New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Study Guide Unit 3 Exam

Star Star Star
8 reviews
by: Stefanie Villiotis

Study Guide Unit 3 Exam BSC 2010

Stefanie Villiotis
GPA 4.0

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

I have taken the class notes as well as the posted notes and created an easy to remember study guide with highlighted critical test questions. (Includes the charts we need to draw on the exam, as w...
General biology
Study Guide
50 ?




Star Star Star
2 reviews
Star Star Star Star Star
Star Star Star Star Star
Hayleigh Martin

Popular in General biology

Popular in Biological Sciences

This 11 page Study Guide was uploaded by Stefanie Villiotis on Wednesday November 18, 2015. The Study Guide belongs to BSC 2010 at Florida State University taught by Dennis in Fall 2015. Since its upload, it has received 249 views. For similar materials see General biology in Biological Sciences at Florida State University.


Reviews for Study Guide Unit 3 Exam

Star Star Star Star Star


Star Star Star Star Star


-Hayleigh Martin

Star Star Star Star Star


Star Star Star Star Star

-Sydni Sands

Star Star Star Star

Yes very helpful


Star Star Star

-Brittany Kotowske


Not working.



it doesnt work



Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 11/18/15
Unit 3 Exam Study Guide - 6 Seminal lines of inquiry: i. Griffith- Demonstrations of bacterial transformation (“r” bacteria became “s” permanently, and called this phenomenon transformation, now defined as a change in genotype and phenotype due to assimilation of foreign or external DNA) ii. Avery McCarty and Macleod- Transforming factor is separable (announced transforming factor -> acidic; extract from nucleus= DNA; “only DNA works transforming harmless bacteria into pathogenic bacteria” & reductionist biology separates carbohydrates, fats, proteins) iii. Hershey and Chase- Transforming factor is DNA (used T2 phage “ S (proteins) or 3P (DNA) to show that transforming factor = DNA) iv. Chargaff- Ratios of DNA within and between species (DNA= Nitrogenous base, sugar, phosphate group; 30.3% of human DNA = A and 26% of bacteria DNA = A A=T and G=C) v. Watson and Crick- DNA structure A=T G=C The relationship between structure and function is manifest in the double helix Semi-conservative model of replication (each daughter strand has 1 old strand) vi. Wilkins and Franklin- DNA Structure vii. Meselson and Stahl- Semi-conservative replication DNA can be Light/Light, Heavy/Light, or Heavy/Heavy DNA Replication i. Enzymes: a. Helicase- untwists the double helix (unzip your genes) b. Topoisomerase- relieves the strain of twisting caused by unwinding (breaks swivels and regions bonds) c. Primase- Makes a primer (a strating point) for synthesis of a DNA strand. (5’ end of leading strand) d. DNA Pol III- polymerizes a new strand of DNA by adding nucleotides i. DNA Pol I- removes RNA and replaces with nucleotides e. Ligase- ligates or joins together nucleic acids ii. Replication begins at sites called origins of replication, where 2 DNA strands are separated, opening up a replication “bubble” iii. Replication proceeds in both directions from each origin, until entire molecule is copied. iv. At the end of each replication bubble is a replication fork, a Y-shaped region where new DNA strands are elongating. a. Helicases are enzymes that untwist double helix at replication forks b. Single-strand binding protein binds to and stabilizes single- stranded DNA until it can be used as a template c. Topoisomerase corrects “overwinding” ahead of replication forks by breaking, swiveling, and rejoining DNA strands. d. DNA polymerases cannot initiate synthesis of a polynucleotide; they can only add nucleotides to the 3’ end. The initial nucleotide strand is a short RNA primer, made by primase. Antiparallel Elongation a. Enzymes called DNA polymerases catalyze the elongation of new DNA at a replication fork b. Most DNA polymerases require a primer and a DNA template strand c. The rate of elongation is about 500 nucleotides per second in bacteria and 50 per second in human cells d. Each nucleotide that is added to a growing DNA strand is a nucleoside triphosphate e. dATP, for example, supplies adenine to DNA and is similar to the ATP of energy metabolism f. The difference is in their sugars: dATP has deoxyribose while ATP has ribose g. As each monomer of dNTP joins the DNA strand, it loses two phosphate groups as a molecule of pyrophosphate h. The antiparallel structure of the double helix (two strands oriented in opposite directions) affects replication i. DNA polymerases add nucleotides only to the free 3' end of a growing strand; therefore, a new DNA strand can elongate only in the 5' to 3' direction j. Along one template strand of DNA, the DNA polymerase synthesizes a leading strand continuously, moving toward the replication fork Proofreading and repairing DNA i. DNA polymerases proofread newly made DNA, replacing any incorrect nucleotides ii. In mismatch repair of DNA, repair enzymes correct errors in base pairing iii. DNA can be damaged by chemicals, radioactive emissions, X-rays, UV light, and certain molecules (ex: Cigarette smoke) iv. Nucleotide excision repair, nuclease cuts out and replaces damaged stretches of DNA Replicating ends of DNA molecules a. Limitations create problems for linear DNA b. Usual replication machinery provides no completion of 5¢ ends, so repeated rounds of replication produces shorter DNA molecules c. Eukaryotic chromosomal DNA molecules have end nucleotide sequences called TELOMERES d. TELOMERES do not prevent shortening of DNA but postpone erosion of genes near ends of DNA e. Shortening of telomeres is connected to aging f. If chromosomes of germs became shorter in every cell cycle, genes would be missing from gametes g. Telomerase catalyzes lengthening of telomeres in germ cells 2 Flow of Genetic information - Information content of DNA is in the form of specific sequences of nucleotides - DNA inherited by an organism leads to specific traits by dictating PROTEIN SYNTHESIS - Proteins are the links between genotype and phenotype - Gene Expression = DNA directs protein synthesis, includes 2 stages: transcription and translation i. Archibald Garrod = genes dictate phenotypes through enzymes that catalyze specific chemical reactions; symptoms of inherited disease reflect inability to synthesize certain enzyme ii. Beadle and Tatum = exposed bread mold to X-rays creating mutants unable to survive on medium as a result of inability to synthesize certain molecules; used crosses to identify 3 classes of arginine- deficient mutants; one-gene-one-enzyme hypothesis = each gene dictates production of a specific enzyme (now called one-gene-one- polypeptide hypothesis) Transcription and Translation a. RNA is the intermediate between genes and the proteins for which they code b. Transcription is the synthesis of RNA under the direction of DNA c. Transcription produces messenger RNA (mRNA) d. Translation is the synthesis of a polypeptide, which occurs under the direction of mRNA e. Ribosomes are the sites of translation f. In prokaryotes, mRNA produced by transcription is immediately translated without more processing g. In a eukaryotic cell, the nuclear envelope separates transcription from translation h. Eukaryotic RNA transcripts are modified through RNA processing to yield finished mRNA i. A primary transcript is the initial RNA transcript from any gene j. The central dogma is the concept that cells are governed by a cellular chain of command: DNA  RNA  protein Genetic Code i. 20 amino acids; 4 nucleotide bases ii. Codons: Triplets of bases a. The flow of information from gene to protein is based on a triplet code: a series of non-overlapping, three-nucleotide words b. These triplets are the smallest units of uniform length that can code for all the amino acids c. Example: AGT at a particular position on a DNA strand results in the placement of the amino acid serine at the corresponding position of the polypeptide to be produced d. During transcription, one of the two DNA strands called the template strand provides a template for ordering the sequence of nucleotides in an RNA transcript 3 e. During translation, the mRNA base triplets, called codons, are read in the 5’ to 3’ direction f. Each codon specifies the amino acid to be placed at the corresponding position along a polypeptide g. Codons along an mRNA molecule are read by translation machinery in the 5’ to 3’ direction h. Each codon specifies the addition of one of 20 amino acids i. Of the 64 triplets, 61 code for amino acids, 3 triplets are “stop” signals to end translation j. NO CODON SPECIFIES MORE THAN ONE AMINO ACID k. Genes can be transcribed and translated after being transplanted from one species to another ii. Transcription is the DNA-directed synthesis of RNA a. RNA synthesis is catalyzed by RNA polymerase, which breaks DNA strands apart and hooks together RNA nucleotides b. The DNA sequence where RNA polymerase attaches is called the promoter; in bacteria, the sequence signaling the end of transcription is called the terminator iii. The three stages of transcription: a. Initiation  Promoters signal the initiation of RNA synthesis  Transcription factors mediate the binding of RNA polymerase and the initiation of transcription  The completed assembly of transcription factors and RNA polymerase II bound to a promoter is called a transcription initiation complex  A promoter called a TATA box is crucial in forming the initiation complex in eukaryotes b. Elongation  As RNA polymerase moves along the DNA, it untwists the double helix, 10 to 20 bases at a time  Transcription progresses at a rate of 40 nucleotides per second in eukaryotes  A gene can be transcribed simultaneously by several RNA polymerases c. Termination  The mechanisms of termination are different in bacteria and eukaryotes  In bacteria, the polymerase stops transcription at the end of the terminator  In eukaryotes, the polymerase continues transcription after the pre-mRNA is cleaved from the growing RNA chain; the polymerase eventually falls off the DNA i. Eukaryotic cells modify RNA after transcription a. Enzymes in the eukaryotic nucleus modify pre-mRNA before the genetic messages are dispatched to the cytoplasm b. During RNA processing, both ends of primary transcript are altered 4 c. The 5’ end receives modified nucleotide 5’ cap (modified guanine nucleotide) d. The 3’ end gets a Poly-A tail (50-250 adenine nucleotides) i. This facilitates export of mRNA ii. Protects mRNA from hydrolytic enzymes iii. Help ribosomes attach to 5¢ end e. Introns and Exons splicing: i. Noncoding regions = Introns (intervening sequences) ii. All other regions = Exons (eventually expressed, translated into amino acid sequences) iii. Sometimes RNA splicing is carried out by SPLICEOSOMES iv. Spliceosomes = consist of proteins and ribonucleoproteins (snRNPs) that recognize splice sites v. Proteins often have modular architecture consisting of discrete regions called domains vi. Exon shuffling can result in new proteins vii. Ribozymes a. Catalytic RNA that functions as enzyme and splice RNA b. 3 properties of RNA enable it to function as enzyme: 1. Forms three-dimensional structure because of ability to base pair with itself 2. Some bases in RNA have functional groups 3. RNA may hydrogen-bond with other nucleic acid molecules ii. Synthesis of a polypeptide a. A cell translates an mRNA message into protein with the help of transfer RNA (tRNA) b. Molecules of tRNA are not identical: i. Each carries a specific amino acid on one end ii. Each has an anticodon on the other end; the anticodon base- pairs with a complementary codon on mRNA c. A tRNA molecule consists of a single RNA strand that is only about 80 nucleotides long d. Flattened into one plane to reveal its base pairing, a tRNA molecule looks like a cloverleaf e. Because of hydrogen bonds, tRNA actually twists and folds into a three-dimensional molecule f. tRNA is roughly L-shaped g. Accurate translation requires two steps: i. First: a correct match between a tRNA and an amino acid, done by the enzyme aminoacyl-tRNA synthetase ii. Second: a correct match between the tRNA anticodon and an mRNA codon h. Flexible pairing at the third base of a codon is called wobble and allows some tRNAs to bind to more than one codon Building a Polypeptide a. The three stages of translation: i. Initiation ii. Elongation iii. Termination b. All three stages require protein “factors” that aid in the translation process 5 c. Initiation iv. The initiation stage of translation brings together mRNA, a tRNA with the first amino acid, and the two ribosomal subunits v. First, a small ribosomal subunit binds with mRNA and a special initiator tRNA vi. Then the small subunit moves along the mRNA until it reaches the start codon (AUG) vii. Proteins called initiation factors bring in the large subunit that completes the translation initiation complex d. Elongation viii. During the elongation stage, amino acids are added one by one to the preceding amino acid ix. Each addition involves proteins called elongation factors and occurs in three steps: codon recognition, peptide bond formation, and translocation e. Termination x. Termination occurs when a stop codon in the mRNA reaches the A site of the ribosome xi. The A site accepts a protein called a release factor xii. The release factor causes the addition of a water molecule instead of an amino acid xiii. This reaction releases the polypeptide, and the translation assembly then comes apart A ribosome has three binding sites for tRNA: i. The P site holds the tRNA that carries the growing polypeptide chain ii. The A site holds the tRNA that carries the next amino acid to be added to the chain iii. The E site is the exit site, where discharged tRNAs leave the ribosome Completing and Targeting Proteins i. Translation is not sufficient to make a functional protein ii. Polypeptide chains are modified after translation iii. Protein Folding and Post-Translational modifications: a. During and after synthesis, polypeptide chain spontaneously coils and folds into three-dimensional shape iv. Two populations of ribosomes are evident in cells: Free ribosomes (cytosol) = synthesize proteins; and bound ribosomes (attached to ER)= make proteins of endomembrane system and secreted from cell v. POLYPEPTIDE SYNTHESIS ALWAYS BEGINS IN CYTOSOL *** Mutations i. Changes in genetic material of a cell or virus ii. POINT MUTATIONS= chemical changes in one base pair of gene iii. SILENT MUTATIONS= no affect on amino acid produced by a codon because of redundancy of genetic code iv. MISSENSE MUTATIONS= still code for an amino acid but not the right one v. NONSENSE MUTATIONS= change amino acid codon into stop codon, leading to nonfunctional protein vi. FRAMESHIFT MUTATIONS= insertion or deletion of nucleotides alter the reading frame; additions or losses of nucleotide pairs in a gene vii. Mutagens = physical or chemical agents that can cause mutations Gene Expression 6 i. Prokaryotes and Eukaryotes alter gene expression in response to changing environment ii. Gene expression regulates development and is responsible for differences in cell types iii. Bacteria respond to environmental change by regulating transcription iv. A cell can regulate the production of enzymes by feedback inhibition or by gene regulation v. Gene expression in bacteria is controlled by the operon model: 1. Operons a. A cluster of functionally related genes can be under coordinated control by a single on-off “switch” b. The regulatory “switch” is a segment of DNA called an operator usually positioned within the promoter c. An operon is the entire stretch of DNA that includes the operator, the promoter, and the genes that they control d. Repressors **** i. The operon can be switched off by a protein repressor ii. The repressor prevents gene transcription by binding to the operator and blocking RNA polymerase iii. The repressor is the product of a separate regulatory gene iv. The repressor can be in an active or inactive form, depending on the presence of other molecules v. A corepressor is a molecule that cooperates with a repressor protein to switch an operon off vi. For example, E. coli can synthesize the amino acid tryptophan Negative Gene Regulation i. Repressible and Inducible Operons – Two Types of Gene Regulation a. Repressible Operon = ON; binding of a repressor to the operator shuts off transcription. (TRP Operon) b. Inducible Operon = OFF; inducer molecule inactivates the repressor and turns on transcription. (LAC Operon) c. TRP Operon = i. by default, is on; genes for tryptophan synthesis are transcribed. ii. Tryptophan binds to TRP repressor which turns operon OFF iii. Repressor only active in the presence of its corepressor tryptophan; thus the TRP operon is turned off (repressed) if tryptophan levels are high. d. LAC Operon = i. Inducible operon, contains genes for the code for enzymes used in the hydrolysis and metabolism of lactose 7 ii. By itself the lac repressor is active and switches the lac operon off iii. An INDUCER inactivates the repressor to turn the lac operon on e. Repressible and Inducible Operons i. INDUCIBLE enzymes function in CATABOLIC pathways; synthesis is induced by chemical signal ii. REPRESSIBLE enzymes function in ANABOLIC pathways; synthesis is repressed by high levels of the end product. iii. Regulation of TRP and LAC involves negative control of genes because operons are switched off by active repressor - Some operons are also subject to positive control through a stimulatory protein, such as CATABOLITE ACTIVATOR PROTEIN (CAP), an activator of transcription - When glucose is scarce, CAP is activated by binding with cyclic AMP - GLUCOSE IS A PREFERRED FOOD SOURCE OF E.COLI - Activated CAP attaches to the promoter of the lac operon and increases the affinity of RNA polymerase, thus accelerating transcription - When glucose levels increase, CAP detaches from the lac operon, and transcription returns to a normal rate - CAP helps regulate other operons that encode enzymes used in catabolic pathways i. Eukaryotic gene expression can be regulated at ANY stage a. ALL organisms MUST regulate which genes are expressed 24/7 b. In multicellular organisms, gene expression is essential for cell specialization c. Differences between cell types result from differential gene expression, the expression of different genes by cells with the same genome ii. Regulation of Chromatin Structure a. HIGH HETEROCHROMATIN GENES = NOT EXPRESSED b. Chemical modifications to HISTONES and DNA of chromatin influence structure and gene expression c. HISTONE ACETYLATION= acetyl groups attached to + charged lysines in histone tails - Loosens chromatin structure – promotes initiation of transcription - METHYLATION (adding methyl groups) condenses chromatin - PHOSPHORYLATION (adding phosphate groups) NEXT TO METHYLATED AMINO ACIDS can loosen chromatin - HISTONE CODE HYPOTHESIS = specific combinations of modifications help determine chromatin configuration and influence transcription d. DNA METHYLATION - Addition of methyl groups to certain bases of DNA - Methylation can reduce transcription (some species) - Can cause long-term inactivation of genes in cellular differentiation - GENOMIC IMPRINTING -- methylation regulates expression of maternal and paternal alleles of certain genes at start of development EPIGENIC INHERITANCE - Inheritance of traits not involving the nucleotide sequence 8 e. Regulation of Transcription Initiation - Chromatin-modifying enzymes—initial control of gene expression— make region of DNA more or less able to bind the transcription machinery - CONTROL ELEMENTS = segments of noncoding DNA that help regulate transcription by binding certain proteins ^^ critical to precise regulation of gene expression - to INITIATE TRANSCRIPTION, eukaryotic RNA polymerase requires assistance of proteins called TRANSCRIPTION FACTORS - Transcription factors essential for transcription of all protein-coding genes - EUKARYOTES –- high levels of transcription of particular genes depend on control elements interacting with specific transcription factors -Enhancers and Specific Transcription Factors: ********* 1. Proximal control elements located close to promoter 2. Distal control elements (enhancers) are far away from gene; sometimes located in an intron 3. ACTIVATOR = protein that binds to enhancer and stimulates transcription 4. BOUND ACTIVATORS cause mediator proteins to interact with proteins at promoter 5. some transcription factors function as repressors, inhibiting expression of gene 6. some activators and repressors act indirectly by influencing chromatin structure to promote or silence transcription f. Post- Transcriptional Gene Regulation - RNA Processing: a. Alternative RNA splicing—different RNA produced from same primary transcript, depending on which RNA segments are exons or introns - mRNA Degradation a. life span of mRNA in cytoplasm is key to determining protein synthesis b. Eukaryotic mRNA lives longer than Prokaryotic mRNA c. mRNA lifespan determined by sequences in leader/trailer regions - INITIATION of TRANSLATION a. Initiation of translation of mRNAs can be blocked by REGULATORY PROTEINS that bind to sequences or structures of mRNA b. Translation of all mRNAs regulated simultaneously EX: Translation Initiation Factors simultaneously activated in an egg after fertilization - Protein Processing and Degradation a. AFTER TRANSLATION protein processing (cleavage & adding chemical groups) are subject to control b. PROTEASOMES = giant protein complexes that bind molecules and degrade them g. Non-coding RNA role in gene expression - Significant amount of genome can be transcribed into noncoding RNAs 9 - Noncoding RNAs regulate gene expression at 2 points: mRNA TRANSLATION and CHROMATIN CONFIGURATION - miRNA = MICRO RNA = small single-stranded RNA that can bind to mRNA – they can degrade mRNA or block translation - siRNA = SMALL INTERFERING RNA a. Inhibition of gene expression by RNA molecules is called RNA interference (RNAi) b. RNAi caused by siRNA c. siRNA play role in HETEROCHROMATIN FORMATION and block large regions of chromosome d. siRNA and miRNA similar but form different RNA precursors Charts and Diagrams Schematic Model Showing Binding Sites 10 ** Know how to draw this: 3 Amino Acid binding site 5 Hydrogen bonds ’ between paired Anticodon 11


Buy Material

Are you sure you want to buy this material for

50 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Amaris Trozzo George Washington University

"I made $350 in just two days after posting my first study guide."

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.