into chromosomes, spindle fibers and centrioles form, nuclear membrane dissolves - synapsis (alignment) of homologous chromosomes to form a tetrad: align for crossing over ✨ ✨ source of genetic variation Metaphase I - Spindle fibers from centrioles attach to centromere of each chromosome - Homologous chromosomes line up (maternal and paternal chromosomes on opposite sides) - Chromosome pairs line up independently of each other (independent assortment✨) ✨ source of genetic variation Random fertilization = source of genetic variation 2n = possible gametes ~ 70 trillion possible 223 = 8 388 chromosome 608 possible combinations in genetically potential offspring unique gametes per parent! Anaphase I - Spindle fibers shorten, pulling homologous chromosomes apart Recall Telophase I - Spindle fibers dissolve, nuclear membrane begins to form around each set of chromosomes - Each pole of the cell now has a haploid set of chromosomes Meiosis II - Interphase does not occur, no DNA replication before meiosis II - Same process as mitosis but with haploid cells  Spermatogenesis: cytokinesis in males → 4 sperm cells - Spermatogonium matures into primary spermatocyte via mitosis - Occurs continuously - Meiosis begins at puberty - Spermatogenesis occurs in seminiferous tubules in testes Spermatogenesis Oogenesis: cytokinesis in females → 1 ovum + 2 polar bodies + secondary oocyte - Oogonium matures into primary oocyte - Primary oocyte arrested in prophase I until puberty begins - Primary oocyte undergoes unequal cytokinesis, resulting in 1 polar body and secondary oocyte - Secondary oocyte arrested in metaphase II until fertilization - Secondary oocyte finishes meiosis after fertilization, resulting in 1 polar body and zygote Nondisjunction: errors in meiosis Nondisjunction in Meiosis I - Homologous chromosomes fail to separate in anaphase I - Both chromosomes pulled to the same pole - All 4 gametes affected - 2 with extra chromosome - 2 short one chromosome Nondisjunction in Meiosis II - Sister chromatids fail to separate in anaphase II - Chromosome pulled to one pole of the cell - 2 normal gametes, 2 affected gametes - 1 with extra chromosome - 1 short a chromosome Example: Trisomy 21 (Down’s Syndrome) - Results in an embryo with 3 copies of chromosome 21 - Can be nondisjunction from maternal or paternal gametes  Notes: pg. 10 - 11 Textbook: pg. 573 - 580 Reproductive Strategies: Alternation of Generations Animal Life Cycle - Based on a regular pattern of meiosis and mitosis - Humans live and reproduce in the diploid stage Plant Life Cycle - Plants (and some fungi) that exist in distinct haploid and diploid stages throughout their lifecycle. Key Terminology: - Sporophyte = diploid generation of a plant (produces haploid spores) - Gametophyte = haploid generation of a plant (produces gametes) - Gametes fuse (fertilization) to create another sporophyte Fill in the blanks: Asexual Reproduction: Reproduction via Mitosis - Budding: a complete (mini) version of the parent grows out of the parent’s body - Fragmentation: creation of offspring from a fragment (portion) of the parent - Parthenogenesis: unfertilized egg develops into an adult Mendelian Genetics Common Diploma Questions (approx 40% of exam - 24 questions) Cell Division Mendelian Genetics Molecular Genetics Mitosis~2 Crosses~4-5 DNA Replication~2 Meiosis~ 3 Pedigree Protein Synthesis~3 Nondisjunction ~1 Chrom. Mapping Recombination (what if)~1 Life Cycles~1 What if → mode of inheritance  Notes: pg. 2 - 6 Textbook: pg. 586 - 598 Mendelian Genetics Genotype: genes present in an Phenotype: physical expression of individual genes present in an individual  Alleles: different “versions” of the same Genes: heritable sections of DNA. gene. In a pair of homologous Can be protein coding or non-protein chromosomes, one allele is from maternal coding. origin, the other is from paternal origin. Gene Pool: genes present within a population Dominant Allele: an allele that can mask the expression of other variants of the same gene Recessive Allele: an allele that can only be expressed in the absence of a dominant allele  Genotypes - Homozygous Dominant: individual with two copies of the dominant allele (BB) - Homozygous Recessive: individual with two copies of the recessive allele (bb) - Heterozygous: individual with two different alleles (Bb) Gregor Mendel: Father of Modern Genetics “Experiments in Plant Hybridization” Monohybrid Cross: a cross between two organisms with variations in one gene P = parental generation F1 = first filial generation (produced from crossing P generation) F2 = second filial generation (produced from crossing F1 generation)  Try this! 1. Try the following monohybrid cross using a Punnett Square. Find the phenotypic and genotypic ratios. Tt x tt. Assume tall plant height is the dominant allele. 2. Brachydactyly (B) is a dominant trait over normal length fingers (b). A man, heterozygous for brachydactyly, marries a woman with normal length fingers. What are the possible genotypes and phenotypes for their offspring? 2. Brachydactyly (B) is a dominant trait over normal length fingers (b). A man, heterozygous for brachydactyly, marries a woman with normal length fingers. What are the possible genotypes and phenotypes for their offspring? B b b Bb bb b Bb bb 1 brachydactyly: 1 normal 1 Bb : 1 bb Mendel’s Law of Segregation Pairs of alleles (on homologous chromosomes) segregate independently during anaphase I of meiosis. Only one of the two gene copies present in an organism is distributed to each gamete that is made. The allocation of these alleles is random. Mendel’s Law of Independent Assortment Different pairs of alleles (on homologous chromosomes) align independently of each other during metaphase I of meiosis. Test Cross: determining the unknown genotype of an individual expressing the dominant trait by breeding it with a homozygous recessive individual and then examining the offspring Summary of a Test Cross: - If no recessive offspring are observed (all express the dominant allele) then the unknown genotype of the parent is homozygous dominant - If some recessive offspring are observed then the unknown genotype of the parent is heterozygous Test Crossing Sheep - Sheep ranchers like producing white wool (easier to dye and less brittle than black wool) - White wool is dominant, black wool is recessive - If a farmer suspects white sheep in a herd are heterozygous, what should he/she do? Test Crossing Sheep Cross the suspected heterozygotes with a black sheep. If black sheep offspring are produced, the white sheep is confirmed to be heterozygous  Notes: pg. 7 - 10 Textbook: pg. 594 - 597, 601 - 609 Multiple Alleles & Modes of Inheritance Multiple alleles: a gene that has more than 2 alleles. Examples: - Eye colour in Drosophila - Blood types in humans - Fur colour in rabbits Eye colour in Drosophila E1E1, E1E2, E1E3, E1E4 E2E2, E2E3, E2E4 E3E3, E3E4 E4E4 Example: Predict the phenotypic outcome if a honey-coloured heterozygous fly is crossed with a white fly. Another example of multiple alleles: rabbit coat colour C = grey Cch = chinchilla Ch = light grey c = albino C > Cch > Ch > c Incomplete Dominance & Codominance 1. Incomplete Dominance No dominant allele - alleles act together to form an intermediate trait Example: Snapdragons (flowers) Red x White = pink 2. Codominance Both alleles are expressed equally Example: Cattle Red x White = roan (both red and white hair present) Incomplete or Codominance? Incomplete or Codominance? Sickle Cell Anemia = codominance Incomplete or Codominance?  Blood Typing in Humans: Codominance - Surface antigen determines blood type - Controlled by 3 alleles - IA - IB - i - IA and IB are codominant, i is recessive to IA and IB Genotypes and Phenotypes for ABO Blood Types IAIA, I Ai IBIB, I Bi IAI B ii  Calculating Probability in Genetics # chances for an event Probability (P) = # possible events Product Rule = (probability of event one) x (probability for event two) Example: Calculate the probability of producing a daughter who is blood type A from a cross between a Type B mother and a Type A father, given that each of these individuals have a parent with type O. Example: Predict the phenotypic outcome if a red snapdragon (CRCR) is crossed with a pink snapdragon (CRCW) Example: The mating of a white horse (CWCW) and a chestnut coloured horse (CHCH) produces a red roan (CWCH). Predict the phenotypic and genotypic ratios of a mating between two roan horses. Sex - Linked Traits - Genes that are located on the X or Y chromosome, rather than autosomes - Barr body: inactivated X chromosome in cells with more than one X. - In female cells: 1 X chromosome, 1 barr body - In male cells: 1 X chromosome, 1 Y https://www.youtube.com/watch?v=Y9vXhmI5FXM Example: Colour Blindness (X-Linked Recessive)  Notes: pg. 11 - 12 Textbook: pg. 605 - 609 Dihybrid Crosses Dihybrid crosses: Involve two genes (2 different traits) Mendel’s Law of Independent Assortment: the two alleles of one gene segregate independently of the alleles for other genes during gamete formation T = tall plant t = short plant Y = yellow plant y = green plant Determine the gametes of a dihybrid cross: TTYY Determine the gametes of a dihybrid cross: TtYY Determine the gametes of a dihybrid cross: TtYy P = true breeding F1 = produced from crossing individuals from P F2 = produced from crossing individuals from F1 Complete a Punnett Square for the following: TtYy x TtYy Phenotypes: Genotypes: Example: A dwarf, red snapdragon is crossed with a homozygous tall white plant. Determine the phenotype and genotype of the F1 generation (Tall is dominant and red is incompletely dominant to white). Example: a dwarf pink snapdragon is crossed with a heterozygous tall, white snapdragon. Determine the phenotypic and genotype ratios of the offspring. Example: a dwarf pink snapdragon is crossed with a heterozygous tall, white snapdragon. Determine the phenotypic and genotype ratios of the offspring. tCR tCR tCW tCW TCW TtCRCW TtCRCW TtCWCW TtCWCW TCW TtCRCW TtCRCW TtCWCW TtCWCW tCW ttCRCW ttCRCW ttCWCW ttCWCW tCW ttCRCW ttCRCW ttCWCW ttCWCW Example: a dwarf pink snapdragon is crossed with a heterozygous tall, white snapdragon. Determine the phenotypic and genotype ratios of the offspring. tCR tCR tCW tCW Genotypic Ratio: 1 TtCRCW TCW TtCRCW TtCRCW TtCWCW TtCWCW 1 TtCWCW 1 ttCRCW TCW TtCRCW TtCRCW TtCWCW TtCWCW 1 ttCWCW W R W R W W W W W Phenotypic Ratio: tC ttC C ttC C ttC C ttC C 1 tall pink 1 tall white tCW ttCRCW ttCRCW ttCWCW ttCWCW 1 short pink 1 short white  Notes: pg. 13 - 14 Textbook: pgs. 610 - 614 Pedigrees Pedigree = a chart showing the expression of a trait through the generations of a family Eg. Hemophilia in the royal family  Pedigrees can be used to predict genotypes and mode of inheritance (autosomal dominant, autosomal recessive, X-linked recessive and X- linked dominant)  AUTOSOMAL DOMINANT Mode of Inheritance: _____________________________ CLUES: - both males and females are equally affected - trait usually appears in all generations - affected children cannot be born from unaffected parents - affected parents can produce an unaffected child  AUTOSOMAL RECESSIVE Mode of Inheritance: _____________________________ CLUES: - both males and females are equally affected - trait tends to skip generations - unaffected parents may produce affected offspring  X-LINKED RECESSIVE Mode of Inheritance: _____________________________ CLUES: - mostly males affected - no transmission of trait from father to son - trait skips generations - Females can only get condition if father has condition and mother has condition or is a carrier  X-LINKED DOMINANT Mode of Inheritance: _____________________________ CLUES: - mostly males affected - trait does not skip generations - affected sons must have an affected mother - affected daughters may inherit from either parent (if father is affected = all daughters affected) - less common than x-linked recessive If a third phenotype emerges - incomplete or codominance! This could be denoted as different colours, half/half black/white, or an x through - always double check your keys What would the genotype of III-1 be? What would the genotype of III-2 be?  Notes: pg. 15 - 17 Textbook: pg. 600 - 602 Gene Linkage & Chromosome Mapping Interphase - S phase Meiosis I - Prophase I - Linked genes = genes found close together on the same chromosome - the closer the genes are, the more likely they will be inherited together (stay together during crossing over) Not linked - on different types of chromosomes Incomplete Linkage - on same chromosome but far apart Complete linkage - close together on same chromosome - If genes are NOT linked: - 50% parental, 50% recombinant gametes - If genes ARE linked: - less than 50% recombinant gametes Example: Fruit flies may exhibit the following traits: normal wings (N) which are dominant to vestigial wings (n). Yellow body (Y) which is dominant to black body (y). Determine the phenotypic ratio of a cross of two flies from the F1 generation. Phenotypic Ratio: Example: Fruit flies may exhibit the following traits: normal wings (N) which are dominant to vestigial wings (n). Yellow body (Y) which is dominant to black body (y). Determine the phenotypic ratio of a cross of two flies from the F1 generation. NY Ny nY ny Phenotypic Ratio: NY NNYY NNYy NnYY NnYy 9: Normal wings, yellow body 3: Normal wings, black body 3: vestigial wings, yellow body Ny NNYy NNyy NnYy Nnyy 1: vestigial wings, black body nY NnYY NnYy nnYY nnYy ny NnYy Nnyy nnYy nnyy Example: Fruit flies may exhibit the following traits: normal wings (N) which are dominant to vestigial wings (n). Yellow body (Y) which is dominant to black body (y). Determine the phenotypic ratio of a cross of two flies from the F1 generation, assuming these genes are linked. NnYy x NnYy Example: Fruit flies may exhibit the following traits: normal wings (N) which are dominant to vestigial wings (n). Yellow body (Y) which is dominant to black body (y). Determine the phenotypic ratio of a cross of two flies from the F1 generation, assuming these genes are linked. NnYy x NnYy Out of 400 fruit flies, we would expect: NY ny 75% Normal wings, yellow body = 300 flies 25% Vestigial wings, black body = 100 flies NY NNYY NnYy Actual Data: ny NnYy nnyy 282 normal wings, yellow body 9 normal wings, black body 9 vestigial wings, yellow body 92 vestigial wings, black body How is this possible? Chromosome Mapping: based on recombination frequency recombination frequency = # recombinants x 100% total # offspring Note: a recombination frequency > 50% indicates that genes are on different chromosomes (or very far apart) Ex. determine the distance between the wing and colour genes shown below Example: Determine the sequence of genes A, B and C. Crossover frequency between A and B is 12%, B and C is 7% and between A and C is 5%. Determine the sequence of these genes (Map these genes!) Determine the sequence of these genes (Map these genes!) Determine the sequence of genes (map these genes!) Determine the sequence of these genes (map them!) Molecular Biology  Textbook: Pg. 624 - 629 Notes: Pg. 2 - 5 DNA Structure  DNA Deoxyribonucleic Acid Deoxyribose: 5 carbon sugar molecule that helps form the sugar-phosphate backbone Nucleic Acid: biomolecules essential to all living things (comprised of nucleotides - the building blocks of DNA) Key Features of DNA Structure 1. Hydrogen bonds between nitrogenous bases 2. Adenine pairs with thymine, cytosine pairs with guanine Apple Tree Car Garage 3. DNA strands run antiparallel (5’ and 3’ ends in opposite directions) 4. Each nitrogenous base is bonded to a deoxyribose sugar group 5. Sugar groups are bonded to phosphate groups 6. A nucleotide = 1 nitrogenous base + 1 deoxyribose sugar + 1 phosphate DNA Structure DNA Location - 3 Places Nucleus (nDNA): - Packed into chromosomes - Encode proteins for growth, reproduction and maintenance  DNA Location - 3 Places Mitochondria (mt DNA): - Circular DNA separate from main diploid set - Encode proteins needed for oxidative phosphorylation DNA Location - 3 Places Chloroplasts (cpDNA): - Circular (or linear) DNA separate from main set - Mostly encode genes for protein synthesis and photosynthesis  Mitochondria DNA (mtDNA) - Conserved across eukaryotes because of the critical role of cellular respiration - Less effective DNA repair mechanisms than nDNA increase the mutation rate - useful to study evolution: can compare mtDNA of different species to determine phylogeny History of DNA Phoebus Levene (early 1900s): - Discovered that DNA and RNA are composed of nucleotides - Identified by their nitrogen bases: - A: Adenine - T: Thymine - C: Cytosine - G: Guanine History of DNA Erwin Chargaff (1940s): - Found nucleotides are present in equal amounts - Chargaff’s Rule: A = T, C = G Example: Complete the table for the nucleotide composition of the following DNA samples Nucleotide Composition (%) DNA Sample Adenine Thymine Cytosine Guanine 1 16 2 32 3 11 Example: Complete the table for the nucleotide composition of the following DNA samples Nucleotide Composition (%) DNA Sample Adenine Thymine Cytosine Guanine 1 16 16 34 34 2 32 32 18 18 3 39 39 11 11 History of DNA Rosalind Franklin (1950s): - Discovered that DNA is a double stranded helix with consistent spacing between DNA strands History of DNA James Watson & Francis Crick (1950s): - Discovered the precise double-helical, antiparallel structure of DNA DNA looks like a twisted ladder! STRUCTURE: 2 complementary strands BACKBONE: deoxyribose & phosphate RUNGS: 2 nitrogen bases Adenine --- Thymine Guanine --- Cytosine  Textbook: Pg. 630 - 633 Notes: Pg. 6 - 7 DNA Replication  DNA Replication - Occurs during S phase of interphase during the cell cycle in the nucleus - Semi-conservative process: each strand serves as a template for the creation of its complementary strand - 3 steps: initiation, elongation & termination Semiconservative Replication - Each strand on the double helix = template for the synthesis of a new, complementary strand - Requirements: - Free nucleotides - Replication enzymes (DNA polymerase, helicase) - Decondensed DNA strand Step 1: Initiation - Helicases bind to the replication origin on the DNA strand - Helicases cut the strand and “unzip” the double helix - Y-shaped area where DNA is split = replication fork - Unwound region = replication bubble - Newly unwound single strands = templates for replication 2. Elongation - Primase enzymes synthesize a primer sequence telling DNA polymerase where to bind - DNA polymerase adds nucleotides within the replication bubble, creating 2 new strands of DNA (complementary to template strands) - Elongation always occurs in the 5’ to 3’ direction - Leading strand = replicated continuously - Lagging strand = replicated in short fragments, called okazaki fragments - Okazaki fragments are spliced together by DNA ligase Step 3: Termination - Newly synthesized strands automatically rewind into a helix (chemically stable structure) - DNA replication continues until the DNA molecules separate - Replication protein complex dissembles DNA Replication: Key Enzymes Enzyme Function Helicase Unwinds DNA Primase Synthesizes a DNA primer needed for elongation to begin DNA polymerase Adds new nucleotides from the 5’ to 3’ end of template DNA strands DNA ligase Splices together okazaki fragments of the lagging DNA strand  Notes: pg. 11-13 Textbook: pg. 636 - 642 Intro to Protein Synthesis Recall: DNA Replication Human Genome Project  Replication: production of an exact copy of DNA during cell division Transcription: messenger RNA (mRNA) produced from a template segment of DNA The Central Dogma: flow of genetic information Translation: RNA is used as a template to synthesize proteins Translation: amino acid chain (protein) produced from mRNA. Facilitated by tRNA and rRNA Gene Expression What is a gene? ○ Segment of DNA that codes the amino acid sequence of a polypeptide (protein) DNA does not directly control protein synthesis ○ Its information is transcribed into RNA Gene expression/protein synthesis requires 2 steps: 1. Transcription (DNA → mRNA) 2. Translation (mRNA → protein) Composition of Proteins - 20 different amino acids - Determined by DNA sequence - Different combinations of hemoglobin amino acids = different proteins - One gene = one “protein” Immunoglobulin (antibody) - Protein synthesis relies on rRNA, mRNA and tRNA keratin  DNA RNA Location: nucleus and cytoplasm Location: nucleus and cytoplasm (during cell division) Strand number: single stranded Strand number: double stranded Sugar: ribose Sugar: deoxyribose Nitrogen bases: cytosine, guanine, Nitrogen bases: cytosine, guanine, adenine, uracil adenine, thymine 3 Types of RNA involved in protein synthesis  mRNA 3 Types of RNA Cytoplasm and nucleus 1. Messenger RNA (mRNA) - Involved in transcription (1st stage of protein synthesis) - Copies DNA and carries the message from DNA in the nucleus to ribosomes in the cytoplasm - Contains codons encoding for specific amino acids  tRNA & rRNA All RNA 3 Types of RNA Cytoplasm Produced in the nucleus 2. Ribosomal RNA (rRNA) - Component of ribosomes - Helps link specific amino acids together depending on the mRNA sequence 3. Transfer RNA (tRNA) - Carries specific amino acids to the ribosome for protein synthesis - Contains anticodons complementary to codons on mRNA  Textbook: pg. 636 - 642 Notes: pgs. 14-16 Transcription & Translation Transcription has 3 steps: 1. Initiation: ○ RNA polymerase binds to the promoter sequence on the template DNA strand 2. Elongation: ○ DNA strands unwind and mRNA synthesis begins in the 5’→ 3’ direction ○ DNA rewinds after RNA polymerase passes 3. Termination: ○ RNA polymerase reaches termination sequence and detaches ○ mRNA is released to the cytoplasm  pre-mRNA must be processed into mRNA - Introns must be spliced out of pre-mRNA. Exons remain in the final mRNA transcript - Introns = non-protein coding sequences - Exon = protein coding sequences - Processed mRNA is released to the cytoplasm ✨ Note: this RNA splicing process only occurs in eukaryotes ✨ 2. Translation: mRNA → protein Translation has 3 steps: 1. Initiation ○ Ribosome binds to mRNA ○ Start codon (AUG) on mRNA recruits tRNA carrying methionine Initiation = P site, new tRNA recruited to A site  Translation has 3 steps: 2. Elongation ○ Next codon recognized and tRNA is recruited to A site (amino acids in P and A sites form peptide bond) ○ Ribosome shifts down 1 codon (tRNA in P site is now in E site, tRNA in A site is now in P site) ○ tRNA in E site exits and new tRNA recruited to A site Translation has 3 steps: 3. Termination a. Ribosome reaches a stop codon and dissembles, releasing newly formed polypeptide  Notes: pg. 17 - 18 Textbook: pg. 643 - 646 Mutations Mutation = mistake in DNA replication, transcription or translation, can be harmful, beneficial or neutral Mutations occurring in somatic cells (body cells) are not inherited Mutations occurring in germ cells (gametes) can be inherited Mutations can be spontaneous or induced Spontaneous = accidental error Induced = mutagenic agents promoting during DNA replication changes to the DNA sequence  Types of Mutations 1. Point mutation: substitution, deletion or insertion of a single nucleotide ○ Usually does not result in a changed amino acid due to the redundancy of the genetic code ○ Can result in frameshift mutations Types of Mutations 2. Chromosomal mutations: inversion, deletion, duplication or translocation of a chromosomal region Effects of Mutations No change in amino acid sequence, no impact on cell Silent function Change in amino acid sequence, results in changed Missense protein Change in amino acid sequence resulting in a premature Nonsense STOP codon  Notes: Pg. 19 - 22 Textbook: pg. 647 - 661 Genetic Engineering Cloning of a Gene Cloning ○ Production of many identical copies of an organism by some asexual means Gene cloning ○ Production of many identical copies of a single gene Two ways ○ Recombinant DNA ○ PCR (Polymerase Chain Reaction) Genetic Engineering = Recombinant DNA (rDNA) Technology - Building new DNA by manipulating the DNA of organisms - Biotechnologists splice new pieces of DNA (genes) from one chromosome to another - Refinement of traditional breeding practices Recombinant DNA requires four things 1. A plasmid from E. coli 2. Gene of interest 3. Restriction enzymes (for “cutting” DNA) 4. Ligase (for “gluing” the gene to plasmid DNA) Recombinant DNA requires four things 1. A plasmid from E. coli 2. Gene of interest 3. Restriction enzymes (for “cutting” DNA) 4. Ligase (for “gluing” the gene to plasmid DNA) Why is E.coli bacteria used? - Plentiful, cheap and easy to manipulate - Short life span (grow quickly) - Contain a plasmid Plasmid: a small, circular piece of DNA used as a “host” to clone a gene (distinct from chromosomal DNA) How to make rDNA: “Sticky ends” 1. A bacterial plasmid 2. Isolation of the 3. Both DNA samples is opened using a desired gene using are combined in a restriction enzyme the same restriction petri dish or test (scissors) enzyme (scissors) tube 4. Ligase is added to glue DNA samples together 5. Recombinant DNA is inserted back into E. coli. The bacteria replicates, replicating the desired gene product with it. 6. The product is separated and sterilized. Some may be bottled and sold (eg. insulin) Applications of rDNA 1. Agriculture: Genetically Modified Organisms (GMOs) a. Herbicide and pathogen resistance, increased crop yield, better nutrient profiles, environmental stress tolerance Applications of rDNA 2. Pharmaceuticals a. Insulin, antibodies, hormones (eg. HGH), interferons, vaccines Applications of rDNA 3. Gene Therapy: replacing a defective gene with a non-defective gene. Eg.Cystic fibrosis, hemophilia. Issue: How do we get the genes in the body and control them? Evaluating Recombinant DNA  DNA Fingerprinting Humans only use about 2% of their DNA (protein coding) Certain sections of DNA are unique to each individual (about 0.1%) (except twins). Use PCR to amplify portions of DNA Technologies to Analyze DNA Applications in: - Forensics - Detect trace amounts of DNA - First step in DNA fingerprinting and southern blotting - Paternity testing - Genetic research - Gene therapy - Medicine (infectious disease detection, genetic testing, cancer diagnoses) Use of Restriction Enzymes: Gel Electrophoresis To determine crime scene data Who was at the crime?  To identify paternity Who is related to the child?  PCR (Polymerase Chain Reaction) What if we only have a small sample?? We can use PCR (polymerase chain reaction) to amplify the amount of DNA we have 1. Add primers (sequences on either side of the sequence we want to amplify), ○ Taq polymerase (DNA polymerase that works at high temp) ○ Free nucleotides 2. Heat up DNA to 92 degrees to separate 3. Cool to 50 degrees so primers can bind to DNA 4. Heat to 72 degrees to synthesize new DNA 5. Repeat over and over