BIOL360-Genetics-Week6-LectureA-4-4 PDF

Week 6  Student Learning Outcomes 1. Analyze the molecular mechanisms of PCR, including the roles of primers, Taq polymerase, and thermal cycling, and evaluate how these components contribute to DNA amplification. 2. Design an experiment to use gel electrophoresis to separate an...

Week 6  Student Learning Outcomes 1. Analyze the molecular mechanisms of PCR, including the roles of primers, Taq polymerase, and thermal cycling, and evaluate how these components contribute to DNA amplification. 2. Design an experiment to use gel electrophoresis to separate and visualize DNA fragments and justify your choice of molecular markers based on their size and application. 3. Critically evaluate the impact of mutations in the CFTR gene on ion transport and cellular homeostasis and predict how these mutations lead to the clinical symptoms of cystic fibrosis. 4. Compare and contrast the steps involved in Sanger sequencing and next-generation sequencing and discuss how advances in sequencing technologies have transformed genetic research. 5. Interpret the results of a genetic linkage analysis for Huntington’s disease and explain how trinucleotide repeat expansions affect gene function and disease inheritance.  Student Learning Outcomes 6. Develop a molecular strategy to identify VNTRs (Variable Number Tandem Repeats) in human DNA using PCR and defend your approach for its use in forensic or paternity testing. 7. Assess the importance of lipid rafts in signal transduction and propose a hypothesis for how these microdomains could be involved in immune cell activation. 8. Synthesize knowledge of X-linked recessive and X-linked dominant inheritance patterns to predict the outcomes of genetic crosses involving sex-linked traits. 9. Examine the role of genetic markers, such as VNTRs and SNPs (single nucleotide polymorphisms), in tracking disease inheritance, and evaluate their effectiveness in genetic counseling and population studies. 10. Apply knowledge of genetic sequencing technologies to propose a method for identifying novel mutations in cancer genomes and explain how these mutations could drive tumorigenesis.  Introduction to Molecular Genetic Analysis Molecular genetic analysis relies on understanding DNA replication processes. PCR and dideoxynucleotide sequencing are essential techniques for studying genetic variation. PCR amplifies DNA segments, making billions of copies from small amounts of DNA. PCR has applications in evolutionary biology, forensic genetics, and medical research. Biomedical Example: PCR is used in forensic science to match DNA at crime scenes. New insights into the clinical and molecular spectrum of the MADD-related neurodevelopmental disorder  Polymerase Chain Reaction (PCR) Overview PCR is an automated DNA replication process performed in a test tube. The process includes denaturation, primer annealing, and primer extension steps. Taq polymerase, a heat-stable enzyme, synthesizes new DNA during PCR. Each cycle of PCR doubles the number of DNA copies, producing billions after 30 cycles. Biomedical Example: PCR is used in genetic testing for identifying hereditary diseases.  Steps of PCR Reaction Denaturation: The reaction mixture is heated, causing double-stranded DNA to denature. Primer Annealing: Single-stranded DNA primers bind to complementary sequences. Primer Extension: Taq polymerase synthesizes DNA by adding nucleotides to primers. Each cycle of PCR amplifies the target DNA exponentially. Case Study: PCR is used in evolutionary biology to compare ancient and modern DNA.  Question 1 Which enzyme is crucial for the DNA synthesis step in PCR? A) DNA polymerase I B) RNA polymerase C) Taq polymerase D) Reverse transcriptase E) Ligase  Question 1 Which enzyme is crucial for the DNA synthesis step in PCR? A) DNA polymerase I B) RNA polymerase C) Taq polymerase D) Reverse transcriptase E) Ligase Correct Answer: C) Taq polymerase Reasoning: Taq polymerase is heat-stable and can withstand the high temperatures used during PCR, making it ideal for DNA synthesis in this process.  Sanger DNA Sequencing Sanger sequencing involves in vitro DNA replication to read DNA nucleotide sequences. Dideoxynucleotides halt DNA synthesis, producing fragments of various lengths. Each dideoxynucleotide is labeled for detection during sequencing. Fragments are separated using gel electrophoresis, and the DNA sequence is inferred. Biomedical Example: Sanger sequencing was used in the Human Genome Project to determine the sequence of ~22,000 genes. Polymerase Chain Reaction (PCR) and Genetic Analysis PCR amplifies specific DNA regions, making genetic analysis possible even from trace DNA. VNTR markers are used in PCR to analyze genetic variation, such as in paternity tests. Gel electrophoresis separates amplified DNA fragments based on size. PCR and gel electrophoresis are used to compare alleles in codominant inheritance. Case Study: PCR-based VNTR analysis is crucial in forensic identification cases.  Question 2 What is the purpose of gel electrophoresis in genetic analysis? A) To amplify DNA sequences B) To separate DNA fragments by size C) To sequence DNA D) To denature proteins E) To visualize chromosomes  Question 2 What is the purpose of gel electrophoresis in genetic analysis? A) To amplify DNA sequences B) To separate DNA fragments by size C) To sequence DNA D) To denature proteins E) To visualize chromosomes Correct Answer: B) To separate DNA fragments by size Reasoning: Gel electrophoresis separates DNA based on fragment length, with smaller fragments moving faster through the gel.  PCR Applications in Research PCR is used to study DNA sequence similarity across species. Primers from one species can be used to amplify DNA in another species. Amplification occurs even if there are base-pair mismatches between primers and target DNA. Lowering the reaction temperature increases primer annealing during PCR. Biomedical Example: PCR has been applied in cancer research to identify mutations.  Gel Electrophoresis in PCR Analysis Gel electrophoresis separates DNA fragments based on size, allowing visualization. Molecular-weight markers help determine fragment sizes in base pairs. Smaller DNA fragments move faster through the gel compared to larger fragments. PCR products are highly concentrated and can be directly analyzed with gel electrophoresis. Biomedical Example: Gel electrophoresis is used to detect genetic disorders such as cystic fibrosis. Figure Out the Sequence DNA Replication and PCR Limitations PCR can amplify DNA sequences from as little as one molecule of DNA. The length of amplified products is generally limited to under 10 kb. Knowledge of target sequences is necessary to design primers for PCR. PCR has revolutionized the study of evolutionary genetics and personalized medicine. Biomedical Example: PCR is used in prenatal screening for genetic anomalies.  Variable Number Tandem Repeat (VNTR) Markers VNTR markers consist of repeating DNA sequences that vary between individuals. PCR amplifies VNTR regions for genetic comparison in forensic and paternity tests. VNTR markers are codominantly inherited, meaning both alleles are detectable. VNTRs create unique genetic profiles useful for identification purposes. Case Study: VNTR analysis is a core technique in criminal investigations for matching suspects to crime scenes. Genetic Applications of PCR and VNTR VNTR markers provide genetic evidence in legal and familial disputes. PCR amplifies VNTR alleles, which vary in repeat number between individuals. VNTRs are inherited in a codominant fashion, useful for parent-child comparisons. Amplified VNTR regions are visualized through gel electrophoresis. Biomedical Example: VNTR analysis was pivotal in identifying victims of natural disasters through DNA.  Dideoxynucleotide DNA Sequencing Dideoxy sequencing, developed by Frederick Sanger, uses chain-terminating nucleotides. Sequencing reactions terminate when a dideoxynucleotide is incorporated into DNA. Fragments are separated by size, and the DNA sequence is deduced from gel electrophoresis. This method forms the foundation of high- throughput DNA sequencing technologies. Biomedical Example: Sanger sequencing was employed to diagnose inherited genetic conditions.  Question 3 Which of the following statements best describes Variable Number Tandem Repeats (VNTRs)? A) VNTRs are inherited from only one parent B) VNTRs are single nucleotide polymorphisms C) VNTRs consist of repeating sequences of DNA that vary in number between individuals D) VNTRs cannot be amplified by PCR E) VNTRs are the result of chromosomal mutations  Question 3 Which of the following statements best describes Variable Number Tandem Repeats (VNTRs)? A) VNTRs are inherited from only one parent B) VNTRs are single nucleotide polymorphisms C) VNTRs consist of repeating sequences of DNA that vary in number between individuals D) VNTRs cannot be amplified by PCR E) VNTRs are the result of chromosomal mutations Correct Answer: C) VNTRs consist of repeating sequences of DNA that vary in number between individuals Reasoning: VNTRs are regions of DNA with repeating sequences that differ between individuals, making them useful for genetic profiling.  Dideoxynucleotide Sequencing and PCR Dideoxy sequencing utilizes dideoxynucleotides to produce DNA fragments of varying lengths. The absence of an OH group at the 3' carbon in dideoxynucleotides prevents chain elongation. Sequencing gels reveal DNA sequence by the relative position of each fragment. Modern sequencers automate this process for faster, large-scale genome sequencing. Biomedical Example: Sanger sequencing was essential in mapping the BRCA1 gene linked to breast cancer.  Next-Generation Sequencing (NGS) Technologies Next-generation sequencing (NGS) allows for massively parallel sequencing of DNA fragments. DNA is fragmented, and clusters of identical DNA strands are amplified for sequencing. Fluorescently labeled nucleotides identify the sequence as DNA is synthesized. NGS is faster and cheaper than Sanger sequencing, producing billions of base pairs in a day. Biomedical Example: NGS is used to sequence cancer genomes, identifying mutations in tumor DNA.  Illumina Sequencing Technology Illumina sequencing involves attaching DNA fragments to a flow cell for amplification. Each fragment is sequenced by synthesis using fluorescently labeled nucleotides. A laser excites the nucleotides, and the emitted light identifies the added base. Millions of DNA fragments are sequenced simultaneously, increasing throughput. Biomedical Example: Illumina sequencing is used to detect rare genetic variants in population studies.  Third-Generation Sequencing (TGS) Technologies Third-generation sequencing reads long stretches of DNA in real time. TGS does not require DNA amplification, reducing errors caused by PCR. PacBio and Oxford Nanopore are key third- generation sequencing technologies. Nanopores detect electrical signals generated by nucleotides passing through. Biomedical Example: Third-generation sequencing is employed in real-time pathogen detection in clinical settings.  Question 4 Which of the following are characteristics of X-linked recessive inheritance? A) Males are more affected than females B) Females are always affected if they inherit one copy of the mutant allele C) The gene is located on the Y chromosome D) Carrier females can pass the allele to sons E) Daughters of affected males will always inherit the mutant allele F) The trait skips generations G) Affected males inherit the trait from their fathers H) Carrier females have one normal and one mutant allele  Question 4 Which of the following are characteristics of X-linked recessive inheritance? A) Males are more affected than females B) Females are always affected if they inherit one copy of the mutant allele C) The gene is located on the Y chromosome D) Carrier females can pass the allele to sons E) Daughters of affected males will always inherit the mutant allele F) The trait skips generations G) Affected males inherit the trait from their fathers H) Carrier females have one normal and one mutant allele Correct Answers: A), D), E), F), H) Reasoning: X-linked recessive traits are more common in males, and carrier females can pass the allele to their sons. Daughters of affected males will inherit the mutant allele, and the trait often skips generations. Carrier females have one normal and one mutant allele.  Applications of DNA Sequencing in Medicine DNA sequencing technologies enable the identification of genetic mutations linked to diseases. Next-generation sequencing (NGS) has revolutionized personalized medicine by allowing genome-wide scans for disease predisposition. DNA sequencing aids in diagnosing inherited disorders, guiding treatment decisions based on genetic profiles. Whole-genome sequencing is increasingly used in oncology to identify cancer-causing mutations. Case Study: Whole-genome sequencing was used to diagnose a rare genetic disorder in a child, allowing for targeted treatment.  Application of PCR in Medical Genetics A family with a history of cystic fibrosis seeks genetic testing for their newborn. PCR is used to amplify and detect the CFTR gene mutation responsible for cystic fibrosis. The results show that the newborn carries two copies of the mutated allele, confirming the diagnosis. Genetic counseling is provided to the parents to discuss treatment options and future reproductive decisions. Application: This case study demonstrates how PCR can be used in pre-symptomatic diagnosis of genetic disorders.  Question 5 Which of the following are true regarding Huntington's disease and genetic linkage? A) It is an autosomal dominant disorder B) It results from a CAG repeat expansion in the HTT gene C) Genetic linkage analysis can identify carriers of the mutation D) The disease is X-linked recessive E) Genetic counseling is recommended for individuals at risk F) The gene for Huntington's is located on chromosome 21 G) Disease onset is correlated with the number of CAG repeats H) The mutation affects males more than females  Question 5 Which of the following are true regarding Huntington's disease and genetic linkage? A) It is an autosomal dominant disorder B) It results from a CAG repeat expansion in the HTT gene C) Genetic linkage analysis can identify carriers of the mutation D) The disease is X-linked recessive E) Genetic counseling is recommended for individuals at risk F) The gene for Huntington's is located on chromosome 21 G) Disease onset is correlated with the number of CAG repeats H) The mutation affects males more than females Correct Answers: A), B), C), E), G) Reasoning: Huntington's disease is autosomal dominant, caused by a CAG repeat expansion. Genetic linkage analysis helps identify mutation carriers, and genetic counseling is important for families at risk. Disease onset is linked to the number of CAG repeats.  Genetic Linkage in Disease Research Researchers study a family with a history of Huntington's disease, an autosomal dominant disorder. Genetic linkage analysis is used to map the location of the mutation on chromosome. The study reveals that the CAG repeat expansion in the HTT gene correlates with the severity of the disease. This information helps predict disease onset and severity in family members carrying the mutation. Application: Genetic linkage is a powerful tool in identifying disease-causing genes and understanding their inheritance patterns. Quiz 12  Question 1 A research team is using PCR to amplify a DNA sequence from a small biological sample found at a crime scene. After amplification, they run the PCR products on an agarose gel to separate the DNA fragments by size. However, the team notices that no bands appear on the gel after staining. What is the most likely reason why the research team failed to see any DNA bands on the gel? A) They did not add DNA primers to the PCR reaction. B) They used too much DNA in the PCR reaction. C) They forgot to heat the sample during the denaturation phase. D) The PCR reaction was run for too many cycles. E) They used the wrong buffer in the gel electrophoresis chamber.  Question 1 A research team is using PCR to amplify a DNA sequence from a small biological sample found at a crime scene. After amplification, they run the PCR products on an agarose gel to separate the DNA fragments by size. However, the team notices that no bands appear on the gel after staining. What is the most likely reason why the research team failed to see any DNA bands on the gel? A) They did not add DNA primers to the PCR reaction. B) They used too much DNA in the PCR reaction. C) They forgot to heat the sample during the denaturation phase. D) The PCR reaction was run for too many cycles. E) They used the wrong buffer in the gel electrophoresis chamber. Correct Answer: A) They did not add DNA primers to the PCR reaction. Reasoning: Without primers, DNA cannot be amplified during PCR because Taq polymerase requires a primer to initiate DNA synthesis. Other options would either still produce some detectable DNA or result in visible bands under different conditions.  Question 2 A geneticist is studying a population of plants and finds that they have different alleles for a gene that controls flower color. The gene has a VNTR (Variable Number Tandem Repeat) region that can have anywhere from 5 to 15 repeats. PCR is used to amplify the VNTR region, and the alleles are analyzed using gel electrophoresis. What can the geneticist infer from the banding patterns observed on the gel? A) The plants with fewer repeats have larger alleles. B) The plants with more repeats have smaller alleles. C) The number of repeats correlates with the size of the PCR product. D) The PCR products with the same number of repeats will have different sizes. E) VNTR regions are not visible on agarose gels.  Question 2 A geneticist is studying a population of plants and finds that they have different alleles for a gene that controls flower color. The gene has a VNTR (Variable Number Tandem Repeat) region that can have anywhere from 5 to 15 repeats. PCR is used to amplify the VNTR region, and the alleles are analyzed using gel electrophoresis. What can the geneticist infer from the banding patterns observed on the gel? A) The plants with fewer repeats have larger alleles. B) The plants with more repeats have smaller alleles. C) The number of repeats correlates with the size of the PCR product. D) The PCR products with the same number of repeats will have different sizes. E) VNTR regions are not visible on agarose gels. Correct Answer: C) The number of repeats correlates with the size of the PCR product. Reasoning: The more repeats present in the VNTR region, the longer the PCR product will be, resulting in a larger band on the gel. This allows the geneticist to distinguish between alleles based on their size.  Quiz 3 A young boy is diagnosed with hemophilia, a condition characterized by poor blood clotting. His mother is a carrier of the hemophilia gene, but his father does not have the condition. What is the most likely inheritance pattern of hemophilia, based on this family’s history? A) Autosomal recessive B) Autosomal dominant C) Y-linked D) X-linked recessive E) X-linked dominant  Quiz 3 A young boy is diagnosed with hemophilia, a condition characterized by poor blood clotting. His mother is a carrier of the hemophilia gene, but his father does not have the condition. What is the most likely inheritance pattern of hemophilia, based on this family’s history? A) Autosomal recessive B) Autosomal dominant C) Y-linked D) X-linked recessive E) X-linked dominant Correct Answer: D) X-linked recessive Reasoning: Hemophilia is an X-linked recessive condition. The mother, as a carrier, has one affected X chromosome and one unaffected X chromosome. The boy inherits the affected X from his mother, while his Y chromosome comes from his unaffected father.  Quiz 4 Which of the following are applications of next- generation sequencing (NGS) in biomedical research? A) Identifying mutations in cancer cells B) Diagnosing rare genetic diseases C) Detecting chromosomal rearrangements D) Sequencing microbial genomes in infectious diseases E) Visualizing proteins within cells F) Identifying gene expression changes through RNA sequencing G) Detecting large structural variations in the genome H) Mapping the brain’s neural networks  Quiz 4 Which of the following are applications of next-generation sequencing (NGS) in biomedical research? A) Identifying mutations in cancer cells B) Diagnosing rare genetic diseases C) Detecting chromosomal rearrangements D) Sequencing microbial genomes in infectious diseases E) Visualizing proteins within cells F) Identifying gene expression changes through RNA sequencing G) Detecting large structural variations in the genome H) Mapping the brain’s neural networks Correct Answers: A), B), C), D), F), G) Reasoning: Next-generation sequencing is used to identify mutations in cancer, diagnose genetic diseases, detect chromosomal rearrangements, sequence microbial genomes, and identify gene expression changes (RNA sequencing). It can also detect structural variations like deletions or duplications. However, NGS is not typically used for protein visualization or mapping brain networks.  Quiz 5 A research lab is studying the effects of a mutation in the CFTR gene, which causes cystic fibrosis. The team uses PCR to amplify the CFTR gene from DNA samples of patients and unaffected individuals. The mutation affects chloride ion transport, leading to thick mucus in the lungs and digestive system. Researchers want to understand how this mutation affects protein function and its impact on health. 1. Explain the role of the CFTR gene in normal cellular function and why mutations in this gene lead to cystic fibrosis. 2. Describe how PCR is used to detect the CFTR gene mutation in individuals. 3. Discuss the potential treatments for cystic fibrosis and how understanding the genetic mutation informs therapeutic approaches.  Quiz 5 Correct Answer 1: The CFTR gene codes for a chloride ion channel, which regulates the transport of chloride ions across epithelial cell membranes. In cystic fibrosis, mutations in CFTR result in dysfunctional ion transport, leading to thick, sticky mucus in the lungs, pancreas, and other organs. This impairs breathing and digestion. Correct Answer 2: PCR amplifies specific regions of the CFTR gene from a patient’s DNA sample. Primers are designed to flank the mutation site, allowing researchers to detect whether the mutation is present by analyzing the PCR product through gel electrophoresis or DNA sequencing. Correct Answer 3: Treatments include medications that thin mucus, improve lung function, and manage infections. Gene therapy, which aims to correct the CFTR mutation, and CFTR modulator drugs that improve the function of the defective protein, are more targeted approaches. Understanding the specific mutation allows for personalized treatment strategies. Assignment 12 Problem 1: PCR Amplification of a Genetic Marker in a Forensic Case A crime scene investigator has found a small biological sample at the scene of a robbery. They decide to amplify a specific VNTR (Variable Number Tandem Repeat) region using PCR. The VNTR region is known to vary significantly between individuals, making it an ideal genetic marker for identification purposes. The investigator wants to compare the amplified DNA with the DNA profiles of several suspects. You are tasked with analyzing the PCR results to determine whether one of the suspects can be linked to the crime scene. The following questions will guide your analysis of PCR amplification and gel electrophoresis results. 1. What is the purpose of amplifying the VNTR region using PCR in this case? Hint: Think about how the VNTRs differ between individuals and how PCR can reveal this. 2. After PCR, how will gel electrophoresis help you determine if the suspect’s DNA matches the DNA from the crime scene? Hint: Consider how DNA fragments of different sizes move through the gel. 3. What could be some limitations of using VNTR analysis for this forensic case? Hint: Think about potential issues with DNA degradation, contamination, or the statistical likelihood of different individuals having similar VNTR profiles. Problem 1: PCR Amplification of a Genetic Marker in a Forensic Case Answer 1: The purpose of amplifying the VNTR region is to generate a DNA profile that can be compared across different individuals. VNTRs vary in length due to differences in repeat number, making them unique to each person. Answer 2: Gel electrophoresis will separate the amplified DNA fragments by size. If the suspect’s DNA has a similar VNTR profile (same number of repeats) as the crime scene sample, the DNA fragments will produce bands at the same positions on the gel. Answer 3: Limitations include degraded DNA at the crime scene, contamination of samples, or the possibility that different individuals may coincidentally have similar VNTR profiles, making conclusive identification more difficult. Problem 2: The Role of CFTR Gene Mutations in Cystic Fibrosis Cystic fibrosis (CF) is a genetic disorder caused by mutations in the CFTR gene, which encodes a chloride ion channel. In affected individuals, mutations in CFTR lead to faulty chloride transport across epithelial cells, causing thick mucus buildup in the lungs and other organs. A research lab is studying the effects of specific CFTR mutations using PCR to detect the presence of mutations in patient DNA. Your task is to understand how CFTR mutations are detected and analyze how these mutations affect protein function and contribute to the disease. 1. How would PCR be used to detect a mutation in the CFTR gene in a patient’s DNA? Hint: Consider what PCR amplifies and how the resulting DNA can be analyzed. 2. How does the CFTR mutation affect chloride ion transport, and how does this lead to the symptoms of cystic fibrosis? Hint: Think about how ion channels function in normal cells versus in cells with the CFTR mutation. 3. Why is early detection of CFTR mutations important for the management of cystic fibrosis? Hint: Consider the potential benefits of early treatment for genetic disorders. Problem 2: The Role of CFTR Gene Mutations in Cystic Fibrosis Answer 1: PCR can be used to amplify the region of the CFTR gene where the mutation occurs. Primers designed to flank the mutation site would allow researchers to amplify the DNA, and subsequent sequencing or gel electrophoresis would reveal if the mutation is present. Answer 2: The mutation disrupts the function of the CFTR chloride channel, preventing chloride ions from exiting epithelial cells. This leads to water being retained within the cells, resulting in thick mucus in the lungs and digestive organs, which causes the characteristic symptoms of CF. Answer 3: Early detection allows for early intervention, including therapies that can reduce mucus buildup and improve lung function, potentially slowing disease progression and improving quality of life. Problem 3: Genetic Linkage Analysis in Huntington’s Disease Huntington’s disease (HD) is an autosomal dominant neurodegenerative disorder caused by a CAG repeat expansion in the HTT gene on chromosome 4. The number of repeats correlates with the age of onset and severity of the disease. Genetic linkage analysis is used to map the gene and predict whether individuals in affected families will develop HD. You are a genetic counselor working with a family affected by Huntington’s disease. Your job is to analyze the genetic data and provide information about the inheritance pattern and likelihood of disease onset. 1. What is the role of genetic linkage analysis in understanding Huntington’s disease? 1. Hint: Consider how linkage analysis helps map the location of the disease-causing gene. 2. How does the number of CAG repeats in the HTT gene correlate with the severity and onset of Huntington’s disease? 1. Hint: Think about how trinucleotide repeat expansions affect protein function. 3. What ethical considerations should be taken into account when counseling families about genetic testing for Huntington’s disease? 1. Hint: Consider the implications of knowing one’s genetic risk for an untreatable, late-onset disease. Problem 3: Genetic Linkage Analysis in Huntington’s Disease Answer 1: Genetic linkage analysis helps identify the location of the HTT gene by examining the co-segregation of the disease with genetic markers on chromosome 4. This allows researchers to predict who in a family may carry the expanded CAG repeats. Answer 2: The more CAG repeats in the HTT gene, the earlier the onset of Huntington’s disease and the more severe the symptoms. This is due to the abnormal accumulation of the mutant huntingtin protein, which leads to neuronal dysfunction. Answer 3: Ethical considerations include the psychological impact of knowing one’s risk, the potential for discrimination based on genetic information, and respecting individuals’ autonomy in deciding whether to undergo testing. It is important to provide non-directive counseling and support throughout the decision-making process. Week 6  Basic Elements of Pedigree Analysis  Autosomal Inheritance Patterns Autosomal inheritance involves genes located on non- sex chromosomes. Autosomal recessive traits require two copies of the allele to be expressed. Autosomal dominant traits require only one copy of the allele to be expressed. Both males and females are equally likely to inherit autosomal traits, as these traits are not linked to sex chromosomes. Biomedical Example: Cystic fibrosis is an autosomal recessive disorder, requiring both parents to carry the defective allele for the trait to be passed on.  Autosomal Dominant Inheritance Autosomal dominant inheritance occurs when only one copy of a dominant allele is necessary for an individual to express the associated phenotype. Key Characteristics: Both males and females are equally affected, as autosomal genes are located on non-sex chromosomes. An affected individual typically has one affected parent, as only one allele is required to pass on the trait. Autosomal dominant disorders do not skip generations, and every affected person has at least one affected parent. Example: Huntington’s Disease: A neurodegenerative disorder that typically manifests in mid-adulthood. An individual with one copy of the mutated gene (Hh) will develop the disease. Punnett Square Example: In a cross between a heterozygous affected individual (Hh) and an unaffected individual (hh), there is a 50% chance of having an affected child (Hh) and a 50% chance of having an unaffected child (hh). Pedigree Characteristics: The trait appears in every generation. Affected individuals can pass the trait to both sons and daughters. Unaffected individuals do not pass on the trait. Autosomal Dominant Inheritance  Question 1 What is the probability of person IV-2 being affected?  Question 1 What is the probability of person IV-2 being affected? III-1 (mom): aa III-2 (dad): A- (i.e., Aa or AA) Cross: aa x A- For: Expression of A- phenotype At least 50% chance when phenotype is unknown and 100% chance when phenotype is observed.  Autosomal Recessive Inheritance Autosomal recessive inheritance occurs when two copies of a recessive allele are necessary for an individual to express the associated phenotype. Key Characteristics: Both males and females are equally affected, as autosomal genes are located on non-sex chromosomes. Affected individuals usually have parents who are both carriers (heterozygous), but do not exhibit the trait themselves. Autosomal recessive disorders can skip generations, as the trait can be hidden in carriers. Example: Cystic Fibrosis: A genetic disorder that affects the lungs and digestive system. An individual must inherit two copies of the recessive allele (aa) to express the disease, while carriers (Aa) are unaffected. Punnett Square Example: In a cross between two heterozygous carriers (Aa), there is a 25% chance of having an affected child (aa), a 50% chance of having a carrier child (Aa), and a 25% chance of having an unaffected child (AA). Pedigree Characteristics: The trait may skip generations. Affected individuals often have unaffected (carrier) parents. Both male and female offspring can be affected. Autosomal Recessive Inheritance  Question 2 What is the probability of person V-11 being affected?  Question 2 What is the probability of person V-11 being affected? IV-10 (mom): aa IV-11 (dad): AA Cross: aa x AA For: Expression of aa phenotype 0% chance because aa is rare and there is no information linking dad to the recessive allele. (a) Parents who have a dominant phenotype and produce a child with a recessive phenotype (dd) must be heterozygous (Dd). The siblings with the dominant phenotype are D–. (b) Punnett square for the cross of heterozygous parents. Sex-Linked Inheritance  Y-Linked Inheritance Y-linked inheritance occurs when the gene causing the trait or disorder is located on the Y chromosome, meaning only males can inherit and pass on the trait. Key Characteristics: Only males are affected, as only males possess a Y chromosome (XY). The trait is passed directly from father to son, and all male offspring of an affected father will inherit the trait. Females are never affected because they do not have a Y chromosome. Example: Y-Linked Hearing Loss: A rare genetic condition associated with genes on the Y chromosome that can cause male- specific traits or disorders. Affected fathers pass the condition to all of their sons, but none of their daughters. Punnett Square Example: In a cross involving an affected male (XY*) and an unaffected female (XX), all daughters (XX) will be unaffected, and all sons (XY*) will inherit the Y-linked trait from their father. Pedigree Characteristics: Only males are affected. The trait does not skip generations if affected males have sons. There is a direct line of inheritance from father to son. Y-Linked Inheritance  Question 3 What is the probability of person 11 being affected?  Question 3 What is the probability of person 11 being affected? 3 (mom): XX 4 (dad): XYD Cross: XX x XYD For: Expression of YD phenotype 100% chance of person 11 being affected because they inherited the only affected Y chromosome from dad.  X-Linked Recessive Inheritance X-linked recessive inheritance occurs when a recessive gene on the X chromosome causes a trait or disorder, primarily affecting males, who have only one X chromosome (XY). Key Characteristics: Males are more commonly affected because they only need one copy of the recessive allele on the X chromosome to express the trait (since they only have one X chromosome). Females must inherit two copies of the recessive allele (one from each parent) to express the trait because they have two X chromosomes (XX). Carrier females (heterozygous, XRRX) can pass the trait to their sons, who will inherit the disorder if they receive the affected X chromosome. Example: Hemophilia: A well-known X-linked recessive disorder that impairs the body’s ability to clot blood. Males who inherit the affected X chromosome (XhhY) from their carrier mother will express the disorder. Punnett Square Example: In a cross between a carrier mother (X*X) and an unaffected father (XY): 50% chance of having an unaffected daughter (XX) or a carrier daughter (X*X). 50% chance of having an unaffected son (XY) or an affected son (X*Y). Pedigree Characteristics: The trait is more common in males. Carrier females can pass the trait to sons. There is no male-to-male transmission of the disorder, but affected males can pass the carrier status to their daughters. Case Study: Queen Victoria was a carrier of hemophilia, which spread through European royal families, affecting many male descendants. X-Linked Recessive ceInheritance  Question 4 What is the probability of Leopold in generation IV being affected?  Question 4 What is the probability of Leopold in generation IV being affected? III-Beatrice (mom): XXh III-Beatrice’s Husband (dad): XY Cross: XXh x XY For: Expression of Xh in Leopold 50% chance when phenotype is unknown and 100% chance when phenotype is observed. X-Linked Recessive Inheritance  X-Linked Dominant Inheritance Definition: X-linked dominant inheritance occurs when a dominant gene on the X chromosome causes a trait or disorder. The trait is expressed in both males and females, but it tends to be more severe in males. Key Characteristics: Females only need one copy of the dominant allele to express the trait (XDDX). Males who inherit the dominant allele on the X chromosome (XDY) will always express the trait. Affected males will pass the trait to all of their daughters (as they inherit the father’s X chromosome), but none of their sons (as sons inherit the Y chromosome from their father). Example: Fragile X Syndrome: An X-linked dominant disorder that is the most common inherited cause of intellectual disability. Affected females may exhibit a milder form of the disorder due to the presence of one normal X chromosome, while affected males often experience more severe symptoms. Punnett Square Example: In a cross between an affected father (XDY) and an unaffected mother (XX): 100% chance of affected daughters (XDX), as they inherit the dominant allele from the father. 0% chance of affected sons (XY), as sons inherit the father’s Y chromosome, not the X chromosome. Pedigree Characteristics: The trait is expressed in both males and females. Affected males pass the trait to all daughters but no sons. Affected females have a 50% chance of passing the trait to both sons and daughters. Biomedical Example: Fragile X syndrome, an X-linked dominant disorder, is the most common inherited cause of intellectual disability. X-Linked Dominant Inheritance  Question 5 What is the probability of person III-3 being affected?  Question 5 What is the probability of person III-3 being affected? II-3 (mom): XXH II-4 (dad): XY Cross: XXH x XY For: Expression of XH in son 50% chance before phenotype in known and 100% chance once phenotype is observed. Quiz 13  Question 1 In a pedigree where an autosomal dominant disorder is being passed through generations, a father in Generation II is affected, while the mother is unaffected. Their son in Generation III is unaffected and marries an unaffected partner. What is the probability that their child in Generation IV will be affected if we assume the father in Generation II is heterozygous? A) 0% B) 25% C) 50% D) 75% E) 100%  Question 1 In a pedigree where an autosomal dominant disorder is being passed through generations, a father in Generation II is affected, while the mother is unaffected. Their son in Generation III is unaffected and marries an unaffected partner. What is the probability that their child in Generation IV will be affected if we assume the father in Generation II is heterozygous? A) 0% B) 25% C) 50% D) 75% E) 100% Correct Answer: A) 0% Reasoning: Since the son in Generation III is unaffected, he must be homozygous recessive (aa). As a result, no matter whom he marries, he cannot pass the dominant allele, meaning his children will not inherit the disorder.  Question 2 In a family pedigree, the mother is a carrier of an X-linked recessive disorder, and the father is unaffected. They have one daughter who is unaffected and marries an unaffected man. What is the probability that their son will inherit the X-linked recessive disorder? A) 0% B) 25% C) 50% D) 75% E) 100%  Question 2 In a family pedigree, the mother is a carrier of an X-linked recessive disorder, and the father is unaffected. They have one daughter who is unaffected and marries an unaffected man. What is the probability that their son will inherit the X-linked recessive disorder? A) 0% B) 25% C) 50% D) 75% E) 100% Correct Answer: A) 0% Reasoning: Since the daughter is unaffected, she can either be homozygous normal (XX) or a carrier (XrX). In either case, the father of her son is unaffected (XY), so the son cannot inherit the disorder unless the mother passes on a recessive allele. If the daughter is not a carrier, her son will not inherit the disorder.  Question 3 In a pedigree involving a Y-linked disorder, an affected father has several children with an unaffected partner. What is the probability that their son in Generation IV will inherit the Y-linked disorder? A) 0% B) 25% C) 50% D) 75% E) 100%  Question 3 In a pedigree involving a Y-linked disorder, an affected father has several children with an unaffected partner. What is the probability that their son in Generation IV will inherit the Y-linked disorder? A) 0% B) 25% C) 50% D) 75% E) 100% Correct Answer: E) 100% Reasoning: Y-linked traits are passed from father to son via the Y chromosome. Since only males inherit the Y chromosome from their father, all sons of an affected male will inherit the Y-linked disorder.  Question 4 In a pedigree, both parents in Generation II are unaffected but have an affected child in Generation III. The affected individual in Generation III marries an unaffected partner. What is the probability that their child in Generation IV will be affected by the autosomal recessive disorder if the unaffected partner is not a carrier? A) 0% B) 25% C) 50% D) 75% E) 100%  Question 4 In a pedigree, both parents in Generation II are unaffected but have an affected child in Generation III. The affected individual in Generation III marries an unaffected partner. What is the probability that their child in Generation IV will be affected by the autosomal recessive disorder if the unaffected partner is not a carrier? A) 0% B) 25% C) 50% D) 75% E) 100% Correct Answer: A) 0% Reasoning: Since the unaffected partner in Generation III is not a carrier (homozygous dominant, AA), there is no chance of the child inheriting two recessive alleles (aa). Therefore, the child cannot be affected.  Question 5 In a pedigree, a father is affected by an X-linked dominant disorder, while the mother is unaffected. They have a daughter in Generation III. What is the probability that the daughter will be affected? A) 0% B) 25% C) 50% D) 75% E) 100%  Question 5 In a pedigree, a father is affected by an X-linked dominant disorder, while the mother is unaffected. They have a daughter in Generation III. What is the probability that the daughter will be affected? A) 0% B) 25% C) 50% D) 75% E) 100% Correct Answer: E) 100% Reasoning: Fathers pass their X chromosome to all daughters. Since the father is affected by an X-linked dominant disorder, he will pass the dominant allele to his daughter, ensuring that she will inherit the disorder. Assignment 13 Pedigree Construction Assignment Title: Constructing Pedigrees and Probability Analysis for Genetic Diseases Total Points: 10 Due Date: Sunday by 11:59 PM Assignment Description: In this assignment, you will create pedigrees for two different genetic diseases, allowing you to practice reverse pedigree analysis and understand inheritance patterns more deeply. You will act as a genetic counselor by constructing your own pedigrees and explaining the probabilities for affected and carrier individuals. This exercise will help you solidify your understanding of how different inheritance patterns work. Assignment Requirements: 1. Choose Two Genetic Diseases: One must follow autosomal inheritance (either dominant or recessive). The other must follow X-linked inheritance (either dominant or recessive). Ensure that one of the diseases follows a dominant inheritance pattern, and the other follows a recessiveinheritance pattern. 2. Construct Two Pedigrees: For each chosen disease, construct a pedigree spanning four generations. The pedigree should clearly display individuals who are affected, carriers (if applicable), and unaffected. Indicate the genotypes for each individual where possible. 3. Probability Analysis for 4th Generation: For one affected individual and one carrier (if applicable) in the 4th generation of each pedigree, calculate the probability that they will inherit the disease. Provide a clear explanation and justification for the probabilities you have calculated, using the rules of inheritance (dominant/recessive, autosomal/X-linked). 4. Format: Submit your assignment as a PDF or Word document. Include a visual diagram of the pedigrees for both diseases. Provide your written explanation below each pedigree, focusing on the probabilities for the affected and carrier individuals in the 4th generation.