Genetics Exam #1

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Chapter 9: Translation Genetic Code Table

(WILL BE GIVEN ON EXAM)

Chapter 7: DNA Structure Know how to determine the complement of a given single-stranded DNA sequence. *** CONTINUED *** Genetic Analysis 7.1 A portion of one strand of a DNA duplex has the sequence: 5′-ACGACGCTA-3′. Identify the sequence and polarity of the other DNA strand. For this double-stranded DNA fragment, identify the total number of phosphodiester bonds it contains and identify the total number of hydrogen bonds in its base pairs.

* DNA is a double helix composed of single strands that contain complementary base pairs (A pairs with T, and G with C). The complementary strands are antiparallel (i.e., one strand is 5′ to 3′, and its complement is 3′ to 5′). * One phosphodiester bond forms between adjacent nucleotides on each strand of DNA. A-T base pairs (joining the two strands) contain 2 hydrogen bonds, and G-C base pairs contain 3 hydrogen bonds. The complementary sequence is TGCTGCGAT. The polarity of the complementary strand is 3′-TGCTGCGAT-5′. Between the adjacent nucleotides of this fragment there are 8 phosphodiester bonds per strand for a total of 16 phosphodiester bonds. There are 4 A-T bases pairs containing 2 hydrogen bonds each, and 5 G-C base pairs containing 3 hydrogen bonds each, for a total of 8 + 15 = 23 hydrogen bonds in this DNA fragment.

Chapter 7: DNA Structure Really good summary of the structure of DNA * Be able to know and understand all the details in these pictures. - What groups are on the 5' end? 3' end? - Identify the major groove and minor groove. - Know the measurements in angstroms, (and bp when applicable). - Where are the hydrogen bonds? How many are there and where? - Where is the site of phosphodiester bonds?

* DNA is a double helix of antiparallel strands. * Base pairs are on the inside and phosphate groups are on the outside. * One helical turn = 34 angstroms = 3.4 nm = 10.5 base pairs Textbook: DNA is composed of four kinds of DNA nucleotides joined covalently by phosphodiester bonds that link one nucleotide to its neighbors in polynucleotide chains. Two polynucleotide chains come together along their lengths to form a double helix, also called a DNA duplex. The nucleotides in one strand complement the corresponding nucleotides in the partner strand in a specific pattern called "complementary base pairing." The pairs of complementary bases are held together by hydrogen bonds; and while relatively weak in comparison with covalent bonds, these are the forces that bind one strand to the other. A key observation made from Franklin's research was the recognition of two slightly different forms of DNA. These were designated A-form DNA and B-form DNA. B-form DNA was much more common than A-form DNA, and it is now known to predominate in all organisms. A third type of DNA is also known, as we describe at the end of this section. The molecular dimensions of DNA are measured using the unit called an angstrom (Å) or in nanometers (nm). One angstrom is equal to 10^-10 meters, or 1 ten-billionth of a meter, and 1 nm equals one-billionth of a meter, or 10^-9 meters. In B-form DNA, the distance from the axis of symmetry to the outer edge of either sugar-phosphate backbone is 10 Å (1 nm), and the molecular diameter is 20 Å (2 nm) at any point along the length of the helix (Figure 7.7a). The 20-Å molecular symmetry of the double helix was the key observation that told Watson and Crick that DNA structure results from pairing of a purine (A or G) with its complementary pyrimidine (T or C). The purine-pyrimidine base-pair pattern gives each base pair the same dimension. A second key observation derived from Franklin's Photo 51 is that nucleotide base pairs are spaced at intervals of 3.4 Å along DNA duplexes. This tight packing of DNA bases in the duplex leads to base stacking, the slight rotation of adjacent base pairs around the axis of symmetry so that their planes are parallel, imparting a twist to the double helix.

Chapter 7: DNA Replication, PCR, Sequencing Describe the role of NTPs and primers in replication.

* NTPs are used, so as to make the reaction thermodynamically favorable. DNA Polymerase can't start from scratch: a DNA or RNA primer is required. Since the primer is RNA, it will later get degraded and replaced by DNA. Textbook: DNA polymerases are unable to initiate DNA strand synthesis on their own. To perform its catalytic activity, a DNA polymerase requires the presence of a primer sequence, a short single-stranded segment that begins a daughter strand and provides an OH end to which a new DNA nucleotide can be added by DNA polymerase. To satisfy the requirement for a primer, DNA replication in bacteria is initiated by primase, a specialized enzyme that synthesizes a short RNA primer (see step ➌ of Figure 7.14). Measuring just one dozen to two dozen nucleotides in length, RNA primers provide the 3' OH needed for DNA polymerase activity. RNA primers contain the nucleotide base uracil (U) in place of thymine. Consequently, RNA primers cannot remain as part of fully replicated DNA. Thus, although they are essential for allowing DNA polymerase to begin its DNA synthesis, RNA primers are temporary and are removed from newly synthesized DNA strands before replication is completed. Primase enzymes also operate in the initial stages of archaeal and eukaryotic DNA replication. As in bacterial replication, primases in archaea and eukaryotes synthesize a short RNA primer that functions identically to bacterial RNA primers.

Chapter 8: Transcription (Not a learning objective, but should know) What are noncoding RNAs? What are the noncoding RNAs and their functions? How are ncRNAs generally classified?

* Substantial attention is paid to mRNA and its role in coding protein, but increasingly researchers are discovering functional RNAs in eukaryotic cells that do not encode polypeptides. Identified as noncoding RNAs (ncRNAs), these molecules have diverse cellular functions. The functional activities of ncRNAs derive from their ability to bind to other nucleic acids or to proteins to form complexes that carry out essential activities in cells. * In all organisms the two most prominent forms of noncoding RNA are ribosomal RNA (rRNA) and transfer RNA (tRNA). Ribosomal RNA combines with numerous proteins to form the ribosome, the molecular machine responsible for translation. Specific segments of rRNA molecules interact with mRNA to initiate translation. Transfer RNA is the RNA that carries amino acids to the ribosomes for construction of proteins, and it is encoded in dozens of different forms in all genomes. Each tRNA is responsible for binding a particular amino acid that it carries to the ribosome. At the ribosome, a group of nucleotides of a tRNA temporarily base-pair with nucleotides of mRNA. The tRNA deposits its amino acid that is added to the protein chain being produced there. * ncRNAs are generally classified by their length. Those that are longer than about 200 nucleotides are classified as long noncoding RNAs (lncRNAs), and those that are shorter than 200 nucleotides are small regulatory RNAs, also called short noncoding RNAs (sncRNAs). Certain types of ncRNAs vary in length (some are greater than 200 nucleotides, while some are less than 200 nucleotides). These are classified as variable-length noncoding RNAs.

Chapter 8: Transcription Compare the coding and template strands.

* The coding strand is identical to the mRNA created, except U is substituted for T. * The template strand is complementary to the mRNA created, except U is substituted for T. Textbook: Like all RNA polymerases, bacterial RNA polymerase uses one strand of DNA, the template strand, to assemble the transcript by complementary and antiparallel base pairing of RNA nucleotides with DNA nucleotides of the template strand (see Figure 1.9 for a review). The coding strand of DNA, also known as the nontemplate strand, is complementary to the template strand.

GENE EXPRESSION QUIZ A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA) The anticodon loop of the first tRNA that will complement this mRNA is: * 3' UGC 5' * 5' UGC 3' * 5' GGC 3' * 3' GGC 5' * 5' ACG 3'

3' GGC 5' The first tRNA that will complement this RNA will match 5' CCG 3'. It's anticodon will be the reverse complement: 3' GGC 5'.

Chapter 8: Transcription Predict mRNA sequences from DNA and vice versa. The sequence of an RNA transcript is: 5' - GGUUACAUUC - 3' What is the sequence of the coding strand?

5' - GGTTACATTC - 3'

Chapter 9: Translation What is the anticodon for the Thr codon ACC?

5' - GGU - 3'

DNA STRUCTURE QUIZ 5'-AGTC-3' is the "top" strand of a DNA double helix. What is the bottom strand? Write your answer 5' to 3'.

5'-GACT-3'

DNA STRUCTURE QUIZ Which group is typically found at the 5' end of a DNA strand?

DNA STRUCTURE QUIZ Phosphodiester bond formation involves a chemical reaction involving group ______ on one molecule with group ____ on another molecule.

A,B

DNA REPLICATION QUIZ You are setting up a dideoxysequencing reaction. What is the correct ratio of dNTPs (deoxyribonucleotides) to ddNTPs (dideoxyribonucletides)?

Adding more dNTPs than ddNTPs

Chapter 7: DNA Replication, PCR, Sequencing Textbook Explanation Continued Describe semi-conservative replication and explain its significance. * Who is Rosalind Franklin? * Who are Watson and Crick?

Almost immediately after the structure of DNA was identified, three competing models of DNA replication emerged (Figure 7.8). The ➊ semiconservative DNA replication model—which proved to be correct—proposed that each daughter duplex contains one original parental strand of DNA and one complementary, newly synthesized daughter strand. The ➋ conservative DNA replication model predicts that one daughter duplex contains the two strands of the parental molecule and the other contains two newly synthesized daughter strands. Lastly, the ➌ dispersive DNA replication model predicts that each daughter duplex is a composite of interspersed parental duplex segments and daughter duplex segments.

Chapter 8: Transcription Articulate how alternative splicing can increase protein diversity. * How can large eukaryotic genomes express more proteins than there are genes in the genomes? - Name the transcription-associated mechanisms.

Alternative splicing of the same transcript can give rise to different proteins. * Single splicing pathway vs. alternative splicing - NOTE: the order of the exons is not altered. - Introns are usually much larger than exons. Alternative splicing can produce forms of a protein that have different functions. (in pic: upon infection, pattern of splicing was altered) * Alternative splicing results in the inclusion or exclusion of an exon - it does not change their order. * Alternative splicing does not result in the duplication of an exon in the mature mRNA. Textbook: It is common for large eukaryotic genomes to express more proteins than there are genes in the genomes. Three transcription-associated mechanisms can account for the ability of single DNA sequences to produce more than one polypeptide. First, a pre-mRNA can be spliced in alternative patterns in different types of cells. In other words, the same transcript might produce one mature mRNA (and a particular protein) in one type of cell and a different mature mRNA (and a different protein) in another type of cell, This process is called alternative pre-mRNA splicing. Second, alternative promoters can initiate transcription at distinct start points in different cell types, and third, alternative polyadenylationuses different polyadenylation signal sequences in a gene to produce different mRNAs. The CT/CGRP gene produces the same pre-mRNA transcript in many cells, including thyroid cells and neuronal cells. The transcript contains six exons and five introns and includes two alternative polyadenylation sites, one in exon 4 and the other following exon 6. In thyroid cells, CT/CGRP pre-mRNA is spliced to form mature mRNA containing exons 1 through 4, using the first poly-A site for polyadenylation. Translation produces calcitonin, a hormone that helps regulate calcium. In neuronal cells, the same pre-mRNA is spliced to form mature mRNA containing exons 1, 2, 3, 5, and 6. Polyadenylation takes place at the site that follows exon 6, since exon 4 is spliced out as though it were an intron. Translation in neuronal cells produces the hormone CGRP.

Chapter 9: Translation Delineate how proteins are a chain of amino acids. * What are proteins composed of? * Know the structure of an amino acid. * How many different AA's are there? (You don't need to name and categorize them). * What is a protein overall? * How many AA long is a typical protein? * What is a polypeptide? * Describe the structure of a protein. * Describe peptide bond formation. * Identify the peptide bond in a protein. * How are proteins synthesized (in what direction)? * What terminus will the amino acid be added to? * Identify the N terminus and the C terminus.

Amino acids and proteins * Proteins are composed of amino acids (AA's) - The R group side chain is different among different AAs. * There are 20 different AA's. * Protein = AA's linked in a chain by peptide bonds. * A typical protein is about 450 amino acids long. * "polypeptide" is synonymous with "protein". * Although proteins are linear chains of AA's, they fold into complicated 3D structures. * Peptide bond formation: like phosphodiester bond formation, peptide bond formation is a dehydration synthesis (aka condensation) reaction. * Proteins are synthesized N (amino) to C (carboxy). Textbook: = good explanation, don't memorize. Twenty different amino acids are the basic building blocks of polypeptides. All amino acids have features in common and features that are distinct. The distinctive features impart specific characteristics that allow the amino acid to participate in certain chemical reactions or behave in a hydrophilic or hydrophobic manner. In part, the common features allow amino acids to be joined into polypeptides by covalent bond formation between adjacent amino acids in the chain. The distinctive portion of each amino acid is its side chain, known as an R-group, which is also joined to the α-carbon. The R-groups range in complexity from a single hydrogen atom to ringed structures that in themselves contain multiple carbon atoms. Each R-group imparts specific characteristics as shown in Table 9.1. Ten of the amino acids have nonpolar R-groups, meaning they have no charged atoms that can participate in formation of hydrogen bonds with other amino acids. Five other amino acids have polar R-groups that can carry partial charges and can participate in hydrogen bond formation with other amino acids. The five remaining amino acids have electrically charged R-groups: Three are basic and two are acidic. Electrically charged R-groups allow these amino acids to form ionic bonds and hydrogen bonds.

Problem Set #1: DNA Replication, PCR, Sequencing Digestion of DNA polymerase III enzyme with a protease results in two fragments, A & B. The B fragment can still polymerize DNA, but misincorporates nucleotides at much higher rate. What activity is likely located infragment A? A) 3′-to-5′ exonuclease activity B) 5′-to-3′ exonuclease activity C) 3′-to-5′ polymerase activity D) 5′-to-3′ polymerase activity E) 3′-to-5′ helicase activity

Answer: A) 3′-to-5′ exonuclease activity

Problem Set #1: DNA Replication, PCR, Sequencing The bacterial virus SV40 has a double-stranded DNA genome, 5243 base pairs in length. You determine thebase composition and find that it has 24% Guanine residues. What percentage of its genome will consist of Adenine residues? A) 24% B) 26% C) 48% 2D) 52% E) impossible to determine

Answer: B) 26% G-C, A-T * G=24% C=24%, G-C=48% total * 100%-48%=52% * A-T=52%, A=26% T=26%

Problem Set #1: DNA Replication, PCR, Sequencing In Drosophila, the first 14 cell divisions after fertilization take no more than 10 minutes each. E. coli incontrast take ~ 30 minutes to undergo a cell division. What is one difference between DNA replication inbacteria versus Drosophila that could explain this? A) Eukaryotic chromosomes are replicated bidirectionally, while bacterial chromosomes are replicated in onedirection. B) Eukaryotic chromosomes have many origins of replication and replicate bidirectionally, while bacteria haveonly one origin of replication and replicate unidirectionally. C) Bacterial chromosomes are replicated bidirectionally, while eukaryotic chromosomes are replicated in onedirection. D) Eukaryotic chromosomes have many origins of replication, while bacteria have only one origin of replication. E) The process is identical in bacterial and eukaryotic DNA replication.

Answer: D) Eukaryotic chromosomes have many origins of replication, while bacteria have only one origin of replication.

Problem Set #1: DNA Replication, PCR, Sequencing Which of the following might you find in a DNA sequencing reaction but not in a polymerase chain reaction? A) DNA template B) DNA polymerase C) dNTPs D) ddNTPs E) DNA primer

Answer: D) ddNTPs

Problem Set #1: DNA Replication, PCR, Sequencing The bacterial virus M13 phage has a single stranded DNA genome. You determine the base composition andfind that it has 24% Guanine residues. What percentage of its genome will consist of Adenine residues? A) 24% B) 26% C) 48% D) 52% E) impossible to determine

Answer: E) impossible to determine

Problem Set #1: DNA Replication, PCR, Sequencing The complement (bottom strand) of the DNA sequence 5'-AGC-3' is? A. 5'-AGC-3' B. 5'-CGA-3' C. 5'-CTG-3' D. 5'-TCG'-3' E. 5'-GCT-3'

Answer: E. 5'-GCT-3'

Problem Set #1: DNA Replication, PCR, Sequencing Based on the following replication bubble, which of these statements is true? A) X and Y are leading strands, W and Z are lagging strands B) X and Z are leading strands, W and Y are lagging strands C) X and W are leading strands, Y and Z are lagging strands D) W and Z are leading strands, X and Y are lagging strands E) W and Y are leading strands, X and Z are lagging strands

Answer: E) W and Y are leading strands, X and Z are lagging strands

DNA STRUCTURE QUIZ In order to answer "This chemical structure is found in...", which chemical group was most informative?

DNA STRUCTURE QUIZ Which group is, or contains, a purine?

DNA STRUCTURE QUIZ Which group typically participates in complementary base pairing?

Chapter 7: DNA Replication, PCR, Sequencing Compare and contrast DNA replication in cells with PCR. * In cells vs. PCR/Dideoxy sequencing

COMPARE In cells * Need template, primers, DNA polymerase, and nucleotides. PCR/Dideoxy sequencing * Need template, primers, DNA polymerase, and nucleotides. CONTRAST In cell * RNA is used as primer. * Enzymes unwind the DNA duplex. * Replication origins, replication forks, leading strand, lagging strand. * Multiple enzymes,es and other proteins involved. PCR/Dideoxy sequencing * DNA primers are used. * Heat is used to unwind (denature) the DNA. * No rep origins or forks, no leading strand/lagging strand (all strands can be considered leading strands). * Only one enzymes,e needed (DNA polymerase).

Chapter 9: Translation What is the red box around?

His anticodon

Chapter 7: DNA Replication, PCR, Sequencing Demonstrate a basic understanding of dideoxycytidine's sequencing.

Chain-terminating bases are the key to Dideoxy DNA sequencing. Sequencing reactions utilize fluorescent chain-terminating nucleotides. * The chain terminating nucleotides are present at a lower concentration (from color and length, you can identify nt). Textbook: Sanger's DNA sequencing method is known as dideoxynucleotide DNA sequencing or dideoxy DNA sequencing or simply Sanger sequencing. Based on in vitro DNA replication reactions that closely resemble PCR, dideoxy sequencing, like PCR, uses DNA primers and DNA polymerase to replicate new DNA from a single-stranded template. In dideoxy sequencing reactions, the four standard deoxynucleotides (dNTP) of DNA are used in high concentrations, but to them is added a much smaller amount of a dideoxynucleotide triphosphate (ddNTP). Tens of thousands of identical DNA fragments are used in each sequencing reaction. The fragments are generated by cloning. Dideoxynucleotides differ from deoxynucleotides in lacking two oxygen atoms (dideoxy means "two deoxygenated sites") rather than having the usual one deoxygenated site. Recall that dNTPs are deoxygenated at the 2' carbon and have a hydroxyl group (OH) at the 3' carbon. In contrast, ddNTPs have hydrogen (H) atoms rather than hydroxyl groups at both the 2' and 3' carbons (Figure 7.27a). The absence of a hydroxyl group at the 3' carbon in ddNTP prevents the ddNTP from forming a phosphodiester bond, so when a ddNTP is incorporated into a growing strand by DNA polymerase, the synthesis of the strand is terminated at that point (Figure 7.27b). Dideoxy sequencing therefore produces a large number of partial replication products, each terminated by incorporation of a ddNTP at a different site in the sequence. (a) Dideoxynucleotides (ddNTPs) are deoxygenated at both the 2' and 3' carbons and cannot form a phosphodiester bond for the further elongation of DNA. (b) The incorporation of a dideoxynucleotide of cytosine (ddCTP) terminates the replication reaction.

Chapter 7: DNA Structure Identify the parts of DNA and RNA nucleotides. * Know the numbering of carbons. - 5'C vs. 5C * How are DNA and RNA nucleotides different? * What are the two types of nitrogenous bases in DNA and RNA? - Which bases are which? * Understand naming of nucleotides.

DNA and RNA are composed of nucleotides. * A nucleotide has three components: the phosphate group, the ribose sugar, and the nitrogenous base. * Nucleotide = sugar, base, phosphate group. * Nucleoside = sugar + base. "Prime" means the carbon belongs to the sugar. Purines and Pyrimidines are the two types of nitrogenous bases in DNA and RNA * Pyrimidines: Cytosine, Uracil, Thymine. - "Pyromaniacs in Connecticut". * Purines: Guanine and Adenine. - "Pure as Gold" DNA and RNA: Cytosine, Guanine, Adenine. DNA only: Thymine RNA only: Uracil * RNA has a OH group on 2'C, instead of a H (like DNA). - DNA: deoxyadenosine 5'-monophosphate (missing OH group on 2' C) - Think "DNA = deoxyribonucleic acid", "deoxy = OH died" - RNA: adenosine 5'-monophosphate - Think "RNA = ribonucleic acid"

DNA REPLICATION QUIZ Suppose you provide an actively dividing culture of E. coli with radioactive thymine. What would you expect to see if the cell replicates its DNA and divides once in the presence of the radioactive base?

DNA in both daughter cells will be radioactive

Chapter 7: DNA Structure Explain complementary base pairing.

DNA is a double helix of antiparallel strands. The two strands of DNA are held together by hydrogen bonds formed between complementary base pairs (nitrogenous base pairs). * A pairs with T, G pairs with C. * There are two hydrogen bonds between A-T, and there are three hydrogen bonds between G-C. * Hydrogen bonds are non-covalent and relatively weak. They can come apart and come back together. * Lots of hydrogen bonds (think: velcro). Note that a purine (A or G) always pairs with a pyrimidine (C or T or U). In pic... * A pairs with T * G pairs with C * A pairs with T Textbook: DNA is stable as a double helix. The two polynucleotide strands that make up the duplex have a specific relationship that follows two rules: (1) The arrangement of the nucleotides is such that the nucleotide bases of one strand are complementary to the corresponding nucleotide bases on the second strand (A pairs with T, and G pairs with C), and (2) the two strands are antiparallel in orientation (see the opposite-pointing arrows on each side of the diagrams in Figure 7.6). If one strand is, for example, 5'-ATCG-3', then the complementary strand is 3'-TAGC-5'. Complementary base pairing joins a purine nucleotide on one strand to its complementary pyrimidine nucleotide on the other. The chemical basis of such pairing is the formation of a stable number of hydrogen (H) bonds between the bases of the different strands. Hydrogen bonds are noncovalent bonds that form between the partial charges that are associated with hydrogen, oxygen, and nitrogen atoms of the nucleotide bases. Antiparallel strand orientation is essential to the formation of stable hydrogen bonds. In Figure 7.6, notice that the nucleotides in one strand are oriented with their 3' carbon toward the top and their 5' carbon toward the bottom. The complementary nucleotides in the other strand are antiparallel to these; that is, their 5'-to-3' orientations run in the oppositedirection.

Chapter 7: DNA Replication, PCR, Sequencing Demonstrate a basic understanding of dideoxycytidine's sequencing. Genetic Analysis 7.3 QUESTION: From the dideoxy DNA sequencing33 gel shown here, deduce the sequence and strand polarities of the DNA duplex fragment.

DNA polymerase incorporates nucleotides in four parallel reactions. Each reaction mixture includes the four normal DNA nucleotides (dNTPs) and one labeled dideoxynucleotide (ddNTP). Incorporation of a dNTP allows continued strand synthesis, but incorporation of a ddNTP terminates synthesis. The 3′ end of the primer is used to initiate DNA synthesis. The first nucleotide incorporated during synthesis is cytosine, as determined by identifying the location of the smallest synthesized fragment: the "C" lane. The second and third nucleotides are both adenine. The first three nucleotides are therefore 5′-CAA-3′ The synthesized strand is 5′-[primer]-CAATAGCTGAGGAGTCGATTCATGCCGATA-3′ The template DNA strand is 3′-GTTATCGACTCCTCAGCTAAGTACGGCTAT-5′

DNA STRUCTURE QUIZ This chemical structure is found in...

DNA, but not RNA

Chapter 7: DNA Replication, PCR, Sequencing Demonstrate a basic understanding of dideoxycytidine's sequencing. *** CONTINUED TEXTBOOK EXPLANATION ***

Dideoxy DNA sequencing is carried out in four separate reaction mixtures—one for each of the four ddNTPs. Each reaction mixture contains the DNA strand to be sequenced, a single-stranded DNA primer, DNA polymerase, large amounts of each of the four standard nucleotides (dATP, dGTP, dCTP, and dTTP), and a small amount of onedideoxynucleotide: adenine (ddATP), thymine (ddTTP), cytosine (ddCTP), or guanine (ddGTP). Figure 7.28 shows that in each reaction mixture, DNA synthesis terminates at each site where a ddNTP is incorporated into the newly synthesized molecule. Figure 7.28a shows the DNA fragment being sequenced at the top, annealed to the 18-mer primer used to initiate DNA synthesis (18-mer means the primer is 18 nucleotides in length). It also shows that for the "C reaction mixture" (the mixture that includes ddCTP), each location at which a cytosine can be incorporated into the growing chain generates some DNA replication fragments that terminate at that location. Keep in mind that most of the cytosine in the C reaction mixture is the more highly concentrated dCTP, so it is most likely that this nucleotide will be incorporated into the growing chain. If so, replication continues. If, on the other hand, the less concentrated ddCTP is incorporated, as it will be in a small proportion of the replicating molecules, replication terminates. Figure 7.28a shows five different DNA fragment lengths generated by ddCTP incorporation into C reaction mixture products. a) A target region of DNA is located by binding a single-stranded primer of 18 nucleotides (an "18-mer") that carries a 5' label. Replication products terminated by ddCTP each have a different length. (b) Replication products terminated by ddGTP. (c) Termination products generated by ddTTP. (d) Termination products generated by ddATP. Figure 7.28 Full Alternative Text The same process ensues in the three reaction mixtures containing, respectively, ddGTP, ddTTP, and ddATP (Figures7.28b-d). Upon the completion of the four parallel sequencing reactions, there will be, for every nucleotide in the sequence, some partial replication DNA fragments terminating at that nucleotide.

Chapter 7: DNA Replication, PCR, Sequencing Define replication origin, replication fork, leading strand, lagging strand. * What is notable about eukaryotic chromosomes? * How is DNA generally replicated? * How is accuracy ensured during DNA replication? - Explain the process. *** Continued textbook explanation ***

Each strand of parental DNA acts as a template for the synthesis of a new daughter strand of DNA. In E. coli, daughter DNA strands are synthesized at the replication fork by DNA polymerase III (DNA pol III), the principal DNA-synthesizing enzyme (see Figure 7.14, step ➍). DNA pol III begins its work at the 3'-OH end of an RNA primer and rapidly synthesizes new DNA by adding one nucleotide at a time in a sequence that is complementary and antiparallel to the template-strand nucleotides. Pol III requires a template nucleotide to add a new nucleotide to a daughter strand. Enzymes with functions identical to DNA pol III are found in archaea and eukaryotes. Experimental evidence indicates that most of the enzymes participating in DNA replication are part of a large protein complex called a replisome. There is one replisome at each replication fork. Replisomes have numerous components, including, in each replisome, two complete molecules of DNA pol III. One of these DNA pol III molecules carries out the 5'-to-3' synthesis of one daughter strand continuously, in the same direction in which the replication fork progresses. The second pol III in the replisome carries out synthesis of the other daughter strand. The continuously elongated daughter strand is called the leading strand (Figure 7.18). Notice that Figure 7.18 divides the replication bubble into four quadrants. The upper right and lower left quadrants contain leading strands. The daughter strands in the upper left and lower right quadrants shown in Figure 7.18 have a 5'-to-3' direction of elongation that runs opposite to the direction of movement of the replication fork. These daughter strands are elongated discontinuously,in short segments, each of which is initiated by an RNA primer. The discontinuously synthesized daughter strand is called the lagging strand. Thus in Figure 7.18, the lower right and upper left quadrants of the replication bubble contain lagging strands (see also step ➎ of Figure 7.14).

DNA STRUCTURE QUIZ What is group A? (Note: in all questions, the term "ribose sugar" is meant to apply to both oxy- and deoxy- ribose)

Phosphate group

Figure 7.7a shows that one complete helical turn spans 34 Å. This span is occupied by approximately 10.5 base pairs. Figure 7.7b is a space-filling model that illustrates base-pair stacking and the twisting of the sugar-phosphate backbones. Figure 7.7c is a ball-and-stick model illustrating how base pairs twist around the axis of symmetry to create the helical spiral. Base-pair stacking creates two grooves in the double helix, gaps between the spiraling sugar-phosphate backbones that partially expose the nucleotides. The alternating grooves, known as the major groove and minor groove, are highlighted in Figures 7.7b and 7.7c. The major groove is approximately 12 Å wide, and the minor groove is approximately 6 Å wide. The major and minor grooves are regions where DNA-binding proteins can most easily make direct contact with nucleotides along one or both strands of the double helix. In this chapter and in later chapters, we discuss many of the important functions DNA-binding proteins perform, such as regulating the initiation of transcription and controlling the onset and progression of DNA replication. Most of these functions depend on the presence of characteristic sequences of DNA nucleotides. DNA-binding proteins gain access to DNA nucleotides in major and minor grooves of the molecule. B-form DNA, overwhelmingly the most common DNA structure in organisms, has a right-handed twisting of the sugar-phosphate backbone. A-form DNA also has a right-handed twist to the helix. It is more compact than B-form DNA, with about 11 base pairs per complete helical twist, although its diameter is a little greater than that of B-form DNA (Table 7.1). A-form DNA is occasionally detected in cells, and it appears to be particularly common in bacteriophages, where its more compact size makes it functional for packaging of bacteriophage DNA. A-form DNA may be less amenable to binding by DNA-binding regulatory proteins, due to alterations of the major and minor grooves in comparison with B-form DNA. A-form DNA major grooves are deeper and narrower than those of B-form DNA, and its minor grooves are wider and shallower than those of B-form DNA.

Chapter 7: DNA Replication, PCR, Sequencing *** TEXTBOOK EXPLANATION CONTINUED AGAIN *** Demonstrate a basic understanding of dideoxycytidine's sequencing.

Following completion of the ddNTP reactions, the contents of each reaction are loaded into separate lanes of a DNA electrophoresis gel, and the contents undergo separation by their length in base pairs (Figure7.29a). Each DNA fragment in the gel can be radioactively labeled for visualization, allowing the sequence of the newly synthesized strand to be "read" off the gel. Knowing that the smallest DNA fragment migrates the farthest from the origin of migration, and knowing that newly synthesized DNA is elongated in the 5'-to-3' direction and that the primer is located at the 5' end of the sequenced strand, we can identify the consecutive nucleotides by the gel lane in which successively longer DNA fragments are located. Thus, the first incorporated nucleotide after the primer is A (i.e., ddATP is incorporated and terminates replication), followed by another A (ddATP incorporated), followed by T (ddTTP incorporated), and so on. Once the sequenced strand is determined, the complementary strand can be determined, using the knowledge that DNA strands are antiparallel and display complementary base pairing. An example of a dideoxy DNA sequencing gel is shown in Figure 7.29b, and a portion of the sequence is given.

GENE EXPRESSION QUIZ There is a mutation in a bacterial gene "X"'s promoter that changes a nucleotide from a T to an A in a position between the -10 and -35 elements. What does this mutation do (most likely)? * Have little or no effect on the rate of gene X's transcription. * Decrease the rate of gene X's transcription. * Increase the rate of gene X's transcription.

Have little or no effect on the rate of gene X's transcription. Mutations to the -10/-35 elements may increase or decrease gene expression. Base substitutions in between will most likely not effect gene expression.

Chapter 7: DNA Replication, PCR, Sequencing Describe semi-conservative replication and explain its significance. * Who is Rosalind Franklin? * Who are Watson and Crick?

He doesn't ask us to memorize dates and experiments!!! Dr. Rosalind Franklin took Photograph 51, an x-ray crystallography exposure. * Photograph 51 is a key piece of data supporting the model that DNA was a helix. * Although she took the photo, she did not receive appropriate credit from Watson and Crick. - After taking the photo, Rosalind left the photo on her desk. Watson and Crick saw the photo and used information they obtained from seeing the photo to make a model of DNA. "It has not escaped our notice that the specific base-pairing we have proposed immediately suggests a possible copying mechanism for the genetic material" - Watson and Crick DNA Replication * The new strand is not a copy of the old strand. But, it can be used as a template to reproduce a copy of the old strand. * But, if we start with a double-stranded molecule, and make a complement of each strand... we end up with 2 double-helical "daughter" molecules that are identical copies of their "mother". - Each daughter contains one strand from mother and one newly-synthesized strand. - Mom is dead. She has been ripped in half, and lives on only in her daughters. DNA replication is semiconservative! * Messelson and Stahl's experiment demonstrated that DNA replicates by a semiconservative mechanism (rejects other models).

Chapter 8: Transcription Describe promoters, enhancers, silencers; define "transcription factor". * How can the level of gene expression be regulated in prokaryotes? * How can the level of gene expression be regulated in eukaryotes? * What does the promoter determine? * How is transcription initiated? Describe the steps. (not on this fc) - What happens shortly after transcription initiation? * What determines where an mRNA ends? * What is happening to DNA during RNA synthesis?

How does the RNA polymerase know where to start transcription? * RNA polymerase binds to promoter sequences in DNA to initiate transcription. - +1: start site. Prokaryotes: * the level of gene expression can be affected by how similar the -10 and -35 sequences are to the consensus sequence. * The promoter consists of two short sequences. * -10 sequence (aka -10 box) * -35 sequence * The exact sequence in between doesn't matter * The sequence of the -10 and -35 boxes do not need to exactly match the sequence for the lac gene, but they need to be close. (-35: TTTACA, -10: TATGTT) * RNA polymerase binds to the -35 and -10 sequence. - Amino acid residues in the protein use hydrogen bonds, electrostatic interactions, and hydrophobic interactions to recognize these sequences. Enhancers and silencers can regulate gene expression in eukaryotes. * In addition to a promoter, eukaryotic genes can also be regulated by other sequences in the DNA called enhancers and silencers. * These sequences can be close to the promoter, or far upstream or downstream. * Enhancers and silencers have binding sites for specialized proteins called "transcription factors". - Transcription factors bind to promoter sequence and help turn gene on/off, and help RNA polymerase bind to help express genes. The promoter determines where transcription starts and the direction in which it proceeds.

Chapter 9: Translation Read the genetic code and predict protein sequences from DNA or RNA sequences. * How is protein sequence encoded in DNA/RNA? Answer this by answering: - What is the protein coding region of mRNA made up of? * What are these things called, and what do they correspond to? * What does the Universal Genetic Code state? * How many codons are in the universal genetic code? - How are these codons split up? * What signals the start of protein synthesis? * What signals that the end of the coding sequence has been reached, and what does this signal lead to?

How is protein sequence encoded in DNA/RNA? * The protein coding region of an mRNA is made up of non-overlapping nucleotide triplets (codons), each of which corresponds to an amino acid. * There is no punctuation or space between codons. In the pic, UAA is one of the stop codons (blue). When looking at the polypeptide and the mRNA, you can see the codons and the amino acids the codons code for. EX: Codon 1 (AUG) codes for MET, codon 2 (ACA) codes for THR, etc. The Universal Genetic Code * With some minor exceptions, all living organisms on Earth use this same genetic code. * 64 codons total. * 61 codons for 20 amino acids -> redundancy (means that several codons could code for one AA). * ATG (DNA language)/AUG (RNA language) signals the start of protein synthesis with M (M=Met=Methionine) M can also be used internally in proteins. - All proteins start with Methionine. * 3 stop codons signal that the end of the coding sequence has been reached, and translation should be terminated.

Chapter 7: DNA Replication, PCR, Sequencing Demonstrate the understanding of what "5' to 3'" means with regard to DNA replication. * What enzyme makes DNA replication happen? What is used in DNA replication?

How is the new strand synthesized? * The new strand is synthesized one nucleotide at a time. * NTPs are used, so as to make the reaction thermodynamically favorable. * Polymerization is 5' -> 3' - This means new nucleotides are added to 3' OH of growing strand. * DNA Polymerase is the enzyme that makes all this happen. Textbook: In all organisms, DNA polymerase enzymes that are responsible for synthesizing new DNA strands use a template strand to direct the addition of nucleotides to daughter strands in a complementary and antiparallel manner. These new nucleotides are added to the 3' end of the growing daughter strand, and the overall direction of daughter strand elongation is 5' to 3'.

Chapter 7: DNA Structure Know how to determine the complement of a given single-stranded DNA sequence.

If we know the sequence of one strand... Then we can deduce the sequence of the other strand... * A pairs with T * G pairs with C

Chapter 9: Translation (Not a learning objective, but must know) Where are proteins translated? * Where in eukaryotes? * Where in prokaryotes? Is the picture (attached) from a bacterial cell or a eukaryotic cell?

In eukaryotes, mRNA is processed in the nucleus then exported to the cytoplasm where it is translated. - Transcription in nucleus, translation in cytoplasm. In prokaryotes, transcription and translation are coupled (there is no nucleus in bacteria). - Both transcription and translation take place in cytoplasm. The picture is taken of a bacterial cell (transcription and translation are both occurring in the cytoplasm).

Chapter 7: DNA Replication, PCR, Sequencing ** CONTINUED AGAIN ** New Generations of DNA Sequencing Technology (Not included in learning objectives, but included in the section of the book he told us to know) * Benefits of new generations? * Next-Generation Sequencing * Third-Generation Sequencing

Its small size makes it portable and ideal for fieldwork or other settings where the use of larger pieces of equipment is not feasible. Both methods can detect covalently modified nucleotides, including 4mC (4-methylcytosine). Potentially, these systems can also detect DNA nucleotides that have been damaged by ultraviolet light or chemicals.

Chapter 8: Transcription TEXTBOOK EXPLANATION Describe promoters, enhancers, silencers; define "transcription factor". * How can the level of gene expression be regulated in prokaryotes? * How can the level of gene expression be regulated in eukaryotes? * What does the promoter determine? * How is transcription initiated? Describe the steps. - What happens shortly after transcription initiation? * What determines where an mRNA ends? * What is happening to DNA during RNA synthesis?

Like all RNA polymerases, bacterial RNA polymerase uses one strand of DNA, the template strand, to assemble the transcript by complementary and antiparallel base pairing of RNA nucleotides with DNA nucleotides of the template strand (see Figure 1.9 for a review). The coding strand of DNA, also known as the nontemplate strand, is complementary to the template strand. The gene—that is, the stretch of DNA regions that produces an RNA transcript—contains several segments with distinct functions (Figure 8.3). The promoter of the gene is immediately upstream—that is, within a few nucleotides of the 5' start of transcription, which is identified as corresponding to the +1 nucleotide. The promoter is not transcribed. Instead, the promoter sequence is a transcription-regulating DNA sequence that controls the access of RNA polymerase to the gene. The coding region is the portion of the gene that is transcribed into mRNA and contains the information needed to synthesize the protein product of the gene. The termination region is the portion of the gene that regulates the cessation of transcription. The termination region is located immediately downstream—that is, immediately 3' to the coding segment of the gene. Promoters are double-stranded regulatory DNA sequences that bind transcription proteins such as RNA polymerase and direct the RNA polymerase to the nearby start of transcription. RNA polymerase is attracted to promoters by the presence of consensus sequences, short regions of DNA sequences that are highly similar, though not necessarily identical, to one another and are located in the same position relative to the start of transcription of different genes (see Section 7.4 for an introduction to consensus sequences). Although promoters are double stranded, promoter consensus sequences are usually written in a single-stranded shorthand form that gives the 5' -to- 3' sequence of the coding (nontemplate) strand of DNA (Figure 8.5). The most commonly occurring bacterial promoter contains two consensus sequence regions that each play an important functional role in recognition by RNA polymerase and the subsequent initiation of transcription.

Chapter 9: Translation (Not a learning objective, but must know) What are polysomes? * How would researchers figure out which mRNAs are being actively translated?

Many ribosomes are simultaneously on a single mRNA. * These entities are called polysomes. * Researchers can isolate polysomes from cells and figure out which mRNAs are being actively translated. Textbook: = don't need to memorize, just good summary. Polyribosomes are busy translational complexes containing multiple ribosomes that are each actively translating the same mRNA (Figure 9.11). Each ribosome in the polyribosome structure independently synthesizes a polypeptide, markedly increasing the efficiency of utilization of an mRNA.

Chapter 7: DNA Replication, PCR, Sequencing New Generations of DNA Sequencing Technology (Not included in learning objectives, but included in the section of the book he told us to know) * Benefits of new generations? * Next-Generation Sequencing * Third-Generation Sequencing

New generations of DNA sequencing technologies are continuing to be developed. These technologies sequence DNA fragments in parallel, meaning that hundreds of thousands to millions of DNA fragments are sequenced simultaneously. This brings the cost of DNA sequencing down so far that the goal of the "thousand-dollar genome sequence"—that is, the availability of genome sequencing as an affordable component of everyday medicine—is within reach. In the 40 years since Sanger introduced dideoxy DNA sequencing, the process has gotten both faster and cheaper by many orders of magnitude. These advances have been made possible by methods that sequence hundreds of thousands to millions of DNA fragments simultaneously in a reaction, in a process characterized as "massively parallel sequencing." There are several different versions of methods that take a massively parallel approach, but they are all based on a similar elaboration of dideoxy DNA sequencing. Collectively, these advanced methods are identified as next-generation sequencing, or NGS. NGS procedures begin with the fragmentation of genomic DNA. Figure 7.30 illustrates this process as the first step of one version of NGS known as Illumina sequencing. In Illumina sequencing, the DNA is fragmented ➊, tagged with adaptor molecules attached to both ends of each strand ➋, and then denatured for analysis ➌. The adaptor molecules anchor the strand in a later step and may contain a PCR primer. The single-strand fragments are next placed in a flow cell and amplified to produce clusters of identical strands ➍ ➎. The mixture used for amplification contains DNA polymerase, the four dNTPs, and other necessary compounds. The dNTPs of A, T, G, and C are tagged with different fluorescent compounds that emit light in specific wavelengths when excited ➏. After each new nucleotide is incorporated into a growing strand, a laser light excites the fluorescent compound attached to the base, and a photoreceptor records the emission wavelength to identify the intensity ➐ ➑. Software records this information and converts it to identify the nucleotide as either A, T, C, or G ➒.

DNA STRUCTURE QUIZ What is group C?

Nitrogenous base

DNA STRUCTURE QUIZ Which group is, or contains, a pyrimidine?

None of them

DNA STRUCTURE QUIZ What is this chemical structure?

Nucleotide

Chapter 7: DNA Structure Explain how nucleotides are joined in chains to form DNA molecules. * What are they joined by? - Be able to identify one of these. - Understand the name. * What's happening in this reaction?

Nucleotides are linked in a chain by phosphodiester bonds. * Phosphodiester bond formation: a chemical reaction between the phosphate group on one molecule and the sugar on another molecule. * Phosphodiester bond formation is an example of a dehydration synthesis, or condensation, reaction. * A chain of 3 nucleotides linked by 2 phosphodiester bonds is shown in the image attached. In a phosphodiester, the R3 is a hydrogen, and the R1 and R2s are carbons. A phosphodiester is a phosphate ester with two esters (a phosphate (PO4)3- containing two esters). Textbook: DNA-strand formation is catalyzed by the enzyme DNA polymerase. The enzyme catalyzes the formation of a phosphodiester bond between the 3' hydroxyl group of one nucleotide and the 5' triphosphate group of an adjacent nucleotide (Figure 7.6). Two of the three phosphates of the dNTP are removed during phosphodiester bond formation, leaving the nucleotides of a polynucleotide chain in their monophosphate form. The two discarded phosphates are called the pyrophosphate group. As mentioned before, the resulting strand is a polynucleotide chain composed of nucleotides joined by covalent bonds. The pattern of phosphodiester bond formation gives each strand a sugar-phosphate backboneconsisting of alternating sugar and phosphate groups along its length.

Chapter 8: Transcription TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Describe promoters, enhancers, silencers; define "transcription factor". * How can the level of gene expression be regulated in prokaryotes? * How can the level of gene expression be regulated in eukaryotes? * What does the promoter determine? * How is transcription initiated? Describe the steps. - What happens shortly after transcription initiation? * What determines where an mRNA ends? * What is happening to DNA during RNA synthesis?

Often, promoters alone, while necessary, are not sufficient to initiate transcription of eukaryotic genes. In such cases, additional regulatory sequences, and additional transcription-activating proteins, are needed to drive transcription. This is particularly the case for multicellular eukaryotes that have many different types of cells with distinctive patterns of gene expression, including patterns that change as the organisms grow and develop. This type of transcriptional regulation is discussed in Section 13.2. Enhancer sequences are one important group of DNA regulatory sequences that increase the level of transcription of specific genes. Enhancer sequences bind specific proteins that interact with the proteins bound at gene promoters, and together promoters and enhancers drive transcription of certain genes. In many situations, enhancers are located upstream of the genes they regulate; but enhancers can be located downstream as well. Some enhancers are relatively close to the genes they regulate, but others are thousands to tens of thousands of base pairs away from their target genes. Thus, important questions for molecular biologists are: What proteins are bound to enhancers, and how do enhancer sequences regulate transcription of the gene given their different distances from the start of transcription? One answer is that enhancers bind activator proteins and associated coactivator proteins to form a protein "bridge" that bends the DNA and links the transcription complex at the promoter to the activator-coactivator complex at the enhancer (Figure 8.13). The bend produced in the DNA may contain dozens to thousands of base pairs. The action of enhancers and the proteins they bind dramatically increases the efficiency of RNA pol II in initiating transcription, and as a result increases the level of transcription of genes regulated by enhancers. At the other end of the transcription-regulating spectrum are silencer sequences, DNA elements that act to repress transcription of their target genes. Silencers bind proteins that bend DNA in such a way that genes become sequestered in the folded segment and thus are shielded from transcription activation by RNA pol II.

One advantage of this method is its ability to detect covalent modifications of nucleotides, including methylation (we discuss the importance of methylation and other nucleotide modifications to gene expression in Chapter 13). The PacBio sequencer is a large laboratory piece of equipment. It can process DNA reads in the 10- to 50-kb range, and DNA replication is rapid. Initially, this method had an error rate of 10-15% in nucleotide assignment. However, newer techniques that resequence the same molecule multiple times can produce a consensus sequence with substantially fewer errors. The second widely used TGS method analyzes the sequence of single strands of DNA using the electrical signals associated with each nucleotide. Produced by Oxford Nanopore Technologies, this method uses a flow cell made of an electrical resistant membrane with thousands of tiny pores 1 nm in diameter known as nanopores (Figure 7.31b). The nanopores are formed by proteins that puncture the membrane and provide a channel through which the DNA strand to be sequenced will pass. As DNA enters a nanopore, it is denatured by a DNA helicase-like enzyme. A single DNA strand passes through the nanopore and each nucleotide emits electrical signals. Because each nucleotide is a different size and has distinct electrical properties, it is identified as it passes through the nanopore. These nucleotide-specific electrical signatures are collected and interpreted as a nucleotide sequence. Like the PacBio system, the nanopore system can detect covalently modified nucleotides. The nanopore system can also process very long reads of over a megabase. This method initially had an error rate around 15% due to the rapid rate of DNA passage through nanopores. Recent improvements to the method have slowed the rate of DNA passage and thus reduced the error rate. Third-generation sequencing is much faster and cheaper than next-generation sequencing. There is no need to clone DNA fragments for sequencing in TGS, and very long sequence reads are more readily completed with TGS. Comparing the two TGS methods, the PacBio system is laboratory-based, while the Oxford Nanopore system is a small unit ranging in size from a flash drive to a cellphone.

Chapter 7: DNA Replication, PCR, Sequencing Outline the process of PCR and delineate PCR reaction components. * What does PCR stand for? What does that mean? * What does PCR require? * What does PCR do? * Is PCR conservative, semiconservative, or not conservative? * What are the steps of PCR? * Why do we need PCR?

PCR has revolutionized medicine, forensics, and experimental biology. * Kary Mullis, shared Nobel Prize in Chemistry in 1993 for inventing PCR. He currently lives in Corona Del Mar. PCR = polymerase chain reaction. * It's DNA replication in a test tube (polymerase reaction). * The number of copies of the molecule that's being replicated doubles each cycle ("chain" reaction). Applications of PCR * Pre-implantation genetic diagnosis as an application of PCR. * Forensics (the Innocence Project). * Screening blood products for diseases. * Viral infections in wild monkeys by collecting feces. * Sequencing the Neanderthal genome. * Test for presence of SARS-CoV-2 (virus that causes COVID-19). The polymerase chain reaction (PCR): DNA replication in a test tube. * Requires a sequence-specific primer - you have to know at least part of the DNA sequence. (Because you need a primer that will anneal to the complementary bp) * Requires a special polymerase (that won't denature upon heat). * In PCR, the primer is made of DNA, not RNA, so they don't get removed. * Is semiconservative. * Doubles the amount of DNA with each cycle. PCR involves repeated cycles of polymerization to amplify DNA ----------------------- DONT MEM TEMPS------------ (DENATURATION -> ANNEALING -> EXTENSION) REPEAT * Denaturation: - Temp: 95 degrees Celsius. - What happens: All the DNA duplexes melt into single strands. * Annealing: - Temp: 50 degrees Celsius or so (depends on primer sequences). - What happens: The primers anneal (bind) to complementary sequence on the template strands. * Extension: - Temp: 72 degrees Celsius. - What happens: The polymerase does its thing. Why do we need PCR? * Because we need more DNA. * Because we want to measure the size of a piece of target DNA.

Chapter 9: Translation (Not a learning objective, but must know) What are the four levels of organization of polypeptides?

Polypeptides have four levels of organization (Table 9.2). The polypeptide primary structure is the sequence of amino acids contained in the polypeptide. The differences in the order of amino acids and in the lengths of polypeptides (the number of amino acids they contain) are effectively limitless. There are billions of possible amino acid sequences. At the same time, the specific order of amino acids in any given polypeptide is critical to its proper folding and functioning. Polypeptide secondary structure consists of certain common configurations adopted by portions of polypeptides, owing primarily to hydrogen bonds and ionic interactions that form between amino acids. Hydrogen bond formation causes amino acids with polar R-groups to align with one another. This can result in local bending or twisting of the polypeptide into one of two possible structures: An α-helix (alpha helix) is a twisted coil of amino acids stabilized by hydrogen bonds between partially charged R-groups; a β-pleated sheet (beta-pleated sheet) is a roughly 130-degree bend created when hydrogen bonding between amino acids induces a segment of a polypeptide to fold. A polypeptide's tertiary structure is the three-dimensional structure of the folded polypeptide as a whole. Polypeptides that are active are in their tertiary structure. Some polypeptides are capable of assuming two or more somewhat different tertiary structures. These may include an active structure and an inactive structure, or other combinations. A range of interactions involving the R-groups—hydrogen bonds, covalent bonds, ionic interactions, and hydrophobic interactions—produce the overall shape of the protein.

Chapter 9: Translation (Not a learning objective, but must know) What are the four levels of organization of polypeptides? *** CONTINUED ***

Primary, secondary, and tertiary structures of polypeptides are interdependent—the primary structure leads to certain secondary structure possibilities and these, in turn, lead to the formation of the one or more possible tertiary structures of a polypeptide. But some proteins in their active form consist of two or more polypeptides, and this level of organization is called the quaternary structure. Proteins that have two or more polypeptides (and therefore a quaternary structure) are often described as multimers. The individual polypeptides of a multimer may be identical or may be different. A protein composed of four identical polypeptides, for example, can be called a homotetramer, whereas a four-polypeptide protein that contains two or more different polypeptides can be identified as a heterotetramer. Table 9.2 summarizes these four levels of polypeptide structure for the red blood cell protein hemoglobin—a heterotetrameric protein that is responsible for carrying oxygen.

Chapter 8: Transcription Describe promoters, enhancers, silencers; define "transcription factor". * How can the level of gene expression be regulated in prokaryotes? * How can the level of gene expression be regulated in eukaryotes? * What does the promoter determine? * How is transcription initiated? Describe the steps. - What happens shortly after transcription initiation? * What determines where an mRNA ends? * What is happening to DNA during RNA synthesis?

Prokaryotes: RNA polymerases do not require a primer to initiate transcription. 1. The RNA polymerase core enzyme and sigma subunit bind to -10 and -35 promoter consensus sequences. * -10 consensus sequence: TATAAT * -35 consensus sequence: TTGACA 2. DNA unwinds near the transcription start site to form the open promoter complex. 3. RNA polymerase holoenzyme initiates transcription and begins RNA synthesis. The sigma subunit dissociates shortly after transcription initiation, and the core enzyme continues transcription. Termination sequences determine where an mRNA ends. 4. The core enzyme synthesizes until it encounters the termination sequence. As RNA synthesis progresses, the DNA duplex unwinds to allow the template strand to direct RNA assembly. The duplex closes following synthesis.

Chapter 8: Transcription Detail the parts of a mature mRNA; compare and contrast bacterial and eukaryotic mRNAs.

Prokaryotic vs. Eukaryotic mRNAs * Each eukaryotic mRNA encodes a single protein. * Prokaryotic mRNA can be polycistronic (code for multiple proteins). The primary transcript is processed into mRNA in eukaryotes. * 5' - CAP and polyA tail (prokaryotes lack these). * Eukaryotes have mRNA splicing - Introns are usually much larger than exons. - Introns are spliced out and discarded. - The processed transcript: * 5' cap, exons, polyA tail. Prokaryotes: the level of gene expression can be affected by how similar the -10 and -35 sequences are to the consensus sequence. Enhancers and silencers can regulate gene expression in eukaryotes. * In addition to a promoter, eukaryotic genes can also be regulated by other sequences in the DNA called enhancers and silencers. * These sequences can be close to the promoter, or far upstream or downstream. * Enhancers and silencers have binding sites for specialized proteins called "transcription factors". For Eukaryotes: Mature mRNAs contain 5′ and 3′ untranslated regions (UTRs, in prok. too), a modified base at the 5′ end called 'cap', a polyA tail at the 3′ end, and a protein-coding region in between. * 5' cap and polyA tail make the mRNA easily identifiable to make into protein. They help with translation of protein. Tail is put on after transcription by PolyA polymerase

GENE EXPRESSION QUIZ Which of the following is true about promoter sequences? * They are found upstream of the transcription site. * They are the places that DNA polymerase lands, and they tell DNA polymerase where to start transcribing. * They can be read in either direction. * They are identical for all genes in the genome.

Promoter sequences are found upstream of the transcription site. Promoter sequences are where RNA polymerase lands, and are what directs where RNA polymerase needs to start transcription. Promoter sequences are read in only one direction. Promoters can be different from one gene to another.

Chapter 8: Transcription Describe the process of transcription. * Compare DNA and RNA. * Describe polarity and sequence. * Which is the coding strand? Which is the template strand? * Where will the next nucleotide be added? * Which nucleotide will be added next? (Use lower pic to answer) - A, G, C, T, or U?

RNA is synthesized by RNA polymerase using DNA as a template. (requires a template) * RNA strands are synthesized 5' -> 3', just like DNA strands are. * Each incoming RNA nucleotide (oxynucleotide triphosphate) undergoes base pairing with the DNA template. Some differences between DNA and RNA. * DNA is composed of dNMPs and is synthesized from dNTPs. * RNA is composed of NMPs and is synthesized from NTPs. * In RNA, uracil is used instead of thymine (i.e. UMP instead of dTMP). DNA is copied into messenger RNA before it 'becomes' protein. * This is called transcription. * The DNA coding strand and the mRNA transcript have the same polarity and sequence, substituting U in mRNA for T in DNA. * The next nucleotide will be added onto 3' end of the mRNA?? (green) * G will be the next nucleotide to be added (mRNA is identical to the DNA coding strand except T is swapped for U). Textbook: Both DNA and RNA are nucleic acids. They are composed of nucleotide building blocks that are joined together by phosphodiester bonds to form polynucleotide strands. One principal difference between their molecules is the stable single-stranded structure of RNA versus the double-stranded structure of DNA. A second difference is the frequently encountered folding of RNA molecules. Many RNA molecules adopt folded secondary structures by complementary base pairing of segments of the molecule as part of the process by which they become functional. The RNA nucleotides, like those of DNA, are composed of a five-carbon sugar, a nucleotide base, and one or more phosphate groups. Each RNA nucleotide carries one of four possible nucleotide bases. But RNA nucleotides differ chemically from DNA nucleotides in two critical ways. The first difference concerns the identity of the RNA nucleotide bases. The purines adenine and guanine in RNA are identical to the purines in DNA. Likewise, the pyrimidine cytosine is identical in RNA and DNA. In RNA, however, the second pyrimidine is uracil (U) rather than the thymine carried by DNA. The four RNA ribonucleotides (A,U,G,C) are shown in Figure 8.1.

Chapter 8: Transcription TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN Describe promoters, enhancers, silencers; define "transcription factor". * How can the level of gene expression be regulated in prokaryotes? * How can the level of gene expression be regulated in eukaryotes? * What does the promoter determine? * How is transcription initiated? Describe the steps. - What happens shortly after transcription initiation? * What determines where an mRNA ends? * What is happening to DNA during RNA synthesis?

RNA polymerase holoenzyme initiates transcription through a process involving two steps. In the first step, the holoenzyme makes an initial loose attachment to the double-stranded promoter sequence and then binds tightly to it to form the closed promoter complex (➊ in Foundation Figure 8.6). In the second step, the bound holoenzyme unwinds approximately 18 bp of DNA around the -10 consensus sequence to form the open promoter complex ➋. Following formation of the open promoter complex, the holoenzyme progresses downstream to initiate RNA synthesis at the +1 nucleotide on the template strand of DNA ➌. Upon reaching the +1 nucleotide, the holoenzyme begins RNA synthesis by using the template strand to direct RNA assembly. The holoenzyme remains intact until the first 8 to 10 RNA nucleotides have been joined. At that point, the sigma subunit dissociates from the core enzyme, which continues its downstream progression (➌ in Foundation Figure8.6). The sigma subunit itself remains intact and can associate with another core enzyme to transcribe another gene. Downstream progression of the RNA polymerase core is accompanied by DNA unwinding ahead of the enzyme to maintain approximately 18 bp of unwound DNA ➍. As the RNA polymerase passes, progressing at a rate of approximately 40 nucleotides per second, the DNA double helix re-forms in its wake. When transcription of the gene is completed, the 5' end of the RNA trails off the core enzyme ➎.

Replication Origins - sites where DNA synthesis initiates in cells. Replication forks circled in red! Eukaryotic Chromosomes have Multiple Replication Origins * Electron micrograph showing multiple origins of replication on a Drosophila chromosome. * The human genome contains more than 10,000 origins, about 50,000 base pairs apart from each other. DNA is generally replicated in both directions at once. Replication Fork Textbook: Experimental evidence clearly demonstrates that DNA replication is most often bidirectional, progressing in both directions from a single origin of replication in bacterial chromosomes and from multiple origins of replication in eukaryotic chromosomes. In 1963, John Cairns reported the first evidence of a single origin of DNA replication in E. coli. Based on Cairns's evidence, it appeared that once replication gets underway in bacteria, there is expansion around the origin of replication, forming a replication bubble (Figure 7.10). The image shown in the figure is similar to the type of result Cairns obtained, and shows two regions known as replication forks at either end of the replication bubble. A replication bubble expands bidirectionally from an origin of replication, and active DNA synthesis takes place at each replication fork. (a) The arrows indicate replication bubbles, which are expanding bidirectionally. Different-sized replication bubbles indicate different replication start times. (b) Replication bubbles from multiple origins (upper) expand bidirectionally (middle) and merge, ultimately forming two sister chromatids (lower). Regardless of differences in the timing of initiation at the multiple origins of replication on a eukaryotic chromosome, each of the replication bubbles emanating from an origin of replication expands toward the others to eventually merge, resulting in the replication of all of the DNA in each eukaryotic nucleus by the end of S phase (Figure 7.13b). The end products of replication of each eukaryotic chromosome are a pair of identical DNA duplexes that are sister chromatids. The sister chromatids will remain joined through and will be separated at anaphase of the upcoming M phase.

Research subsequent to Franklin's initial discovery of A-form DNA identifies the level of hydration of DNA as the principal factor determining its formation. Dehydration converts B-form DNA to the A form, and it is thought that by assuming the A form, DNA is better protected from damage under desiccation conditions. Such conditions occur in certain bacterial species. The third form of DNA, Z-form DNA, was discovered by Robert Wells and colleagues in 1970, and its structure was determined by Andrew Wang, Alexander Rich, and colleagues in 1979. Z-form DNA is quite different from A-form and B-form DNA in that it has a left-handed twist that gives the sugar-phosphate backbone a zigzag appearance—hence the name Z-form—and other structural differences as well (see Table 7.1). No definitive biological significance has been identified for Z-form DNA; however, it occurs in cells and is particularly common near the start sites for gene transcription. Studies of Z-form DNA have identified certain DNA sequences that are associated with Z-form DNA formation. These too occur most frequently near the starting points of gene transcription. Study of the human genome reveals numerous such sequences where Z-form DNA is detected. Human chromosome 22 appears to have a number of transcription start sequences where Z-form DNA can occur.

DNA STRUCTURE QUIZ What is group B?

Ribose sugar

Chapter 9: Translation Genetic Analysis 9.2 The following segment of DNA encodes a polypeptide containing six amino acids. DNA triplets encoding the start codon (AUG) and a stop codon are included in the sequence. 5′-... CCCAGCCTAGCCTTTGCAAGAGGCCATATCGAC ...-3′ 3′-... GGGTCGGATCGGAAACGTTCTCCGGTATAGCTG ...-5′ * Write the sequence and polarity of the mRNA encoded by this gene. * Determine the amino acid sequence of the polypeptide, and identify the N- and C-terminal ends of the polypeptide. * If a base-substitution mutation changes the first transcribed G of the template strand to an A, how will this alter the polypeptide?

Scanning both DNA strands in their 3′-to-5′ direction identifies a single 5′-ATG-3′ sequence. The sequence is on the lower strand in the diagram beginning with the seventh nucleotide from the right. Since just one DNA triplet encoding a start codon is present, a scan of the strand at the correct distance from the start codon finds a 5′-TAG-3′ triplet sequence encoding a UAG stop codon: 3′-GGGTCG GAT CGGAAACGTTCTCCG GTA TAGCTC-5′ The mRNA sequence is: 5′-AUG GCC UCU UGC AAA GGC UAG-3′ The polypeptide sequence is: N-Met-Ala-Ser-Cys-Lys-Gly-C Substituting the first transcribed G->A on the template strand alters the second codon of mRNA by changing GCC->GUC and substitutes valine (Val) for alanine (Ala) in the second position of the polypeptide sequence.

Chapter 9: Translation How is an mRNA translated into protein? * Explain the role of ribosomes, tRNAs, and amino acyl tRNA synthetases in translation. - Know what they are, their components, and their functions. * How is mRNA read (what direction)? * What is a "charged" tRNA?

Some key players in translation: * Ribosomes * tRNAs * Aminoacyl tRNA synthetases Ribosomes: * Ribsomes = 'machine' composed of 3-4 rRNAs and over 50 proteins. - large and small subunits. * Have channels that hold the RNA and nascent polypeptide (the polypeptide being made). * Perform three key functions: 1. Bind mRNA and identify the start codon for translation. - mRNA is read from 5' -> 3' 2. Help bring about complementary pairing between mRNA codons and tRNA anti-codons. 3. Catalyze peptide bond formation between amino acids. Transfer RNAs (tRNAs): pair codons with amino acids. * pair amino acid with codon in mRNA. * tRNAs are covalently attached to AA. * The anticodon binds to codon in mRNA. * In pic, for example, the anticodon (purple) would be the anticodon for M and then the amino acid covalently bound to tRNA would be M. Aminoacyl tRNA synthetases: enzymes that 'recognize' both the anticodon and the cognate (correct) amino acid, and then attach the correct amino acid to the tRNA. * In pic, Glutamate is the AA and the anticodon stem includes the anticodon for glutamate. The "start" codon is AUG and codes for Methionine. * A tRNA attached to an amino acid is called a "charged" tRNA. * In pic, we see mRNA and the codon AUG. We also see tRNA and the anticodon 3' - UAC -5'. The anticodon of tRNA will form complementary base pairs with the codon of mRNA.

Chapter 9: Translation TEXTBOOK EXPLANATION CONTINUED How is an mRNA translated into protein? * Explain the role of ribosomes, tRNAs, and amino acyl tRNA synthetases in translation. - Know what they are, their components, and their functions. * How is mRNA read (what direction)? * What is a "charged" tRNA?

Textbook. : = don't memorize, marked as skim (good summary). Polypeptide assembly is orchestrated by ribosomes, which are ribonucleoprotein "machines" containing multiple molecules of ribosomal RNA (rRNA) and dozens of proteins. Ribosomes of all organisms are composed of two subunits that assemble into a ribosome as translation begins. Ribosomes bind mRNA and provide an environment for complementary base pairing between mRNA codon sequences and the anticodon sequences of tRNA. (See Section 1.3 for a basic review of translation.) Figure 9.2 encapsulates the essential elements of translation. Ribosomes translate mRNA in the direction, beginning with the start codon and ending with a stop codon. At each triplet codon, complementary base pairing between mRNA and tRNA determines which amino acid is added to the nascent (growing) polypeptide. The start codon and stop codon define the boundaries of the translated segment of mRNA. The resulting polypeptides have an N-terminal (amino-terminal) end corresponding to the end of mRNA and a C-terminal (carboxyl-terminal) end that corresponds to the end of mRNA. Ribosomes perform three essential tasks: 1. Bind messenger RNA and identify the start codon where translation begins. 2. Facilitate the complementary base pairing of mRNA codons and tRNA anticodons that determines amino acid order in the polypeptide. 3. Catalyze peptide bond formation between amino acids during polypeptide formation. The small ribosomal subunit contains a channel to hold the mRNA. In addition, there is a channel in the large subunit through which the nascent polypeptide is extruded from the ribosome Transfer RNA molecules are transcribed from tRNA genes. Recall that the three-dimensional structure of tRNAs features a CCA terminus at the end of tRNA molecules as the site of attachment of an amino acid (see Figure 8.28). Each tRNA carries only one of the 20 amino acids, and correct charging of each tRNA is crucial for the integrity of the genetic code.

Chapter 7: DNA Structure Why do complementary base pairs consist of one purine and one pyrimidine? Why are the strands of the double helix antiparallel and not parallel?

Textbook: Two questions about DNA frequently come up in discussions of DNA molecular structure. Why do complementary base pairs consist of one purine and one pyrimidine? Why are the strands of the double helix antiparallel and not parallel? The presence of one purine and one pyrimidine per base pair is a matter of molecular symmetry. Were the molecule to be composed of two paired purines (both double-ringed) the base pair would measure more than 20 Å across. Conversely, if the base pair was two pyrimidines (both single-ringed), the measurement would be much less than 20 Å. This would give the molecule an irregular diameter that might make it more difficult to package in cells and nuclei, or might make the binding of DNA-binding regulatory proteins more difficult. The reason why DNA strands are antiparallel to one another and not parallel is a matter of hydrogen bond formation. For hydrogen bonds to form, the negative charge of an oxygen or nitrogen must occur opposite the positive charge of a hydrogen. This occurs when complementary base pairs align in antiparallel strands. If a purine and a pyrimidine were aligned in parallel strands, positively charged hydrogens would be opposite one another, as would negatively charged nitrogens and oxygens. These repelling forces would prevent hydrogen bond formation.

Chapter 7: DNA Structure Textbook Explanation Identify the parts of DNA and RNA nucleotides. * Know the numbering of carbons. - 5'C vs. 5C * How are DNA and RNA nucleotides different? * What are the two types of nitrogenous bases in DNA and RNA? - Which bases are which? * Understand naming of nucleotides.

Textbook: A DNA nucleotide has three components: (1) a deoxyribose sugar, (2) one of four nitrogenous bases, and (3) up to three phosphate groups (Figure 7.5). Deoxyribose is a 5-carbon sugar, with the individual carbons identified as 1', 2', 3', 4' and 5'. An oxygen atom connects the 1' carbon to the 4' to form a five-sided (pentose) ring. The 5' carbon projects outward from the 4' carbon (and from the ring). Three of the carbon molecules have particular functional importance in determining nucleotide type and nucleotide function. A nitrogenous nucleotide base is attached to the 1' carbon by a covalent bond. The nucleotide base is either A, G, T, or C. A hydroxyl group (OH) is attached to the 3' carbon. The hydroxyl group participates in phosphodiester bond formation with the adjacent nucleotide in a DNA strand. Finally, phosphate molecules are attached to the 5' carbon. The image attached shows the four DNA nucleotides in their monophosphate forms. This is the form in which they appear while in a nucleotide chain. Free DNA nucleotides that are not incorporated into a polynucleotide chain are triphosphates; that is, they have three phosphate molecules attached to the 5' carbon. Note that deoxyribose has only a hydrogen atom bound at the 2' carbon, not a hydroxyl (OH) group as at the 3' carbon. This is the basis for naming the sugar deoxyribose (deoxy means "not oxygenated"). The four nitrogenous bases in DNA are of two structural types—a single-ringed form called a pyrimidine, and a double-ringed form called a purine. Cytosine (C) and thymine (T) are pyrimidines, and adenine (A) and guanine (G) are purines. In their monophosphate configurations, the nucleotides that carry the purine bases adenine and guanine are designated, respectively, deoxyadenosine 5'-monophosphate (dAMP) and deoxyguanosine 5'-monophosphate (dGMP); and the nucleotides that carry the pyrimidine bases cytosine and thymine are deoxycytidine 5'-monophosphate (dCMP) and deoxythymidine 5'-monophosphate (dTMP). Collectively, these are identified as the deoxynucleotide monophosphates, or dNMPs, where N can refer to any of the four nucleotide bases. In contrast, free (reactive) DNA nucleotides in their triphosphate configurations are dNTPs

Chapter 7: DNA Replication, PCR, Sequencing Understand DNA Proofreading and its significance (not a learning objective, but supposed to know).

Textbook: A critically important process for ensuring DNA replication accuracy is DNA proofreading, a capability of most DNA polymerases that momentarily stops and reverses replication to remove an incorrect nucleotide and replace it with a correct nucleotide. In E. coli, this proofreading ability resides in the 3'-to-5' exonuclease activity of DNA polymerases. Polymerases like pol III and pol I have a structure resembling an open hand: A "thumb" and "fingers" hold the template and daughter strands in the "palm," where 5'-to-3' polymerase activity is centered (Figure 7.21). When a replication error occurs, the mismatched DNA bases of the template and daughter strands are unable to hydrogen bond properly. As a result, the 3'-OH end of the daughter strand becomes displaced, blocking the further addition of nucleotides and inducing rotation of the daughter strand into the 3'-to-5' exonuclease site at the "heel" of the hand. Several nucleotides, including the mismatched one, are then removed from the 3' end of the daughter strand, after which the daughter strand rotates back to the polymerase site in the palm and replication resumes. Like their counterparts in bacteria, the principal DNA replication polymerases in eukaryotes and archaea also have proofreading ability to help ensure the accuracy of DNA replication.

Chapter 7: DNA Replication, PCR, Sequencing Textbook Explanation Describe semi-conservative replication and explain its significance. * Who is Rosalind Franklin? * Who are Watson and Crick?

Textbook: Considering the importance of DNA throughout the biological world, it was no surprise to discover that the general mechanism of DNA replication is the same in all organisms. This universal process evolved in the earliest life-forms and has been retained for billions of years. As organisms diverged and became more complex, however, an array of differences did develop among DNA replication proteins and enzymes. Despite the diversification of these specific components of DNA replication, three attributes of DNA replication are shared by all organisms: Each strand of the parental DNA molecule remains intact during replication. Each parental strand serves as a template directing the synthesis of a complementary, antiparallel daughter strand. Completion of DNA replication results in the formation of two identical daughter duplexes, each composed of one parental strand and one daughter strand. Watson and Crick concluded with the following observation: It has not escaped our notice that the specific base-pairing we have proposed immediately suggests a possible copying mechanism for the genetic material. Specifically, Watson and Crick recognized that a consequence of complementary base pairing was that nucleotides on one strand of the duplex could be used to guide the ordering of nucleotides on the other strand. Watson and Crick presumed that DNA replication used the nucleotide sequence of each strand to form a new pair of DNA duplexes, hypothesizing that each DNA strand of the original duplex would act as a template for the synthesis of a new daughter strand. Watson and Crick did not know the precise mechanism by which template-based replication took place, however, raising the crucial question of what the exact mechanism of replication might be.

Chapter 8: Transcription TEXTBOOK EXPLANATION Detail the parts of a mature mRNA; compare and contrast bacterial and eukaryotic mRNAs.

Textbook: In the figure, untranslated mRNA at the 5′ end (5′ UTR) and at the 3′ end (3′ UTR) separate the 5′ mRNA end from the start codon and the stop codon from the rest of the mRNA, respectively. Bacterial and eukaryotic RNA transcripts differ from one another in at least two important ways. First, eukaryotic transcripts are more stable than bacterial transcripts. The half-life of a typical eukaryotic mRNA is measured in hours to days, whereas bacterial mRNAs have an average half-life measured in seconds to minutes. A second difference is the presence of introns in eukaryotic genes that are absent from most bacterial genes. Keep in mind that in bacteria the lack of a nucleus leads to coupling of transcription and translation. In eukaryotic cells, on the other hand, transcription takes place in the nucleus, allowing pre-mRNA processing to take place in the cytoplasm and translation to occur at free ribosomes or at ribosomes attached to the rough endoplasmic reticulum in the cytoplasm. In discussing posttranscriptional processing, we highlight three processing steps that are coordinated during transcription to modify the initial eukaryotic gene mRNA transcript, called pre-mRNA, into mature mRNA, the form of mRNA that is translated. These modification steps are (1) 5' capping, the addition of a modified nucleotide to the 5' end of mRNA; (2) 3' polyadenylation, cleavage at the 3' end of mRNA and addition of a tail of multiple adenines to form the poly-A tail; and (3) intron splicing, RNA splicing to remove introns and ligate exons. These processes occur in the nucleus. They occur simultaneously and while transcription is underway. They are also tied to the termination of transcription. Once these three steps are complete, the pre-mRNA has been fully processed into mature mRNA. The mature mRNA is released from the nucleus and makes its way to ribosomes, where translation takes place.

Chapter 7: DNA Replication, PCR, Sequencing Textbook Explanation Outline the process of PCR and delineate PCR reaction components. * What does PCR stand for? What does that mean? * What does PCR require? * What does PCR do? * Is PCR conservative, semiconservative, or not conservative? * What are the steps of PCR? * Why do we need PCR?

Textbook: The polymerase chain reaction (PCR) is an automated version of DNA replication that takes place in a test tube containing a total reaction volume of 20 to 50 microliters (one microliter is one-millionth of a liter). Despite this very small total reaction volume, a typical PCR reaction, beginning with just a few copies of a short, targeted DNA sequence, produces millions of copies of the sequence in a few hours. Polymerase chain reactions are in vitro DNA-replication reactions performed using (1) double-stranded DNA containing the target sequence that is to be copied, (2) a supply of the four DNA nucleotides, (3) a heat-stable DNA polymerase, and (4) two different single-stranded DNA primers (described in the list of steps below). These PCR components are mixed with a buffer solution, and then the automated reaction is run through 30 to 35 three-step "cycles." Each cycle doubles the number of copies of the targeted DNA sequence. The PCR process is generally identified as "amplification," and it is common to speak of "PCR amplification" in reference to the process and of "amplified DNA" as the product of the reaction. PCR reactions are carried out in a device known as a PCR thermal cycler. Thermal cyclers are programmable, allowing the length and temperature of each cycle step to be adjusted to meet the needs of the experimenter. The thermal cycler takes just a few seconds to change temperature between steps. Figure 7.25 illustrates the three steps of a PCR reaction. The steps and functions of each PCR cycle are as follows: ➊ Denaturation. The reaction mixture is heated to approximately 95 degrees, causing double-stranded DNA to denature into single strands as the hydrogen bonds between complementary strands break down. The step duration is usually 1 to 2 minutes. ➋ Primer annealing. The reaction temperature is reduced to between about 45 and 68 degrees to allow primer annealing—the hybridization of the two short, single-stranded DNA primers to complementary sequences bracketing the target sequence. These primers have the same function as RNA primers in DNA replication. They are, as mentioned, short (12 to 24 nucleotides), and one primer binds to each of the denatured DNA strands. Step drtn: 1-2 min

Chapter 9: Translation TEXTBOOK EXPLANATION CONTINUED Describe translation initiation, elongation, and termination. * Initiation: - What is the initiation complex? - What are the tRNA binding sites in the ribosome? What do they do? - Know where things are in terms of which site at each point during translation. * Elongation: - Know where things are in terms of which site at each point during translation. - Know the steps of elongation. - What is peptidyl transferase and what does it do? * Termination: - Know where things are in terms of which site at each point during translation. - Know the steps of termination. * Know why they happen (what caused it), and what results after the step. - What is the 3' UTR?

Textbook: = don't need to memorize, just good summary. Figure attached identifies two segments of the mRNA transcript that do not undergo translation. Between the 5' end of mRNA and the start codon is a segment known as the 5' untranslated region, abbreviated 5' UTR. The region between the stop codon and the 3' end of the molecule is the 3' untranslated region, or 3' UTR. The 5' UTR contains sequences that help initiate translation and the 3' UTR contains sequences associated with transcription termination. Translation initiation in all organisms begins when the small ribosomal subunit binds near the 5' end of mRNA and identifies the start codon sequence. In the next stage, the initiator tRNA, the tRNA carrying the first amino acid of the polypeptide, binds to the mRNA start codon. In the final stage of initiation, the large subunit joins the small subunit to form an intact ribosome, and translation begins. During these stages, initiation factor proteins help control ribosome formation and binding of the initiator tRNA, and guanosine triphosphate (GTP) provides energy. The tRNAs used during translation each carry a specific amino acid and are identified as charged tRNAs. In contrast, a tRNA without an amino acid is uncharged. Specialized enzymes discussed in a later section are responsible for recognizing different tRNAs and charging each one with the correct amino acid. Elongation, the second phase of translation, begins with the recruitment of elongation factor (EF) proteins into the initiation complex. Elongation factors facilitate three steps of polypeptide synthesis: 1. Recruitment of charged tRNAs to the A site 2. Formation of a peptide bond between sequential amino acids 3. Translocation of the ribosome in the 3' direction along mRNA GTP cleavage provides the energy for each step of elongation.

Chapter 8: Transcription (Not a learning objective, but should know) * Know the Central Dogma and what it explains. * What gives us our phenotype? - Describe these things. * What gives us our genotype? * What determines how we function and how we look? * What controls our phenotype? How? - As a general rule, what else can contribute to our phenotype?

The Central Dogma explains how DNA (genotype) becomes a phenotype. * DNA ---transcription----> RNA ----translation ----> protein Proteins give us our phenotype. * Proteins are enzymes, and structural components of cells, etc. * Which proteins and how much we have of them determine how we function and how we look. * DNA controls our phenotype by encoding proteins. * NOTE: This is true as a general rule, but... non-coding RNAs, including tRNAs, rRNA, microRNAs, and lncRNAs, can also contribute to phenotype.

Chapter 9: Translation TEXTBOOK EXPLANATION CONTINUED AGAIN How is an mRNA translated into protein? * Explain the role of ribosomes, tRNAs, and amino acyl tRNA synthetases in translation. - Know what they are, their components, and their functions. * How is mRNA read (what direction)? * What is a "charged" tRNA?

The charging of tRNAs is catalyzed by enzymes called aminoacyl-tRNA synthetases, or, more simply, tRNA synthetases.There are 20 different tRNA synthetases, one for each of the amino acids. To charge an uncharged tRNA, a tRNA synthetase catalyzes a reaction that forms a bond between the carboxyl group of the amino acid and the hydroxyl group of adenine in the CCA terminus. When tRNA is in contact with tRNA synthetase, the tRNA acceptor stem fits into an active site of tRNA synthetase. The active site contains the amino acid that will be added to the tRNA acceptor stem, as well as ATP that provides energy for amino acid attachment.

Chapter 8: Transcription Define "gene expression", and understand that it's regulated in space and time. * What are some general variations relating to gene expression? * What is mRNA? What happens to mRNA? * When we say a gene is being expressed, what do we mean? * What are housekeeping genes? * What are cell-type specific genes? * If you were to quickly summarize gene expression and its variation, what would you say? * What is regulated gene expression? - What does it involve?

The conversion of a gene into its product (generally a protein via an RNA intermediate) is called: GENE EXPRESSION * Some genes function as RNA and are transcribed but not translated. * mRNAs are translated into proteins. GENE EXPRESSION: when we say a gene is being expressed, we mean it is being transcribed and translated into protein. Many genes are not expressed all the time. * Some genes are expressed virtually all the time. * Many genes are only expressed some of the time - in response to external or internal conditions or signals. Many genes are not expressed in all cells. * Each cell of a multicellular organism contains the same DNA = same genes. * But, different cells express different genes (thus different proteins). * Housekeeping genes - expressed virtually in all cells (EX: RNA Polymerases). * Cell-type-specific genes - expressed in only certain cells, e.g. neurons, muscle cells, blood cells. Summary: Most genes are not expressed all the time, or in all cells. * Only a subset of genes are expressed in any given cell type: - cell-type-specific or tissue-specific gene expression. - regulated gene expression. * And only a subset of those are expressed at any given time. - regulated gene expression The regulation of gene expression is very complex, and we hardly touch on it here, but it involves promoters, enhancers, silencers, and transcription factors.

Chapter 8: Transcription (Not a learning objective, but should know) How do researchers verify that a segment of DNA is a functionally important component of a promoter?

The diversity of eukaryotic promoters begs an important question: How do researchers verify that a segment of DNA is a functionally important component of a promoter? The research has two components; the first, outlined in Research Technique 8.1, is discovering the presence and location of DNA sequences that transcription factor proteins will bind to. The second component involves mutational analysis to confirm the functionality of the sequence. Researchers produce many different point mutations in the DNA sequence under study and then compare the level of transcription generated by each mutant promoter sequence with transcription generated by the wild-type sequence. Figure 8.12 shows a synopsis of promoter mutation analysis from an experiment performed by the molecular biologist Richard Myers and colleagues on a mammalian β-globin gene promoter. These researchers produced mutations of individual base pairs in TATA box, CAAT box, and GC-rich sequences, and of nucleotides between the consensus sequences, to identify the effect of each individual mutation on the relative transcription level of the gene. The bars in the figure indicate the impact of base substitution mutations of individual base pairs in and around the consensus sequences of the promoter. A relative transcription level of 1.0 represents the wild-type promoter; thus, a bar that is lower than 1.0 indicates a decrease in transcription level, and a bar that is higher than 1.0 indicates an increased level of transcript The researchers found that most base-pair mutations in the three consensus regions significantly decreased the transcription level of the gene, and they found two base substitutions in the CAAT box region that significantly increased transcription. In contrast, mutations outside the consensus regions had nonsignificant effects on transcription level. These results show the functional importance of specific DNA sequences in promoting transcription and confirm a functional role in transcription for TATA box, CAAT box, and GC-rich sequences. Notice that the sequences of these regulatory regions in this particular gene differ slightly from the consensus sequences shown. The precise regulatory sequence of any gene may vary slightly from con

The elongation cycle continues until one of the three stop codons, UAG, UGA, or UAA, enters the A site of the ribosome. There are no tRNAs with anticodons complementary to stop codons, so the entry of a stop codon into the A site is a translation-terminating event. Release factors (RF) bind a stop codon in the A site (Figure 9.10 ➊). The catalytic activity of RFs releases the polypeptide bound to tRNA at the P site ➋. Polypeptide release causes ejection of the RF from the P site and leads to the separation of the ribosomal subunits ➌.

DNA REPLICATION QUIZ A single strand of DNA, 20 nucleotides long, with the sequence 5'-AAAAACCCCCGGGGGTTTTT-3' is in a test tube. What else needs to go in the test tube to so that we end up with a piece of double stranded DNA, 20 base pairs long, with the above sequence comprising one of the two strands?

The primer 5'-AAAAAC-3' DNA polymerase deoxynucleotide triphosphates

Chapter 8: Transcription TEXBOOK EXPLANATION CONTINUED Describe the process of transcription. * Compare DNA and RNA. * What is mRNA? Function? * Describe polarity and sequence. * Which is the coding strand? Which is the template strand? * Where will the next nucleotide be added? * Which nucleotide will be added next? (Use lower pic to answer) - A, G, C, T, or U?

The structure of uracil is similar to that of thymine, but notice, by comparing the structure of uracil in Figure 8.1 with that of thymine in Figure 7.5, that thymine has a methyl group at the 5 carbon of the pyrimidine ring, whereas uracil does not. In all other respects, uracil is similar to thymine, and when uracil undergoes base pairing, its complementary partner is adenine. The second chemical difference between RNA and DNA nucleotides is the presence of the sugar ribose in RNA rather than the deoxyribose occurring in DNA. The ribose gives RNA its name (ribonucleic acid). Compare the ribose molecules shown in Figure 8.1 with deoxyribose in Figure 7.5, and notice that ribose carries a hydroxyl group (OH) not found in deoxyribose at the 2' carbon of the ring. Except for this difference, ribose and deoxyribose are identical, having a nucleotide base attached to the 1' carbon and a hydroxyl group at the 3' carbon. The similarity of the sugars of RNA and DNA leads to the formation in RNA of phosphodiester bonds between nucleotides of a strand and to a sugar-phosphate backbone that is identical to that of DNA. RNA-strand phosphodiester bond formation takes place by the same general mechanism as found in DNA (Figure 8.2). RNA is synthesized from a DNA template strand using the same purine-pyrimidine complementary base pairing described for DNA, except that in RNA, adenine pairs with uracil rather than thymine. RNA polymerase enzymes catalyze the addition of each ribonucleotide to the 3' end of the nascent strand, forming a phosphodiester bond between the 5' carbon of one nucleotide and the 3' carbon of the adjacent nucleotide, eliminating two phosphates (the pyrophosphate group) from the incoming ribonucleotide triphosphate in the process, just as in DNA synthesis. * Messenger RNA (abbreviated mRNA) is a type of single-stranded RNA involved in protein synthesis. mRNA is made from a DNA template during the process of transcription. The role of mRNA is to carry protein information from the DNA in a cell's nucleus to the cell's cytoplasm (watery interior), where the protein-making machinery reads the mRNA sequence and translates each three-base codon into its corresponding amino acid in a growing protein chain.

Chapter 7: DNA Structure Understand the role of hydrogen bonds in complementary base pairing.

The two strands of DNA are held together by hydrogen bonds formed between complementary base pairs (nitrogenous base pairs). * Hydrogen bonds are non-covalent and relatively weak. They can come apart and come back together. Hydrogen bonds are non-covalent bonds. Thus, the two strands of DNA are held together non-covalently - they can separate (melt) at high enough temperatures, or when "unzipped" by specialized enzymes.

Chapter 8: Transcription TEXTBOOK EXPLANATION CONTINUED Articulate how alternative splicing can increase protein diversity. * How can large eukaryotic genomes express more proteins than there are genes in the genomes? - Name the transcription-associated mechanisms.

The use of alternative promoters occurs when a gene contains more than one upstream sequence that can bind transcription factors and initiate transcription at different transcription start sites. Similarly, alternative polyadenylation is possible in those genes that contain more than one polyadenylation signal sequence that can activate 3' pre-mRNA cleavage and polyadenylation. Alternative promoters and alternative polyadenylation are driven by the variable expression of transcriptional or polyadenylation proteins in a cell-type-specific manner, and the processes generate characteristic mature mRNAs and distinctive proteins in specific cells. The result is that transcription of a given gene may lead to the production of several different mature mRNAs in different types of cells, and to their translation into distinct proteins in each of those cell types.

Chapter 7: DNA Structure Identify the components of DNA and RNA. * What are the necessary properties for the carrier of genetic info? - What is the carrier of genetic info?

These are the necessary properties for the carrier of genetic information. 1. Complex (must be able to "code" for 20 amino acids plus punctuation). 2. Stable (so it doesn't fall apart in cell). 3. Able to be replicated. 4. Mutable (changeable, so we can evolve). In life as we know it, the carrier of genetic information is either DNA or RNA. * Double-Stranded DNA genomes: Most life, except some viruses. * Single-Stranded DNA genomes: Parvoviruses, Circoviruses... * Double-Stranded RNA genomes: Rotaviruses... * Single-Stranded RNA genomes: Coronaviruses, polio virus, coxsackie virus, rhinoviruses, Hep A virus... * ssRNA -> dsDNA -> ssRNA genomes: Retroviruses - At one point in their life cycle they were ssRNA, and then dsDNA etc. DNA and RNA are composed of nucleotides.

Chapter 8: Transcription TEXTBOOK EXPLANATION CONTINUED AGAIN Describe promoters, enhancers, silencers; define "transcription factor". * How can the level of gene expression be regulated in prokaryotes? * How can the level of gene expression be regulated in eukaryotes? * What does the promoter determine? * How is transcription initiated? Describe the steps. - What happens shortly after transcription initiation? * What determines where an mRNA ends? * What is happening to DNA during RNA synthesis?

These consensus sequences are located upstream from the +1 nucleotide (the start of transcription) in a region flanking the gene where the nucleotides are denoted by negative numbers and are not transcribed. At the -10 position of the E. coli promoter is the Pribnow box sequence, or the -10 consensus sequence, consisting of 6 bp having the consensus sequence 5'-TATAAT-3'. The Pribnow box is separated by about 25 bp from another 6-bp region, the -35 consensus sequence, identified by the nucleotides 5'-TTGACA-3'. The nucleotide sequences that occur upstream, downstream, and between these consensus sequences are highly variable and contain no other consensus sequences. Thus, in a functional sense, the -10 (Pribnow) and -35 consensus sequences are important because of their nucleotide content, their location relative to one another, and their location relative to the start of transcription. In contrast to the consensus sequences themselves, the nucleotides between -10 and -35 are important as spacers between the consensus elements, but their specific sequences are not critical. Natural selection has operated to retain strong sequence similarity in consensus regions and to retain the position of the consensus regions relative to the start of transcription. The effectiveness of evolution in maintaining promoter consensus sequences is illustrated by comparison with the sequences between and around -10 and -35 which are not conserved and which exhibit considerable variation. In addition, the spacing between the sequences and their placement relative to the +1 nucleotide is stable. RNA polymerase is a large molecule that binds to -10 and -35 consensus sequences and occupies the space between and immediately around the sites. Crystal structure models show that the enzyme spans enough DNA to allow it to contact promoter consensus regions and reach the +1 nucleotide. Once bound at a promoter in this fashion, RNA polymerase can initiate transcription. RNA pol II recognizes and binds to promoter consensus sequences in eukaryotes with the aid of proteins called transcription factors (TF). The TF proteins bind to promoter regulatory sequences and influence transcr. initiation by interacting, directly or indirectly w/ RNA pol

GENE EXPRESSION QUIZ What do DNA and RNA polymerase have in common? * They both make nucleic acids in a 3' to 5' fashion. * They both catalyze the formation of phosphodiester bonds. * They both incorporate dNTPs into a growing nucleic acid molecule (RNA or DNA).

They both catalyze the formation of phosphodiester bonds. Both RNA and DNA polymerase work in a 5' to 3' direction and catalyze phosphodiester bonds. DNA polymerase uses dNTPs, RNA polymerase uses NTPs.

Chapter 7: DNA Replication, PCR, Sequencing ** CONTINUED ** New Generations of DNA Sequencing Technology (Not included in learning objectives, but included in the section of the book he told us to know) * Benefits of new generations? * Next-Generation Sequencing * Third-Generation Sequencing

This process repeats itself very rapidly as nucleotides are added to the strands. The result is a sequence for the fragments in each cluster. In this manner, next-generation sequencing identifies the sequence of a DNA strand "by synthesis" rather than "by chain termination" (the approach in dideoxy sequencing). Well over 40 years have passed since Sanger first outlined his dideoxy nucleotide DNA sequencing method, and DNA sequencing technology has entered its third generation. DNA sequencing technology is now focused on reading long stretches of DNA sequence and on obtaining a sequence in real-time from single DNA molecules. In dideoxy DNA sequencing, the sequence of DNA is read at a length of up to a thousand bases at a time using electrophoretic gel-based analysis. NGS methods discarded electrophoretic gel analysis in favor of assessing the sequence of many DNA molecules simultaneously. This so-called massively parallel approach is capable of reading millions of pieces of DNA simultaneously in blocks of a few hundred nucleotides at a time. Previous DNA sequencing methods required that individual DNA fragments be sequenced and then computationally assembled by joining contiguous fragments to produce a complete DNA sequence. Beginning in the mid-2000s, third-generation sequencing (TGS) technologies were introduced to provide real-time sequencing of long stretches of single DNA molecules. TGS methods can read DNA lengths in the 10- to 100-kb range and do not require the same kind of sequence assembly as NGS. Two TGS methods have emerged over the last decade. Both methods analyze single DNA molecules in real time. The first method, introduced by Pacific Biosciences (PacBio), uses a small structure called a flow cell, which contains thousands of tiny wells. Each well contains a single DNA fragment that undergoes replication using an engineered DNA polymerase that is anchored at the bottom of the well (Figure 7.31a). Using fluorescently labeled nucleotides excited by a light source, each newly incorporated nucleotide emits a pulse of light. Each emission is recorded, and the nucleotide in that position is assigned.

Chapter 9: Translation What is the red box around?

Thr codon

Chapter 7: DNA Replication, PCR, Sequencing Explain how nucleotides are polymerized to form DNA molecules.

Through phosphodiester bond formation! Nucleotides polymerize by chemically linking a phosphate group at the 5' position of one nucleotide to the hydroxyl group at the 3' position of the next nucleotide, eliminating a water molecule (condensation polymerization), catalyzed by DNA polymerase.

Chapter 8: Transcription TEXBOOK EXPLANATION CONTINUED AGAIN Describe the process of transcription. * Compare DNA and RNA. * Describe polarity and sequence. * Which is the coding strand? Which is the template strand? * Where will the next nucleotide be added? * Which nucleotide will be added next? (Use lower pic to answer) - A, G, C, T, or U?

Transcription is the synthesis of a single-stranded RNA molecule by RNA polymerase. It is most clearly understood and described in bacteria, and E. coli is the model experimental organism from which the majority of our knowledge of bacterial transcription has been derived. In this section, we examine the four stages of transcription in bacteria: (1) promoter recognition and identification, (2) the initiation of transcript synthesis, (3) transcript elongation, and (4) transcription termination. The end product of transcription is a single-stranded RNA that is complementary and antiparallel to the template DNA strand. The transcript has the same 5' -to-3' polarity as the coding strand of DNA, the strand complementary to the template strand. The coding strand and the newly formed transcript also have identical nucleotide sequences, except for the presence of uracil in the transcript in place of thymine in the coding strand. For this reason, gene sequences are written in 5' -to-3' orientation as single-stranded sequences based on the coding strand of DNA. This allows easy identification of the mRNA sequence of a gene by simply substituting U for T.

Chapter 9: Translation Describe translation initiation, elongation, and termination. * Initiation: - What is the initiation complex? - What are the tRNA binding sites in the ribosome? What do they do? - Know where things are in terms of which site at each point during translation. * Elongation: - Know where things are in terms of which site at each point during translation. - Know the steps of elongation. - What is peptidyl transferase and what does it do? * Termination: - Know where things are in terms of which site at each point during translation. - Know the steps of termination. * Know why they happen (what caused it), and what results after the step. - What is the 3' UTR?

Translation can be divided into three phases: * Initiation * Elongation * Termination Initiation: * The initiation complex (includes small ribosomal subunit and initiator Met-tRNA (pre-bound)) binds to 5' cap on mRNA. * The initiator complex scans along the mRNA 5' to 3' until it finds an AUG. * Large subunit then binds. * At this point, the initiating methionine tRNA is in the P (peptidyl) site. tRNA binding sites in the ribosome * P (peptidyl) site - holds the tRNA to which the growing polypeptide chain is attached. * A (acceptor) site - binds tRNA carrying the next amino acid to be added. * E (exit) site - empty tRNAs leave from here after their amino acid has been added. Elongation * New tRNA enters the A (acceptor) site. * Peptide bond is formed between amino acids in the P and A site. * Ribosome moves down one codon: - tRNA in the P site moved to E site, tRNA in the A site moved to P site. * tRNA in E site exits. * Next tRNA enters the A site; repeat. - "Repeat": -> Peptide bond is formed between amino acids in the P and A site, etc. 1. tRNA recruitment * Charged tRNA accompanied by EF1 (elongation factor 1) enters and binds at the A site. - ACC = Thr - CAU = His 2. Peptide bond formation * Peptidyl transferase forms a peptide bond between amino acids in the P and A sites. Peptide chain moves to the A site. * Ribosome slides over to the next codon. 3. Translocation * Elongation factors translocate the ribosome; the uncharged tRNA is released to the E site, and a new tRNA is recruited to the A site. * Ribosome moves along mRNA. Termination 1. Release-factor recruitment: release factors are recruited when a stop codon occurs at the A site. 2. Polypeptide release: eRF fills the A site, triggering the release of polypeptide by hydrolysis of GTP. 3. Ribosome dissociation and mRNA release. * No tRNA for a STOP codon, release factor binds instead. * The portion of the mRNA after the stop codon is the 3' UTR.

Chapter 7: DNA Replication, PCR, Sequencing Question: Of the 6 bases shown, which one of them was most recently added to the growing strand?

What we mean when we say DNA synthesis proceeds 5' -> 3' is this: New bases are added onto the 3' end of the growing strand.

Chapter 8: Transcription Predict mRNA sequences from DNA and vice versa. The sequence of an RNA transcript is: 5' - GGUUACAUUC - 3' What is the sequence of the template strand?

coding: 5' - GGTTACATTC - 3' template: 5' - GAATGTAACC - 5'

Chapter 8: Transcription Delineate introns from exons, and explain the role of splicing.

mRNA splicing * Introns are usually much larger than exons. * Introns are spliced out and discarded (they don't code for protein). * The processed transcript: - 5' cap, exons, polyA tail. mRNA splicing: * Occurs at sites determined by consensus sequences. * Requires multiple proteins. * Takes place in the nucleus. * Most of the RNA is discarded in mammals. * Splicing pattern can vary under different conditions (alternative splicing). Splicing links up exons from within a given gene, not exons from different genes. Textbook: Most eukaryotic genes contain two kinds of segments. One kind, the exons, become part of mature mRNA and encode segments of proteins. The other kind, the introns, are intervening segments that separate exons. Introns are removed from pre-mRNA by processes that excise the introns and splice together the exons.

Chapter 7: DNA Replication, PCR, Sequencing Textbook Explanation ***CONTINUED AGAIN*** Outline the process of PCR and delineate PCR reaction components. * What does PCR stand for? What does that mean? * What does PCR require? * What does PCR do? * Is PCR conservative, semiconservative, or not conservative? * What are the steps of PCR? * Why do we need PCR?

or individual genes. The requirement for primer sequence information can be satisfied by informed guesses about the sequences likely to occur at primer binding sites or by using primers from one species to amplify similar sequences in another species. For example, a biologist wanting to study DNA-sequence similarity between species could use a pair of primers that amplify a Drosophila gene to examine the human genome for a related gene. There may be one or more base-pair mismatches between the Drosophila primers and the human DNA sequences they bind to, but the mismatches need not prevent primer annealing if the temperature of the PCR reaction is lowered during step 2 of the reaction. The lower temperature can increase the stability of hybridization of the primers and their target sequences enough to allow the former to prime the PCR amplification. The PCR process selectively amplifies DNA fragments ranging in size from a few dozen base pairs to several thousand base pairs in length. The fragments generated are almost all of the same double-stranded target region. So highly concentrated are the results of PCR amplification that they can be analyzed directly using gel electrophoresis (see Chapter1 for a discussion of this method). Gel electrophoresis separates fragments of DNA by their sizes in base pairs. Recall from our discussion in Chapter 1 that DNA fragments containing fewer base pairs move more quickly in the electrical separation field than fragments with more base pairs. This means that smaller fragments, with higher electrophoretic mobility, migrate farther from the origin of migration than do fragments with more base pairs. The use of molecular-weight size markers (in gel electrophoresis DNA fragments containing known numbers of base pairs) allows researchers to determine the size of DNA fragments of unknown length by comparing their migration with that of the known size markers.

Chapter 9: Translation The anticodon is a part of the...

tRNA

Chapter 7: DNA Replication, PCR, Sequencing Textbook Explanation ***CONTINUED*** Outline the process of PCR and delineate PCR reaction components. * What does PCR stand for? What does that mean? * What does PCR require? * What does PCR do? * Is PCR conservative, semiconservative, or not conservative? * What are the steps of PCR? * Why do we need PCR?

➌ Primer extension. Raising the temperature of the reaction to 72 degrees Celsius allows primer extension, during which a specialized DNA polymerase known as Taq polymerase synthesizes DNA, beginning at the 3' end of each primer. Taq polymerase, described in more detail below, synthesizes new DNA at the rate of about 1000 bp per minute. This step duration is usually 3 to 5 minutes. Figure 7.26 shows the two important features of the locations of PCR-primer binding. First, the primers bind just outside of the target region for amplification, and, second, the primers bind to opposite complementary strands. This primer binding pattern ensures that the target region will be copied during the PCR procedure, and it establishes the 5' and 3' boundaries of the amplified PCR products that will be produced by the procedure. Each complete PCR cycle doubles the number of copies of the target DNA sequence, so beginning with a single copy of double-stranded target sequence, completion of the first PCR cycle produces two copies of the target sequence, completion of the second cycle produces four copies, completion of the third cycle, eight copies, and so on. After 30 PCR cycles the yield would be 2^30 or more than 1 billion copies of the target sequence, and completion of 36 cycles could yield more than 68 billion copies of the target sequence. Taq polymerase is named for the thermophilic bacterial species Thermus aquaticus. This bacterium lives in hot springs at near-boiling temperatures and has evolved heat-stable proteins that remain active at these temperatures. The heat stability of Taq DNA polymerase is important to the efficiency of PCR, since step 1 of a PCR cycle raises the reaction temperature to near boiling. The DNA polymerases of most organisms are not heat stable, and they denature and become inactive at temperatures above about 45 degrees Celsius PCR has an enormous variety of applications, but it also has limitations, the most important of which are (1) the necessity of having some knowledge of the sequences needed for primers and (2) the difficulty of producing amplification products longer than 10 to 15 kb. In most cases, the length limitations on PCR restrict its use to the study of selected DNA segments

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