BIOL 1105 Test 3-2 PDF

Lesson 12 READ 3’ to 5’ WRITE 5’ to 3’ 12.1- Gene Expression Overview, Prokaryotic Transcription Distinguish between transcription and translation - Transcription: DNA===>RNA - RNA polymerase binds to a DNA t...

Lesson 12 READ 3’ to 5’ WRITE 5’ to 3’ 12.1- Gene Expression Overview, Prokaryotic Transcription Distinguish between transcription and translation - Transcription: DNA===>RNA - RNA polymerase binds to a DNA template strand - Synthesizes RNA complementary to the DNA - Occurs in the nucleus of eukaryotic cells and the cytoplasm of prokaryotic cells - Translation: RNA===>proteins - mRNA used to synthesize proteins - Takes place in ribosome where rRNA and tRNA are used to assemble amino acids into a protein chain based on the mRNA sequence List the functions of different types of RNA 1. Messenger RNA (mRNA): Intermediate that carries the genetic code from the DNA to the ribosome, where it directs protein synthesis 2. Ribosomal RNA (rRNA): forms the core structure of ribosomes and catalyzes peptide bond formation between amino acids during protein synthesis 3. Transfer RNA (tRNA): reads mRNA sequence and brings the correct amino acids to the ribosome for protein synthesis 4. Small nuclear RNA (snRNA): splices the pre-mRNA in eukaryotic cells, where it helps remove non-coding sequences (introns) 5. MicroRNA (miRNA): regulates gene expression by binding to specific mRNA molecules, inhibiting their translation or promoting their degradation Summarize the initiation, elongation, and termination phases of transcription 1. Initiation a. RNA polymerase binds to the promoter region of a gene b. Sigma subunit helps RNA polymerase position correctly at the transcription start site (+1) c. RNA polymerase then unwinds the DNA double helix d. RNA synthesis begins with the incorporation of the first nucleotide complementary to the DNA template 2. Elongation a. RNA polymerase moves along the DNA template strand, synthesizing RNA in the 5’ to 3’ direction b. Adds ribonucleotides that are complementary to the DNA strand  c. Forms an RNA strand that is almost identical to the DNA coding strand (uracil replaces thymine) 3. Termination a. RNA polymerase reaches a terminator sequence in the DNA b. The RNA transcribed from the terminator forms a “hairpin structure” which pauses the RNA polymerase c. RNA polymerase dissociates from the DNA template, releasing the newly synthesized RNA molecule 12.2 - Eukaryotic Transcription Compare eukaryotic transcription and prokaryotic transcription RNA polymerase - Eukaryotic - Have multiple types of RNA polymerase (at least 3) - Each polymerase transcribes different kind of RNA - RNA polymerase II synthesizes mRNA, which is responsible for encoding proteins  - Prokaryotic - Have a single type of RNA polymerase that handles the synthesis of all types of RNA Promoters and Initiation - Eukaryotic - Genes have more complex promoters with specific sequences - Binding of RNA polymerase to the promoter involves a series of proteins called transcription factors - Essential to recruit RNA polymerase to the promoter and initiate transcription - Prokaryotic - Promoters are simpler - RNA polymerase can bind more directly to the promoter sequences (with the help of a sigma factor) - No need for additional proteins Termination - Eukaryotic - Poorly understood - Specific termination sequences are less clearly defined compared to prokaryotes - Prokaryotic - More straightforward - Well-defined terminator sequences with either form… - Hairpin loops in the RNA - Other factors to signal the end of transcription RNA Processing - Eukaryotic - Primary transcripts (pre-mRNA) undergo extensive processing before becoming mature RNA - Addition of 5’ cap to protect mRNA from degradation and aid in ribosome binding during translation - Addition of 3’ poly-A tail which also protects the transcript from degradation - Splicing to remove non-coding regions (introns) and join the coding regions (exons) together - Alternative splicing allows a single gene to produce multiple mRNA variants and, therefore, different proteins  - Prokaryotic - mRNA does not undergo significant modifications - Transcription results in mature mRNA that is immediately ready for translation Simultaneous Translation - Eukaryotic - Transcription occurs in nucleus - Translation occurs in cytoplasm - These processes are spatially and temporally separated - Prokaryotic - Transcription occurs in cytoplasm - Translation occurs in cytoplasm - These processes can occur simultaneously Review anticodons 12.3 - Translation Describe the critical features of the genetic code 1. Codons a. Sequences of three nucleotides in the mRNA b. Each one specifies a particular amino acid or stop signal during translation 2. 64 possible Codons a. Only 20 amino acids however b. Redundancy of the genetic code—multiple codons can specify the same amino acid 3. Start and stop codons a. AUG is the start codon which specifies the amino acid methionine  i. First codon in the coding sequence for almost all proteins in both eukaryotes and prokaryotes b. UUA, UAG, and UGA are the three stop codons that do not code for amino acids, but instead signal the termination of translation 4. Universality a. Most organisms use the same codons to specify the same amino acids 5. Reading frames a. mRNA can be read in 3 possible reading frames depending on where translation starts b. Only one reading frame will produce the correct protein  6. Wobble pairing a. Some tRNA molecules can recognize more than one codon due to wobble pairing—less strict base-pairing rules at the third position of the codon i. Allows one tRNA to pair with multiple codons for the same amino acid Summarize the initiation, elongation, and termination of translation Initiation - Prokaryotic - Small ribosomal subunit binds to a sequence on the mRNA called the ribosome-binding sequence (RBS) - First initiator tRNA, carrying a modified form of methionine (N-formyl-methionine), binds to the P-site of the ribosome and pairs with the starting codon AUG - Large ribosomal subunit joins the complex, completing the initiation phase and positioning the initiator tRNA at the P-site, with the A-site open for the next tRNA - Eukaryotic - Small ribosomal subunit bind to the 5’ cap of the mRNA and scans along the mRNA to find the start codon AUG - The initiator tRNA bring in normal methionine (not the modified form in eukaryotes) - Large ribosomal subunit binds and the initiation complex is formed Elongation 1. Ribosome moves along the mRNA in the 5’ to 3’ direction 2. New tRNA molecules, each carrying a specific amino acid, enter the A-site based on the codon being read by the ribosome 3. A peptide bond is formed between the amino acid on the tRNA at the A-site and the growing polypeptide chain attached to the tRNA at the P-site a. Catalyzed by rRNA in the large ribosomal subunit 4. Ribosome shifts one codon down the mRNA 5. tRNA that was in the P-site moves to the E-site and is released, while the tRNA from the A-site moves to the P-site 6. This allows the A-site to accept the next incoming tRNA Termination 1. Elongation continues until ribosome encounters a stop codon 2. No tRNA that matches a stop codon. Instead, a release factor protein bind to the stop codon in the A-site 3. Release factor triggers the release of the completed polypeptide chain from the tRNA in the P-site 4. After the polypeptide is released, the ribosome disassembles into its small and large subunits, completing the process of translation 12.4 - Gene expression, phenotype, and the effects of mutation Connect gene expression to the production of functional molecules and the effect they have on an organism’s phenotype - Gene expression: the process by which the information encoded in a gene is used to produce a functional molecule, typically a protein or RNA  - These molecules carry out essential cellular functions and their activity directly impact an organism’s phenotype (observable characteristics or traits) - Hemoglobin and sickle cell anemia - Protein composed of four subunits - A single nucleotide change in the gene encoding one subunit causes a single amino acid substitution(genotype) - This change alters the structure, causing it to form long chains instead of the normal globular shape - Chains change the shape of red blood cells from round to sickle-shaped - Phenotype: sickle cell anemia - Genetically modified pig - Gene encoding for fluorescence in jellyfish (genotype) - Results in the production of a protein that gives the pig yellow pigmentation in its skin (phenotype) List the different types of a mutation and predict the effects of each 1. Point mutations a. Base substitution: one nucleotide is substituted for another in the DNA sequence i. Silent mutation: new codon still codes for the same amino acid, so there is no effect on the protein’s structure or function ii. Missense mutation: base substitution causes a different amino acid to be incorporated into the protein (sickle cell anemia example) iii. Nonsense mutation: base substitution creates a stop codon in the middle of the gene, resulting in a truncated and often nonfunctional protein 2. Frameshift mutations a. Insertion or deletion mutations: addition or removal of one or more bases changes the reading frame of the gene, leading to a completely different sequence of amino acid downstream of the mutation i. Often results in severely altered and nonfunctional protein 3. Chromosomal mutations a. Deletions: segment of chromosome is lost, potentially removing one or more genes i. Can result in loss of gene function b. Duplications: section of the chromosome is copied, leading to extra copies of certain genes i. Can lead to overexpression of genes, which may disrupt normal cellular function c. Inversions: segment of chromosome is reversed i. Can disrupt gene function if it breaks a gene or alters its regulation d. Translocations: segment of chromosome moves to a new location, which may result in the fusion of two genes i. Can lead to abnormal gene function or the production of novel proteins ***note: while mutations can have negative effects (diseases, non functional proteins), they also play a key role in evolution by introducing genetic variation which is essential for species adaptation and survival over time  Lesson 13 13.1 - Transcriptional Regulation Identify the critical point at which gene expression is commonly controlled - Primarily controlled through transcriptional regulation - First and most crucial opportunity for cells to regulate gene expression - Most common form in both prokaryotes and eukaryotes - Depends on whether RNA polymerase can interact with the DNA and initiate transcription - occurs at a specific region of the DNA called the promoter which is located upstream of the transcription start site for the gene that they regulate - Prokaryotes: multiple genes can be controlled by a single promoter - Eukaryotes: each gene has its own promoter sequence - Core promoter region: contains the necessary binding sites for RNA polymerase; if it is removed or altered, RNA polymerase cannot bind, and transcription cannot start (halting gene expression at the transcriptional stage) - Regulatory proteins: bind to other regions of the promoter and either promote or inhibit RNA polymerase binding Distinguish between positive and negative control of transcription Positive Control - Regulatory protein called an activator binds to a specific site on the DNA to promote or enhance transcription - Help RNA polymerase bind to the promoter region - Preface operons: allow prokaryotes to respond quickly to environment because all the proteins that are required for a cellular response can be made by activating transcription on a single promoter (all related genes are grouped together) - Examples: in the lac operon (operon=cluster of genes which have related functions), the CAP protein serves as an activator - It binds to CAP binding site on the DNA when glucose is absent - Response to environmental cues that indicate the presence of lactose and absence of glucose Negative Control  - Regulatory protein called a repressor bind to DNA on the site called the operator and blocks transcription - Prevents RNA polymerase from accessing the DNA and starting transcription - Example: in the lac operon, the lac repressor binds to the operator in the absence of lactose, preventing the transcription of the genes involved in lactose metabolism Distinguish between activator and repressor proteins in transcription -Two types of transcription factors in regulating gene expression 1. Activator Proteins a. Promote transcription b. Positive control of gene expression 2. Repressor proteins a. Inhibit transcription b. Negative control of gene expression Additional information Lac operon: produces the proteins required for lactose metabolism has both positive and negative control Lac repressor inhibits when lactose is absent Cap activator protein promotes transcription when glucose is absent Top left:present Bottom left: allolactose binds to lac repressor Top right: absent Bottom right: lac repressor binds to lac operon Left transcription is: induced Right transcription is: repressed Trp operon Trp present: binds to trp repressor and there will be no transcription Trp not present: no molecule there to bind to repressor so rna polymerase can freely transcribe 13.2 - Eukaryotic transcriptional regulation Contrast control by induction and control by repression 1. Induction - A process where a molecule in the environment stimulates transcription - An environmental molecule binds to a regulatory protein (repressor), causing it to release the DNA, allowing transcription to proceed - Example: Lac opreron - Allolactose (metabolite of lactose) binds to the lac repressor, causing it to release the DNA - Stimulates transcription when glucose is absent 2. Repression - A process where a molecule in the environment inhibits transcription - An environmental molecule binds to a regulatory protein which then binds to the operator region of the DNA and blocks transcription - Example: trp operon - Tryptophan binds to the trp repressor, enabling it to bind to the operator and block transcription *Summary: induction “turns on” transcription when needed while repression “turns off” transcription when unnecessary Compare the control of gene expression in prokaryotes and eukaryotes 1. Prokaryotes - Primarily in response to environmental changes and is often rapid - Operons allow multiple genes to be controlled under one promoter - Gene regulation is straightforward: activators and repressors act near the promoter - Example 1: lac operon - Regulated by the availability of lactose (inducer) and glucose (repressor) - Example 2: trp operon - Regulated by tryptophan, which represses transcription when abundant 2. Eukaryotes - More complex due to larger genomes and the involvement of multiple transcription factors - General transcription factors: required for basal transcription at the core promoter - Specific transcription factors: enhance transcription and can bind to distant enhancer elements - Enhancers are the binding site of the specific transcription factors (activators) - Chromatin structure plays a major role in accessibility of genes for transcription - Example: in eukaryotes, specific transcription factors may bind to enhancers, looping the DNA to bring them into proximity with the core promoter and influence transcription *Summary: prokaryotic regulation is mostly about quick environmental responses, while eukaryotic regulation involves multiple layers of transcription factors and chromatin modifications Describe how chromatin structure may regulate gene expression - DNA is packaged into chromatin to fit with the cell - Nucleosomes (DNA wrapped around histone proteins) are the basic unit - Heterochromatin: tightly packed chromatin, transcriptionally inactive - Euchromatin: loosely packed chromatin, transcriptionally active - Histone modifications 1. Acetylation: adds acetyl groups to histones, loosening the interaction between DNA and histones, making DNA more accessible for transcription (promotes gene expression) 2. ATP-dependent chromatin remodeling: enzymes can slide or remodel nucleosomes, making regions of DNA accessible or inaccessible for transcription *Summary: looser chromatin allows for active transcription, while tighter chromatin restricts access to DNA, limiting gene expression 13.3 - Post-transcriptional regulation Summarize the different methods of post-transcriptional regulation. -Post-transcriptional regulation controls gene expression after transcription but before translation 1. mRNA splicing: removing introns from the primary transcript a. alternative splicing: allows a single gene to produce multiple proteins by splicing exons in different combinations b. Example: i. in the thyroid, mRNA produces calcitonin ii. In the hypothalamus, the same transcript produces CGRP through different exon combinations 2. Regulation of mRNA export: a. mRNA must exit the nucleus to be translated in the cytoplasm b. Only a small fraction (5%) of mRNA is exported to the cytoplasm c. Limiting mRNA export is a way to prevent its translation 3. Inhibition of Translation a. General inhibition: limiting the availability of proteins required for translation can inhibit translation of all mRNAs in the cytoplasm b. Specific inhibition: proteins can selectively bind to a specific mRNA and prevent its translation 4. Regulation by small RNAs (microRNAs/miRNAs and small interfering RNAs/siRNAs) a. Both small RNAs prevent mRNA translation by binding to complementary mRNA sequences b. miRNAs are produced from self-complementary RNA strands that form hairpin loops c. siRNAs are derived from long double stranded RNA, often introduced from external sources (ex. viruses) Explain how small RNAs may affect gene expression 1. miRNAs a. Derived from the organism’s DNA b. Fold into a hairpin structure to form double-stranded RNA, later processed into single-stranded miRNA 2. siRNAs a. Typically originate from long double-stranded RNA or external sources b. May arise from viral infection or experimental methods 3. Both a. Processed by enzymes into short, single-stranded RNAs b. Form a RISC (RNA-induced silencing complex) c. Bind to complementary regions on mRNA i. miRNAs bind to the 3’ untranslated region (UTR) ii. siRNAs bind to various parts of the mRNA d. Result in mRNA degradation or inhibition of translation, preventing protein production Explain how proteins may be targeted for degradation -Through post-translational regulation which controls the lifespan and activity of proteins through the processes of… 1. Ubiquitination: a process where a protein is tagged with ubiquitin molecules, marking it for degradation a. Enzyme: ubiquitin ligase attaches ubiquitin to the protein using ATP b. Polyubiquitination: a chain a ubiquitins is added, marking the protein for destruction  2. Proteasome Degradation a. Proteasome: a cylindrical complex that breaks down ubiquitinated proteins into smaller polypeptides b. The ubiquitin molecules are removed before degradation and reused by the cell -This is important for maintaining protein quality, degrading old, damaged, or unnecessary proteins, and regulating protein activity within the cell Lesson 14 14.1-Cell Reproduction Compare binary fission and mitosis 1. Binary fission a. Occurs in prokaryotes b. Single circular chromosome is replicated c. Cell elongates, moving chromosomes to opposite ends of the cell d. Septum forms at the cell’s midline  e. Cell splits into two genetically identical daughter cells i. Mitochondria and chloroplasts in eukaryotic cells also divide by binary fission, supporting the endosymbiotic theory 2. Mitosis a. Occurs in eukaryotes b. Nuclear division c. Involves linear chromosomes d. Duplicated chromosomes attach to the mitotic spindle (made of microtubules) e. Spindle aligns and segregates chromosomes into two daughter cells f. Followed by cytokinesis (cytoplasmic division) i. Animal cells: use actin-based contractile ring ii. Plant cells: form a cell plate due to the rigidity of the cell wall 14.2-Chromosome Structure Summarize the structure and packaging of eukaryotic chromosomes - Structure - continuous, double-stranded DNA molecule  - In humans: each chromosome contain approximately 140 million nucleotides - Chromatin: the material that makes up chromosomes, a complex of DNA and proteins - Heterochromatin: tightly packed, not expressed - Euchromatin: less tightly packed, where genes are located and actively expressed - DNA Packaging 1. Nucleosomes a. First level of packaging b. DNA wraps around histone proteins, forming bead-like structures c. Nucleosomes are separated by short stretches of DNA, creating a “beads on a string” structure (about 10 nm fiber) 2. Solenoid a. 10 nm fiber condenses into a thicker, more compact structure (30 nm) b. Typical packing state during interphase (non-dividing phase of cell cycle) 3. Mitotic Chromosome a. During mitosis, chromatin fibers condense further into the X-shaped structures familiar in dividing cells Other notes: - Before replication, each chromosome is a single DNA molecule - After replication, each chromosome consists of two identical sister chromatids - Held together by cohesin proteins until separated during mitosis - Each chromatid has a centromere (point of constriction) and a kinetochore (protein complex where the spindle fibers attach during mitosis) - Karyotype: the full set of chromosomes in an organism, arranged by size, number, and other characteristics - Human: 23 pairs of chromosomes (46 total in diploid cells) (homologous to each other) - Number of chromosomes is not necessarily related to the organism’s complexity - Haploid (n): one complete set of chromosomes (ex. gametes) - Diploid (2n): two sets of chromosomes, one from each parent, found in most somatic cells - Homologous chromosomes: a pair of chromosomes where one chromosome comes from the mother and the other from the father (two genetically similar chromosomes from each parent are termed homologues) - When homologues replicate they produce sister chromatids 14.3-The Cell Cycle and Stages of M Phase Diagram the eukaryotic cell cycle 1. G1 Phase (growth 1) 2. S Phase (synthesis) 3. G2 Phase (growth 2) 4. M Phase (Mitosis and cytokinesis) Summarize the events that take place during interphase and during M phase Interphase 1. G1 Phase a. Primary growth phase b. Cell grows in size, duplicates organelles, and synthesizes macromolecules necessary for later stages c. Cell carries out its regular functions d. Longest Phase  2. S Phase a. DNA replication occurs b. Cell’s genome is duplicated, resulting in two identical copies of each chromosome (sister chromatids) 3. G2 Phase a. Cell continues to grow and makes final preparations for division b. Centrosomes, which organize mitotic spindle, are duplicated c. Cell checks for any DNA replication errors and repairs them before entering mitosis (important for preventing cancerous cell growth) d. Organelles replicate e. Microtubules organize M Phase (PPMAT + cytokinesis) 1. Prophase a. Chromosomes condense into visible structures b. Mitotic spindle forms c. nuclear envelope begins to disassemble 2. Prometaphase a. nuclear envelope fully breaks down; nuclear envelope breakdown (NEBD) b. Chromosomes attach to the spindle microtubules via their kinetochores 3. Metaphase a. Chromosomes align at the cell’s equator (metaphase plate) b. sister chromatids are attached to spindle poles on opposite sides 4. Anaphase a. sister chromatids are pulled apart toward opposite poles b. Microtubules shorten, separating the chromatids 5. Telophase a. Chromosomes decondense b. nuclear envelope reforms around the separated chromosomes c. Cleavage furrow (animal cells) or cell plate (plant cells) begins to form for cytoplasmic division 6. Cytokinesis a. Cytoplasm is divided between the two daughter cells b. Cleavage furrow or cell plate is fully formed - At the end of M phase, two genetically identical daughter cells are produced, each entering G1 to begin the cycle again 14.4-Cell Cycle Control Explain the role of checkpoints in cell cycle control - checkpoints serve as regulatory points that ensure the cell proceeds through the stages of growth and division correctly - They can halt the cycle if issues are detected, allowing cells to pause and repair problems before moving forward - Apoptosis: programmed cell death 1. G1/S Checkpoint a. Determines whether the cell is ready to enter the S phase and begin DNA replication b. Cell checks for sufficient nutrients, growth factors, and whether it has grown large enough c. Critical for ensuring the cell has the resources needed to support DNA replication and the survival of daughter cells 2. G2/M Checkpoint a. Assess if DNA replication is complete and checks for DNA damage b. If errors or incomplete replication are detected, the cell pauses to allow repairs before entering mitosis c. Prevents passing defective or incomplete genetic information to daughter cells 3. Spindle checkpoint a. Ensures all chromosomes are properly aligned at the metaphase plate and attached to the spindle fibers b. If chromosomes are not correctly aligned, the cell will not proceed to anaphase to prevent errors in chromosome segregation c. Prevents chromosome nondisjunction, where daughter cells could receive too many or too few chromosomes, which is often fatal to the cell Explain how cancer may result from failure of cell cycle control - Cancer can develop when the regulatory mechanisms controlling the cell cycle are disrupted - Often occurs due to mutations in genes responsible for cell cycle control  - Leads to uncontrolled cell growth and division 1. Proto-Oncogenes and Oncogenes a. Proto-oncogenes: normal genes that promote cell growth and division b. When mutated they become oncogenes, which are permanently activated or overexpressed c. Oncogenes push the cell through the cycle even when it should stop i. Leads to uncontrolled cell proliferation d. Gain-of-function mutation: only one copy of the gene need to be mutated to promote cancer 2. Tumor suppressor genes a. Normally act as brakes on the cell cycle b. Detect problems like DNA damage and stop the cycle at checkpoints c. When mutated, these checkpoints fail i. Allows damaged or abnormal cells to continue dividing d. Loss-of-function mutation: both copies of the gene need to be inactivated for the cell to lose its ability to halt division Lesson 15 15.1-Cell Division: Meiosis Explain the role of meiosis in sexual reproduction - It reduces the chromosome number in cells by half to form gametes from germ-line cells (sperm/egg) - Gametes from two individuals of the opposite sex can fuse together, producing a zygote that has a complete set of chromosomes - Genetic diversity - Independent assortment: when homologous chromosome pairs align at the metaphase plate (during metaphase I), the orientation of each pair is random  - Paternal and maternal chromosomes can align in any order - Crossing over: (explained in question 2) - Meiosis alternates between diploid (2n) and haploid (n) states - In most animals, the diploid state dominates, but meiosis reduces diploid germ cells to haploid gametes Describe the importance of synapsis, crossing over, and reductive division during meiosis - Important for genetic recombination Synapsis - Occurs during Prophase I of Meiosis I when homologous chromosomes (one from each parent) align tightly and pair up - This forms a structure called the synaptonemal complex - Allows for crossing over to happen Crossing over  - During synapsis (prophase I), homologous chromosomes exchange genetic material - Occurs between non-sister chromatids - Introduces new combinations of alleles into gametes - Chiasmata: crossover points - Hold homologous chromosomes together until they are separated during Anaphase I - Significantly increases genetic diversity Reductive Division - Meiosis I is called reductive division - It reduces the chromosome number from diploid (2n) to haploid (n) - Important for maintaining the correct chromosome number across generations - If this process didn’t occur, chromosome numbers would double with each generation - Homologous chromosomes are separated - Ensuring each gamete receives only one chromosome from each pair Summarize the behavior of chromosomes during both meiotic divisions Meiosis I (also known as Reduction Division) 1. Prophase I: Homologous chromosomes pair up through synapsis, and crossing over occurs 2. Metaphase I: Homologous chromosomes align randomly at the metaphase plate (independent assortment) 3. Anaphase I: Homologous chromosomes are pulled apart to opposite poles, but sister chromatids remain attached 4. Telophase I: Two haploid cells are formed, each containing one set of chromosomes (with sister chromatids still connected) Meiosis II (Equational Division) 1. Prophase II: Chromosomes condense again without undergoing DNA replication 2. Metaphase II: chromosomes align at the metaphase plate 3. Anaphase II: sister chromatids are finally separated and pulled to opposite poles 4. Telophase II: four haploid cells (gametes) are produced, each containing one chromatid from each chromosome Compare and contrast mitosis and meiosis 1. Purpose a. Mitosis: to create identical cells for growth, repair, or asexual reproduction b. Meiosis: to produce genetically diverse gametes for sexual reproduction 2. Number of Divisions a. Mitosis: 1 (PMMAT) b. Meiosis: 2 (Meiosis I and Meiosis II) 3. DNA replication a. Mitosis: occurs once before division b. Meiosis: occurs once before meiosis I, but not between meiosis I and II 4. Daughter cells a. Mitosis: two identical diploid daughter cells b. Meiosis: four genetically diverse haploid cells 5. Genetic variation a. Mitosis: none b. Meiosis: high (due to crossing over and independent assortment) 6. Homologous Chromosomes a. Mitosis: do not pair up or exchange genetic material b. Meiosis: pair up during prophase I synapsis, and crossing over Meiosis only occurs in the germ-line cells to produce gametes Reduction division Highlight all vocabulary words in notes!!!!! 1 chromosome 2 copies after meiosis I Lesson 16 16.1-Mendelian Crosses Explain the advantages of Mendel's experimental system Mendel’s choice of pea plants as his model for studying genetics provided several advantages: Distinct Traits with Two Variations: Traits (e.g., flower color) had two clear variations, like purple or white, making inheritance patterns easier to observe. Quick Growth and Short Generation Time: Pea plants mature rapidly, allowing Mendel to observe several generations in a relatively short time. Large Sample Size: Pea plants produce many seeds, enabling Mendel to gather large datasets and observe consistent inheritance ratios. Controlled Fertilization: Pea plants can self-fertilize or be cross-fertilized, giving Mendel control over plant mating and the ability to test pure (true-breeding) vs. hybrid offspring. Predict and evaluate the outcomes of mono-hybrid crosses In a monohybrid cross, Mendel studied inheritance patterns of one trait with two variations. Here’s a breakdown: Parent Generation (P): True-breeding plants with homozygous genotypes, e.g., purple (PP) and white (pp) flower colors. F1 Generation: Cross between P plants produced all heterozygous (Pp) offspring with the dominant phenotype (purple flowers). F2 Generation: Self-fertilization of F1 plants led to: ○ Phenotypic Ratio: 3:1 (75% purple : 25% white) ○ Genotypic Ratio: 1:2:1 1/4 homozygous dominant (PP) 1/2 heterozygous (Pp) 1/4 homozygous recessive (pp) Key Concept: Dominant traits mask recessive ones in heterozygous genotypes, but recessive traits reappear in the F2 generation in a predictable ratio. Explain how Mendel's Principle of Segregation can be explained by Meiosis Mendel’s Principle of Segregation describes how alleles for each trait separate during gamete formation, which is explained by meiosis: Meiosis I: Homologous chromosomes (one from each parent) separate, resulting in gametes that carry only one allele for each gene. Gamete Formation: Heterozygous individuals (e.g., Aa) produce gametes with either A or a alleles, not both. Fertilization: During fertilization, gametes from each parent combine, reuniting the alleles in offspring, producing various genotype combinations (AA, Aa, aa). Meiosis Connection: The segregation of alleles observed in Mendel’s experiments mirrors the separation of chromosomes during meiosis, explaining how each parent passes on one allele per trait to the offspring. Key Terms to Remember True-breeding: Organisms that consistently pass down a specific trait (homozygous). Homozygous: Two identical alleles for a trait (e.g., PP or pp). Heterozygous: Two different alleles for a trait (e.g., Pp). Genotype: Genetic makeup of an organism (e.g., PP, Pp, or pp). Phenotype: Observable physical traits (e.g., purple or white flowers). Principle of Segregation: Mendel’s conclusion that allele pairs separate during gamete formation, explaining trait inheritance patterns. 16.2-Punnett Squares and Di-hybrid Crosses Explain how Mendel’s Principle of Segregation can be explained by meiosis. A di-hybrid cross examines the inheritance of two different traits, each with two alleles. Here’s how to predict outcomes using a Punnett Square: Example Traits: Seed shape (round, wrinkled) and seed color (yellow, green). ○ Round (R) and Yellow (Y) are dominant traits. ○ Wrinkled (r) and Green (y) are recessive traits. Use Punnett Squares to Predict and evaluate the outcomes of di-hybrid crosses 1. Determine Parent Genotypes: Cross between homozygous dominant (RRYY) and homozygous recessive (rryy) plants: ○ The F1 generation is heterozygous (RrYy) for both traits. 2. Gamete Formation: ○ Each F1 plant (RrYy) can produce four types of gametes: RY, Ry, rY, ry. ○ Use the First, Inner, Outer, Last (FOIL) mnemonic to ensure all gametes are represented. 3. Create the Punnett Square: ○ Draw a 4x4 grid for the di-hybrid cross. ○ List the four possible gametes (RY, Ry, rY, ry) from each parent on the top row and first column. 4. Fill in the Punnett Square: ○ Each box represents the genotype of an F2 offspring based on the alleles from each parent’s gamete. ○ This cross predicts a 9:3:3:1 phenotypic ratio in the F2 generation: 9 round yellow seeds (dominant for both traits) 3 round green seeds (dominant for shape, recessive for color) 3 wrinkled yellow seeds (recessive for shape, dominant for color) 1 wrinkled green seed (recessive for both traits) Key Concept: The 9:3:3:1 ratio indicates the inheritance patterns for two independently assorting traits. Mendel’s Principle of Independent Assortment describes how alleles for different traits segregate independently of each other during gamete formation. Understand the Principle of Independent Assortment Independent Behavior of Alleles: ○ In a di-hybrid cross, alleles for one gene (e.g., seed color) assort independently of alleles for another gene (e.g., seed shape). ○ Thus, traits like seed color and shape are inherited separately, leading to new combinations in offspring. Connection to Meiosis: ○ During metaphase I of meiosis, homologous chromosome pairs (each carrying alleles for different traits) align independently of each other. ○ This independent alignment results in each gamete receiving one allele from each gene independently, supporting the random combination of traits. Predicted Outcomes: Using a Punnett Square for a di-hybrid cross, the 9:3:3:1 ratio observed in the F2 generation shows how each pair of alleles segregates independently, supporting Mendel’s findings. Summary: The Principle of Independent Assortment explains that alleles for different genes segregate independently into gametes, leading to genetic diversity. Key Terms to Remember Punnett Square: A tool to predict offspring genotypes and phenotypes in genetic crosses. Dihybrid Cross: A cross between two organisms that are both heterozygous for two traits. FOIL Method: Mnemonic to remember gamete combinations in a di-hybrid cross. Independent Assortment: The concept that alleles of different genes separate independently during gamete formation. 9:3:3:1 Ratio: Expected phenotypic ratio in the F2 generation for a di-hybrid cross with independent assortment. 16.3-Pedigrees Interpret pedigrees for dominant and recessive traits Pedigrees are graphical tools used to track traits or diseases through multiple generations. Understanding the symbols and the inheritance patterns is essential for interpreting whether a trait is dominant or recessive. 1. Symbols in Pedigrees Circles represent biological females. Squares represent biological males. Shaded shapes indicate individuals who express the trait (phenotype). Unshaded shapes represent individuals who do not express the trait. Half-shaded shapes (if present) indicate carriers who do not express the recessive trait but carry one recessive allele. Single horizontal lines between a male and female indicate mating. Double horizontal lines between a male and female indicate mating between related individuals. Vertical lines and brackets connect parents to their offspring. 2. Interpreting Dominant Trait Pedigrees Key Characteristics of Dominant Traits:  Affected individuals have at least one dominant allele (e.g., B). Dominant traits typically appear in every generation (i.e., they do not skip generations). If one parent has the dominant trait, at least some offspring are likely to express the trait. Example Interpretation: Generation 1: ○ Male 1 is affected, so he has at least one dominant allele (e.g., B). ○ Female 1 is unaffected, so she must have two recessive alleles (e.g., bb). Generation 2: ○ Male 1 and Female 1 have three children. ○ One child is unaffected, indicating they are bb, meaning they did not inherit the dominant allele. ○ The other two children are affected, meaning they each have at least one dominant allele (Bb), inherited from their affected parent. Confirmation of Genotypes: ○ Since the unaffected child inherited a recessive allele (b) from each parent, we can conclude that the father in Generation 1 is heterozygous (Bb). 3. Interpreting Recessive Trait Pedigrees Key Characteristics of Recessive Traits:  Affected individuals must be homozygous recessive (e.g., aa). Recessive traits can skip generations since heterozygous individuals (carriers) do not show the trait but can pass the recessive allele to their offspring. In pedigrees, recessive traits are more likely to appear if both parents are carriers, especially in cases of mating between related individuals. Example Interpretation: Generation 1: ○ Individuals do not show the trait, so they are either homozygous dominant (AA) or heterozygous (Aa). Generation 2: ○ Several individuals are carriers (heterozygous, Aa), as indicated by half-shaded symbols. ○ No individual expresses the trait, meaning all individuals with the recessive allele are heterozygous and do not show the phenotype. Generation 3: ○ Affected individuals are born to carrier parents. ○ Both parents in this case are heterozygous (Aa), so there is a chance their offspring could inherit two recessive alleles (aa) and express the trait. Generation 4: ○ More affected individuals may appear due to the presence of the recessive allele passed through carriers. Key Points for Interpretation 1. Dominant Traits: ○ Generally appear in every generation. ○ At least one parent of an affected individual will also be affected. ○ Genotype of affected individuals: BB or Bb (heterozygous or homozygous dominant). 2. Recessive Traits: ○ Often skip generations. ○ Two unaffected parents can produce affected offspring if both are carriers. ○ Genotype of affected individuals: aa (homozygous recessive). 16.4-Non-Mendelian Inheritance Patterns 1. Patterns of Inheritance That Do Not Fit Mendel's Model Mendel’s model of inheritance is based on traits controlled by single genes with two alleles in a clear dominant-recessive relationship. However, many inheritance patterns do not follow these rules. Examples of Non-Mendelian Inheritance Patterns: Multiple Alleles: ○ Some genes have more than two alleles within a population, although an individual can only inherit two alleles. ○ Example: The ABO blood type system, which includes three alleles: IA, IB, and i. Codominance: ○ In codominance, both alleles in a heterozygote are fully expressed, resulting in a phenotype that shows both traits distinctly. ○ Example: Blood type AB, where both IA and IB alleles express A and B antigens on red blood cells. Incomplete Dominance: ○ In incomplete dominance, the heterozygous phenotype is intermediate between the two homozygous phenotypes. ○ Example: Flower color in certain plants, where crossing red and white flowers results in pink offspring. Though the pink flowers appear blended, alleles remain discrete, and future offspring can show both red and white colors. Polygenic Inheritance: ○ Traits controlled by multiple genes that contribute to a continuous range of phenotypes, often showing a bell-shaped distribution. ○ Example: Human height, where many genes impact overall height, and environmental factors can also influence the outcome. Pleiotropy:  ○ When one gene affects multiple traits or phenotypes. ○ Example: Cystic fibrosis, where a mutation in one gene causes various symptoms across the respiratory, digestive, urinary, and reproductive systems. Epistasis: ○ The expression of one gene is dependent on the presence of one or more 'modifier genes.' ○ Example: Coat color in Labrador Retrievers, where one gene determines pigment color (black or brown) while another gene determines the distribution of color (or lack thereof). Homozygous recessive alleles in the second gene cause the dog to have a yellow coat. Environmental Influence: ○ Some traits are influenced by environmental factors in addition to genetic factors. ○ Example: Coat color in Siamese cats is influenced by temperature, where warmer body parts are light-colored and cooler areas are darker due to heat-sensitive enzyme activity. 2. Genetic Inheritance Patterns of Blood Typing Blood typing, particularly the ABO blood group system, is a classic example of multiple alleles and codominance in genetics. ABO Blood Group System: Alleles Involved: ○ There are three alleles for the blood type gene: IA, IB, and i. ○ IA and IB are codominant, meaning they both fully express their traits when together. ○ i is recessive and does not produce any surface sugars on blood cells. Genotypes and Corresponding Blood Types: ○ Type A Blood: Genotypes: IAIA or IAi Only A-type sugars (antigens) are present on red blood cells. ○ Type B Blood: Genotypes: IBIB or IBi Only B-type sugars (antigens) are present on red blood cells. ○ Type AB Blood: Genotype: IAIB Both A and B antigens are present due to codominance, meaning both alleles contribute equally to the phenotype. ○ Type O Blood: Genotype: ii No surface sugars are present, as both alleles are recessive. Rh Factor: The Rh factor is another blood antigen, separate from the ABO system, controlled by two alleles: ○ R (dominant): Produces the Rh antigen. ○ r (recessive): Does not produce the Rh antigen. ○ Phenotypes: Rh+ individuals have at least one dominant R allele (RR or Rr). Rh- individuals are homozygous recessive (rr). These inheritance patterns show how blood types and other traits can exhibit multiple alleles, codominance, and dominance-recessive relationships, extending Mendel’s initial model of single-gene inheritance. Lesson 17 17.1-Sex Chromosomes Explain how sex chromosomes are important for sex determination Sex chromosomes (X and Y) are crucial for determining the sex of an individual in many species, including humans and fruit flies. Mechanism of Sex Determination: In Drosophila (Fruit Flies): ○ The number of X chromosomes determines sex. ○ Females have two X chromosomes (XX). ○ Males have one X chromosome and one Y chromosome (XY). ○ During fertilization, a female fly contributes an X chromosome, while a male can contribute either an X or Y chromosome, resulting in an XX (female) or XY (male) combination.  In Humans: ○ The presence of the Y chromosome determines sex. ○ Females are XX, and males are XY. ○ The Y chromosome contains a critical gene called SRY (Sex-determining Region Y), which triggers the development of male characteristics. Without the Y chromosome, the default development is female. ○ Like in fruit flies, human females contribute an X chromosome in their eggs, while males contribute either an X or a Y in their sperm. Sex Chromosome Contributions During Meiosis and Fertilization: During meiosis, females produce only X-containing eggs, while males produce sperm with either an X or a Y chromosome. X-bearing sperm fertilizing an egg results in a female (XX). Y-bearing sperm fertilizing an egg results in a male (XY). Describe the inheritance pattern of sex-linked traits Sex-linked traits are associated with genes located on the sex chromosomes, typically the X chromosome. These traits follow unique inheritance patterns because males and females have different combinations of X and Y chromosomes. X-Linked Inheritance Patterns: Example Trait in Drosophila: Eye color in fruit flies, studied by T.H. Morgan. ○ The gene responsible for eye color is located on the X chromosome. ○ Red eyes are dominant (X^R), while white eyes are recessive (X^w). Patterns in Crosses: 1. Crossing a White-Eyed Male (X^wY) with a Red-Eyed Female (X^RX^R): ○ F1 Generation: All offspring have red eyes, as the red allele (X^R) is dominant. Females are heterozygous (X^RX^w), carrying both red and white alleles. Males inherit only the red allele (X^R) from the mother, so they have red eyes (X^RY). 2. Crossing F1 Red-Eyed Offspring with Each Other: ○ F2 Generation: A portion of male offspring have white eyes (X^wY) because they inherit the white allele from the heterozygous mother. All white-eyed individuals are male because they lack a second X chromosome to mask the recessive allele. Female offspring can carry the white-eye allele if they are heterozygous (X^RX^w), but they will still show red eyes due to the dominant red allele. Key Points in X-Linked Inheritance: Males inherit only one X chromosome from their mother, meaning any recessive X-linked trait (e.g., white eyes in Drosophila, colorblindness in humans) will be expressed in males if present. Females have two X chromosomes, so they can be carriers of a recessive X-linked trait without expressing it if they are heterozygous. Female Expression of X-Linked Traits: If a female inherits two recessive alleles, one from each parent, she will express the recessive trait. Examples of X-Linked Traits in Humans: Hemophilia: A disorder affecting blood clotting. Colorblindness: Affects the ability to perceive colors and is more common in males due to the single X chromosome inheritance pattern. 17.2-Non-Mendelian Inheritance Explain the epigenetic mechanisms of dosage compensation and genomic imprinting Epigenetic mechanisms influence gene expression without altering the DNA sequence itself. Two important examples are dosage compensation and genomic imprinting. Dosage Compensation Dosage compensation is the process by which organisms equalize the expression levels of genes on the X chromosome between males and females, despite females having two X chromosomes and males having only one. Mechanism in Humans: In human females, dosage compensation occurs through X-chromosome inactivation, where one X chromosome in each cell is randomly inactivated. (especially important in heterozygous individuals) ○ The inactivated X chromosome condenses into a structure called a Barr body. ○ This process ensures that females, like males, have only one active X chromosome in each cell, balancing protein levels. ○ Genetic Mosaicism: Heterozygous females for X-linked traits are genetic mosaics, meaning different cells may express different alleles depending on which X chromosome is inactivated. ○ Example: Calico cats, where patchy fur coloration results from random X-inactivation of genes controlling fur color. Some patches show dark fur, and others show orange, based on which X chromosome is inactivated in each cell. Genomic Imprinting Genomic imprinting is another epigenetic mechanism where genes are "imprinted" (inactivated) based on parental origin. Mechanism: ○ Imprinting affects only certain genes in mammals and flowering plants. ○ Genes are inactivated depending on whether they were inherited from the mother or father, meaning that only one active copy of the gene is present in the offspring. ○ Maternal Imprinting: The maternal copy of the gene is inactivated. ○ Paternal Imprinting: The paternal copy of the gene is inactivated. ○ Implication: The offspring will express only the non-imprinted (active) copy of the gene, affecting how certain traits are inherited and expressed. 2. Inheritance Patterns of Chloroplast and Mitochondrial DNA The inheritance of DNA from mitochondria and chloroplasts is an example of non-Mendelian inheritance, as it does not follow the patterns seen with nuclear DNA. Mitochondrial DNA Inheritance (Maternal Inheritance) Mechanism: ○ Mitochondria are passed down from the mother, as the zygote receives its cytoplasm (including mitochondria) from the egg cell. ○ As a result, mitochondrial DNA is inherited only from the mother. ○ Pattern: All offspring of an affected mother will inherit the mitochondrial trait. Males with the trait cannot pass it on to their offspring, as mitochondria are not inherited from sperm. Example: ○ In human pedigrees, all affected individuals (both male and female) can trace the trait back to the maternal line. No affected males pass the mitochondrial trait to their children. Chloroplast DNA Inheritance in Plants Mechanism:  ○ Chloroplasts, like mitochondria, are typically inherited from only one parent, usually the mother in many plant species. ○ This maternal inheritance results in all offspring expressing chloroplast traits inherited from the mother. Species Variation: ○ In most plants, chloroplast inheritance follows a maternal pattern, but some species may exhibit paternal or biparental chloroplast inheritance. Summary Dosage Compensation: Balances gene expression on the X chromosome between sexes through mechanisms like X-chromosome inactivation in females (e.g., calico cats). Genomic Imprinting: Selective inactivation of genes based on parental origin, affecting gene expression in offspring. Mitochondrial and Chloroplast DNA: Follows maternal inheritance, as these organelles are typically passed from the mother only, leading to distinct inheritance patterns for mitochondrial traits in animals and chloroplast traits in plants. This guide highlights how epigenetic and organelle inheritance patterns extend beyond Mendel’s rules, showing the complexity of genetic inheritance. 17.3-Genetic Disorders Describe how mutations and nondisjunction can lead to human genetic disorders Mutations Mutations are changes in the DNA sequence that can lead to altered or nonfunctional proteins, impacting cellular functions and sometimes resulting in genetic disorders. Example: Sickle-Cell Anemia ○ Cause: A single base substitution in the gene encoding hemoglobin. ○ Effect: This mutation changes hemoglobin’s shape and function, impairing oxygen transport and altering the shape of red blood cells, resulting in the characteristic “sickle” shape. ○ Inheritance Pattern: Individuals homozygous for the sickle-cell allele exhibit symptoms, while heterozygous individuals generally do not. However, heterozygous individuals have increased resistance to malaria, providing a selective advantage in regions where malaria is common. Nondisjunction Nondisjunction is the failure of chromosomes (either homologous chromosomes or sister chromatids) to separate correctly during meiosis, leading to an abnormal number of chromosomes, known as aneuploidy. Types of Aneuploidy: ○ Monosomy: Loss of one chromosome (e.g., Turner Syndrome, with XO chromosome pattern). ○ Trisomy: Gain of one chromosome (e.g., Trisomy 21 or Down Syndrome, with an extra chromosome 21).  Examples of Disorders Caused by Nondisjunction: ○ Down Syndrome (Trisomy 21): Caused by an extra copy of chromosome 21. Results in physical and developmental challenges. Increased risk is associated with maternal age.  ○ Klinefelter Syndrome (XXY): Extra X chromosome in males, leading to male physical traits with some female characteristics and often reduced fertility. ○ Turner Syndrome (XO): Missing X chromosome in females, leading to underdeveloped sex organs and possible developmental delays. ○ Jacob’s Syndrome (XYY): Extra Y chromosome in males, typically resulting in increased stature with some potential for decreased cognitive abilities. Discuss the advantages and disadvantages of various types of genetic testing Genetic testing methods vary in terms of timing, accuracy, and potential risks. Common methods include amniocentesis, chorionic villi sampling, and pedigree analysis. Advantages of Genetic Testing Early Diagnosis and Risk Assessment: ○ Genetic counseling and testing can provide individuals and families with information on the likelihood of inheriting or passing on genetic disorders, helping in family planning.  ○ Identifies carriers of genetic mutations, such as those in the BRCA1 and BRCA2 genes, which are associated with an increased risk of breast cancer. In-Utero Genetic Testing: ○ Amniocentesis and Chorionic Villi Sampling (CVS) allow detection of genetic abnormalities in the fetus, providing early diagnosis and planning options for parents. Personalized Medicine: ○ Genetic testing helps in tailoring treatment options for individuals with specific genetic mutations, such as those at higher risk for certain cancers. Disadvantages of Genetic Testing Risk of Miscarriage: ○ Amniocentesis involves inserting a needle into the amniotic cavity to collect fetal cells, which carries a small risk of miscarriage. It is generally performed around the fourth month of pregnancy. ○ Chorionic Villi Sampling (CVS), performed as early as five weeks into pregnancy, carries a risk of injury to the embryo as well as possible miscarriage. Timing Limitations: ○ Amniocentesis cannot be performed until later in pregnancy, around the fourth month, limiting the time available for decision-making based on the results. ○ CVS can be done earlier but carries higher risks, presenting parents with difficult choices. Ethical Considerations and Psychological Impact: ○ The knowledge of carrying or potentially passing on genetic mutations can create emotional and ethical dilemmas for individuals and families. ○ Early detection of genetic disorders in a fetus raises complex ethical questions regarding pregnancy continuation or termination. Summary Mutations (e.g., sickle-cell anemia) and nondisjunction (e.g., Down Syndrome, Klinefelter Syndrome) lead to a variety of human genetic disorders through changes in DNA sequence or chromosome number. Genetic Testing methods like amniocentesis and CVS provide critical information but have associated risks, including miscarriage and ethical concerns. Genetic counseling helps families navigate the risks and benefits of these testing options. This guide highlights the complex relationship between genetic mutations, inheritance patterns, and the technologies available to predict or detect genetic disorders. 17.4-Genetic mapping Describe how genetic recombination and linkage analysis can be used to determine distance between genes Genetic Recombination: Crossing Over: During prophase I of meiosis, homologous chromosomes can exchange segments in a process called crossing over. This event occurs between genes on the same chromosome and produces recombinant chromosomes, which contain alleles from both parents. Recombinant Gametes: The gametes formed from these recombinant chromosomes are termed recombinant gametes, as they carry a mix of alleles not seen in either parent. Linkage and Gene Distance:  Linked Genes: Genes that are located close together on the same chromosome tend to be inherited together and are termed linked. Linked genes show fewer recombinant phenotypes because crossing over between closely spaced genes is less frequent. Measuring Recombinant Frequency: The frequency of recombinant phenotypes in offspring reflects the physical distance between genes: ○ Higher Recombination Frequency: When two genes are farther apart on a chromosome, there is a greater likelihood of crossing over occurring between them, leading to a higher percentage of recombinant phenotypes in the offspring. ○ Lower Recombination Frequency: When two genes are close together, crossing over is less likely, resulting in fewer recombinant phenotypes. Calculating Gene Distance: Test Cross Example: By performing a test cross and counting recombinant offspring, geneticists can determine the recombination frequency and thus the distance between genes. ○ Example: In Drosophila, a test cross between doubly heterozygous F1 progeny and homozygous recessive parental types produced 1,000 total offspring, of which 180 were recombinant. The recombination frequency is calculated as follows: Recombination Frequency = (Recombinant Offspring / Total Offspring) × 100 = (180 / 1,000) × 100 = 18% ○ Map Units and Centimorgans: Each 1% recombination frequency represents a distance of one map unit, or centimorgan (cM). Therefore, the distance between the two gene loci in this example is 18 centimorgans. Linkage Analysis and Gene Mapping: Linkage analysis, through test crosses and calculation of recombination frequencies, provides a method for mapping gene locations on chromosomes by establishing the relative distances between genes based on recombination rates. Identify current strategies used to construct genetic maps Genetic mapping has evolved with advances in molecular biology, allowing the use of molecular markers to map genes, especially in humans where traditional test crosses are not feasible. 1. Molecular Markers for Human Genetic Mapping Short Tandem Repeats (STRs): ○ STRs are short sequences of DNA (2-4 bases) that repeat in tandem and vary in number between individuals. ○ STRs serve as molecular landmarks on the genome because their repeat numbers are unique to individuals and heritable. Single-Nucleotide Polymorphisms (SNPs): ○ SNPs are single-base variations in the DNA sequence, which occur at specific loci in the genome.  ○ SNPs occur at relatively low frequency but are abundant enough to provide a dense coverage of markers across the genome, aiding in precise gene mapping. 2. Genetic Mapping Using Recombination and Molecular Markers Linkage Mapping with STRs and SNPs: ○ STRs and SNPs are used as reference points, or “markers,” to locate genes within the genome. ○ By examining co-inheritance patterns of these markers with known genetic traits, researchers can identify the approximate location of genes associated with those traits. Gene Disruption and Mutant Phenotypes (in Model Organisms): ○ In organisms like fruit flies, researchers can experimentally disrupt genes to observe resulting phenotypes and create genetic maps. This strategy is valuable in model systems but is not suitable for humans. 3. Benefits of Current Strategies in Human Genetic Mapping Improved Resolution: ○ The high density of STRs and SNPs across the human genome allows for more detailed mapping than was possible using phenotypic traits alone. Application in Genome-Wide Association Studies (GWAS): ○ SNPs and STRs are widely used in GWAS to identify genetic variants linked to complex traits and diseases, advancing personalized medicine and risk assessment. Summary Genetic Recombination and Linkage Analysis: Used to determine distances between genes by measuring recombination frequencies, providing relative positioning of genes on chromosomes. Current Mapping Strategies: STRs and SNPs have enhanced genetic mapping capabilities, especially in humans, allowing researchers to construct detailed genetic maps and perform association studies. This guide highlights how recombination and molecular markers are essential tools in modern genetics, allowing for precise genetic mapping and insights into the genetic basis of traits and diseases.