Lehrer BIOSC 0160 Exam 1 Study Guide PDF

Lehrer BIOSC 0160 Exam 1 Study Guide Experiments that demonstrated that the genetic material is contained in the DNA (13.1) ​ circumstantial evidence: ○​ DNA is located in the nucleus and chromosomes ○​ somatic cells have twice as much DN...

Lehrer BIOSC 0160 Exam 1 Study Guide Experiments that demonstrated that the genetic material is contained in the DNA (13.1) ​ circumstantial evidence: ○​ DNA is located in the nucleus and chromosomes ○​ somatic cells have twice as much DNA as reproductive cells ○​ amount of DNA differs between species ​ three experiments: ○​ Frederick Griffith’s experiment ​ bacterial transformation is caused by a certain biomolecule ○​ Oswald Avery’s experiment ​ bacterial transformation is caused by DNA ○​ Alfred Hershey and Martha Chase’s experiment ​ DNA, and not protein, is associated with genetic material ​ Griffith’s experiment ○​ hypothesis: material in dead bacterial cells can genetically transform living bacterial cells ○​ Griffith used two strains of pneumococcus: the S strain (virulent, kills mice) and the R strain (not virulent, does not kill mice) ○​ when he injected mice with heat-killed S-type pneumococcus cells, the mice did not die ○​ when he injected mice with a mixture of heat-killed S-type cells and living R-type cells, the mice died ○​ he concluded that some of the R-type cells were converted to S-type cells through bacterial transformation ○​ conclusion: a chemical substance from one cell is capable of genetically transforming another cell  ○​ ​ Avery’s experiment ○​ hypothesis: the chemical nature of the transforming substance from pneumococcus is DNA ○​ Avery treated dead S-type cells with different enzymes, including RNase (which destroys RNA), protease (which destroys proteins), and DNase (which destroys DNA) in separate mixes ○​ each mix was then added to R-type cells ○​ the R strain became virulent in the mixes treated with RNase and protease, but not the mix treated with DNase, meaning that DNA had to be present in order for bacterial transformation to occur ○​ conclusion: because only DNase destroyed the transforming substance, the transforming substance must be DNA ○​  ○​ ​ Hershey and Chase’s experiment ○​ hypothesis: either component of a bacteriophage, DNA or protein, might be the hereditary material that enters a bacterial cell to direct the assembly of new viruses ○​ Hershey and Chase studied bacteriophage T2, which infects the bacterium E. coli ○​ the virus injects materials into the bacterial cell to replicate itself ○​ in their experiment, they labelled viral DNA and viral proteins with radioactive isotopes ​ DNA was labeled with ³²P, and proteins were labeled with ³⁵S ○​ the viruses infected bacteria and then the bacterial cells were separated from the rest of the culture with a centrifuge ○​ ³²P was found associated with the pellet (containing bacteria), while ³⁵S was found associated with the supernatant (containing viruses) ○​ conclusion: DNA, not protein, enters bacterial cells and directs the assembly of new viruses ○​ ○​  ○​ DNA structure (13.2) ​ DNA is a polymer of nucleotides and consists of a phosphate group, a deoxyribose sugar, and a nitrogenous base ○​ ​ DNA is supercoiled within the cell ○​ ​ Erwin Chagraff found that in a DNA strand, A + G = T + C ○​ the amount of A = the amount of T ○​ the amount of G = the amount of C ​ Rosalind Franklin used X-ray diffraction and discovered that DNA takes the shape of a double helix with 10 nucleotides in each turn, along with other physical properties ​ Watson and Crick built the DNA structure using Chagraff and Franklin’s observations ​ 5 defining features of the DNA molecule: ○​ 1. it is a double-stranded helix with a sugar-phosphate backbone on the outside and base pairs on the inside ​ phosphate groups repel each other and therefore face the outside ○​ 2. the strands are held together by complementary base pairing, where A pairs with T and C pairs with G ​ the stacked nitrogenous bases are weakly attracted to each other, creating more stability in the structure of DNA ○​ 3. the strands are antiparallel, meaning they run in opposite directions ​ a phosphate group is attached to the 5’ carbon of the deoxyribose sugar, while a hydroxyl group is attached to the 3’ carbon of the deoxyribose sugar ​ ○​ 4. DNA is a right-handed helix  ​ ○​ 5. DNA has major and minor grooves (caused by the uneven spacing of the sugar-phosphate backbone) ​ in the grooves, the outer edges of the nitrogenous bases are exposed, allowing additional hydrogen bonds to form ​ ​ functions/properties of DNA: ○​ stores genetic information in its base pairs ○​ is susceptible to mutations (permanent changes in the sequence of nucleotides, thereby changing the information encoded) ○​ is replicated in the cell using complementary base pairing ○​ codes for a phenotype such as a protein DNA replication (13.3) ​ there were three replication patterns of DNA that we thought were possible ○​ semiconservative - each parent strand serves as a template ○​ conservative - parent strands are templates but do not contribute to the daughter strand ○​ dispersive - parent strands are templates but daughter strands contain parts of the parent strand and the new molecule ○​ ​ Matthew Meselson and Franklin Stahl’s experiment ○​ hypothesis: DNA replicates semiconservatively (as opposed to conservatively or dispersively) ○​ Meselson and Stahl used the heavy non-radioactive isotope of nitrogen called ¹⁵N to grow a culture of E. coli; they grew another culture of E. coli in the isotope ¹⁴N ○​ when DNA extracts were taken from the E. coli cultures, they were combined and centrifuged in a salt solution that formed a density gradient ○​ initially, two separate bands formed in the centrifugation tube; one represented the DNA from the ¹⁵N culture (it was denser so it formed towards the bottom), and one represented the DNA from the ¹⁴N culture (formed towards the top) ○​ afterwards, they grew another ¹⁵N E. coli culture and transferred the bacteria to a ¹⁴N medium; they allowed the bacteria to continue growing ○​ after one generation, intermediate DNA was observed (had a density that was between the heavier and lighter DNA), and this could only be explained by semiconservative replication ○​ after two generations, half of the DNA was intermediate and half was light; again, this could only be explained by semiconservative replication  ○​ conclusion: this pattern could have only been observed if each DNA molecule contains a template strand from the parental DNA, so DNA replication is semiconservative ○​ ​ 3 main steps of DNA replication: ○​ initiation - double helix is unwound by DNA helicase, separate strands are stabilized by single-stranded binding proteins, and each strand is primed with RNA primers synthesized by a primase ○​ elongation - synthesis of new strand by adding complementary dNTPs in direction 5’ to 3’, catalyzed by DNA polymerase; the reaction releases 2 phosphates from each dNTP, releasing energy for the reaction ○​ termination - end of synthesis; RNA primers are replaced by DNA via DNA polymerase, and the fragments are connected by DNA ligase ​ DNA replication starts with the binding of the pre-replication complex at the origin of replication (ori) ○​ the complex contains several proteins like the DNA polymerase and it binds to specific sequences of nucleotides on the ori ​ DNA helicase unwinds the double helix, while single-stranded binding proteins keep the template strands separated ​ DNA polymerase needs an RNA primer (short single strand of RNA) to begin replication ○​ the primer is synthesized by primase ​ once the primer is synthesized, DNA polymerase starts on the 3’ end of the primer, building in the 5’ to 3’ direction and stopping at the end of the template strand ○​ DNA polymerase is stabilized by a sliding DNA clamp ​ the leading strand grows continuously forward at the 3’ end, while the lagging strand grows discontinuously backwards ○​ one primer is needed for the leading strand, while multiple primers are needed for the lagging strand ○​ short strands formed as a part of the lagging strand are known as Okazaki fragments ​ in the lagging strand, a different DNA polymerase replaces the RNA primers with DNA, and then DNA ligase connects the Okazaki fragments ​ ​ telomeres - highly repeated sequences at the end of the linear chromosomes in many eukaryotes ○​ recognized as breaks in the DNA strands ○​ special proteins bind to the telomeres, so they are not recognized as breaks in the chromosome  ○​ ​ chromosomes lose part of their telomeres with each cell division ○​ after the RNA primer is removed, the lagging strand does not have a 3’ to extend to, so the fragment is removed ○​ after too many cell divisions, genes are lost and the cell dies ○​ some cells avoid losing telomeres using an enzyme called telomerase, which contains an RNA sequence used as a template and adds missing parts of the telomeres ​ expressed in stem cells, germ cells, and 90% of human cancers ​ Reparation of errors in DNA (13.4) ​ errors in DNA sequences: ○​ DNA polymerases occasionally insert incorrect bases and create mismatches between base pairs (typically 1 incorrect base per 100,000 bases replicated)  ○​ bases can change spontaneously ○​ bases can be damaged by external mutagen agents (e.g. radiation, chemicals) ○​ mismatches between pairs/damaged bases, if not repaired, can lead to mutations ​ DNA sequence correcting mechanisms: ○​ proofreading mechanism - during replication, the replication complex proteins excise the incorrect nucleotide and DNA polymerase III adds the correct base pair ​ ○​ mismatch repair mechanism - right after replication, mismatch proteins excise the incorrect nucleotide and some neighboring ones, DNA polymerase I adds new nucleotides, and a ligase repairs the nick ​ ○​ excision repair mechanism - during the life of the cell, mismatch proteins excise the incorrect nucleotide and some neighboring ones, DNA polymerase I adds new nucleotides, and a ligase repairs the nick ​ mismatches during the life of the cell can be a result of radiation or chemical exposure/damage from the environment or to the DNA itself; these damages are not associated to replication  ​ Experiments that showed that genes code for proteins (14.1) ​ Archibald Garrod: “Do genes determine enzymes?” ○​ Garrod observed alkaptonuria (black urine) in children whose parents were often first cousins (sharing ⅛ alleles) ​ he believed alkaptonuria was caused by a recessive mutant allele ○​ he isolated homogentisic acid (accumulates in urine and turns it black) and noticed that it had a similar structure to tyrosine (amino acid) ​ he hypothesized that homogentisic acid was a breakdown product of tyrosine and the enzyme that was needed to break down homogentisic acid was not being produced in kids with alkaptonuria; a normal human allele was needed to synthesize the enzyme ○​ conclusion: one gene codes for one enzyme  ○​ ​ George Wells Beadle and Edward Lawrie Tatum: “Genes do determine enzymes” ○​ Beadle and Tatum grew wild-type Neurospora fungus in a minimal medium that allowed for fungal growth ​ Neurospora is a model organism; it is haploid (all alleles are expressed, no heterozygotes) and it has simple genomes ○​ they exposed the fungi to X-rays (which mutate genes), and some strains could no longer grow on the minimal medium ○​ they hypothesized that the X-rays caused mutations in the genes that code for the production of enzymes; these enzymes are needed to synthesize additional nutrients required for the strains to grow ○​ for each mutant strain, they found a single compound that supported its growth when added to the medium, suggesting that mutations have simple effects ○​ conclusion: confirmed Garrod’s hypothesis that one mutated gene impacted one specific enzyme ​ Adrian Srb and Norman Horowitz’s experiment ○​ hypothesis: each gene determines an enzyme in a biochemical pathway ○​ they isolated Neurospora strains that could not survive without the amino acid arginine in their growth medium ○​ they added specific compounds hypothesized to be in the arginine biosynthetic pathway to the medium, allowing them to identify a series of steps in the pathway  ○​ ○​ ​ the idea of a one-gene/one-enzyme relationship has been modified because many proteins like enzymes are composed of more than one polypeptide chain; it is more correct to say one gene corresponds to one polypeptide Central dogma of molecular biology (14.2) ​ central dogma of molecular biology describes information flow from genes to proteins ​ includes: ○​ DNA replication ○​ transcription - DNA is copied into a complementary RNA sequence  ○​ translation - RNA sequence is used to create the amino acid sequence of a polypeptide ​ exceptions: viruses having RNA instead of DNA as genetic material (undergoing RNA replication), retroviruses copying DNA from RNA and using host’s transcriptional machinery to make more RNA, telomerase using reverse transcription using an RNA template Transcription: from DNA to RNA (14.3) ​ DNA serves a template for its own replication and as a template for the synthesis of RNA ○​ based on complementarity but uracil (U) replaces thymine (T), and RNA has a ribose sugar rather than deoxyribose ​ occurs within the nucleus in eukaryotes ​ 3 steps of transcription: ○​ initiation: RNA polymerase recognizes a promoter ​ one strand of DNA serves as the template ​ promoter - specific DNA sequence that tells RNA where to start transcription, which strand to use, and the direction of synthesis ​ binding proteins direct RNA polymerase onto the promoter (sigma factors in prokaryotes/transcription factors in eukaryotes) ○​ elongation: RNA polymerase adds nucleotides in the 5’ to 3’ direction ​ RNA polymerase unwinds the DNA ​ reads 3’ to 5’ and adds 5’ to 3’ ○​ termination: RNA polymerase reaches the termination site and releases the DNA ​ resulting RNA is complementary to the template DNA and similar to the coding strand except for U replacing T  ○​ ​ DNA and RNA polymerases both synthesize in the 5’ to 3’ direction and use (d)NTPs as substrates; however, RNA polymerase can unwind DNA unlike DNA polymerase, RNA polymerase does not require a primer while DNA polymerase does, and RNA polymerase does not proofread Processing of mRNA in eukaryotes (14.4) ​ differences between prokaryotic and eukaryotic gene expression  ○​ ​ nucleic acid hybridization: ○​ ​ nucleic acid hybridization in eukaryotes leaves stretches of DNA that are not hybridized due to missing a complementary sequence to RNA (these stretches are called introns) ​ introns are transcribed in the nucleus (pre-mRNA) but are also spliced out in the nucleus (mRNA) ​ ​ small nuclear ribonucleoproteins (snRNPs) bind to pre-mRNA to facilitate splicing ​ after cuts are made between an upstream exon (upstream = an exon towards the 5’ end) and intron, the intron forms a closed loop and the 3’ OH group of the upstream exon reacts with the 5’ phosphate of the downstream exon (downstream = an exon towards the 3’ end)  ○​ ​ both ends of pre-mRNA are also modified in the nucleus ○​ a cap is added at the 5’ end to export mature mRNA to the cytosol, facilitate the binding of mRNA to the ribosome for translation, and protect the mRNA from being digested by ribonucleases ○​ a poly-A tail is added to the 3’ end to export mRNA and it contributes to mRNA recognition/stability in the cytosol ​ The genetic code: from mRNA to polypeptide (14.5) ​ discovery of the genetic code: “How can 4 bases, A, U, G, and C, code for 20 different amino acids?” ○​ a triplet code (3-letter codons) was considered likely; 4 ✕ 4 ✕ 4 = 64 codons, which is more than enough to encode 20 amino acids ​ Marshall W. Nirenberg and J. Heinrich Matthaei hypothesized that an artificial mRNA containing only one repeating base will direct the synthesis of a protein containing only one repeating amino acid ○​ they synthesized RNA called poly U that only contained uracil (with UUU codons) and discovered the codons code for phenylalanine through radioactive marking ○​ ​ the genetic code is redundant (multiple codons can code for the same amino acid), unambiguous (each codon only codes for one amino acid), and nearly universal (shared by almost all organisms) ○​ codons are read 5’ to 3’, and there are 64 codons in total ​ 61 sense codons - code for amino acids ​ 1 start codon - initiation signal for translation and codes for Met ​ 3 stop/nonsense codons - termination signals for translation  ​ mRNA translation (14.6) ​ three types of RNA involved in transcription and translation: ○​ messenger RNA (mRNA) - determines sequence of amino acids in the polypeptide ○​ ribosomal RNA (rRNA) - forms part of the ribosome (which reads mRNA and synthesizes the polypeptide) ○​ transfer RNA (tRNA) - binds specific amino acids and carries them to the ribosome where it recognizes mRNA sequences ​ tRNA carries a specific amino acid that corresponds to a mRNA codon ○​ the anticodon is complementary to the mRNA codon ​ ​ wobble base pairing - interaction of bases in the third position of the codon (3’ end of codon/5’ end of anticodon) is not that specific ○​ e.g. all codon pairs that start with the same two bases but end with C or U code for the same amino acid ○​ allows there to be 40 tRNAs for 61 amino acids ○​ ​ translation occurs in the ribosome ○​ made of many proteins and rRNA, and has a small and large subunit ○​ three tRNA binding sites: A, P, and E ​ A - binding site for tRNA bound to an AA ​ P - binding site for tRNA bound to the polypeptide being built ​ E - binds “empty” tRNA before it exits the ribosome ○​ small subunit validates the correct interaction between mRNA and tRNA  ○​ ​ 3 steps of translation: ○​ 1. initiation - the small ribosomal subunit binds to its recognition sequence on mRNA (AUG); tRNA with the amino acid methionine binds to the AUG start codon; the large ribosomal subunit joins the initiation complex and the tRNA occupying the P site  ​ ○​ 2. elongation (occurs in a 5’ to 3’ direction and builds from the N terminus to the C terminus) - the anticodon of an incoming tRNA binds to the codon at the A site; the methionine is linked to the next amino acid by the large subunit; the ribosome shifts down a codon to move the uncharged tRNA to the E site, while the tRNA carrying the polypeptide chain moves to the P site, opening up the A site; a new charged tRNA enters the A site and the uncharged tRNA is released from the E site; the polypeptide is transferred to the amino acid on the A site tRNA; the ribosome shifts down one codon and moves the tRNA with the polypeptide to the P site, while opening up the A site; the process repeats  ​ ○​ 3. termination - a stop codon enters the A site; a release factor binds to the stop codon; the release factor allows hydrolysis (breaking) of the bond between the polypeptide and the tRNA in site P; the polypeptide is released  ​ ​ polysomes can accelerate the rate of protein synthesis (multiple ribosomes translating the same mRNA to make many copies of the same polypeptide chain)  ○​ Post-translational modification of polypeptides (14.7) ​ translation may require post-translational edits and/or transportation ○​ polypeptides could serve functions in cells but may also be needed in different organelles or outside the cell ​ signal sequences (or signal peptides) are short stretches of amino acids that tell the polypeptides where it belongs ○​ they bind to receptor proteins on the outer membrane of the corresponding organelle ○​ if there is no signal sequence, the polypeptide stays in the cytosol ​ a signal sequence of around 20 hydrophobic amino acids at the N terminus directs to the rough endoplasmic reticulum for additional processing ​ post-translational modifications: ○​ phosphorylation - catalyzes by protein kinases; changes protein conformation and can expose active site ○​ glycosylation - adds sugars to create glycoproteins; essential for protein transport to different organelles ○​ proteolysis - cutting of peptide chain; catalyzed by enzymes called proteases; needed by HIV due to its large proteins ○​