Microbial Genetics Exam 1 Review PDF

1) Differences Between Bacterial and Eukaryotic Genomes ​ Size & Structure: Bacterial genomes are smaller and circular, while eukaryotic genomes are larger and linear. Plasmids: Bacteria often carry plasmids, which provide extra genes for antibiotic resistance or virulence. ​...

1) Differences Between Bacterial and Eukaryotic Genomes ​ Size & Structure: Bacterial genomes are smaller and circular, while eukaryotic genomes are larger and linear. Plasmids: Bacteria often carry plasmids, which provide extra genes for antibiotic resistance or virulence. ​ Introns & Operons: Bacterial genomes lack introns and contain operons (groups of genes transcribed together), whereas eukaryotic genes have introns and are usually individually regulated. ​ Compaction: Bacterial chromosomes are highly compacted using histone-like proteins, while eukaryotic DNA is wrapped around histones to form chromatin. ​ eukaryotic has a nucleus’s, bacterial dna is in nucleoid ​ one origin of replication in bacterial genome, and multiple origins of replication in eukaryotic 2) Genome Size & Bacterial Lifestyles: After sequencing the genome of several bacterial species, it was observed that the size of the genome varies and is related to the bacterial lifestyle (free-living, facultative pathogen, obligate pathogen/symbiont). Please describe which bacterial lifestyle has the largest genome and which has the smallest genome. Also provide an evolutionary explanation for the differences in genome size among these lifestyles. ​ Largest Genomes: Free-living bacteria have the largest genomes because they need to encode a wide variety of genes to adapt to diverse environments. ​ Smallest Genomes: Obligate pathogens or symbionts have the smallest genomes because they rely on their host for nutrients, allowing them to lose many genes over time. ​ Evolutionary Explanation: Genome reduction occurs in host-dependent bacteria due to gene loss over time. ​ Free-living bacteria maintain a larger genome to encode enzymes and transporters for independent survival. ​ Why did the obligate bacteria lose the genes and have a smaller genome? 3) Regarding DNA synthesis, in which end (5’ or 3’) of an existing sequence does the DNA polymerases enzyme add new nucleotides? Explain the reason why nucleotides are added to this specific end. DNA Polymerase Adds Nucleotides to the 3’ End ​ DNA polymerases add nucleotides to the 3’ hydroxyl (OH) end of an existing strand. This is because nucleotides are added via phosphodiester bonds, requiring a free OH group on the 3’ end. A phosphate is added to the hydroxyl group. The synthesis direction is 5’ → 3’. 4) Differences & Similarities Between In Vivo DNA Replication & PCR Similarities: ​ Both require DNA polymerase, template DNA, and nucleotides. ​ Both follow the 5’ → 3’ synthesis direction. Differences: Feature​ In Vivo DNA Replication​ ​ ​ ​ ​ PCR Enzyme​ DNA polymerase III​ ​ ​ ​ ​ Taq polymerase Initiation​ Requires primase & RNA primer​ Uses pre-designed DNA primers Separation​ Helicase opens DNA​ ​ ​ Heat denaturation Location​ Inside the cell​ ​ ​ ​ ​ In a test tubei What​ ​ Whole genome ​ ​ ​ ​ Replicates a smart part Temp​ ​ Body temp:37 celcius​​ ​ Recycles thru three temps: 95,98 PCR steps: Initiation(94-98), denaturation (50-65), extension (72) What do you need for PCR Reaction: -​ DNA template -​ Primers/Short DNA fragments that you design -​ Buffer/ Mg cofactor -​ Enzyme (DNA pol) -​ dnTPs (A,T,C,G all nucleotides) 5) Cis-Acting Elements & Trans-Acting Factors: Several regulatory elements are deployed by the cell to regulate gene expression. Define what are cis-acting elements and trans-acting factors. Provide at least one example from each during the process of transcription. ​ Cis-acting elements: INTRAMOLECULAR, Regulatory DNA sequences located on the same DNA molecule as the gene they control. ○​ Example: Promoters like the -10 (TATAAT) and -35 regions, Regulator ​ Trans-acting factors: COMES FROM ANOTHER MOLECULE, Proteins or RNAs that regulate gene expression and can act at a distance. ○​ Example: Sigma factors, which bind to promoters and recruit RNA polymerase 6) Coupled Transcription & Translation in Bacteria: Regarding transcription and translation, what does it mean when we say that transcription and translation are coupled in bacteria? Why does this coupling not occur in eukaryotes? ​ In bacteria, transcription and translation occur simultaneously because there is no nucleus and both processes can take place in the cytoplasm. ○​ Ribosomes bind to mRNA while it is still being transcribed. ​ In eukaryotes, transcription happens in the nucleus, and mRNA must be processed and exported before translation occurs in the cytoplasm. 7) Role of Sigma Factor in Transcription: The RNA polymerase holoenzyme in bacteria consists of five core subunits and a sigma factor. Describe the molecular function of the sigma factor during transcription. Explain how bacteria utilize sigma factors to regulate gene expression. ​ Sigma factor is part of the RNA polymerase holoenzyme and helps recognize promoters. ​ Once RNA polymerase binds, sigma factor dissociates after transcription starts. ​ Bacteria regulate gene expression by using alternative sigma factors:  ○​ σ70 (RpoD): Housekeeping genes. ○​ σ32 (RpoH): Heat shock response. ○​ σ54 (RpoN): Nitrogen metabolism. ​ what is a regalom? 8) Promoter Strength: Some genes in the bacterial genome are expressed at a higher rate compared to others. One of the mechanisms leading to this difference relies on the strength of the promoter. Describe the concept of promoter strength and the molecular mechanism behind this effect ​ The strength of a promoter refers to how efficiently RNA polymerase binds to it and initiates transcription. ​ Strong promoters have sequences closer to the consensus (TATAAT at -10 and TTGACA at -35). ​ If mutations weaken these sequences, transcription is less frequent. 9) Polycistronic mRNA in Bacteria: Explain what polycistronic mRNAs are and why genes are transcribed in such an organization, and why such organization is beneficial to bacteria. ​ Definition: Polycistronic mRNA in bacteria is a single mRNA encoding multiple proteins. ​ Why?: Bacteria group functionally related genes into operons, allowing coordinated regulation. ​ Benefit: Efficient gene regulation and faster protein production. 10) Degeneracy of the Genetic Code: The genetic code defines the rules by which information in DNA and RNA is translated into proteins. It is described as degenerate. Explain why the genetic code is considered degenerate and how it provides advantages to the cell. ​ The genetic code is degenerate, meaning multiple codons code for the same amino acid. ○​ Example: Proline (Pro) is encoded by CCU, CCC, CCA, and CCG. ​ Advantages: ○​ Reduces mutation effects, evolutionary flexibility ○​ Allows wobble base pairing, improving translation efficiency. ○​ Translation efficiency, less tRNAs 11) Shine-Dalgarno Sequence: TRANSLATION: Describe the function and general nucleotide composition of the ribosome-binding site, also known as the Shine-Dalgarno sequence. ​ Definition: A ribosome-binding site (AGGAGG) located upstream of the start codon in bacterial mRNA. ​ 5-10 nucleotides from start codon ​ Function: Aligns the ribosome with the start codon (AUG) for translation initiation. Recruits ribosome to start translation at the 16s. ​ gonna react with the 16s ribosomal subunit and gonna start translation with AUG  ​ Location – The Shine-Dalgarno sequence is 5–10 nucleotides upstream of the start codon (AUG) in bacterial mRNA. ​ Sequence Composition – It is rich in purines (AGGAGG is a common consensus sequence), but slight variations exist across different bacterial species. ​ Function – It pairs with a complementary sequence in the 16S rRNA of the 30S ribosomal subunit, aligning the ribosome correctly for translation initiation. ​ Initiation Process – Once aligned, the ribosome begins translation at the AUG start codon. 12) Trans-Translation & tmRNA: Transfer-messenger RNAs (tmRNAs) play a crucial role in the process of trans-translation. Explain how trans-translation works and why it is important for bacterial cells ​ Problem: If an mRNA lacks a stop codon, the ribosome stalls. ​ Solution: tmRNA (transfer-messenger RNA) rescues stalled ribosomes by: ​ have mRNA, tmRNA will come and insert 1 or 2 alanine and start adding small peptide tag, and this tag will be recognized by proteins and cut it. 1.​ Binding to the A-site of the ribosome. 2.​ Adding a peptide tag to the incomplete protein. 3.​ Directing it for degradation by proteases. 13) Stanley Falkow’s Molecular Koch’s Postulates: Describe the three major points of the molecular Koch’s postulates established by Stanley Falkow and explain why these postulates are important for studies of gene function. ​ Postulate 1: The gene must be associated with pathogenic bacteria. ​ Postulate 2: If the gene is inactivated, the bacteria must become less virulent. ​ Postulate 3: Restoring the gene should restore virulence. ​ Importance: These postulates help scientists determine the role of specific virulence genes in pathogens. ​ edward Lewis doing complementation test: two flies with a. White eye and bred them and saw offspring was white or red. If white, mutation on same gene. If red eye, mutations on two diff genes. Found complementation. Study Tips 1.​ Use Flashcards: Quiz yourself on key terms (e.g., operons, sigma factors, tmRNA). 2.​ Draw Diagrams: Illustrate DNA replication, transcription, and translation. 3.​ Practice Questions: Explain concepts out loud or to a friend. 4.​ Use Mnemonics: DNA polymerase III: “3 strikes, you’re out” (main enzyme, 3’-5’ exonuclease proofreading). 5.​ Sigma factors: “σ70 runs the show” (housekeeping genes). Here are mnemonics and diagrams to help with key concepts from the review. I’ll focus on confusing or detailed topics and include visual explanations where helpful. 1) DNA Replication: Leading vs. Lagging Strand Mnemonic: “5 to 3 is the key, but lagging is free!” ​ ​ DNA polymerase only adds nucleotides 5’ → 3’. ​ ​ The leading strand is continuous (one piece). ​ ​ The lagging strand is discontinuous (Okazaki fragments). Diagram 5' -------------> 3' (Leading Strand - Continuous) 3'