Practice 10 structured short- and long-answer medical questions on Molecular Biology,Molecular Genetics & Microbial-Biochemistry LAQs & SAQs. Review...
Answer: - **Gene structure:**Eukaryotic genes are composed of **exons** (coding sequences) and **introns** (non-coding sequences), with introns being removed during RNA splicing. - **Promoter region:**Upstream of the coding sequence lies a **promoter** containing specific sequences (e.g., the **TATA box**) essential for the initiation of transcription by RNA polymerase II. - **Regulatory elements:**Eukaryotic genes are also controlled by **enhancers**, **silencers**, and **insulators**, which regulate the level, timing, and tissue-specific expression of genes. - **Chromatin packaging:**DNA is wrapped around **histone proteins** to form **nucleosomes**, enabling compaction into chromatin; gene accessibility depends on the degree of chromatin condensation (euchromatin vs. heterochromatin). - **Post-transcriptional modification:**After transcription, primary RNA undergoes **capping**, **polyadenylation**, and **splicing** to form mature mRNA before translation. - **Ingestion of oocysts:**Consuming food or water contaminated with mature oocysts from cat feces. - **Ingestion of tissue cysts:**Eating undercooked or raw meat containing tissue cysts (especially from pigs, sheep, or goats). - **Congenital transmission:**Vertical transmission from an infected mother to her fetus across the placenta during pregnancy. - **Mutation:**Spontaneous changes in DNA introduce new alleles into a population, altering genotype frequencies. - **Natural selection:**Differential survival and reproduction favor certain genotypes over others, leading to changes in allele frequencies. - **Genetic drift:**Random fluctuations in allele frequencies, especially in small populations, can lead to loss or fixation of alleles. - **Gene flow (migration):**Movement of individuals between populations introduces new genetic material, affecting allele frequencies. - **Non-random mating:**Mating preferences (such as assortative or disassortative mating) can alter genotype distributions by increasing homozygosity or heterozygosity. - **Denaturation (94–98°C):**Double-stranded DNA is heated to separate into two single strands. - **Annealing (50–65°C):**Short DNA primers bind (anneal) to complementary sequences on the single-stranded templates. - **Extension (72°C):**Taq DNA polymerase synthesizes new DNA strands by adding nucleotides to the primers. - **Importance:**Taq DNA polymerase is **thermostable**, meaning it remains active at the high temperatures used in PCR, allowing continuous DNA synthesis without denaturation of the enzyme. - **Source:**It is derived from **Thermus aquaticus**, a thermophilic bacterium found in hot springs. - **Mediating virus entry:**Virus receptors on host cells enable **attachment and entry** of viruses, initiating infection. - **Tropism determination:**The type and distribution of receptors determine the **host range** and **tissue specificity** of the virus (e.g., HIV uses CD4 receptors found on T-helper cells). - **Disease pathogenesis:**Viral binding to specific receptors can influence **disease severity** and **clinical manifestations**. - **Therapeutic targets:**Virus-receptor interactions are targeted in **drug development** (e.g., receptor blockers to prevent viral entry). - **Vaccine development:**Knowledge of receptors aids in **vaccine design** by focusing immune responses on blocking virus attachment.
Answer: - **Chromatin remodeling** opens DNA (via HATs and remodeling complexes). - **Core promoter**: Usually has a **TATA box** (~25 bp upstream). - **Pre-Initiation Complex (PIC)**:**TFIID** binds TATA (TBP subunit). - **TFIIA**, **TFIIB**, **TFIIF** (brings RNA Pol II), **TFIIE**, and **TFIIH** assemble. - **TFIIH**:**Helicase**: Unwinds DNA. - **Kinase**: Phosphorylates RNA Pol II CTD → activates transcription. - **Promoter clearance**: RNA polymerase II escapes the promoter and starts RNA synthesis. ---
Answer: - **RNA polymerase II** synthesizes RNA 5’ → 3’, reading DNA 3’ → 5’. - Uses **ribonucleotide triphosphates (NTPs)**. - **Co-transcriptional processing**:**5’ Capping** (7-methylguanosine). - **Splicing** (introns removed, exons joined). - **Elongation factors** help polymerase move efficiently. ---
Answer: - **Polyadenylation signal (AAUAAA)** triggers termination. - **Cleavage** of RNA transcript downstream. - **Poly-A tail** (~200 As) added at 3’ end. - **RNA polymerase II** disengages from DNA. ---
Answer: - **5’ cap**: Protects RNA, aids translation. - **Splicing**: Removes introns; can allow **alternative splicing**. - **Poly-A tail**: Stabilizes RNA, helps export from nucleus. --- QUICK MEMORY TIP: **"CRP → PIC → TFIIH → Cap-Spin-Tail"**(Chromatin Remodeling → Pre-Initiation Complex → TFIIH activation → Capping, Splicing, Tail addition) --- --- **Protein Translation (8 Marks)** **Definition:**Translation is the process of **protein synthesis** where the **ribosome** reads the **mRNA codon sequence** and assembles **amino acids** into a polypeptide chain. --- ### **Stages of Translation** **1. Initiation** - mRNA binds to the **small ribosomal subunit**. - **Initiator tRNA** (carrying methionine) binds to the **start codon (AUG)**. - **Large ribosomal subunit** joins → functional ribosome formed. **2. Elongation** - Ribosome reads mRNA **one codon at a time**. - tRNA delivers **correct amino acids** based on codon-anticodon matching. - **Peptide bonds** form between amino acids. - Ribosome shifts (translocates) along the mRNA. **3. Termination** - Ribosome reaches a **stop codon** (UAA, UAG, UGA). - **Release factors** bind → polypeptide chain is released. - Ribosomal subunits dissociate. --- **Role of tRNA** - **tRNA** molecules bring amino acids to ribosome. - Each tRNA has:**Anticodon loop** (binds mRNA codon). - **Amino acid** attachment at 3′ end. --- **Location** - **Prokaryotes**: Cytoplasm (free ribosomes). - **Eukaryotes**: Cytoplasm and rough endoplasmic reticulum (RER). --- **Summary Line:** Translation = mRNA is decoded by ribosomes → tRNA delivers amino acids → polypeptide chain forms. --- **Explanation:** - **Albinism** is caused by a **recessive allele** ("a"). - Normal skin color is controlled by the **dominant allele** ("A"). - For two normal-skinned parents to have an albino child:Both parents must be **heterozygous** (Aa). - They carry one normal ("A") and one albino ("a") gene. - Their genotypes are **Aa × Aa**. - The albino child’s genotype must be **aa** (homozygous recessive). --- **Punnett Square:** --- **Probability of Albino Child:** - From the Punnett square:1/4 (25%) chance of **aa** (albino) per pregnancy. Thus, the **probability that the fifth child will be albino = 25%** or **1/4**. **Key Points:** - Each organism carries **two alleles** for each gene (one from each parent). - During **gamete formation** (meiosis), these two alleles **separate (segregate)**. - Each gamete receives **only one allele** for each gene. - Fertilization restores the **pair of alleles** (one from each parent). - Segregation is **random** — either allele is equally likely to go into any gamete. - Explains the **3:1 ratio** seen in monohybrid crosses (dominant to recessive traits). --- **Example:** - A pea plant with genotype **Tt** (tall) will produce gametes:50% with **T** (tall allele). - 50% with **t** (short allele). --- **Summary Line for Quick Memory:** In Mendel’s Principle of Segregation, **alleles separate during gamete formation**, ensuring each gamete carries only one allele for each trait. ### Development of a Live-Attenuated Vaccine: - **Isolate the Pathogen**: Obtain the virus or bacterium causing the disease. - **Attenuation**: Weaken the pathogen through culturing in non-human cells or genetic modification to reduce virulence. - **Safety Testing**: Test in animals to ensure no disease is caused but immunity is stimulated. - **Immune Response**: The weakened pathogen stimulates the immune system to produce antibodies and memory cells. - **Production**: Grow the pathogen in large quantities, purify, and prepare the vaccine. - **Clinical Trials**: Conduct phased human trials for safety, efficacy, and optimal dosing. - **Approval**: Regulatory bodies approve the vaccine after successful trials. - **Examples**: MMR, Yellow Fever, Polio (oral). **Summary**: Live-attenuated vaccines are made by weakening a pathogen to stimulate immunity without causing disease. Mechanism Underlying the Immune Protective Ability of Live-Attenuated Vaccines: - **Mimics Natural Infection**: The weakened pathogen in the vaccine replicates in the body, mimicking a natural infection without causing disease. - **Activation of Innate Immune Response**: Upon entry, the immune system detects the pathogen using pattern recognition receptors (PRRs) on immune cells like macrophages and dendritic cells, triggering an innate immune response. - **Adaptive Immune Response**:**Humoral Immunity**: B cells are activated, produce antibodies specific to the pathogen, and provide protection against future infections. - **Cell-Mediated Immunity**: T cells (both helper and cytotoxic) are activated to recognize and destroy infected cells. - **Memory Formation**: The immune system creates memory B cells and T cells, which "remember" the pathogen and respond more rapidly and effectively if exposed again in the future. - **Long-Term Immunity**: Because the vaccine contains a weakened but replicating form of the pathogen, it provides long-lasting immunity, often for years or even life. **Summary**: Live-attenuated vaccines stimulate both humoral and cell-mediated immune responses, leading to the creation of memory cells, which provide long-term protection against future infections. - **X-Inactivation (in females):** One of the two X chromosomes in female cells is randomly inactivated to equalize X-linked gene expression between males (XY) and females (XX). - **Upregulation of X in males:** In males, the single X chromosome is slightly upregulated to match the expression level of X-linked genes in females. - **X Duplication (in some species):** In some species, like marsupials, the male's X chromosome is duplicated to balance gene expression between the sexes. Here are four key areas where chromosomal studies are important in laboratory diagnostics: - **Genetic Disorders Diagnosis:** Chromosomal studies help diagnose genetic conditions like Down syndrome (trisomy 21), Turner syndrome (45,X), and Klinefelter syndrome (47,XXY) through karyotyping or FISH. - **Cancer Diagnosis and Prognosis:** Chromosomal abnormalities, such as translocations (e.g., Philadelphia chromosome in chronic myelogenous leukemia), are critical for diagnosing cancers and determining prognosis. - **Prenatal Screening:** Chromosomal analysis (e.g., amniocentesis or CVS) is used to detect chromosomal abnormalities in a fetus, such as trisomies or other congenital disorders. - **Infertility and Miscarriage Investigation:** Chromosomal studies can identify structural or numerical chromosomal abnormalities in couples with recurrent miscarriages or unexplained infertility, guiding appropriate management. These areas are vital for diagnosing and managing various genetic and health conditions. - **Skin color** - **Height** - **Eye color** - The Hardy-Weinberg Equilibrium principle states that allele frequencies in a population remain constant from one generation to the next in the absence of evolutionary influences, such as mutation, natural selection, genetic drift, and migration. This can be mathematically represented by the equation: - where p and q are the frequencies of the two alleles in the population. **Explain, using illustrations, how DNA replicates in a semi-conservative manner. (2 marks)****Explanation:** During DNA replication, each of the two strands of the DNA double helix serves as a template for the formation of a new complementary strand. After replication, each daughter DNA molecule consists of one original (parental) strand and one newly synthesized strand. This is called **semi-conservative replication** because each new DNA molecule conserves one of the original strands.**Illustration:**The DNA double helix unwinds.Each original strand serves as a template for a new complementary strand.The result is two DNA molecules, each with one old strand and one new strand. - **DNA polymerase:** Adds complementary nucleotides to the growing DNA strand during replication. It also has proofreading abilities to correct errors. - **DNA ligase:** Seals the nicks between Okazaki fragments on the lagging strand by forming phosphodiester bonds, ensuring the integrity of the newly synthesized DNA strand. - **DNA helicase:** Unwinds the DNA double helix by breaking the hydrogen bonds between the base pairs, creating single-stranded templates for replication. Transcription is the process by which an RNA molecule is synthesized from a DNA template. RNA polymerase binds to a specific region of DNA called the promoter, unwinds the DNA, and synthesizes a complementary RNA strand. This RNA strand is a copy of the coding sequence of the gene, which will later be translated into a protein. **List and explain the function of the three major RNA molecules present in the cell. (3 marks)** - **mRNA (messenger RNA):** Carries the genetic information from DNA in the nucleus to the ribosome, where it is used as a template for protein synthesis. - **tRNA (transfer RNA):** Delivers specific amino acids to the ribosome during translation, matching its anticodon with the codon on the mRNA. - **rRNA (ribosomal RNA):** Forms the structural and functional components of the ribosome, catalyzing the assembly of amino acids into protein chains during translation. Post-translational modification primarily occurs in the **endoplasmic reticulum (ER)** and the **Golgi apparatus**. The **Minimal Inhibitory Concentration (MIC)** is the lowest concentration of an antibiotic that inhibits visible bacterial growth. The following methods can be used to determine MIC: - **Broth Dilution Method:**Prepare a series of test tubes with varying concentrations of the antibiotic in a broth medium. - Inoculate each tube with a standard amount of the bacterial isolate. - After incubation, observe the lowest concentration where no visible bacterial growth is present. This is the MIC. - **Agar Dilution Method:**Prepare agar plates with different concentrations of the antibiotic. - Inoculate the plate with the bacterial isolate. - After incubation, examine the lowest concentration that prevents bacterial growth. This is the MIC. - **E-test (Gradient Method):**A strip containing a gradient of the antibiotic is placed on an agar plate inoculated with the bacterial isolate. - After incubation, the MIC is determined by reading the point where the growth inhibition intersects the strip. Acyclovir is selectively toxic to cells infected by **alpha-herpesviruses** due to its activation mechanism and viral specificity: - **Viral Thymidine Kinase Activation:**Acyclovir is a **nucleoside analogue**. It requires phosphorylation by the viral enzyme **thymidine kinase**, which is produced by the herpesvirus. - Once phosphorylated, acyclovir becomes acyclovir triphosphate, which inhibits viral DNA replication. - **Selective Incorporation into Viral DNA:**Acyclovir triphosphate competes with the natural nucleoside, guanosine, for incorporation into the viral DNA. - Once incorporated into the DNA, it causes chain termination, preventing further DNA replication. - **Minimal Host Cell Activity:**Human cells do not possess significant thymidine kinase activity, so acyclovir is not activated in uninfected cells. This ensures that acyclovir primarily targets infected cells, making it selectively toxic to herpesvirus-infected cells. --- **Development of Multidrug Resistance in Bacteria (10 marks)** - **Spontaneous Mutations:**Mutations in bacterial DNA can occur naturally during replication. These mutations may confer resistance to one or more antibiotics by altering the target sites or enzymes involved in drug metabolism. - For example, mutations in the gene encoding the target of the antibiotic can lead to reduced drug binding or drug inactivation, rendering the antibiotic ineffective. - **Horizontal Gene Transfer (HGT):****Conjugation:** Bacteria can transfer resistance genes through direct contact, especially via plasmids, between bacterial cells. This can spread resistance quickly within a population. - **Transformation:** Bacteria can uptake naked DNA from their environment that contains resistance genes, incorporating it into their genome. - **Transduction:** Bacteriophages (viruses that infect bacteria) can transfer resistance genes between bacterial species during infection. - **Selective Pressure from Overuse/Misuse of Antibiotics:**Overprescription or incorrect use of antibiotics creates a selective pressure that favors the survival of bacteria with mutations or resistance genes. - Bacteria resistant to one antibiotic survive and reproduce, passing on their resistance. When multiple antibiotics are used inappropriately, this can lead to the development of resistance to multiple drugs. - **Efflux Pumps and Enzyme Production:**Some bacteria develop **efflux pumps** that actively expel antibiotics from the cell, reducing the drug concentration inside the bacterium to sub-lethal levels. - Bacteria can also produce **enzymes** (e.g., beta-lactamases) that degrade antibiotics before they can exert their effect, leading to resistance. - **Biofilm Formation:**Many bacteria form biofilms on surfaces, which protect them from antibiotic penetration and immune system attacks. Biofilms act as a physical barrier, reducing the effectiveness of antibiotics and contributing to chronic infections that are difficult to treat. - **Cross-Resistance:**Some bacteria develop resistance to one class of antibiotics, and this can result in cross-resistance to other structurally or functionally similar antibiotics. This can occur due to the shared mechanisms of resistance, such as efflux pumps or mutations in common bacterial targets. - **Fitness Cost and Compensatory Mechanisms:**Some MDR strains may initially have a fitness cost associated with resistance. However, through compensatory mutations, bacteria may adapt to maintain resistance without incurring a significant fitness disadvantage, allowing them to thrive in antibiotic-treated environments. --- **Q.** b) The Notch mutation is a deletion on the X chromosome of *Drosophila melanogaster*. Female flies heterozygous for Notch have an indentation on the margin of their wings; Notch is lethal in the homozygous and hemizygous conditions. The Notch deletion covers the region of the X chromosome that contains the locus for white eyes, an X-linked recessive trait. Give the phenotypes and proportions of progeny produced in the following crosses:i) A red-eyed, Notch female is mated with a white-eyed male.*(2½ marks)*ii) A white-eyed, Notch female is mated with a red-eyed male.*(2½ marks)*iii) A white-eyed, Notch female is mated with a white-eyed male.*(2½ marks)*iv) Describe how dosage compensation of X-linked genes in *Drosophila melanogaster* occurs.*(2½ marks)* * - **Female (red-eyed, Notch heterozygous):** Xᴺ Xᴺ (Notch mutation) - **Male (white-eyed, Notch):** Xᴡ Y (white eyes are X-linked recessive) **Progeny:** - **Females:** Xᴺ Xᴡ (red eyes, Notch mutation) - **Males:** Xᴺ Y (red eyes, Notch mutation) **Phenotypic Proportions:** - 100% of the offspring will be **red-eyed** (since red eyes are dominant to white), and all will have the **Notch indentation** on their wings (since all will inherit one X chromosome with the Notch deletion). - **Female (white-eyed, Notch):** Xᴡ Xᴺ (Notch mutation) - **Male (white-eyed, Notch):** Xᴇᴡ Y **Progeny:** - **Females:** Xᴡ Xᴇ (white-eyed Notch females). - **Males:** Xᴇᴡ Y. White-eyed notches in males - **Female (white-eyed, Notch):** Xᴡ Xᴺ (Notch mutation) - **Male (red-eyed, normal):** Xᴺ Y (red eyes are dominant) **Progeny:** - **Females:** Xᴄᴺ Xᴡ (red eyes, Notch mutation) - **Males:** Xᴡ Y (white eyes, Notch mutation) **Phenotypic Proportions:** - 50% will be **red-eyed females** (with Notch mutation). - 50% will be **white-eyed males** (with Notch mutation). In **Drosophila melanogaster**, **dosage compensation** occurs through **X-chromosome doubling** in males to balance gene expression between males and females. In males (XY), the single X chromosome is **hyperactivated**, meaning it transcribes genes at a higher rate than the two X chromosomes in females (XX). This compensates for the difference in the number of X chromosomes, ensuring that both sexes produce similar amounts of X-linked gene products. The three major DNA conformations are **A-DNA**, **B-DNA**, and **Z-DNA**: --- **1. B-DNA (Biological DNA)** - **Structure:** Right-handed helix, 10 base pairs (bp) per turn. - **Helix Diameter:** ~2 nm. - **Base Tilt:** Nearly perpendicular to helix axis. - **Environment:** Forms under high hydration (physiological conditions). - **Importance:** Most common form in living cells; Watson-Crick model is based on B-DNA. --- **2. A-DNA** - **Structure:** Right-handed helix, 11 bp per turn. - **Helix Diameter:** ~2.6 nm (wider and shorter than B-DNA). - **Base Tilt:** Bases are tilted away from the helix axis (~20°). - **Environment:** Forms under dehydrated conditions (low humidity). - **Importance:** Often found in double-stranded RNA, DNA-RNA hybrids. --- **3. Z-DNA** - **Structure:** Left-handed helix, 12 bp per turn. - **Helix Diameter:** ~1.8 nm (narrow and elongated). - **Base Tilt:** Zig-zag backbone conformation (hence "Z"-DNA). - **Environment:** Forms in regions rich in alternating purine-pyrimidine (e.g., GC repeats), and under high salt or negative supercoiling. - - **Importance:** May play a role in gene regulation and chromatin remodeling. **Q. 24** A geneticist doctor, while walking through the corridor of Kenyatta National Hospital late one night, saw an enclave of gnomes. He was struck by the color of their beards, both male and female, which were blue. One week later, he spied another enclave of gnomes at Cost General Hospital, all with yellow beards. He collected a few yellow-bearded gnomes from Cost General Hospital and brought them back to Kenyatta National Hospital. Later, he discovered that several of the yellow-bearded and blue-bearded gnomes had mated. The offspring of these matings are all green-bearded. Assuming that the gnomes were true breeding for this trait: - Explain TWO (2) possible explanations for these results.*(2 marks)* - ii) Assuming that beard color is controlled by a single locus, give the genotypes of yellow-bearded, blue-bearded, and green-bearded gnomes. *(1 mark)* - iii) Assuming that a pathway of two distinct enzymes encoded by two different genes controls beard color, give the genotypes of yellow-bearded, blue-bearded, and green-bearded gnomes. *(Use A and O to designate the wild-type alleles and a and o to designate the loss-of-function alleles.)* - *(1 mark)* - iv) When two F1 green-bearded gnomes mate, they produce 64 green-bearded, 27 blue-bearded, and 22 yellow-bearded gnomes. Propose a hypothesis and test it against the Mendelian 1:2:1 genotypic ratio. - **Incomplete Dominance**: The blue and yellow beard colors may represent two different alleles that exhibit incomplete dominance. In this case, mating a blue-bearded gnome with a yellow-bearded gnome could result in offspring with an intermediate phenotype, which is green-bearded. - **Co-dominance**: Another possibility is co-dominance, where both the blue and yellow alleles are expressed simultaneously. The combination of both alleles in the offspring might result in a new phenotype, such as green-bearded gnomes, due to the mixing or co-expression of the blue and yellow pigments. - **Yellow-bearded gnomes**: **YY** (homozygous dominant) - **Blue-bearded gnomes**: **bb** (homozygous recessive) - **Green-bearded gnomes**: **Yb** (heterozygous) - **Yellow-bearded gnomes**: **AA OO** (homozygous for both wild-type alleles of the two genes) - **Blue-bearded gnomes**: **aa OO** or **AA oo** (homozygous for loss-of-function alleles in either the first or second gene) - **Green-bearded gnomes**: **Aa Oo** (heterozygous for both genes, which would lead to a partial function of both enzymes and result in a green beard color) - **Hypothesis**: The green-bearded color results from the heterozygous combination of two alleles, **Aa Oo**, where both genes have incomplete dominance or co-dominance. When two green-bearded gnomes (heterozygous **Aa Oo**) mate, the offspring follow Mendelian inheritance patterns. - **Test Against the 1:2:1 Genotypic Ratio**:The observed phenotypic distribution (64 green-bearded, 27 blue-bearded, and 22 yellow-bearded) closely matches a Mendelian 1:2:1 ratio.Green-bearded gnomes (genotype **Aa Oo**) are the heterozygous offspring, and they show the dominant traits of both alleles in an intermediate form. - Blue-bearded gnomes (genotype **aa OO** or **AA oo**) and yellow-bearded gnomes (genotype **YY OO**) represent the homozygous recessive and homozygous dominant combinations, respectively.The phenotypic ratio (64:27:22) suggests a genetic basis of two traits segregating independently, consistent with a **1:2:1 ratio** typically observed for incomplete dominance or co-dominance in a dihybrid cross.Based on the given offspring distribution, this hypothesis holds true, and the results are consistent with Mendelian genetics. **Introduction to PCR** Polymerase Chain Reaction (PCR) is a technique used to amplify specific segments of DNA. It was developed by Kary Mullis in 1983 and involves repeated cycles of DNA denaturation, primer annealing, and DNA extension using a heat-stable enzyme called Taq polymerase. This process produces millions of copies of a DNA segment, enabling the analysis of very small amounts of DNA. **Key Steps in PCR** - **Denaturation**: DNA is heated to 94-98°C, separating the double strands. - **Annealing**: The temperature is lowered to 50-65°C, allowing primers to bind to the target DNA. - **Extension**: The temperature is raised to 72°C, where Taq polymerase synthesizes the complementary strand. **Applications of PCR in Pharmaceutical and Medical Studies** - **Genetic Disease Diagnosis**PCR can identify mutations in genes (e.g., **BRCA1** for breast cancer), enabling early diagnosis of genetic disorders. - Used for newborn screening to detect genetic conditions such as cystic fibrosis. - **Pathogen Detection**PCR is essential for detecting viruses (e.g., **HIV**, **SARS-CoV-2**), bacteria (**TB**, **E. coli**), and fungi from clinical samples. - Allows early detection of infections even with low amounts of pathogen DNA. - **Pharmacogenomics**PCR can detect genetic variations that affect drug metabolism (e.g., **CYP450** gene variants) and predict patient responses to medications, enabling personalized treatment plans. - **Cancer Diagnosis and Monitoring**PCR detects mutations in cancer-related genes (**p53**, **EGFR**), aiding in early cancer detection and monitoring treatment progress. - Used in detecting minimal residual disease (MRD) in leukemia patients. - **Viral Load Quantification**In HIV and Hepatitis infections, PCR quantifies viral load, guiding therapy decisions and monitoring treatment effectiveness. - **Vaccine Development**PCR ensures the presence of desired genetic sequences in vaccine candidates, ensuring safety and efficacy. - Can detect viral or bacterial sequences to confirm the absence of contaminants. - **Forensic Medicine**PCR is used for DNA profiling in paternity testing, criminal investigations, and disaster victim identification by analyzing biological samples. - **Gene Therapy and Genetic Engineering**PCR verifies the successful incorporation of therapeutic genes in gene therapy and gene editing projects like **CRISPR**. - Ensures genetic modifications are accurate and safe. **Conclusion** PCR is a powerful tool with widespread applications in medical diagnostics, pharmaceutical research, genetic studies, and therapeutic development. Its ability to amplify small DNA samples enables significant advancements in disease detection, personalized medicine, and molecular biology. --- **b) Synthesis of Peptidoglycan** --- **Introduction** - **Peptidoglycan** is a rigid, mesh-like polymer forming the bacterial cell wall. - It provides mechanical strength and protects against osmotic lysis. - It is made of repeating units of **N-acetylglucosamine (NAG)** and **N-acetylmuramic acid (NAM)**, cross-linked by short peptide chains.
Answer: - **UDP-NAG** is synthesized from glucose. - UDP-NAG is converted into **UDP-NAM** by the addition of a lactyl group. - A **pentapeptide chain** (commonly containing L-alanine, D-glutamic acid, meso-diaminopimelic acid [or L-lysine], and two D-alanines) is sequentially attached to UDP-NAM.
Answer: - UDP-NAM-pentapeptide is linked to **bactoprenol phosphate** (a lipid carrier) in the membrane, forming **Lipid I**. - UDP-NAG then adds to Lipid I, forming **Lipid II** (NAM-pentapeptide + NAG).
Answer: - **Bactoprenol** carries the Lipid II complex across the membrane to the outer side (periplasmic side in Gram-negative bacteria).
Answer: - **Transglycosylase enzymes** catalyze the linking of the NAG of the new unit to the NAM of the existing chain via β-1,4-glycosidic bonds. - This extends the glycan strands (the sugar backbone).
Answer: - **Transpeptidase enzymes** (also called Penicillin-Binding Proteins, PBPs) form peptide bonds between the peptide chains attached to NAM residues. - This cross-linking provides strength and rigidity to the wall. - Example: Bond between the third amino acid (often diaminopimelic acid or lysine) and the fourth D-alanine of another chain, releasing one D-alanine (energy source). --- **Clinical Relevance** - **β-lactam antibiotics** (e.g., penicillin) inhibit transpeptidases, preventing cross-linking and weakening the cell wall, leading to bacterial lysis. - **Vancomycin** binds to D-Ala-D-Ala residues, inhibiting both transglycosylation and transpeptidation. **Q. 26** The dominant ability to taste PTC is due to the allele **T**. Among a sample of individuals from a population, **150 could detect** the taste of PTC and **65 could not**. Let **T = normal (taster)** and **t = non-taster** allele. *(Show all your calculations.)*Using the above information, determine: a) Calculate the **allele frequencies** of **T** and **t**.*(6 marks)* b) Determine the **percentage of heterozygotes** in the population.*(2 marks)* c) Find the **number of people** in the population who are **normal homozygotes** (i.e., tasters who are not carriers).*(2 marks)* --- - **Mutation in the HBB Gene**:Sickle cell anemia is caused by a **point mutation** in the **HBB gene** located on chromosome 11. - **Specific Base Substitution**:The mutation involves a **single nucleotide substitution** where **adenine (A) is replaced by thymine (T)** in the DNA sequence. - **Amino Acid Change**:This nucleotide change alters the codon, leading to the substitution of **glutamic acid** (hydrophilic) with **valine** (hydrophobic) at the **sixth position** of the β-globin chain of hemoglobin. - **Formation of Abnormal Hemoglobin (HbS)**:The mutant hemoglobin is known as **hemoglobin S (HbS)**, which tends to polymerize under low oxygen conditions. - **Red Blood Cell Deformation**:Polymerization of HbS distorts red blood cells into a **sickle shape**, making them less flexible, pr