Cell Signaling Past Paper PDF

Flipped classroom. Cell signaling Part A. Six MCQ (20 mins working in pairs, 30 mins discussion) 1. Cell signaling can be divided into four major categories depending on how far the signaling molecule is from the receptor and where the receptor is located....

Flipped classroom. Cell signaling Part A. Six MCQ (20 mins working in pairs, 30 mins discussion) 1. Cell signaling can be divided into four major categories depending on how far the signaling molecule is from the receptor and where the receptor is located. What is paracrine signaling? a. A cell targets a distant cell to die b. A cell targets a distant cell to survive c. A cell targets a nearby cell d. A cell targets itself e. A cell targets a nearby cell by direct contact 2. Which chemical modification can serve as a “molecular switch”? a. Acetylation b. Hydroxylation c. Methylation d. Oxidization e. Phosphorylation 3. Which of the following is a major consequence of activation of phospholipase C (PLC) by the Gq trimeric GTPase? a. Elevation of intracellular cAMP levels, leading to the activation of protein kinase A b. Elevation of PIP3 levels in the plasma membrane, leading to the activation of protein kinase B c. Elevation of intracellular Ca2+ levels, leading to the activation of protein kinase C d. Elevation of IP3 in the plasma membrane, leading to the activation of protein kinase D e. Elevation of intracellular cGMP levels, leading to the activation of protein kinase G 4. What two cell-surface receptors are represented in the two simplified diagrams below (from left to right)? a. TGFβ receptor and TNF receptor b. Cytokine receptor and TNF receptor c. TNF receptor and TGFβ receptor d. TGFβ receptor and cytokine receptor e. Cytokine receptor and TGFβ receptor 5. Growth factors stimulate RTK receptors. One of the most important RTKs is EGFR that is involved in cancer. Which GTPase is activated by EGFR? a. Rab b. Ran c. Ras d. Rho e. Rik 6. Alzheimer's disease is thought to be caused by the abnormal build-up of proteins in and around brain cells. It is believed that these abnormal protein deposits are related with the Delta-Notch pathway. How does the expression of Delta on the surface of a cell activate the expression of certain genes in the nucleus of its neighboring cell? a. Delta binding activates Notch, which activates a transcriptional activator through the JAK–STAT pathway. b. Delta binding leads to the proteolytic cleavage of Notch and inhibition of its activity as a transcriptional repressor c. Delta binding leads to the stabilization of a cytoskeleton- associated transcriptional activator. d. Delta binding releases the extracellular part of Notch, which serves as a ligand for the nearby cells. e. Delta binding releases the intracellular tail of Notch, which enters the nucleus and activates transcription.  Part B. Four critical thinking questions (20 mins working in pairs, 30 mins discussion) 7. Nuclear receptors are regulated through the binding of intracellular proteins. What role do intracellular receptors play in relation to steroid hormones, and more specifically, the regulation of gene expression? Describe your response in 2–3 sentences. 8. Describe a signaling pathway that results to the increase of the intracellular Ca2+. 9. Outline the main events that they take place in phototransduction in response to a flash of light. 10. The cell-mediated immune response in multicellular organisms is dependent on the activation and interaction of proteins, known as cytokines, and their cytokine receptors. These receptors are phosphorylated and influence downstream transcription regulation in the cell. What are the main steps of this pathway? Answers 1. Answer: C 2. Answer: E 3. Answer: C 4. Answer: E 5. Answer: C 6. Answer: E 7. Answer: Nuclear receptors bind to specific DNA sequences upstream of the gene that is regulated by the specific ligand. Meanwhile, steroid hormones, such as cortisol, Vitamin D, ecdysone, and thyroid hormones bind to intracellular nuclear receptors. The binding of the hormone (ligand) triggers a conformational change in the receptor, allowing the receptor to interact with regulatory proteins. These interactions then result in the expression of genes or the inhibition of gene expression. 8. Answer: The activated GPCR stimulates the plasma-membrane-bound phospholipase C-β (PLCβ) via a G protein called Gq. The α subunit and βγ complex of Gq are both involved in this activation. Two second messengers are produced when PI(4,5)P2 is hydrolyzed by activated PLCβ. Inositol 1,4,5- trisphosphate (IP3) diffuses through the cytosol and releases Ca2+ from the ER by binding to and opening IP3-gated Ca2+- release channels (IP3 receptors) in the ER membrane. 9. Answer: The product of light activation, initiates the visual phototransduction second messenger pathway by stimulating the G-protein transducin (Gt), resulting in the liberation of its α subunit. This GTP-bound subunit in turn activates a cGMP phosphodiesterase. The cGMP phosphodiesterase hydrolyzes (breaks down) cGMP, lowering its local concentration so it can no longer activate cGMP-dependent cation channels. This leads to the hyperpolarization of photoreceptor cells, transmitting a signal to the optic nerve. 10. Answer: Cytokines bind to cytokine receptors and are associated with either one or multiple JAKs (JAK1, JAK2, JAK3, and Tyk2). The binding of a cytokine to its receptor leads to a conformational change in the JAKs, leading to the phosphorylation of the tyrosine kinase domain in enzyme. The phosphorylated JAKs then phosphorylate tyrosines on the cytoplasmic tails of cytokine receptors. The JAKs also phosphorylate the STAT proteins, which dissociate from the receptor to form dimers that enter the nucleus to control gene expression. Flipped classroom. Transcription and Translation Part A. Six MCQ (20 mins working in pairs, 30 mins discussion) 1. The sequence of a region of DNA around the 5′ end of a gene in Escherichia coli is shown below. The −10 hexamer and the transcription start site are highlighted. What would be the sequence of the first 10 nucleotides of the mRNA transcribed from this gene? 5′…GCGCTTGGTATAATCGCTGGGGGTCAAAGAT…3′ a. AUAUUAGCGA b. CCAGUUUCUA c. UAUAAUCGCT d. GGGGUCGCUA e. GGUCAAAGAU 2. A primary mRNA transcript with three exons is depicted below. Which mature mRNA product of this transcript is a result of exon skipping? a. b. c. d. e. 3. Mature mRNAs in eucaryotes have 5’cap and 3’ polyA at their ends. This is not the case for bacterial mRNAs. What is the role of the 3’ poly-adenylation (poly-A)? a. It’s related with alternative splicing b. It attracts mRNA splicing factors c. It attracts ribosomes d. It protects mRNA from degradation e. It’s a signal for translation termination 4. An elongating ribosome is bound to appropriate tRNAs in both the A and the P sites and is ready for peptidyl transfer. What happens next? a. The carboxyl end of the polypeptide chain is released from the P-site tRNA and joined to the free amino group of the amino acid linked to the A-site tRNA. b. The amino end of the polypeptide chain is released from the P- site tRNA and joined to the free carboxyl group of the amino acid linked to the A-site tRNA. c. The carboxyl end of the amino acid is released from the A-site tRNA and joined to the free amino group of the polypeptide chain linked to the P-site tRNA. d. The amino end of the amino acid is released from the A-site tRNA and joined to the free carboxyl group of the polypeptide chain linked to the P-site tRNA. 5. Both ribosomal subunits (small and large) assemble at the start codon of mRNA. The start codon amino acid is usually not the first amino-acid in mature proteins since it is removed after translation. Which amino-acid is encoded by the start codon? a. Pro b. Gly c. Met d. Leu e. Ala 6. UTRs are found at the 5’ and 3’ ends of eucaryotes’ mRNAs. What UTR means? a. Unrecognized Region b. Untranslated Region c. Untranscribed Region d. Unregulated Region Part B. Four critical thinking questions (20 mins working in pairs, 30 mins discussion) 7. Are there any differences between procaryotes and eucaryotes in transcription process? 8. What is the role of general transcription factors? 9. What is the role of snRNPs? 10. Why alternative splicing is important? Flipped classroom. Transcription and Translation Answers 1. Answer: E 2. Answer: B 3. Answer: D 4. Answer: A 5. Answer: C 6. Answer: B 7. Answer: Procaryotes don’t have introns (no need for RNA splicing) and mRNA 5’/3’ modifications. Procaryotes’ genes have different promoter elements. Procaryotes’ mRNAs proceed directly to translation (in eucaryotes they have first to exit the nucleus). 8. Answer: General transcription factors (GTFs) are proteins that help eukaryotic RNA polymerases find transcription start sites and initiate RNA synthesis (see details in the reading material). 9. Answer: The snRNPs recognize the conserved sequences within introns and quickly bind these sequences once the pre-mRNA is made and initiate splicing. 10. Answer: Alternative splicing allows for multiple proteins to be generated from a single region of DNA. Flipped classroom. DNA repair and DNA transposition Part A. Five MCQ (20 mins working in pairs, 30 mins discussion) 1. DNA chemical damages occur naturally or due to environmental factors. DNA damages cause changes in the structure of the genetic material and prevents the replication mechanism from functioning and performing properly. Which spontaneous lesion in DNA occurs most frequently in a mammalian cell? a. Depurination b. Cytosine deamination c. Guanine oxidation d. Guanine alkylation e. Depyrimidination 2. DNA repair mechanisms must be accurate and reliable. However, some DNA repair mechanisms are less reliable than others and they are activated when the cell must repair multiple DNA errors in a short time. Which DNA repair mechanism can introduce DNA sequence errors? a. Base excision repair b. Nucleotide excision repair c. Direct chemical reversal d. Homologous recombination e. Nonhomologous end joining 3. Ionizing radiation can cause DNA double-strand breaks. These DNA lesions can trigger carcinogenesis mechanisms. Which DNA repair pathway can accurately repair a double-strand break? a. Base excision repair b. Nucleotide excision repair c. Direct chemical reversal d. Homologous recombination e. Nonhomologous end joining 4. Antibiotic resistance is a major challenge in Medicine today. It is estimated that resistance itself caused 1.2 million deaths in 2019. What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics? a. DNA-only transposons b. Retroviral-like retrotransposons c. Nonretroviral retrotransposons d. Site-specific recombinases 5. SARS-CoV-2 is a [+]-strand RNA virus. Knowing this, what can we infer? a. RNA can immediately be translated to protein by the cell machinery. b. RNA needs to be used as a template to first make complementary RNA before translation can occur. c. A genome must be integrated into the host cell’s genome by reverse transcriptase before proteins can be expressed. d. Enzymes use a cut-and-paste mechanism to integrate its genome into the host’s chromosome. Part B. Three critical thinking questions (20 mins working in pairs, 30 mins discussion) 6. What kind of DNA lesions do tobacco benzopyrenes and UV light cause? Which DNA repair mechanism can repair these lesions? 7. Indicate whether each of the following DNA lesions is typically repaired via the base excision or nucleotide excision repair pathway. a. Base excision b. Nucleotide excision __ 1. Deaminated cytosines __ 2. Depurinated residues __ 3. Thymine dimers __ 4. Bulky guanine adducts 8. Upon heavy damage to the cell’s DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. What can these backup polymerases do? Flipped classroom. DNA repair and DNA transposition Answers 1. Answer: A 2. Answer: E 3. Answer: D 4. Answer: A 5. Answer: A 6. Answer: Bulky nucleotide lesions and various pyrimidine dimers. Nucleotide excision repair. 7. Answer: 1. A 2. A 3. B 4. B 8. Answer: Create mutations even on undamaged DNA Flipped classroom. DNA repair and DNA transposition Part A. Five MCQ (20 mins working in pairs, 30 mins discussion) 1. DNA chemical damages occur naturally or due to environmental factors. DNA damages cause changes in the structure of the genetic material and prevents the replication mechanism from functioning and performing properly. Which spontaneous lesion in DNA occurs most frequently in a mammalian cell? a. Depurination b. Cytosine deamination c. Guanine oxidation d. Guanine alkylation e. Depyrimidination 2. DNA repair mechanisms must be accurate and reliable. However, some DNA repair mechanisms are less reliable than others and they are activated when the cell must repair multiple DNA errors in a short time. Which DNA repair mechanism can introduce DNA sequence errors? a. Base excision repair b. Nucleotide excision repair c. Direct chemical reversal d. Homologous recombination e. Nonhomologous end joining 3. Ionizing radiation can cause DNA double-strand breaks. These DNA lesions can trigger carcinogenesis mechanisms. Which DNA repair pathway can accurately repair a double-strand break? a. Base excision repair b. Nucleotide excision repair c. Direct chemical reversal d. Homologous recombination e. Nonhomologous end joining 4. Antibiotic resistance is a major challenge in Medicine today. It is estimated that resistance itself caused 1.2 million deaths in 2019. What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics? a. DNA-only transposons b. Retroviral-like retrotransposons c. Nonretroviral retrotransposons d. Site-specific recombinases 5. SARS-CoV-2 is a [+]-strand RNA virus. Knowing this, what can we infer? a. RNA can immediately be translated to protein by the cell machinery. b. RNA needs to be used as a template to first make complementary RNA before translation can occur. c. A genome must be integrated into the host cell’s genome by reverse transcriptase before proteins can be expressed. d. Enzymes use a cut-and-paste mechanism to integrate its genome into the host’s chromosome. Part B. Three critical thinking questions (20 mins working in pairs, 30 mins discussion) 6. What kind of DNA lesions do tobacco benzopyrenes and UV light cause? Which DNA repair mechanism can repair these lesions? 7. Indicate whether each of the following DNA lesions is typically repaired via the base excision or nucleotide excision repair pathway. a. Base excision b. Nucleotide excision __ 1. Deaminated cytosines __ 2. Depurinated residues __ 3. Thymine dimers __ 4. Bulky guanine adducts 8. Upon heavy damage to the cell’s DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. What can these backup polymerases do? Flipped classroom. DNA repair and DNA transposition Answers 1. Answer: A 2. Answer: E 3. Answer: D 4. Answer: A 5. Answer: A 6. Answer: Bulky nucleotide lesions and various pyrimidine dimers. Nucleotide excision repair. 7. Answer: 1. A 2. A 3. B 4. B 8. Answer: Create mutations even on undamaged DNA Flipped classroom. Cancer Genetics Part A. Six MCQ (20 mins working in pairs, 30 mins discussion) 1. Mutations can lock the GTPases in the active state, and such mutations in the small GTPase Ras are particularly common in some forms of cancer. What is the most common mechanism behind the conversion of Ras into Ras oncogenes? a. Gene amplification b. Regulatory mutation c. Chromosome rearrangement d. Deletion in coding sequence e. Point mutation in coding sequence 2. Mutations in two important cancer-critical genes, encoding p53 and Rb, respectively, are commonly found in cancers. What type of mutations are these expected to be? a. Loss-of-function mutations in both genes b. Loss-of-function mutation in p53 and gain-of-function mutation in Rb c. Gain-of-function mutation in p53 and loss-of-function mutation in Rb d. Gain-of-function mutations in both genes 3. Colorectal carcinogenesis is a multistep procedure. Usually, a standard set of genes are found mutated in sporadic colorectal tumors. Which gene is usually first mutated in colorectal carcinogenesis? a. K-Ras b. β-Catenin c. Apc d. p53 e. MLH 4. Familial adenomatous polyposis (FAP) is a rare cancer syndrome. Which body organ is affected? a. Liver b. Kidney c. Eye d. Colon e. Skin 5. Lynch syndrome, inherited breast cancer and familial adenomatous polyposis, are among the most known hereditary cancer syndromes. What percentage (%) of cancer cases are hereditary? a. 1-2 b. 5-10 c. 20-30 d. 30-40 e. 40-50 6. A 20-year-old woman has been just diagnosed with breast cancer. Her mother, maternal grandmother, and her aunt (mother’s sister) were also diagnosed with breast cancer at the same age. Which is the mutated gene (most probably)? a. TP53 or RB b. APC or RAS c. MLH1 or MSH2 d. BRCA1 or BRCA2 e. TP53 or APC Part B. Four critical thinking questions (20 mins working in pairs, 30 mins discussion) 7. One of the most common hereditary conditions associated with colorectal cancer is hereditary nonpolyposis colorectal cancer (HNPCC), or Lynch syndrome. How would you describe the molecular pathomechanism of HNPCC? 8. Distinguish between driver mutations and passenger mutations in cancer. How many driver mutations might be expected in cancers? 9. What is the effect of RAS gene mutations on Ras protein? 10. Tumors gradually acquire mutations to evolve from benign to malignant lesions. Because that takes some time, cancer is primarily a disease of aging. So, how can childhood cancers be explained? Flipped classroom. Cancer Genetics Answers 1. Answer: E 2. Answer: A 3. Answer: C 4. Answer: D 5. Answer: B 6. Answer: D 7. Answer: HNPCC is associated with the existence of mutations in one copy of several genes related with the DNA mismatch repair system. Afflicted individuals may suffer a loss or inactivation of the remaining gene copy during their lives, resulting in an elevated spontaneous mutation rate. Cells that are genetically unstable are more likely to become malignant. The risk level is high without any increase in the number of colorectal polyps. HNPCC cells are difficult to detect due to their normal or almost normal appearance. 8. Answer: As normal cells evolve to become cancer cells, they pick up many somatic changes—both genetic and epigenetic. A small subset of the genetic changes, known as driver mutations, result in altered expression of genes that confer a growth advantage to their descendants. They are positively selected and causally implicated in cancer development. The remainder are passenger mutations. Whereas there can be many thousands of passenger mutations in a cell, the number of driver mutations per cell is small. Massively parallel genome sequencing studies have suggest that most breast cancers have 6 or less driver mutations to which must be added epigenetic changes such as silencing of tumor suppressor alleles (and also the possibility of tumor initiation). The number of driver mutations also depends on the type of tumor. Less driver mutations might be expected, for example, in tumors of blood cells (which are not constrained by being part of a solid tissue) and childhood tumors. 9. Answer: When stimulated by upstream signaling molecules, wild type RAS proteins interact with guanine nucleotide exchange factors to replace GDP with GTP, resulting in an activated protein conformation. RAS activity is terminated by interaction with GTPase activating protein, which stimulates the GTPase activity of the protein and converts GTP back to GDP, thereby restoring the inactive form of RAS. Mutations in RAS inhibit the GTPase activity and lock the protein in the active GTP bound conformation, activating proteins that promote cell growth and cell proliferation. 10. Answer: In self-renewing tissues the cells contain DNA that has progressively accumulated mutations through multiple DNA replication cycles in progenitor cells (errors in DNA replication and post-replicative DNA repair are frequent causes of mutations). Pediatric tumors often occur in tissues that are not self-renewing, and such tumors typically have fewer mutations than adult tumors. However, leukemias and lymphomas, which are diseases of self-renewing blood cells, can also often develop early in life. Here the precursor cells are already mobile and invasive and are thought to require fewer DNA changes than in solid tumors, in which the tumor cells require additional mutations to confer these biological capabilities. Another possibility is that childhood cancers can arise from an initiating mutation that arises in embryonic stems cells. These cells in the embryo resemble cancer cells—they are poorly differentiated and rapidly dividing. If they receive a cancer- predisposing mutation, they are much more likely to develop into tumors at an early stage than more differentiated cells with the same mutation. TBL1 – Protein structure and Function 1. A cancer researcher has purified from breast cancer tissue cultures a multisubunit extracellular protein that has several interchain disulfide bonds. He wants to study further this protein in its primary structure. Which of the following chemicals he must use to eliminate the disulfide bonds? a. NaCl, a salt b. SDS, an ionic detergent and denaturing agent c. H2O2, an oxidizing reagent d. Tris, a buffering agent e. DTT, a reducing agent 2. Fibril proteins are very important for bones, tendons, and skin. Which is the chief fibril protein of those structures? a. Actin b. Collagen c. Laminin d. Myosin e. Porphyrin 1 TBL1 – Protein structure and Function 3. Cells degrade non-functional proteins or those that are not needed anymore. The ‘degradation machine’ of the cell is proteasome. How these proteins are recognised by the cell to be transferred to the proteasome? a. By a specific phosphatase. b. By the E1 enzyme. c. By the proteasome complex. d. By an arginine residue in the target protein. e. By an E2–E3 ligase. 4. DNA mutations can cause amino acid substitutions that can disrupt α helices and β sheets, the major regular folding patterns in proteins. These structures are essential for the higher conformation structures of proteins. Which chemical bonds are essential for stabilizing these secondary structure elements? a. Disulphide bonds between amino acid side chains b. Hydrogen bonds between amino acid side chains c. Hydrogen bonds between N—H and C=O groups of the polypeptide backbone d. Van der Waals forces between amino acid side chains e. Van der Waals forces between N—H and C=O groups in the polypeptide backbone 2 TBL1 – Protein structure and Function 5. Proteins often renature into their original conformations after they have been unfolded by denaturing solvents. What that implies? a. The cell does not need molecular chaperones for survival. b. Each protein folds into several different conformations inside the cell. c. The final folded structure of a protein is usually NOT the one with the lowest free energy. d. The information needed to specify the three-dimensional shape of a protein is encoded in its amino acid sequence. e. The information needed to specify the three-dimensional shape of a protein is encoded in its secondary structure. 6. For each of the following cartoon representations from left to right, indicate whether the protein structure is composed of: only α helices (1), α helices plus parallel β sheets (2), α helices plus antiparallel β sheets (3) Your answer would be a three-digit number composed of digits 1 to 3 only, e.g. 221. Images created in PyMOL, from PDB entries 1STU, 2LPC, 1SA8, and 3BOB. Correct answer: 312 3 TBL2 – Cytoskeleton and cell junctions 1. Cytoskeleton functions as "railroad tracks," providing support and directionality inside the cell. If proteins are sent in the wrong place inside cell, this may cause severe cell disfunction. Which motor protein is responsible for transporting “cargo” (organelles, vesicles etc) on the cytoskeleton, from the centre of the cell towards the periphery? a. kinesin b. dynein c. myosin d. actin e. cadherin 2. Epidermolysis bullosa is a rare skin blistering disease. Not any treatment exists for this disease and patients can die prematurely because of severe infections. Mutations in which filament protein is the cause of this disease? a. α-tubulin b. β-tubulin c. actin d. keratin e. lamin 1 TBL2 – Cytoskeleton and cell junctions 3. Macrophages are very important cells of our immune system, since they can engulf and destroy bacteria. These cells can move to any direction. Which cytoskeleton protein is related with cell mobility? a. α-tubulin b. β-tubulin c. actin d. keratin e. lamin 4. Progeria is a severe rare inherited disease, where patients develop accelerated aging. Unfortunately, not any treatment exists for progeria and premature death is unavoidable. Which protein is related with this disease? a. α-tubulin b. β-tubulin c. actin d. keratin e. lamin 2 TBL2 – Cytoskeleton and cell junctions 5. Disruption of the adherens junctions or defects in the associated proteins are associated with a variety of diseases including inflammatory bowel disease, disorders of the skin and hair and cancer. Which cytoplasmic filaments bind to adherens junctions? a. Keratin b. Actin c. Lamin d. Microtubules e. Myosin 6. A blistering disease is a condition in which there are fluid-filled skin lesions. Blistering diseases are often caused by lesions in the cell-cell junctions. Which autoimmune-blistering disease of the skin and mucous membranes is caused by autoantibodies reducing desmosomal adhesion between epithelial cells? a. Atopic dermatitis b. Pemphigus c. Ichthyosis d. Acne e. Epidermolysis bullosa 3 TBL1 – Genetic variation and inheritance 1. Mitochondrial DNA is not inherited the same way as nuclear DNA. Consider a family with five members: parents, son, daughter, maternal grandmother. Who has a different mitochondrial DNA in this family? a. The son b. The father c. The mother d. The daughter e. The maternal grandmother 2. Susceptibility to drug and alcohol addiction is partially genetic. For example, the human dopamine receptor DRD2 is associated with cocaine and alcohol abuse; individuals carrying an A nucleotide, instead of the more common C nucleotide, at a certain position within the sixth intron of the DRD2 gene are at a significantly increased risk of abuse. This variation in this sequence is an example of what? a. SNP b. CNV c. Indel d. Inversion e. Translocation 1 TBL1 – Genetic variation and inheritance 3. One medically important trait that demonstrates co-dominant expression is the ABO blood group system, important in blood transfusion and tissue transplantation. Which combination of parental genotypes can give a child belonging in any blood group (A or B or AB or O)? a. AB x AB b. AB x AO c. AB x BO d. AO x BO e. AB x OO The following pedigrees represent the inheritance of a genetic disease inside a family. Black symbols represent the affected members of the family. 4. Which pedigree represents most likely an autosomal recessive disease? A. 5. Which two pedigrees are NOT compatible with the inheritance of an X-linked recessive disease? B. and E. 6. Which pedigree could represent the inheritance of a mitochondrial disease? B. 2 TBL2 – Chromosomal anomalies 1. X chromosome inactivation occurs when more than one X chromosome is present in a somatic cell. This phenomenon is observed in most mammalian species. Why does this occur? a. To cause phenotypic differences between the sexes b. To avoid nondisjunction of sex chromosomes c. To allow a higher rate of expression from the active X chromosome d. To achieve dosage compensation 2. A 44 years-old pregnant woman worries about her fetus. She thinks that due to her advanced age, the probability for her child to have Down syndrome is very high. At what mother’s age the probability for a Down syndrome childbirth starts to be higher? a. 25 b. 35 c. 45 d. 50 e. 60 3. A prenatal genetic diagnosis showed that the embryo had a chromosomal abnormality related with Cri-du-chat syndrome. The embryologist explained to the pregnant woman that Cri-du-chat syndrom is a rare chromosomal syndrome, characterized by mental disability, delayed development, and distinctive facial features. Which is the causal chromosomal abnormality for Cri-du-chat syndrome? a. a deletion at the 5p arm b. an insertion at the 5p arm c. a deletion at the 6p arm d. an insertion at the 6p arm e. an isochromosome 1 TBL2 – Chromosomal anomalies 4. An 11 years-old boy has been diagnosed with Klinefelter syndrome. Klinefelter syndrome is caused by an extra X chromosome in males. However, the geneticist explained to the parents that their son belongs in the 15% of Klinefelter cases, where only some cells of the body of the patient are affected. How do we call this phenomenon in genetics? a. chimera b. hemizygosity c. heterogeneity d. mosaicism e. plasticity 5. In a high percentage of abortions, a chromosomal abnormality has been detected. In about 50% of those cases, the abnormality is related with a chromosomal trisomy. Which are the most common types of autosomal trisomy that survive to birth in humans? a. 21, 16, 13 b. 21, 18, 14 c. 21, 19, 13 d. 21, 17, 14 e. 21, 18, 13 6. Chromosomal deletions or duplications can be the cause of severe syndromes. However, some of those chromosomal abnormalities are more frequent than others. Which chromosomal region is frequently involved with pathogenic deletions or duplications? a. 1q11 b. 5q11 c. 8q11 d. 14q11 e. 22q11 2 TBL1 – Cell cycle 1. The antitumor drug paclitaxel promotes the assembly of microtubules and inhibits their depolymerization. What is the consequence of this effect on the cell cycle? a. The induction of cell-cycle progression b. The arrest of cell-cycle progression at M phase c. The promotion of chromosome segregation in mitosis d. The blockage of S-phase progression e. The suppression of cell death 2. Several studies have shown that overexpression of cyclins is associated with carcinogenesis mechanisms. M-cyclins activate Cdks that do what? a. Cdks that stimulate chromosome duplication b. Cdks that trigger progression through Start c. Cdks that stimulate entry into mitosis d. Cdks that promote DNA replication e. Cdks that remain at elevated levels until mitosis 1 TBL1 – Cell cycle 3. Mutations in BRCA1 gene is a cause of inherited breast cancer in women. The BRCA1 encoded protein is involved in cell-cycle checkpoints that check for DNA integrity and any DNA damages. In the following schematic diagram of a typical eukaryotic cell cycle, choose two major time points (among A to E) at which the cell-cycle control system normally arrests the cycle if DNA damage is detected. a. B and D b. A and B c. C and D d. B and C e. A and C 2 TBL1 – Cell cycle 4. Cells that are continuously mitotically active are highly vulnerable to carcinogenesis. Which of the following cell populations in our body is among the ones with the highest mitotic index? a. Neurons b. Hepatocytes c. Red blood cells d. Intestinal cells e. Skeletal myocytes 5. TP53 gene is considered one of the most important genes in cancer. It is found mutated in about 50% of cancers. How p53 protein (product of TP53 gene), or the proteins that activates, arrest the cell cycle when DNA damage is detected? a. Activate Cdk-cyclin complexes b. Inactivate Cdk-cyclin complexes c. Activate Anaphase Promoting Complex (APC) d. Inactivate Anaphase Promoting Complex (APC) e. Inactivate p21 3 TBL1 – Cell cycle 6. Cyclin B1, a key cell-cycle regulatory protein in vertebrates, is mostly cytosolic before mitosis. Early in mitosis, however, the protein is phosphorylated by certain protein kinases and consequently accumulates in the nucleus. How can phosphorylation bring about nuclear accumulation of this protein? a. Phosphorylation within the nuclear localization signal inhibits the function of the signal. b. Phosphorylation within the nuclear export signal enhances the function of the signal. c. Phosphorylation within the nuclear export signal interferes with the function of the signal. d. Phosphorylation elsewhere on the protein enhances binding to cytosolic proteins. 4 TBL2 – Cell death 1. Pro-apoptotic proteins of the Bcl-2 family can induce apoptosis. How is apoptosis activated by these proteins? a. By activating nucleases and proteases inside the cell b. By activating the surface death receptors of the cell c. By creating openings in the inner mitochondrial membrane d. By creating openings in the outer mitochondrial membrane e. By direct activation of executioner caspases 2. Another way that cytotoxic lymphocytes kill infected cells is by activating apoptosis cell surface receptors. Which are those receptors? a. Bcl2 receptors b. Caspase receptors c. Fas receptors d. IGF receptors e. TGF receptors 1 TBL2 – Cell death 3. In the intrinsic pathway of apoptosis, the mitochondria releases a protein into the cytosol, which binds to the adaptor protein Apaf1. Apaf1 then oligomerizes into a wheel-like assembly called an apoptosome, which recruits initiator caspase-9 proteins. Which is this protein? a. Bad b. Bim c. Cyt c d. FADD e. Fas 4. Necrosis can be described as a pathological process of cell death which could have been resulted from infections, hypoxia, trauma or toxins. What is the main hallmark of necrotic cells? a. Digestion by caspases b. Phagocytosis by neighbor cells c. Release of mitochondrial proteins d. Shrinkage and membrane blebbing e. Swelling and bursting 2 TBL2 – Cell death 5. Over-activation of some caspases can lead to excessive programmed cell death (apoptosis). This is seen in several neurodegenerative diseases where neural cells are lost, such as Alzheimer's disease. How are executioner caspases get activated? a. By cleavage b. By dephosphorylation c. By dimerization d. By phosphorylation e. By tetramerization 6. During development, apoptosis must be strictly under control. Some cells must die and some others not. Dysregulation of this process can lead to severe developmental diseases. Which group of extracellular signal molecules inhibit apoptosis during development? a. Anti-death factors b. Apoptotic factors c. Death factors d. Developmental factors e. Survival factors 3 TBL1 – Genetics of multifactorial diseases and GWAS 1. Genome-wide association studies (GWAS) are widely used today for investigating the genetic component of multifactorial diseases. In which type of studies are GWAS used most? a. Case-control studies b. General population studies c. Patient studies d. Sibling studies e. Twin studies 2. If geneticists intend for their genomic studies to benefit all groups of people, why might it be important to ensure that there is adequate representation from all ethnic groups in human genome databases? a. Because statistical power is needed b. Because otherwise useful genetic variation could be missed c. Because of ethical issues d. Because human populations genetically differ by 20% 3. Alzheimer's disease is a progressive neurologic disorder that causes the brain to shrink (atrophy) and brain cells to die. APOE gene E alleles have been associated with the disease. Which APOE genotype predispose more for the disease? a. E2/E2 b. E3/E3 c. E4/E4 d. E2/E3 e. E3/E4 4. Many congenital malformations are multifactorial. Which is the most common facial/head congenital malformation? a. Cleft lip b. Facial dysplasia c. Microcephaly d. Hypertelorism e. Hypotelorism 1 TBL1 – Genetics of multifactorial diseases and GWAS 5. Studies of identical and fraternal twins suggest up to 80% of variation in height is genetic. Thousands of genetic variants appear to influence this human trait. What kind of genetic trait is human height? a. Chromosomal b. Epigenetic c. Quantitative d. Qualitative e. Monogenic 6. The most informative studies on how and to what degree heredity and the environment influence human traits are based on twins’ observations. Empirical risk ratios have been calculated through these studies, a valuable tool in genetic consultation. Which kind of twin studies can give the most useful information about the contribution of environmental and genetic factors in multifactorial (complex) diseases? a. dizygotic twins of different gender b. dizygotic twins of same gender c. dizygotic twins separated at birth d. monozygotic twins e. monozygotic twins separated at birth 2 TBL2 – Cytogenetics methods 1. A pediatrician suspects Williams syndrome for an 1-year old child. Williams syndrome is a rare genetic condition characterized by unique physical features, delays in cognitive development and potential cardiovascular problems. The syndrome is related with a deletion on chromosome 7. Which is the most rapid and cost-effective genetic test to confirm or not the diagnosis? a. CMA b. FISH c. Karyotype analysis d. MLPA e. Sanger sequencing 2. CMA analysis allows for the exploration of all 46 human chromosomes in a single test. This is highly valuable for chromosomal syndromes’ investigation in children, especially in cases of uncertain clinical diagnosis. Which chromosomal anomaly cannot be detected by CMA? a. Balanced translocation b. Deletion c. Insertion d. Monosomy e. Unbalanced translocation 3. A man had a blood test for his annual check-up. Blood test unexpectedly showed elevated white blood count, elevated platelet count and immature granulocytes. His physician sent him to the Oncology Department where he was tested positive for the Philadelphia chromosome, the hallmark of Chronic Myeloid Leukemia (CML). How is Philadelphia chromosome being detected? a. CMA b. Karyotype analysis c. MLPA d. Next Generation Sequencing e. Sanger Sequencing 1 TBL2 – Cytogenetics methods 4. A pediatrician suspects that the physical and the mental development of a 3-year- old girl is not normal. The doctor is not sure for the syndrome. The family visited a clinical geneticist that he couldn’t put a diagnosis as well. What genetic test, most probably, the clinical geneticist has suggested for this girl? a. CMA b. FISH c. Karyotype analysis d. MLPA e. Sanger sequencing 5. Karyotyping is the process by which a karyotype is prepared from photographs of chromosomes, in order to determine the chromosome complement of an individual, including the number of chromosomes and any abnormalities. Photographs are taken from dividing cells. In which mitosis stage do we photograph the chromosomes for karyotyping? a. Anaphase b. Interphase c. Metaphase d. Prophase e. Telophase 6. About 90% of Duchene or Becker muscular dystrophy cases carry a small size deletion that involves 10 exons or less, 26% of which carry a single-exon deletion. Which method is most appropriate to use for the diagnosis of those cases? a. CMA b. FISH c. Karyotype analysis d. MLPA e. Sanger sequencing 2 Molecular biology techniques CSIUNic GEMD-101 Spring 2024 Dr Chloe Antoniou, PhD 1  The importance of DNA sequencing Medicine: Detect genes associated with hereditary and acquired diseases. Develop techniques such as gene therapy Determining sequences of bacteria and viruses to develop treatments and vaccines Forensics: Paternity and maternity identification Identifications of biological material at crime scenes Evolutionary and genealogical studies Identification of missing persons. Agriculture: Increase in productivity of crops Enhancing resistance of crops towards pests Increase nutritional value and quality of crops 2  Types of DNA mutations 3 Image from: http://schoolbag.info/chemistry/mcat_biochemistry/40.html Standard nomenclature of mutations Numbering of nucleotides Ogino et al., 2007 4 Example of standard mutation nomenclature Ogino et al., 2007 5  DNA sequencing by the Sanger method (dideoxy sequencing) In Sanger sequencing, 4 PCR reactions are performed with each containing a mixture of dNTP’s and one dideoxy nucleotide (ddNTP). Frederick Sanger Nobel Prize in Chemistry 1980 for developing methods in DNA sequencing. When a ddNTP is incorporated by the DNA polymerase, elongation is terminated. Nobel Prize in Chemistry 1958 “for his work on the structure of proteins, especially that of insulin". 6 An overview of “Sanger Sequencing” 7 Question 1 In the Sanger method for DNA sequencing, a small amount of a ddNTPs are added to the sequencing reaction along with a larger amount of the corresponding dNTP. What result would be observed if the dNTP’s were omitted? Answer If dNTP’s are omitted, then when the first complementary base is encountered in the template, ddNTP will be added, and polymerization will stop. Therefore only one band will be seen in the sequencing gel for each nucleotide. 8 A modern approach to Sanger sequencing 9  G = Black Question 2 C = Blue A = Green T = Red Compare Patient 1 with the Wild Type 1 sample and Compare Patient 2 with the Wild Type 2 sample. Wild Type 1 Wild Type 2 AGCGATG GCAATCG 1. What is the DNA sequence in each panel? 2. Do the patients have any mutations? If yes, what are they? 3. Are the patients homozygous or Patient 1 Patient 2 heterozygous for the mutations? GCA A/G TCG AGC G/A ATG 4. What would be your next step once you have this result? 10  Restriction endonucleases Restriction endonucleases recognize specific base sequences in double- helical DNA and cleave, at specific places, both strands of that duplex. The recognition sequences of restriction enzymes are almost always palindromic. Note: The word palindrome is derived from the Greek “palindromos” (running back again). In linguistics, this refers to a word, sentence, or verse that reads the same from right to left as it does from left to right. 11  DNA fingerprinting DNA fingerprinting is a forensic technique used to identify individuals by characteristics of their DNA. It may involving the use PCR followed by the use of restriction enzymes or the determination of sequence repeats. Many people may have the same number of repeats in a certain region of their DNA or share some of the differences observed in DNA amongst individuals. However, the probability of two people sharing the same number of repeats or sequences at several Image from: https://www.ted.com/topics/dna regions in the DNA is very small. 12  Whose suspect’s DNA was found on the crime scene? Case 1: Case 2: 13  Paternity testing Paternity testing depends on the following principle: He is likely to be the father. He is not the father. 14  The Paternity Index (PI) describes the probability of paternity. It is calculated as X/Y, where X is the probability that the alleged father is, in fact, the father and Y is the probability that any randomly selected man of the same race is the father. The product of the individual PIs is the Combined Paternity Index (CPI) which is ultimately used to calculate the Probability of Paternity seen on paternity test reports. 15 Frank H. Stephenson, Calculations for Molecular Biology and Biotechnology 2010.  The genomes of organisms ranging from bacteria to multicellular eukaryotes have been sequenced The genome sequence of the bacterium Haemophilus influenzae was determined in 1995. The genomic DNA was sheared randomly into fragments that were then sequenced. Computer programs assembled the complete sequence by matching up overlapping regions between fragments. The H. influenzae genome comprises 1.8 million bp and encodes approximately 1,740 proteins. The first eukaryotic genome to be completely sequenced was that of baker’s yeast, Saccharomyces cerevisiae, in 1996. The yeast genome comprises approximately 12 million bp and encodes >6,000 proteins. The Human Genome Project was declared complete in April 2003. The genome size is 3,234.83 Mb, only 2% is coding DNA. Estimated 20,000-25,000 coding genes. 16 Image from US National Human Genome Research Institute 17  “Next-generation” sequencing (NGS) NGS is a high-throughput sequencing method. Such methods allow for the simultaneous sequencing of thousands/millions of relatively short DNA sequences. NGS enables the rapid determination of a whole genome sequence. Individual genome sequences will provide information about genetic variation within populations and may usher in an era of personalized medicine, when these data can be used to guide treatment decisions. 18 An interesting read on human chimerism and the importance of DNA sequencing http://abcnews.go.com/Primetime/shes-twin/story?id=2315693 https://en.wikipedia.org/wiki/Lydia_Fairchild 19  Recombinant product technology Composite DNA molecules made of covalently linked segments from two or more sources are called recombinant DNAs. In a review by Sanchez-Garcia et al., 2016, it is stated that as of 2015, there are over 400 recombinant products (peptides and proteins) and another 1300 undergoing clinical trials. Commonly used method to study the structure and function of proteins. This involves expressing and purifying proteins/peptides of mostly human origin in mammalian, yeast or bacterial cells. Eukaryotic cells are quite challenging to work with and so bacterial cells have the advantage of having an easier to work with DNA. However bacterial cells lack the ability to perform more elaborate processes such as post translational modifications. 20 Cloning human DNA into bacteria 21 Site directed mutagenesis After cloning a gene, one can introduce mutations in order to study their effects on a proteins structure or function. 22  Carrying out mutational studies on proteins provides insight into each function Each data point on this graph is a different Each data point on this graph is a different mutation on a specific protein. Via analysis mutation on a specific protein. Via analysis of of these mutations, a model of how a plant these mutations, the mechanism of binding of protein called phototropin 2, changes each two proteins was elucidated. Lack of binding structure as a response to light. This also between these two proteins is involved in had many implications in the field of protein several hereditary types of hemolytic anemias. engineering. Figure from Antoniou, Lam & Fung 2008. Figure from Zayner, Antoniou, French, Hause & Sosnick, 2013 Yes that is me!! 23  Using probes to detect specific DNA or RNA sequences A probe is a single strand of DNA that can hybridize (base-pair) with a complementary sequence on another single-stranded polynucleotide composed of DNA or RNA. The probe must contain a label so that it can detect a complementary DNA or RNA. The label may be radioactive (so it can be detected by autoradiography) or a chemical that can be identified, for example, by fluorescence. 24 Different types of blotting techniques In addition combinations of blotting techniques exist: Southwestern blotting which is used to detect DNA-binding proteins. Northwestern blotting which is which is used to detect RNA-binding proteins. 25  Detection of mutations by allele-specific oligonucleotide probes An oligonucleotide probe is synthesized that is complementary to a region of DNA that contains a mutation. A probe is made for the normal DNA and a different one for the mutant allele. àIf the mutant probe binds to a sample of DNA, the sample contains DNA from a mutant allele. If the normal probe binds, the sample contains DNA from a normal allele. If both probes bind, then both are present (individual is heterozygous). 26 Question 4 Cystic fibrosis is an autosomal recessive disorder which results from mutations on the cystic fibrosis transmembrane conductance regulator (CFTR) protein. Over two thirds of cystic fibrosis patients carry a mutation which results in a deletion of phenylalanine 508 of the CFTR protein. As a result that protein cannot fold probably. Since this protein is a Cl- channel this leads to dysregulation of epithelial fluid transport. The figure below shows data obtained from a family (father, mother and 3 children) in which cystic fibrosis is suspected. Purple indicates a positive signal. Assuming the wild type allele is designated as C and the recessive allele is c, what is the genotype of each member? Individual Genotype Father Cc Mother Cc Child 1 CC Child 2 Cc 1 2 3 Child 3 cc 27  Quantifying gene expression Either mRNA or protein levels can be measured using a variety of techniques. Some techniques are just detection assays, others give relative expression and others give absolute expression. Measuring mRNA: qPCR/qrt-PCR Measuring protein: Western blotting, ELISA, immunolocalization 28  Quantitative (qPCR) and quantitative reverse transcription PCR (qrt-PCR) Technique based on PCR which allows the determination of the mRNA (qrt-PCR) or DNA (qPCR) concentration. RNA molecules are purified from a tissue or a cell culture. It is important that no DNA be present in the preparation. Two DNA primers that specifically match the mRNA of interest are added, along with reverse transcriptase, DNA polymerase, and dNTP’s. The first round of synthesis is the reverse transcription of the RNA into DNA using one of the primers. Next, is the amplification of that DNA strand by PCR. 29  The quantitative part of this method relies on a direct relationship between the rate at which the PCR product is generated and the original concentration of the mRNA species of interest. Chemical dyes are added to the PCR that fluoresce only when bound to double-stranded DNA, therefore a simple fluorescence measurement can be used to track the progress of the reaction The red sample has a higher concentration The qrt-PCR technique is relatively fast and simple to of the mRNA being measured than does perform in the laboratory; it is currently the method the blue sample, since it requires fewer PCR cycles to reach the same half-maximal of choice for accurately quantifying mRNA levels from concentration of double-stranded DNA. any given gene. Based on this difference, the relative amounts of the mRNA in the two samples Can be done in a high throughput manner. can be precisely determined 30  Applications of qPCR and qrtPCR: 1. Detect changes in concentration of mRNA of interest when a mutation(s) is present. 2. Detect changes in concentration of mRNA of interest with and without drug treatment. Control: mRNA whose concentration is not affected by the specific drug. 3. Detect changes in concentration of mRNA of interest in different cell lines/tissues. Control: mRNA whose concentration remains the same in the cell lines compared. 4. Detect changes in concentration of mRNA of interest at different time points. Control: mRNA whose concentration is not affected during the different time points. 5. Detect and measure the viral loads in conditions such as AIDS. 6. Measure DNA concentration e.g after purification. 31  Western blotting A technique used to detect a specific protein from a mixture. Proteins separated by electrophoresis, blotted onto membrane and components detected using specific antibodies. Protein may be quantified by comparison to internal standards. 32  A typical result of a Western blot. Lane 1: Molecule weight marker Lane 2: No inhibition of Bcl-xL expression Lane 3: Inhibition of Bcl-xL expression Lane 4: Inhibition of Bcl-xL expression a-Tubulin is an example of a protein used as a standard in Western blots Lane 1 2 3 4 since its expression remains unchanged under the experimental conditions used. Bands can be quantified and compared when applicable. 33  Enzyme – Linked Immunosorbent Assay (ELISA) Similar in principle with Western blotting but performed in solution instead of on a gel. Target proteins need to be immobilized on the well. àExample: Wells are coated in a protein called avidin. The target protein of interest has biotin attached (proteins can now be easily biotinylated). Biotin and avidin have very high affinity and so the protein of interest is attached onto the well. Washing, antibody and visualization steps are similar to those in western blotting. High throughput as compared to Western blotting which is low throughput. Higher sensitivity than Western blotting (i.e. can detect lower concentrations of proteins). Used also in antibody detection for medical purposes. 34 Example: ELISAs, Western Blots and PCRs are used in HIV testing. ELISA is used to screen individuals for antibodies to the HIV virus. The test has high sensitivity but lower specificity. A positive ELISA is confirmed by a Western blot. In some cases, a PCR is performed using primers against the HIV genome. If person is infected, a PCR product is detected. The PCR is positive much earlier after infection than the other techniques. Also newborns of HIV positive mothers will always be ELISA and Western blot positive, so PCR’s are the method of choice. 35  Immunolocalization An antibody-based technique used to identify the location of molecules or structures within cells or tissues. Applications: 1. Observe localization of a protein of interest in cells/tissues and monitor changes. Examples cyclin B in protein degradation lecture. 2. Specifically label cellular compartments. Mitochondria labeled in red, Staining of the nucleus, nucleolus and actin in green and the nucleus in blue. cytoplasm. Antibodies specific for proteins in each compartment are labeled with different fluorescent proteins (i.e. different colors) 36 Practice question 1 !"! #$%&'%&()(*'%(+&&,(-$,.&/(01(0*&(23&'0$1,(14('(,&5(3&%03$20$1,(&,678&(%$0&9('%($,/$2'0&/($,( 0*&('%%12$'0&/(4$:;3&"(?(5'%(/$:&%0&/(5$0*(@21A!9(',/('(F1;0*&3,(+-10(5'%(/1,&(;%$,:(0*&('='$-'+-&( B31+&"(?(2'33$&3(413(0*$%(/$%13/&3(51;-/(/$%B-'7(5*$2*(+',/%(1,(0*&(F1;0*&3,(+-10G( ( ?"! !"!(',/(!"H(.+( I"! !"!(.+( J"! !"!(.+(',/(K"L(.+( #"! !"!9(!"H9(',/(K"L(.+(( @"! !"H(.+( ( ( 38 Week 2 – Probabilities in Genetics GEMD-101 Fall 2024 Dr Chloe Antoniou, PhD 1 1. Identify the mode of inheritance and provide at least 2 reasons for your answer. Autosomal recessive X-linked (most probably) but can’t rule out autosomal recessive  Autosomal recessive Autosomal dominant (most probable) but can’t rule out recessive 2. A couple have 5 children, 2 girls and 3 boys. The mother is a carrier of the hemophilia allele, an X-linked disorder. She passes the hemophilia allele on to 2 of the boys who died in childhood and 1 of the daughters is also a carrier. Both daughters have 3 children (2 boys and 1 girl) without hemophilia. The carrier daughter has one son with hemophilia. One of the non-carrier daughter's sons has children with a woman who is a carrier and they have twin daughters. What is the percent chance that each twin daughter will also be a carrier? 50% of each twin being a carrier 3. A 27-year-old woman (indicated by the arrow) presents with a family history of a genetic disorder. She states that her maternal uncle and a male cousin died early in life from complications of the disease, and her older brother has also been recently diagnosed with the same disease. Her family pedigree is shown below. Assuming that her husband is genotypically normal, what is the probability that this woman’s child will have the disease? (Tip: give only one possible answer based on the most likely mode of inheritance). If mother is not a carrier à 0% of having children with the disease If mother is a carrier and father of children is unaffected à 50% daughters will be carriers à 50% sons will have the disease; 50% healthy The woman’s genotype and her child’s phenotype are independent events => rule of multiplication: Probability (affected child AND carrier mother) = = Probability (carrier mother) x Probability (affected child) = ½ x ¼ =1/8= 12.5%. 4. A healthy 30-year-old woman comes to the physician with her husband for genetic counselling. She states that she has two siblings who died from an autosomal recessive disease and that there is no other family history of the disease that she is aware of. What is the risk that this woman is a heterozygous carrier of the disease-causing mutation? (Tip: given the fact the she is healthy). Siblings have the disorder, parents normal (no family history) Autosomal recessive => Parents were heterozygous carriers of the disease allele ð Aa x Aa = AA, Aa, Aa, aa. ð Probability (Aa) = 2/3 GEMD-101 Transcription & Translation tutorial Chloe Antoniou, PhD 1. A 2-week-old girl is brought to the physician because of vomiting and a yellow appearance. Her temperature is 37°C (98.6°F), pulse is 138/min, and respirations are 29/min. Physical examination shows a lethargic infant with jaundice and hepatomegaly. Serum studies show an elevated total bilirubin level. Northern blot analysis shows a normal size and amount of galactose-1-phosphate uridyltransferase (GALT) mRNA, but biochemical studies do not show observable enzyme activity. Which of the following is the best explanation for this patient's condition? A. Gene deletion B. Nonsense mutation C. Premature transcription termination sequence in the DNA D. Promoter mutation E. RNA splicing mutation 2. DNA mismatch repair protein Msh2, a human analogue of the MutS protein found in bacteria, has been found to be altered many different types of cancer and is responsible for 40% of the cases of hereditary nonpolyposis colorectal cancer (HNPCC). A scientist working on a sample from a patient with HNPCC discovers that the tumour cells of one patient do not carry any mutations in the exons of the gene encoding for Msh2, but Western blot analysis shows that the amount of Msh2 protein in these cells is much lower than the amount of protein in normal cells. Suggest two possible reasons that can result in a reduced amount of the Msh2 protein in cancer cells, other than mutations on the exons of the gene encoding the Msh2 protein. Answer: 1. methylation of the promoter region of the gene that results to lower transcription of the Msh2 mRNA. 2. Mutation in an intron affecting proper splicing of the protein. 3. Figure 1 shows a schematic diagram of gene Z which codes for protein Z. The numbers indicate the nucleotide position in the sequence. Figure 1: Schematic diagram of gene Z. The numbers indicate the nucleotide position in the sequence. Figure 3 shows a Northern Blot of the mRNA produced from gene Z. Figure 3: Northern blog of gene Z. (a) Briefly explain using the two diagrams above, how the two bands on the Northern blot of gene Z arise (b) What is the size of the largest possible protein that can be produced from this gene? Justify your answer. Answer: (a) The two bands are approximately 900 bp and 500 bp. The 900 bp band is a product of exons 1, 2 and 3 while the 500 bp band is a product of exon 1 and 3 due to alternative splicing that can take place. (b) 300 amino acids as the longest mRNA transcript is 900 bp.