Bio93 Midterm 2 Review - Student Copy PDF

MIDTERM 2 REVIEW BIO 93 TRIO F24 BioSci Peer Tutors Victoria Vo, Sophie Cohen, Charlotte Phan, Hannah Song  Lecture 10: ATP ADP Hydrolysis of ATP is exergonic Hydrolysis o...

MIDTERM 2 REVIEW BIO 93 TRIO F24 BioSci Peer Tutors Victoria Vo, Sophie Cohen, Charlotte Phan, Hannah Song  Lecture 10: ATP ADP Hydrolysis of ATP is exergonic Hydrolysis of ATP can be coupled with endergonic reactions to make them spontaneous hydrolysis Phosphate Group Energy stored in the bonds of the phosphate group; repulsion between H2O negatively charged phosphate groups can easily be broken down to release energy ENERGY  Poll Question! Which one of these processes DO NOT require the use of energy? a) Endocytosis and exocytosis b) Activation of TKR (Tyrosine Kinase Receptors) c) Activation of GPCR (G-protein coupled receptors) d) Na+/K+ pump e) Hydrolysis of ATP  Allosteric Site: a binding site for Lecture 10: Enzyme activator/inhibitor elsewhere on the enzyme besides the active site Catalytic protein Greatly affected by pH and Temperature 37ºC is optimal for human enzyme General Structure Inhibition of Enzymes Competitive Inhibition Noncompetitive Inhibition Allosteric site e Steps st rat Sub 1. Substrate bind to enzyme at active Inhibitor Enzyme site 2. Induced fit Enzyme Substrate 3. Enzyme perform catalytic activities Inhibitor 4. Releases the product  Enzyme-Catalyzed Reaction Lecture 10: Enzyme - Lowered activation energy - ΔG does not change - Enzyme is not consumed in a reaction - Speeds up reaction Transition State Poll Question Which of the following PAIRS of statements are true? a) Noncompetitive inhibitors bind to the allosteric site b) Enzyme inhibitors increases the activation energy of a reaction c) Competitive inhibitors bind to both allosteric and active site d) Although substrates can bind to the allosteric site, it mostly binds to the active site e) Inhibitors and activators can bind to the allosteric site or active site, but substrates can only bind to the active site  Lecture 11: Cellular Respiration Intermediate Step Glycolysis Pyruvate Oxidation Citric Acid Cycle Oxidative Phosphorylation No ATP made Active transport via MPC 2 ATP 2 ATP 2 pyruvate molecules H+ H+ H+ H+ H+ 1 glucose molecule 2CO2 2CO2 2 acetyl coA 2 FADH2 H2O 2 NADH 6 NADH 26-28 ATP  Poll Question Which steps of cellular respiration include substrate level phosphorylation? a) Glycolysis and Pyruvate Oxidation b) Pyruvate Oxidation and Kreb Cycle c) The ETC and Chemiosmosis d) Glycolysis and the Citric Acid Cycle Poll Question At what step of cellular respiration does the body diverge to fermentation if Oxygen is not present? a) Before glycolysis b) Before pyruvate oxidation c) Before the Kreb cycle d) Before Oxidative Phosphorylation  Lecture 11: Fermentation Lactic acid fermentation - By muscle cells - Lactate can be exported to 2 pyruvates Reduction the liver and be reformed by NADH to glucose - Generate 2 ATP/glucose 2 lactate - No CO2 released No Oxygen Reduction Glycolysis by NADH Release of CO2 2 acetaldehyde 2 ethanol Oxygen Present Alcohol Fermentation - By bacteria and yeast - Release 2 CO2/glucose - 2 ATP/glucose Cellular Respiration Lecture 12 Photosynthesis Introduction: 6CO2 + 6H2O —----> C6H12O6 + 6O2 sunlight Site of photosynthesis: Chloroplast! Photosynthesis is a REDOX rxn!! H2O oxidized → O2 2 parts: 1) Light reactions and 2) calvin cycle CO2 reduced → sugar How is light absorbed? Pigments in thylakoid mem. have light absorbing Converting solar E to chemical E porphyrin ring Pt.1: Light Reaction Absorption spectrum directly correlates to Nadph action and ATP spectrum used up in calvin cycle chemiosmosis Pt. 2: Calvin Cycle Rubisco attaches C of CO2 to RuBP Incorporation of CO2 into RuBP Input: CO2 NADPH ATP Output: NADP+ ADP + Pi G3P Reduction to G3P Poll Question If ATP synthase on the thylakoid membrane was inhibited, what would happen to the pH of the thylakoid space? a) Increase b) Decrease c) Stay around same d) First increase, then decrease Poll Question If rubisco became denatured in Bob’s stroma, which of the following stages in the calvin cycle would be disrupted? a) Carbon reduction b) Carbon fixation c) RuBP regeneration d) G3P to glucose process  Lecture 13 The Molecular Basis of Inheritance GRIFFITH EXPERIMENT: R strain (nonvirulent) → mouse lives S strain (virulent) → mouse dies Heat-killed S strain → mouse lives Mix of heat-killed S strain and living R strain → mouse dies BACTERIAL TRANSFORMATION occurs as R cells take up chemical info from heat-killed S cells HERSHEY & CHASE EXPERIMENT Proteins were stained with radioactive sulfur Nucleic Acids were stained with radioactive phosphorus Radioactive phosphorus was found in bacteria pellet, meaning that DNA carried genetic info for bacteria to make new viruses DETERMINED DNA IS HEREDITARY MATERIAL - Radioactive pellet DNA STRUCTURE DNA Nucleotides # of adenine = # of thymine # of cytosine = # of guanine A pairs with T; C pairs with G ○ Purine must pair with pyrimidine! A and T have two hydrogen bonds C and G have three hydrogen bonds DNA STRUCTURE WILKINS AND FRANKLIN X-ray crystallography → determined structure of DNA WATSON AND CRICK DNA is double helix DNA is made out of two strands Double helix has hydrophobic interior (bases) and hydrophilic exterior (phosphate backbone) POLL QUESTIONS! In a single strand of DNA, guanine comprises 27%. How much thymine is present? a. 73% b. 36.5% c. 27% d. Impossible to determine POLL QUESTIONS! What is the arrow pointed towards in this image? a. Origin of replication b. Replication bubble c. Replication fork d. Answers A and C POLL QUESTIONS! Which image displays the correct DNA replication method? a. b. c. POLL QUESTIONS! Which of the following DNA molecules has a site that is most likely to be an origin of replication? a. CTGACCTTACGATCAGGCCC GACTGGAATGCTAGTCCGGG b. AACCCGGCGTACGTGGCCAA TTGGGCCGCATGCACCGGTT c. GCTAGAAATTATTAGTCTGG CGATCTTTAATAATCAGACC  Lecture 14 DNA Replication and Repair DNA REPLICATION DNA is semiconservative (DNA contains 1 strand of parent DNA and 1 strand of daughter DNA) DNA replicates in the 5’ → 3’ direction Replication starts at ORIGINS OF REPLICATION ○ TATA box DNA SYNTHESIS ALWAYS OCCURS FROM THE 5’ TO 3’ DIRECTION DNA Polymerase can only add new nucleotides to 3’ hydroxyl group of existing DNA strand 3’ end of DNA strand has free hydroxyl group → essential for chemical rxn that adds new nucleotides to chain PROTEIN FUNCTION Helicase Unwinds double helix at replication forks by breaking H bonds SSBP (Single-strand Binding Protein) Stabilizes single stranded DNA Topoisomerase Relieves overwinding ahead of replication fork Primase Synthesizes RNA primer at 5’ of leading strand and 5’ of Okazaki fragment DNA Polymerase III Adds DNA nucleotides to synthesize a new DNA strand DNA Polymerase I Removes RNA primer, replacing them with DNA nucleotides Ligase (glue!) Joins Okazaki fragments of lagging strand and 3’ end of DNA to primer DNA PROOFREADING & REPAIR Mismatch pair of DNA - repair enzymes correct errors in base pairing Nucleotide Excision Repair - a nuclease cuts out and replaces damaged stretches of DNA POLL QUESTIONS! During DNA replication in a human cell, helicase begins malfunctioning. What effect would this have on the replication process? a. The replication fork would continue to progress, but the leading strand would be synthesized at a slower rate b. Primase would synthesize primers, but they would not bind to DNA due to the lack of exposed single strands c. DNA polymerase III would be unable to synthesize new DNA strands because the parental DNA remains double stranded POLL QUESTIONS! Peter the Anteater is testing DNA proofreading and repair mechanisms in a lab. He introduces a mismatch base during synthesis. If the mismatch repair mechanism is functioning normally, what is the most likely outcome? a. The mismatch will be recognized and corrected by repair enzymes, restoring the correct sequence b. DNA ligase will detect the mismatch and remove the incorrect nucleotide c. The mismatch will remain uncorrected, potentially leading to a permanent mutation POLL QUESTIONS! A scientist is studying DNA synthesis in cells with a mutation in DNA polymerase III, which prevents it from functioning properly. What would likely happen to the replication process? a. DNA synthesis on the leading strand would stop, but the lagging strand synthesis would continue as normal b. DNA synthesis would start at the origin but would be unable to elongate effectively, resulting in incomplete DNA strands c. The Okazaki fragments on the lagging strand would be directly joined together without further processing. POLL QUESTIONS! Peter the Anteater is conducting an experiment - he has created four new drugs and has curated the following graph showcasing the results of his experiment. What does Drug C most likely inhibit? a. Topoisomerase b. Helicase c. DNA polymerase III d. Nuclease Session Evaluation Form https://forms.gle/BScPeafpb5Kaangp8 Lecture 15 Transcription  Gene Expression - Protein links genotype and phenotype - Genes produce proteins via transcription and then translation - Transcription is the production of mRNA from DNA - Translation is formation of a protein sequence based off the mRNA. This occurs at ribosomes - 3 base pairs code for an amino acid (codon)  Prokaryotic vs Eukaryotic Prokaryotic - Lacks nuclei, transcription and translation are coupled - Ribosomes attach to mRNA molecule while transcription is still in progress Eukaryotic - transcription occurs in the nucleus - translation occurs mainly at ribosomes in the cytoplasm - primary transcript is modified during RNA processing before the finished mRNA is exported to the cytoplasm  Initiation Elongation - RNA polymerase binds to - RNA polymerase builds an promoter region, found near RNA molecule out of the beginning of a gene complementary nucleotides, - RNA polymerase separates making a chain that grows the DNA strands, providing from 5' to 3' the single-stranded template needed for transcription.  Termination - terminators sequences signal that the RNA transcript is complete - transcript is released from the RNA polymerase - Created pre-mRNA in eukaryotic cells Question The following diagram represents the flow of genetic information from DNA to protein. Which of the following describes an event that occurs during Step 1 in the diagram? a. tRNAs and rRNAs are used to add amino acids to a drawing polypeptide chain b. pre-mRNAs are processed to include only certain exons c. pre-mRNAs are produced by RNA polymerase using a template molecule d. mRNAs are translated into pre-mRNA in the cytosol Question Coding Strand: 3’ A T G C G A C T 5’ Non-coding Strand: 5’ T A C G C T G A 3’ Which of the following represents the correct mRNA molecule that will result from the transcription of the DNA molecule? A. 5’ A U G C G A C U 3’ B. 5’ A G U C G C A U 3’ C. 5’ U C A G C G U A 3’ D. 5’ U A C G C U G A 3’ Lecture 16 Translation  RNA Processing Prokaryotes RNA transcripts act as mRNAs right away Eukaryotes needs modifications to the pre-mRNA ○ a) both ends of primary transcript altered Functions? Facilitate mRNA export + protect RNA + help attachment of ribosomes ○ b) RNA splicing (by spliceosomes)  Structure: 3D folding of single RNA via tRNA H-bonds Function: each tRNA carries a specific amino acid depending on its anticodon aminoacyl-tRNA synthetase matches each tRNA to its specific amino acid IMPORTANT: each tRNA is specific to its amino acid, but the anticodon for that amino acid can pair with multiple codons on the mRNA because of the wobble effect” Anticodon read 3’-5’ since mRNA is 5’-3’  Translation: 3 steps 1) Initiation Small subunit 1st binds to initiator tRNA Then binds to 5’ cap till AUG Anticodon H-bonds with AUG Ribosome notes: 2 subunits ribosomal RNA  Translation cont. Several ribosomes translate 1 mRNA 2) elongation molecule: polyribosomes Codon recognition in A-site Peptide bond formation ○ Peptide chain passed to A-site Translocation: A→ P, P→ E 3) Termination A site accepts release factor Adds H2O instead of amino acid to chain to promote hydrolysis Ribosomal subunits and other components dissociate  Location SRP protein binds to signal peptide SRP binds to receptor at ER membrane pore SRP detaches Signal-cleaving enzyme cuts signal peptide at the pore receptor Polypeptide folds into final 3D structure inside ER lumen Point mutations: one base-pair change in a gene Point mutation 2 types: Nucleotide-pair substitution: substitution of Insertions and deletions: causes frameshift different nucleotides for 1 base pair mutation 1. Silent mutations *more disastrous effect (especially when it’s 2. Missense mutations not a group of 3) *one amino acid change that can be good, neutral, or bad 3. Nonsense mutations *STOP this NONSENSE Pollev Q Why is there on average 35-45 tRNAs for 61 codons? a) tRNAs are not involved in protein synthesis, so the number does not correlate with the amino acid count b) There are multiple anticodons for each amino acid c) Each tRNA can recognize multiple codons for the same amino acid due to wobble pairing d) There’s only one codon for each amino acid Pollev Q What kind of mutation occurred for sickle cell anemia? a) Nonsense b) Missense c) Silent d) insertion  Lecture 17 Cell Division Pt. 1 Where do we see cell division? Reproduction: when a unicellular prokaryotic cell like an amoeba divides Tissue renewal: replacement and repair of cells ○ ex) wound healing Growth and development: enables multicellular eukaryotic organisms to develop from a single-cell fertilized egg Chromosome structure A chromosome is a single DNA molecule made up of proteins wrapped around by strands of DNA Sister chromatids are identical copies of a chromosome Centromeres are platforms for kinetochores (protein complex attachment point for spindle microtubules) Cell cycle: 2 main phases 1) Interphase G1: cell grows physically larger, copies organelles, synthesizes necessary macromolecules S: DNA replication G2: more cell growth + mitosis preparation  Cell Cycle cont. to M Phase Chromosome Nuclear Chromosomes Sister Daughter Division of condensation envelope line up chromatids nuclei form cytoplasm fragments metaphase pulled nuclear Nucleoli plate apart envelopes fragments Microtubules and nucleoli Animals: attach to cleavage furrow Mitotic spindles kinetochores start extending Plant cells: cell Nonkinetochor plate e microtubules interact Pollev Q Given this diagram, during anaphase are chromosomes “reeled in” or “walk in”? a) Reeled in b) Walk in Pollev Q Given this diagram, during anaphase are chromosomes “reeled in” or “walk in”? a) Reeled in b) Walk in-dynein in the kinetochore “walks” toward neg. end of microtubule Pollev Q In what phase does chromosomes condense? a) Prophase b) Prometaphase c) Metaphase d) G2 phase  Lecture 18 Cell Division Pt. 2 Checkpoints G1 checkpoint - Cell size, nutrients, growth factors, DNA damage - Hormones and signals trigger CDK4/6 and Cyclin D + CDK2 and Cyclin E - These complexes work together to remove Rb protein (an inhibitor) - Removal of Rb = bypass the G1 checkpoint G2 Checkpoint - DNA damage, DNA replication completeness MPF - Sufficient amount of MPF complexes can bypass 1) Hormones trigger production of cyclin in G2 the G2 checkpoint 2) Accumulation of cyclin binds to Cdk 3) Cdk + cyclin forms MPF (mitosis promoting M Checkpoint factor) - Ensures kinetochores are attached to 4) Enough MPF = pass G2 5) After passing G2, cyclin gets degraded but chromosomes at metaphase plate during Cdk remains in the cell prometaphase Proto-oncogene and Oncogene - Proto-oncogene: normal gene that plays a role in regulating cell division and growth - encode proteins that function to stimulate cell division, inhibit cell differentiation, and halt cell death - Oncogene: mutated version of proto-oncogene, cause cells to divide abnormally and rapidly (cancer cells) - increased cell division, decreased cell differentiation, and inhibition of cell death Hallmarks of cancer cells Loss of Anchorage Dependence Loss of Density Dependent Inhibition Cancer cells can metastasize to other parts of the Cancer cells can divide uncontrollably beyond a body through the circulatory system monolayer of cells, causing clumps and cell masses  Growth hormone Regulating Genes of Cell Division nucleus ras G T Cell Function: regulates cell division by P producing the Ras protein Ras Gene Mutation of Ras (point mutation): result in hyperactive Ras protein, even without growth factors Tumor Function: Gene that code for proteins that initiate DNA repair, regulate the cell cycle, and can trigger apoptosis when cells are detected to be damaged Suppressor Genes (TSG) Healthy colon Small benign polyp Large benign adenoma Malignant tumor p53 “Guardian of the Genome” The final safeguard to tumor development Loss of TSG APC Activation of ras Loss of p53 Loss of TSG DCC Meiosis 1. Each homologous chromosome (one from each Chromosome replication parent) duplicate to form 2 sets of chromosomes, with 2 sister chromatids Homologous chromosome 2. Meiosis 1 a. Formation of chiasmata b. Crossover and separation of homologous chromosomes c. Formation of 2 non identical daughter cells Meiosis 1 3. Meiosis 2 a. Separation of sister chromatids of each chromosome b. Formation of 4 net genetically distinct Meiosis 2 daughter cells Genetic Variations Independent Assortment Crossover of Chromosome Random Fertilization Definition Definition Definition The genetic exchange Pairs of homologous Random combination of between non-sister chromosomes can line up egg and sperm can be chromatids of the same independently at the fertilized at anytime homologous chromosome metaphase plate PollEv Which of the following is the correct definition of a homologous chromosome? a) A chromosome not directly involved in determining the sex of an offspring b) A pair of sister chromatid of the same chromosome c) A pair of chromosome that have the same length and gene loci d) The chromosome that is responsible for determining the sex of the offspring QUESTION If a cell bypasses the G1 checkpoint without repairing damaged DNA, what could be the consequence? a. The cell will replicate damaged DNA, increasing the risk of mutations b. The cell will revert to the G0 phase c. The cell will enter apoptosis immediately d. The cell will continue to cytokinesis QUESTION Which protein complex is crucial for the cell to pass from the G2 checkpoint into mitosis? a. Cyclin D-Cdk4 complex b. Cyclin E-Cdk2 complex c. Maturation Promoting Factor (MPF) d. p53 protein Session Evaluation Form https://forms.gle/BScPeafpb5Kaangp8