Biology Midterm 3, Biology 1207 Midterm 4 Review

2. Be familiar with Griffith's transformation experiment as in what did he do, observe and conclude? Which strain was transformed into what? What effect did this have on the mouse?

*1928 - Frederick Griffith was working with Streptococcus pneumoniae bacteria *Two strains of S. pneumoniae: **Strains that secrete polysaccharide capsules look smooth (S) and infections are fatal in mice (cause pneumonia). **Strains that do not secrete capsules look rough (R) and infections are not fatal in mice *The capsule shields the bacteria from the immune system, so they survive in the blood *Smooth strains (S) with capsule are fatal; rough strains (R) without capsule are not *If mice are injected with heat-killed type S, they survive (because bacteria are dead) *However, mixing live R with heat-killed S kills the mouse **Blood is found to contain living type S bacteria! **Known as transformation *nTransformation = a change in genotype and phenotype due to the assimilation of a foreign substance (now known to be DNA) by a cell.

11. Who proposed the correct structure of DNA in 1953? Whose and what evidence suggested DNA was a helical structure with a uniform diameter? 1

*1953, James Watson and Francis Crick, proposed the structure of the DNA double helix *Watson and Crick used Linus Pauling'smethod of working out protein structures using simple ball-and-stick models *Rosalind Franklin'sX-ray diffraction results were crucial evidence, suggesting a helical structurewith uniform diameter

7. What are the three stages of transcription? In bacteria, what causes RNA polymerase to recognize the promoter region? What is an open complex and where does it form? What protein influences the ability of RNA polymerase to transcribe genes?

*3 stages: initiation-> elongation-> termination *In bacteria, sigma factor causes RNA polymerase to recognize promoter region *open complex: A separation between the two DNA strands that occurs near the promoter during transcription; also called a transcription bubble. *Initiation is complete when an open complex of 10-15 base pairs is made near the promoter. *Sigma factor is released and RNA polymerase slides along the DNA in a way that maintains an open complex as it goes *Transcription factors are proteins that influence the ability of RNA polymerase to transcribe genes.

33. What recognizes any one of the three stop codons in the mRNA at the end of translation? What do these proteins do? What happens as a result?

*3 stop codons - UAA, UAG, UGA **Recognized by release factors 1.Release factor binds to stop codon at the A site 2.Bond between polypeptide and tRNA in the P site is hydrolyzed to release polypeptide 3.Ribosomal subunits and release factors disassociate *The three dimensional structure of a release factor mimics the structure of tRNAs, which allows it to fit in the A site. Bacteria have two different termination factors that recognize stop codons (RF1 and RF2), whereas eukaryotes have only one (eRF).

11. What are the three components in the CRISPR-Cas type II system and what does each do?

*Adaption: The process of adaptation (also called spacer acquisition) occurs after a bacterial cell has been exposed to a bacteriophage. The proteins encoded by the Cas1 and Cas2 genes form a complex that recognizes the bacteriophage DNA as being foreign and cleaves it into small pieces. A piece of bacteriophage DNA, usually between 20 and 50 bp in length, is inserted into the Crispr gene. The mechanism of insertion is not entirely understood. The newly inserted piece of bacteriophage DNA is called a spacer because it acts as a space between adjacent repeats. The different spacers found in the Crispr gene of modern bacterial species are derived from past bacteriophage infections. Each spacer provides a bacterium with defense against a particular bacteriophage. Once a bacterial cell has become adapted to a particular bacteriophage, it will pass this trait on to its daughter cells. By cleaving the bacteriophage into pieces, the adaptation phase can protect a bacterial cell, because it cuts up the bacteriophage DNA and thereby inactivates the phage. However, a more effective way of destroying phages is provided by the expression and interference phases of this system. *Expression: If a bacterial cell has already been adapted to a bacteriophage, a subsequent infection by that phage will result in the expression phase, in which the system gets ready for action by expressing the Crispr, tracr, and Cas9 genes (Figure 13.8c). The Crispr gene is transcribed from a single promoter and produces a long ncRNA called pre-crRNA, which contains several repeat sequences separated by spacers. The gene encoding the tracrRNA is also transcribed, which produces many molecules of tracrRNA. As mentioned earlier, tracrRNA is also an ncRNA. A region of the tracrRNA is complementary to the repeat sequences of the pre-crRNA. Several molecules of tracrRNA base-pair with the pre-crRNA. The pre-crRNA is then cleaved into many small molecules, now called crRNAs. Each crRNA is attached to a tracrRNA. A region of the tracrRNA is recognized by the Cas9 protein. The tracrRNA acts as a guide that causes the tracrRNA-crRNA complex to bind to a Cas9 protein. *Interference After the tracrRNA-crRNA-Cas9 complex has formed, the bacterial cell is ready to destroy the bacteriophage DNA. This phase is called interference because it resembles the process of RNA interference described earlier in this chapter. Each spacer within a crRNA is complementary to one of the strands of a bacteriophage DNA. Therefore, the crRNA acts as a guide that causes the tracrRNA-crRNA-Cas9 complex to bind to that bacteriophage DNA . After binding, the Cas9 protein functions as an endonuclease that breaks both strands in the bacteriophage DNA. This cleavage inactivates the phage and thereby prevents phage proliferation. Since discovering the CRISPR-Cas system in bacteria and archaea, researchers have been able to modify certain components of this system and use them to mutate genes in living cells

19. What is a secondary messenger in the cell? How is it produced? What are some other secondary messengers in the cell? What shuts off the cAMP signal?

*After first messengers bind to receptors such as GPCRs, many signal transduction pathways lead to the production of second messengers—small molecules or ions that relay signals inside the cell. The signals that result in second messenger production often act quickly, in a matter of seconds or minutes, but their duration is usually short. Therefore, such signaling typically occurs when a cell needs a quick and short cellular response. *Mammalian and plant cells make several different types of G protein α subunits. One type of α subunit binds to adenylyl cyclase, an enzyme in the plasma membrane. This interaction stimulates adenylyl cyclase to synthesize cyclic adenosine monophosphate (cyclic AMP, or cAMP) from ATP. cAMP is an example of a second messenger. *Examples: cyclic AMP, cyclic GMP, inositol trisphosphate, diacylglycerol, and calcium *Signaling that involves second messengers is typically of short duration. When the signaling molecule is no longer produced and its concentration falls, a larger percentage of the receptors are not bound by their ligands. When a ligand dissociates from a GPCR, the GPCR becomes deactivated. Intracellularly, the α subunit hydrolyzes its GTP to GDP, and the α subunit and β/γ dimer reassociate to form an inactive G protein. The amount of cAMP decreases due to the action of an enzyme called phosphodiesterase, which converts cAMP to AMP. As the cAMP level falls, the regulatory subunits of PKA release cAMP, and the regulatory and catalytic subunits reassociate, thereby inhibiting PKA. Finally, enzymes called protein phosphatases are responsible for removing phosphate groups from proteins, which reverses the effects of PKA

22. What catalyzes the addition of an amino acid to the 3' end of the tRNA? What is another name for a "charged tRNA" and what does this mean? Does the cell have only one aminoacyl-tRNA synthetase? How many distinct aminoacyl-tRNA synthetases does each cell make?

*Aminoacyl-tRNA synthetase: **Catalyzes attachment of amino acids to 3' end of tRNA; One for each of 20 different amino acids **Reactions result in tRNA with amino acid attached (charged tRNA or aminoacyl tRNA) **Ability of aminoacyl-tRNA synthetase to recognize appropriate tRNA has been called the "second genetic code" *Cell has 20 different aminoacyl-tRNA synthetases. Example: alanyl-aminoacyl-tRNA synthetase recognizes alanine and attaches this amino acid to all tRNAs with alanine anticodons.

1. What did Archibald Garrod observe in his patients and what relationship did he propose between heredity and human characteristics? How did Beadle and Tatum use common bread mold to develop the hypothesis: one gene, one enzyme?

*Archibald Garrod first proposed in 1908 a relationship between genes and the production of enzymes *Studied patients with metabolic defects, or "inborn errors of metabolism" *Alkaptonuria **Patient's body accumulates abnormal levels of homogentisic acid (alkapton) **Recessive pattern of inheritance *Hypothesis: "Disease is due to missing enzyme" nThe nature of the genetic material was completely unknown at the time *Early 1940s - Beadle and Tatum became aware of Garrod's work *They worked on Neurospora crassa, common bread mold *Minimum requirements for growth **Carbon source (sugar), inorganic salts, and biotin **Neurospora can synthesize everything else it needsfrom those precursors (amino acids and vitamins) *8Hypothesized that a mutation might cause a change in an enzyme required for the synthesis of an essential molecule, such as a vitamin or amino acid. *"One gene, one enzyme" **Multiple mutants required arginine to grow **Could they grow if supplemented with precursors instead? **Fell into three groups based on requirements **Conclusion - Supports one gene, one enzyme hypothesis

12. With regard to RNA processing, what is one big difference between where RNA is processed in bacteria versus eukaryotes? What are the events that are included in eukaryotic RNA processing (3 events)?

*Bacteria has only one form of RNA polymerase that transcribes all genes, but bacteria have more than one type of sigma factor that can recognize different promoters. *Bacterial mRNAs can be translated immediately (lack introns). *Eukaryotic mRNAs are made in a longer pre-mRNA form that requires processing into mature mRNA **Introns - transcribed but not translated, between exons **Exons - coding sequence found in mature mRNA *RNA Splicing - removal of introns *Other modifications - addition of tails and caps *Three forms of RNA polymerase **RNA polymerase II - transcribes mRNA **RNA polymerase I and III - transcribes nonstructural genes for rRNA (I) and tRNA (III). **RNA polymerase II requires 5 general transcription factors to initiate transcription

16. In bacteria, how does the ribosome know where to associate with the mRNA? What is the start codon usually? What are the three stop codons?

*Bacterial mRNA has 5' ribosomal-binding site *Start codon usually AUG = Methionine *Typical polypeptide is a few hundred amino acids in length *Stop codons (aka Termination codons) **UAA, UAG or UGA

1. What are general features of cell communication?

*Cells of all living organisms both respond to incoming signals and produce outgoing signals. *Process of cells detecting and responding to signals in the extracellular environment *Coordinates activities in a multicellular organism *Conformational changes of a receptor lead to a response inside the cell *Signals cause a cell to die - apoptosis

3. What is the Central Dogma of Molecular Biology and who proposed it? What is an "exception" to the Central Dogma observed by RNA viruses (retroviruses)?

*Central dogma of molecular biology *Francis Crick - 1958 *DNA-> RNA-> Protein *Transcription **Produces a transcript (RNA copy) of a gene **This messenger RNA (mRNA) specifies the amino acid sequence of a polypeptide (structural gene) *Translation **Process of synthesizing specific polypeptide on a ribosome using the mRNA template **Information "translated" from nucleotide sequence into amino acid sequence. *Proposed by Francis Crick 1958. In addition to transcription and translation, in eukaryotes RNA processing occurs between transcription and translation, so the product of transcription is called a pre-mRNA before it is processed and leaves the nucleus. Prokaryotes - both processes occur in the cytoplasm. Eukaryotes - transcription and RNA processing occurs in nucleus, while translation occurs in the cytosol.

20. What is the function of DNA helicase, DNA topoisomerase, single-stranded binding proteins, DNA polymerase, DNA ligase.

*DNA helicase **Uses ATP to separate strands at the open ori (binds to one strand, moves 5' to 3') *DNA topoisomerase **Relieves kinking ahead of replication fork *Single-strand binding proteins **Keeps strands open as templates *DNA polymerase **Covalently links nucleotides **Deoxynucleoside triphosphates *DNA ligase **covalently attaches adjacent Okazaki fragments in the lagging strand

19. Where does DNA replication begin? How many origins of replication are found in bacteria? In eukaryotes? How many replication forks are there at an origin of replication?

*DNA replication starts at origins of replication, which are sequences in the DNA that are recognized by replication proteins *In bacterial cells there is a single origin of replication. In eukaryotic cells which have much greater amount of DNA to replicate, there are hundreds to thousands of origins of replication. *2 replication forks

10. What are the three steps of the CRISPR-Cas system of defense in bacteria? Be sure to understand what occurs in each of these steps.

*Defense mechanism occurs in three phases: •Adaptation - occurs after bacterial cell has been exposed to a bacteriophage •Expression - subsequent bacteriophage infection results in the expression of a CRISPR RNA. Interference - bacteriophage DNA is destroyed

10. Does the cell always use the same DNA strand as the template for transcription? In all cases, however, what is the direction of RNA synthesis and in what direction is the DNA template read? Make sure you can generate the sequence and polarity of the sequence of mRNA if given a DNA template strand of a particular sequence and polarity.

*Direction of transcription and which DNA strand used varies among genes *In all cases, synthesis of RNA transcript is 5' to 3' and DNA template strand reads 3' to 5

14. What are the 8 structural features of DNA?

*Double stranded *Antiparallel strands *Right-handed helix *Sugar-phosphatebackbone *Bases on the inside *Stabilized by H-bonding *Specific base-pairing *~10 bp per helical turn

15. What happens to the EGF receptors when they bind one molecule of EGF? What chemical modification take place on these receptors? What is this called?

*EGF=epidermal growth factor stimulates cell growth, proliferation and differentiation *Binding of EGF causes subunits to dimerize and phosphorylate each other on tyrosines within the receptors=Autophosphorlation *Mutation in genes that encode proteins in this pathway->cell to be hyperactive, divide uncrontrollably= cancer, psoriasis, or eczema *EGF= "mitogen"= agent that causes cell to divide

12. Be sure that you are able to determine the % of all three nucleotides in a double stranded DNA molecule if given the % of one base. Who made the first observation that %A=%T and %C=%G?

*Erwin Chargoff analyzed base composition of DNA from many different species *Results consistently showed amount of adenine (A) = amount of thymine (T) amount of cytosine (C) = amount of guanine (G) *Chargoff's Rules q%T = %A and %G = %C

1. What are the four criteria necessary for genetic material?

*Four criteria necessary for genetic material: 1.Information to construct organism 2.Replication accurately to preserve information 3.Transmission from parent to offspring 4.Variation to explain differences within and between species

9. What do all G-protein coupled receptors have as a repeated structural domain? What are the five steps that take place when a G-protein coupled receptor is activated (beginning with the ligand binding to the receptor as the first step)? Following its initial activation, how does a G protein become inactivated?

*G-proteins = able to bind guanosine tripphosphate (GTP) and guanosine diphosphate (GDP). *GPCRs typically contain seven transmembrane segments that wind back and forth through the plasma membrane. *% steps: **1.Ligand binds receptor **2.Activated receptor binds to G protein **3.Releases GDP and binds GTP instead **4.GTP causes G protein to dissociate **5.α subunit and β/γ dimer interact with other proteins in a signaling pathway *For the G-protein to return to its inactive state, the alpha subunit that bound to GTP must hydrolyze GTP to GDP and Pi. After this occurs, the alpha and beta/gamma subunits reassociate to form an inactive G protein complex.

5. What is a gene? What do structural genes encode? What are three examples of non-structural genes?

*Gene: **An organized unit of DNA sequences that enablesa segment of DNA to be transcribed into RNA and ultimately results in the formation of a final functional product (which is either RNA or a polypeptide). **Structural Gene - encodes a polypeptide. **Other genes code for RNA itself as its final product (from non-structural genes) ***Transfer RNA (tRNA) - translates mRNA into amino acids ***Ribosomal RNA (rRNA) - part of ribosomes ***micro RNAs (miRNAs) - regulate translantion

15. In translation, what is the genetic code mean, the codons, and what does it mean that the code is degenerate? What are three possible types of information a codon can represent?

*Genetic code - sequence of bases in an mRNA molecule (nearly universal) *Read in groups of three nucleotide bases or codons (64 codons) 43 or 64 possible codons *Most codons specify a particular amino acid **Also Start and Stop codons *Degenerate code - more than one codon can specify the same amino acid

3. How was these observations explained?

*Genetic material had been transferred from the heat-killed type S bacteria to the living type R bacteria *This gave the type R bacteria the capsule-secreting trait and was passed on to their offspring *What was the biochemical basis of this transforming principle? At the time there was no way to know.

16. What grooves are formed in the double helix of DNA and what function is associated with the larger (major) groove and why?

*Grooves are revealed in the space-filling model *Major groove **Proteins bind to affect gene expression **Sequence can be "read" in major groove. nMinor groove **Narrower *Transcription Factors bind in the major groove

28. In prokaryotes, how does the small ribosomal subunit know to bing to the 5'end of the mRNA? What is special about this sequence so the ribosome can recognize it? Where is the start codon relative to this place? What tRNA recognizes the start codon? Then what associates with the small ribosome complex? Once the complete ribosome is made, what occupies the P site at the end of initiation?

*Initiation in bacteria **mRNA binds to small ribosomal subunit facilitated by ribosomal-binding sequence (RSB) at 5' end of mRNA **Ribosomal-binding sequence is complementary to sequence in 16s RNA in small ribosomal subunit. **Start codon is a few nucleotides downstream of RBS **Initiator tRNA recognizes start codon (AUG) in mRNA **Large ribosomal subunit associates **At the end, the initiator tRNA is in the P site

13. What is a spliceosome and what is it composed of? What are snRNPs? What is alternative splicing and what effect does it have on the proteome? Does it have an effect on the genome? What is a ribozyme and to what two RNA species does it apply?

*Introns found in many eukaryotic genes **Most structural genes have one or more introns *Spliceosome - removes introns precisely **Composed of snRNPs (small nuclear RNA + proteins) *Alternative splicing - splicing can occur more than one way to produce different products **Increases size of proteome while minimizing the size of the genome. *rRNA and tRNA are self-splicing **They are ribozymes - RNAs that can catalyze reactions, removal of introns

23. What is the difference between the leading and lagging strands of DNA during DNA replication? How are the RNA primers removed? How are the adjacent DNA fragments joined?

*Leading strand **Synthesized in one long molecule **Primase makes single RNA primer **DNA pol adds nucleotides in a 5' to 3' direction as it moves forward *Lagging strand **Synthesized 5' to 3' but as Okazaki fragments **Okazaki fragments consist of RNA primers plus DNA **Subsequently connected to each other to form a continuous strand. *In both strands **RNA primers are removed by DNA polymerase and replaced with DNA **DNA ligase joins adjacent DNA fragments DNA polymerase III synthesizes DNA of both the leading and lagging strand. To complete the synthesis of the Okazaki fragments within the lagging strand, three additional events occur: the removal of the RNA primers (DNA pol I - digests linkages in a 5' to 3' direction), the synthesis of DNA in the area where the primers have been removed (by DNA pol I), and the covalent joining of adjacent fragments of DNA (DNA ligase).

21. What effect does epinephrine have on liver and muscle cells? What pathway does it use to bring about this effect? What are the enzymes that affect glycogen synthesis and breakdown? How are they regulated in this pathway by epinephrine signaling?

*Let's explore a signal transduction pathway in which the GPCR recognizes the hormone epinephrine (also called adrenaline). This hormone is sometimes called the fight-or-flight hormone. Epinephrine is produced when an individual is confronted with a stressful situation and helps the individual deal with a perceived threat or danger. First, epinephrine binds to its receptor and activates a G protein . The α subunit then activates adenylyl cyclase, which catalyzes the production of cAMP from ATP. One effect of cAMP is to activate protein kinase A (PKA), which is composed of four subunits: two catalytic subunits that phosphorylate specific cellular proteins, and two regulatory subunits that inhibit the catalytic subunits when they are bound to each other. cAMP binds to the regulatory subunits of PKA. The binding of cAMP separates the regulatory and catalytic subunits, which allows each catalytic subunit to be active. *When PKA becomes active, it phosphorylates two enzymes—phosphorylase kinase and glycogen synthase. Both of these enzymes are involved with the metabolism of glycogen, which is a polymer of glucose used to store energy. **When phosphorylase kinase is phosphorylated, it becomes activated. The function of phosphorylase kinase is to phosphorylate another enzyme in the cell called glycogen phosphorylase, which then becomes activated. This enzyme causes glycogen breakdown by phosphorylating glucose units at the ends of a glycogen polymer, which releases individual glucose-phosphate molecules from glycogen: Glycogenn+Pi Glycogenphosphorylase−−−−−−−−−−→Glycogenn−1+Glucose-phosphateGlycogenn+Pi→ Glycogenphosphorylase Glycogenn−1+Glucose-phosphate where n is the number of glucose units in the glycogen polymer. **When PKA phosphorylates glycogen synthase, the function of this enzyme is inhibited rather than activated. The function of glycogen synthase is to make glycogen. Therefore, the effect of cAMP is to prevent glycogen synthesis. *Taken together, the effects of epinephrine in skeletal muscle cells are to stimulate glycogen breakdown and inhibit glycogen synthesis. This provides these cells with more glucose molecules, which they can use for the energy needed for muscle contraction. In this way, the individual is better prepared to fight or flee.

5. What is a ligand and what does it bind to? What does this binding cause in the receptor?

*Ligand = signaling molecule (a steroid or a protein) *Binding occurs when the ligand and receptor happen to collide in the correct orientation with enough to form a ligand-receptor complex. *Binds noncovalently to receptor with high specificity *Binding and release is rapid *Ligand binding causes conformational change in receptor - transmits the signal across the membrane *Once a ligand is released, the receptor reverts andbecomesinactive again

10. In ligand-gated ion channels, what effect does ligand binding have? What effect does this have on the cell? What is one function of such a channel in animal cells?

*Ligand-gated ion channels are a third type of cell surface receptor found in the plasma membranes of animal, plant, and fungal cells. When signaling molecules (ligands) bind to this type of receptor, the ion channel opens and allows the flow of ions through the membrane, changing the concentration of the ions in the cell *In animals, ligand-gated ion channels are important in the transmission of signals between neurons and muscle cells and between two neurons. In addition, ligand-gated ion channels in the plasma membrane allow the influx of Ca2+ into the cytosol. Changes in the cytosolic concentration of Ca2+ often play a role in signal transduction.

18. What was the experiment that differentiated between the three proposed DNA replication mechanisms and who carried out this experiment? How did they distinguish the correct replication mechanism from the others? Be sure you can do the same.

*Meselson and Stahl differentiate among the three proposed DNA replication mechanisms **Nitrogen comes in a common light form (14N) and a rare heavy form (15N) **Grew E. coli in medium with 15N to label, then switched to medium with 14N, collecting samples after each generation **Original parental strands would be 15N while newly made strands would be 14N **Separated DNA by centrifugation after each replication. *Conclusion: Semiconservative DNA replication **Added the DNA sample after each round of replication on top of a solution of cesium chloride (CsCl) and then centrifuged. A CsCl gradient is made in the tube, and the DNA molecules (remain double stranded) travel to a point in the gradient that is equal to their density. Heavier DNA molecules will travel lower in the tube than lighter DNA molecules. After one round of replication, all of the DNA exhibited a half-heavy or intermediate density. The conservative model of replication was disproved after the first round of replication. After the second round of replication, both light and half-heavy DNA bands were observed. If the dispersive mechanism was correct all the DNA strands would have been ¼ heavy after two generations. Instead, as predicted by semi-conservative replication, ½ of the DNA molecules were light and ½ of the DNA molecules were half-heavy.

7. What is the difference between miRNAs and siRNAs?

*MicroRNAs (miRNAs) are ncRNAs transcribed from genes normally found in the eukaryotic genome •Play key roles in regulating gene expression, particularly during embryonic development •miRNA is partially complementary to mRNAs •Estimated that 60% of human protein-encoding genes regulated by miRNAs *Small-interfering RNAs (siRNAs) come from exogenous sources - viruses that infect a cell or from researchers •Usually a perfect match •Thought to prevent viral infections; also important molecular biology tools

6. Understand the mechanism (steps) by which miRNA and siRNA are made in the cell and used for regulating translation via RNA interference.

*MicroRNAs (miRNAs) are ncRNAs transcribed from genes normally found in the eukaryotic genome •Play key roles in regulating gene expression, particularly during embryonic development •miRNA is partially complementary to mRNAs •Estimated that 60% of human protein-encoding genes regulated by miRNAs *Small-interfering RNAs (siRNAs) come from exogenous sources - viruses that infect a cell or from researchers •Usually a perfect match •Thought to prevent viral infections; also important molecular biology tools *RNA interference found in most eukaryotic species *RNA that promotes RNAi can come from two sources •MicroRNAs (miRNAs) Small-interfering RNAs (siRNAs) *Both miRNAs and siRNAs are cut by dicer, releasing a 20 to 25bp dsRNA that associates with the RNA-induced silencing complex (RISC) *Then RISC inhibits translation or degrades the mRNA

2. What is the basis of some modifications to the "one gene, one enzyme" hypothesis today? What observation is responsible for the "one gene, one polypeptide" hypothesis of a gene to be updated to "one gene, one gene product"?

*Modification of "one gene, one enzyme" hypothesis *Enzymes are only one category of cellular proteins - genes also encode other proteins *Also, some proteins are composed of several polypeptides that work together for one function nex: Hemoglobin composed of 4 polypeptides *"One gene, one polypeptide" *Some genes encode RNA that function solely as RNA (for example, tRNA, rRNA, siRNA) = "one gene, one gene product" (RNA or protein).

9. How do ncRNAs function in the cause of neurological and cardiovascular diseases?

*Neurological disorders •Many miRNAs essential for proper development and functioning of the nervous system •Abnormal levels of miRNA expression associated with nearly all neurological disorders in which they've been investigated *Cardiovascular disease •Abnormalities in miRNA levels have been linked to several cardiovascular diseases •miR-1 associated with the development of arrhythmias •Formation of arterial plaques associated with abnormal expression of several miRNAs

10. What bond links the nucleotides together along a strand of DNA or RNA? What type of bond is this? What is the backbone of the DNA or RNA made of? In what orientation is DNA usually always written?

*Nucleotides arecovalently bonded *Phosphodiester bond - phosphate group links two sugar alcohol groups *Backbone - formed from phosphates and sugars **Negatively charged *Bases project away from backbone *Written 5' to 3' *ex:5' - TACG - 3'

25. What determines the overall shape of the ribosome? What are the three discrete sites for tRNA binding and polypeptide synthesis?

*Overall ribosome shape determined by rRNA *Discrete sites for tRNA binding and polypeptide synthesis *E site - Exit site *P site - Peptidyl site *A site - Aminoacyl site

5. In the Hershey and Chase experiment, why was T2 bacteriophage chosen in this experiment? How did they differentially label DNA versus protein? Where did most of the 32P go after the labeled virus infected the bacterium?

*Phage coat=all protein: head/capsid, sheath, tail fibers, base plate *Inside head of phage=DNA *Allow phage to infect bacteria with "genetic infomation"-> seperate using blender->centrifuge *After blending (separated the phage coats from the bacterial cells), the mixture was centrifuged so that the heavier bacterial cells would form a pellet at the bottom of the tube, while the lighter phage coats would remain in the supernatent. They measured the amount of radioactivity with a geiger counter. 80% of the 35-S was found in the supernatent (phage coats) Only 35% of the 32-P was found in the supernatent. Most of the 32-P (65%) was found in the bacterial cells. Did not use the labels in the same experiment, but just compared the labels in two different experiments.

23. Do eukaryotes have only one kind of ribosome? Where are these different ribosomes found and how do they differ from one another?

*Prokaryotes have one kind of ribosome *Eukaryotes have distinct ribosomes in different cellular compartments *Composed of large and small subunits **Each subunit is assembled of many different proteins and one or more RNA molecules *Structural differences between prokaryotes and eukaryotes exploited by antibiotics to inhibit bacterial ribosomes only

26. What are some differences between prokaryotic and eukaryotic DNA?

*Prokaryotes: **Closed circular DNA **Binary Fission **No telomeres **One origin of replication **No organelles - ***No Mitochondria ***No Chloroplasts *Eukaryotes **Linear DNA **Mitosis **Telomeres **Multiple origins of replication **Extranuclear DNA: ***Mitochondrial Genome ***Chloroplast Genome

6. What are the four parts of a gene and what are their functions?

*Promoter = where RNA polymerase binds *Terminator = where RNA polymerase falls off the template DNA. *Transcribed region usually begins with a Methionine. *Regulatory sequences are upstream of the start site of transcription and influence the binding of RNA polymerase to the promoter.This is by the binding of regulatory proteins to sequence specific sites in the regulatory region that bind transcription factors or repressors that influence the binding of RNA polymerase to the promoter

20. How does cAMP affect the ability of Protein Kinase A to phosphorylate target proteins? What are three PKA targets?

*Protein Kinase A (PKA) **Regulatory subunit binds cAMP - active catalytic subunit is released *PKA catalytic subunit phosphorylates target proteins *PKA targets include **enzymes **structural proteins **transcription factors

13. What category of receptor is a "receptor tyrosine kinase" and why? What type of signaling molecules do these receptors recognize. Be sure you remember that EGF binds to a receptor tyrosine kinase which is an enzyme-linked receptor.

*Receptor tyrosine kinases are a category of enzyme-linked receptors that are found in all animals and also in choanoflagellates, which are the protists that are most closely related to animals. However, they are not found in bacteria, archaea, or other eukaryotic species. (Bacteria do have receptor histidine kinases, and all eukaryotes have receptor serine/threonine kinases.) The human genome contains about 60 different genes that encode receptor tyrosine kinases that recognize various types of signaling molecules such as hormones. *Recognize various types of signaling molecules **Growth factor - hormone that acts to stimulate cell growth or division *3 parts of the signal transduction pathway are: **1. relay proteins (i.e. adaptor proteins) activate a protein kinase cascade; **2. the protein kinase cascade phosphorylates proteins in the cell such as transcription factors; **3 the phosphorylated transcription factors stimulate gene transcription

20. What are the components of translation? Why does the cell try to regulate whether or not a gene will be expression prior to translation?

*Requires many components **mRNA **tRNA **Ribosomes (including rRNAs) **translation factors *Most cells use a substantial amount of energy on translation

4. Name and describe six functions associated with ncRNAs.

*Scaffold - bind to multiple components such as proteins, act as scaffold for formation of a complex *Guide - guide one molecule to a specific location in the cell *Alteration of protein function or stability - binds to a protein, can affect ability of the protein to be a catalyst, ability of the protein to bind other molecules, protein stability *Ribozyme - RNA molecule with catalytic function *Blocker - prevents or blocks a cellular process from happening *Decoy - recognize other ncRNAs and sequester them, preventing them from working

27. What are telomeres and what problem do they present in DNA replication? How and by what is this problem solved?

*Short nucleotide sequences repeated at the ends of chromosomes in eukaryotes **= GGGTTA *Specialized form of DNA replication only in eukaryotes in the telomeres *Telomere at 3' does not have a complementary strand and is called a 3' overhang *DNA polymerase cannot copy the tip of the strand with a 3' end **No place for upstream primer to be made *If this replication problem were not solved, linear chromosomes would become progressively shorter *Telomerase enzyme attaches many copies of DNA repeat sequence to the ends of chromosomes Telomerase (RNA and protein). RNA part of enzyme has complementary sequence to telomere. This allows telomerase to bind to the 3' overhang region of the telomere. The RNA sequence beyond the binding site functions as a template, allowing telomerase to synthesize a 6-nucleotide sequence at the end of the DNA strand. The enzyme then moves to the new end of this DNA strand and attaches another 6 nucleotides to the end. This occurs many times, thereby greatly lengthening the 3' end of the DNA in the telomeric region. This lengthening provides an upstream site for an RNA primer to be made. DNA polymerase then synthesizes the complementary DNA strand. In this way, the progressive shortening of eukaryotic chromosomes is prevented. 90% of all types of human cancers, express a high level of telomerase. *Shortening of telomeres is correlated with cellular senescence *Telomerase function is reduced as an organism ages *90% of all types of human cancers have high levels of telomerase

22. What are two advantages of a cAMP signaling pathway?

*Signal Amplification: Amplification of the signal involves the synthesis of many cAMP molecules, which, in turn, activate many PKA proteins. Likewise, each PKA protein phosphorylates many target proteins in the cell to promote a cellular response. *Speed: A second advantage of second messengers such as cAMP is speed. Because second messengers are relatively small and water-soluble, they can diffuse rapidly through the cytosol. For example, Brian Bacskai and colleagues studied the response of neurons to a signaling molecule called serotonin, which is a neurotransmitter that binds to a GPCR. In humans, low serotonin is believed to play a role in depression, anxiety, and other behavioral disorders. To monitor cAMP levels, neurons grown in a laboratory were injected with a fluorescent protein that changes its fluorescence when cAMP is made. As shown schematically in the drawing on the right in, such cells made a substantial amount of cAMP within 20 seconds after the addition of serotonin.

18. What effect does signal binding have on GPCR? (Made sure you know that this stands for G-protein coupled/linked receptor). What does GPCR activate in this state? What does this cause to happen? What does Galpha-GTP go on to do? What does this activated enzyme do? Know that epinephrine binds to its G-protein linked receptor and when it does adenylyl cyclase is activated which goes on to produce cAMP.

*Signal binding to GPCR induces shape change *GPCR activates G protein to bind GTP *Causes dissociation of β/g dimer freeing Gα-GTP *Gα-GTP then activatesadenylyl cyclase *Adenylyl cyclaseconverts ATP tomanycAMP(plus Pi)

8. What are the two roles of SRP RNA in co-translational sorting?

*Signal recognition particle (SRP) - a protein-RNA complex (1 ncRNA and 1 to 6 proteins) that recognizes the ER signal sequence *Two roles for SRP RNA: •Provides a scaffold for the binding of SRP protein(s) •After SRP binds to the SRP receptor in the ER membrane, the SRP RNA stimulates proteins within the SRP and SRP receptor to hydrolyze GTP

5. Be sure to understand how ncRNAs function as: component of Telomerase, Transfer RNA (tRNA), Ribosomal RNA (rRNA), microRNAs and siRNAs, Signal recognition in co-translational sorting.

*Telomerase-TERC facilitates the binding of telomerase to the telomere and acts as a template for DNA replication. *tRNA-tRNA molecules recognize mRNA codons during translation and carry the appropriate amino acid. *rRNA-rRNAs are components of ribosomes, which are the site of polypeptide synthesis. Ribosomes contain an ncRNA that acts as a ribozyme by catalyzing peptide bond formation. *microRNAs and siRNAS-miRNAs and siRNAs regulate the expression and degradation of mRNAs. *SIngal recognition in co-translational sorting-In bacteria, SRP directs some polypeptides to the plasma membrane. In eukaryotes, it directs polypeptides to the endoplasmic reticulum.

18. What is one important function of the start codon with regard to reading the nucleotide sequence into a series of amino acids?

*The start codon, which is AUG, is the site where translation begins. AUG specifies the amino acid methionine. The start codon is only a few nucleotides from the ribosomal-binding site. Beyond this, a large portion of an mRNA functions as a coding sequence—a region that specifies the linear amino acid sequence of a polypeptide. A typical polypeptide is a few hundred amino acids in length. The coding sequence consists of a series of codons. *To translate a nucleotide sequence of mRNA into an amino acid sequence, recognition occurs between mRNA and transfer RNA (tRNA) molecules

15. What is the polarity of one strand to the other in a double helix of DNA? How many nucleotides are in each strand per complete turn of the helix?

*The two strands of the double helix run in anti-parallel (i.e. opposite) directions, with the 5' end of one strand adjacent to the 3' end of the other strand. The double helix has a right-handed twist, (rather than the left-handed twist that is often represented incorrectly in popular media) *There are ~10 nucleotides in each strand per complete turn of the helix.

11. In eukaryotes, what are the three forms of RNA polymerase. Which RNA polymerase synthesizes mRNA in eukaryotes?

*Three forms of RNA polymerase **RNA polymerase II - transcribes mRNA **RNA polymerase I and III - transcribes nonstructural genes for rRNA (I) and tRNA (III). **RNA polymerase II requires 5 general transcription factors to initiate transcription *RNA polymerase II synthesizes mRNA. In eukaryotes, the mRNA molecules always code for one protein, whereas in prokaryotes, many mRNAs code for several proteins

25. What are the three mechanisms that enhance the accuracy of DNA replication?

*Three mechanisms for accuracy 1.Hydrogen bonding between A and T, and between G and C is more stable than mismatched combinations 2.Active site of DNA polymerase is unlikely to add incoming nucleotide if pairs mismatched 3.DNA polymerase can proofread to remove mismatched pairs **DNA polymerase backs up and digests linkages between bad matches, then tries again **Other DNA repair enzymes as well

29. What are the two differences in translation initiation in eukaryotes compared to prokaryotes? Make sure you know the sequence of events for the initiation and elongation steps of translation in prokaryotes.

*Two eukaryotic differences in initiation **Instead of a ribosomal-binding sequence, mRNAs have guanosine cap at 5' end ***Recognized by cap-binding proteins that promote binding of small ribosomal subunit. **Position of start codon more variable (optimal sequence) ***In many cases, first AUG codon used as start codon

14. Other than splicing, what are two other modifications that are included in RNA processing? What is the function of each modification?

*Two of them, called capping and tailing, are modifications to the ends of the pre-mRNA. As described later, these add a 5′ cap and a polyA tail. *Capping **Modified guanosine attached to 5' end **7-methylguanosine - 5'cap **Needed for mRNA to exit nucleus and bind ribosome *Poly A tail **100-200 adenine nucleotides added to 3' end **Increases stability and lifespan in cytosol **Not encoded in gene sequence, added enzymatically

27. What three elements assemble during the being of translation initiation? What help is required? How is this process energized?

*mRNA, the first tRNA and ribosomal subunits assemble *Requires help of ribosomal initiation factors *Also requires input of energy (GTP hydrolysis), initiation factors hydrolyze GTP to provide energy

3. How do ncRNAs bind to DNA or RNA? What can ncRNAs also bind to (other than to nucleic acid)? How do they bind to these structures?

*ncRNAs can bind to DNA or RNA via complementary base pairing *Base pairing allows ncRNAs to affect DNA replication, transcription, and translation *ncRNAs can bind to proteins or small molecules *Stem-loop structures in ncRNAs can bind to pockets on the surface of proteins, or multiple stem-loops may form a binding site for a small molecule

25. What is apoptosis? What is one hormone that can induce apoptosis and how does it do so? How does it reduce inflammation?

*programed cell death *Cell shrinks and forms rounder shape due to destruction of nucleus and cytoskeleton *Plasma membrane forms blebs - irregular extensions that break away *Role of Apoptosis: during development, it is used to sculpt the tissues and organs during development (making fingers from paddle); remove cells that have become damaged by age, environmental mutagens or viruses before they can replicate and give rise to cancerous tumors. *Prednisolone - synthetic steroid used to treat inflammation *Adrenal cortex isinvolved with stressresponses *Prednisolone decreases number of adrenal cortex cells by inducingapoptosis *Effect of prednisolone can be blocked with ACTH which stimulates cell division. *ACTH = adrenocorticotropic hormone Apoptosis can be signaled via an extrinsic pathway that begins with the activation of death receptors on the surface of the cell. Immune system may target a cell for destruction via the binding of a signaling molecule to a death receptor, causing the receptors to aggregate in a trimer. This causes a conformational change in the intercellular domain of the receptors = death domain. This causes the binding of adaptor proteins, which then bind to a procaspase, together which make a death-inducing signaling complex (DISC). The procaspase is cleaved to become a caspase which functions as a protease. The caspase is released from the DISC and it initiates may other caspases in the cell that are directly responsible for digesting intracellular proteins and causing the cell to die (digesting proteins, components of the intracellular matrix, DNA). *prednisone is an anti-inflammatory and immunosuppressive drug that is used to treat a wide variety of disorders, including asthma and rheumatoid arthritis. When taken into the body, it is converted to prednisolone by the liver.

8. Most enzyme-linked receptors are what kind of enzymes? What is activity associated with this type of enzyme? What do the extracellular domain and intracellular domain do?

*protein kinases: enzymes that transfer a phosphate group from ATP to specific amino acids in a protein *Enzyme-linked receptors typically have two important domains: an extracellular domain, which binds to a signaling molecule, and an intracellular domain, which has a catalytic function (Figure 9.6a). When a signaling molecule binds to the extracellular domain, a conformational change is transmitted through the membrane-embedded portion of the protein and affects the conformation of the intracellular catalytic domain. In most cases, this conformational change causes the intracellular catalytic domain to become functionally active.

19. What is the function of the tRNA? What is the relationship between the anticodon and the codon?

*tRNA **Anticodon - 3 RNA nucleotide part of tRNA molecule **Allows binding of tRNA to mRNA codon in the ribosome *tRNA serves as the translator or intermediary between an mRNA codon and an amino acid. *Anticodon in tRNA is complementary to a codon in mRNA

4. What is transcription and what is translation? Why were these terms used to describe these processes?

*transcription, produces an RNA copy of a gene, also called an RNA transcript (Figure 12.3). The term transcription literally means the act of making a copy. Most genes, which are termed protein-encoding genes, produce an RNA molecule that contains the information to specify a polypeptide with a particular amino acid sequence. This type of RNA is called messenger RNA (abbreviated mRNA), because its function is to carry information from the DNA to cellular components called ribosomes. As discussed later, ribosomes play a key role in the synthesis of polypeptides. *The process of synthesizing a specific polypeptide on a ribosome is called translation. The term translation is used because a base sequence in an mRNA is "translated" into an amino acid sequence of a polypeptide.

To which atom is the nitrogenous base covalently linked to the deoxyribose sugar? 5' C 5' N 3' C 1' C 3' OH

1' C

21. What are three common features of tRNA?

1. Cloverleaf structure **Db stranded - H-bonding **Single stranded - loop 2. Anticodon in loop 3. Acceptor stem for amino acid binding at 3' end

3. What are the five types of signals relayed between cells? Make sure that if given a description of one of these types of signaling, you can identify it correctly from the others.

1. Direct Intercellular Signaling-In a multicellular organism, cells adjacent to each other may have contacts, called cell junctions, that enable them to pass ions, signaling molecules, and other materials between the cytosol of one cell and the cytosol of the other (Figure 9.3a). For example, cardiac muscle cells, which cause your heart to beat, have intercellular connections called gap junctions that allow the passage of ions needed for the coordinated contraction of these cells. 2.Contact-Dependent Signaling-Not all signaling molecules diffuse from one cell to another. Some molecules are bound to the surface of a cell and provide a signal to other cells that make contact with the surface of that cell. In the case of contact-dependent signaling, one cell has a membrane-bound signaling molecule that is recognized by a receptor on the surface of another cell. This type of cell-to-cell communication occurs, for example, when portions of neurons (nerve cells) grow and make contact with other neurons. This is important for the formation of the proper connections between neurons. 3.Autocrine Signaling-In autocrine signaling, a cell secretes signaling molecules that bind to receptors on its own cell surface and on the surfaces of neighboring cells of the same cell type, stimulating a response (Figure 9.3c). What is the purpose of autocrine signaling? It is often important for groups of cells to sense cell density. When cell density is high, the concentration of autocrine signals is also high. In some cases, such signals inhibit further cell growth, thereby limiting cell density. 4. Paracrine Signaling-In paracrine signaling, a specific cell secretes a signaling molecule that does not affect that cell but instead influences the behavior of target cells in close proximity. Paracrine signaling is typically of short duration. Usually, the signal is broken down too quickly to be carried to other parts of the body and affect distant cells. A specialized form of paracrine signaling occurs in the nervous systems of animals. Neurotransmitters—molecules made in neurons that transmit a signal to an adjacent cell—are released at the end of a neuron and traverse a narrow space called the synapse. The neurotransmitter then binds to a receptor in a target cell. 5.Endocrine Signaling-In contrast to the previous mechanisms of cell-to-cell communication, endocrine signaling occurs over relatively long distances. In both animals and plants, molecules involved in long-distance signaling are called hormones. They usually last longer than signaling molecules involved in autocrine and paracrine signaling. In mammals, endocrine signaling involves the secretion of hormones into the bloodstream, which may affect virtually all cells of the body, including those that are far from the cells that secrete the signaling molecules. In flowering plants, hormones move through the plant vascular system and also move through adjacent cells. Some hormones are even gases that diffuse into the air. Ethylene, a gas given off by plants, plays a variety of roles, such as accelerating the ripening of fruit.

8. What are structural differences between RNA and DNA (there are two). What type of base is uracil (purine or pyrimidine)?

1. RNA contains the sugar ribose, while DNA contains the slightly different sugar deoxyribose (a type of ribose that lacks one oxygen atom) 2. RNA has the nucleobase uracil while DNA contains thymine. *Uracil is a purine base

12. What are three common themes in signal transduction?

1. Starts with the signal molecule binding to the receptor which induces a conformational change in the receptor 2. The activated conformation of the receptor stimulates / activates first member of the signal transduction pathway 3.Signal transduction often involve protein kinase cascades or second messengers

1. Which of the following choices will be affected by a cell containing two nonfunctional copies of p53? I. Apoptotic pathways II. DNA repair pathways III. Ability to arrest the cell cycle a. I and III b. I, II and III c. I and II d. II only

1. Which of the following choices will be affected by a cell containing two nonfunctional copies of p53? I. Apoptotic pathways II. DNA repair pathways III. Ability to arrest the cell cycle a. I and III b. I, II and III c. I and II d. II only

30. What happens during translation elongation? What is responsible for the correct aminoacyl tRNA associating with the ribosome? How is this process of binding tRNA to the A site energized? What is the tRNA in the P site called, and what is the tRNA in the A site called?

1.Aminoacyl tRNA brings a new amino acid to the A site *Binding occurs due to codon / anticodon recognition *Elongation factors hydrolyze GTP to provide energy to bind tRNA to A site *Peptidyl tRNA is in the P site *Aminoacyl tRNA is in the A site 2.2.Peptide bond is formed between the amino acid at the A site and the growing polypeptide chain *The polypeptide is removed from the tRNA in the P site and transferred to the amino acid at theA site *Called the peptidyl transfer reaction *rRNA catalyzes peptide bond formation (the ribosome is a ribozyme) 3.Movement or translocation of the ribosome toward the 3' end of the mRNA by one codon *Shifts tRNAs at the P and A sites to the E and P sites, respectively *The next codon is now at the A spot qUncharged tRNA exits from E spot

22. What does DNA polymerase require to start replicating a bare template strand? What is required for this step? What direction does DNA polymerase add nucleotides? What part of the nucleotides are directly involved in the polymerization of DNA?

1.DNA polymerase cannot begin synthesis on a bare template strand **Requires a primer to start **DNA primase (an RNA polymerase)makes the primer from RNA **The RNA primer is removed and replaced with DNA later 2.DNA polymerase only works 5' to 3'

28. Describe the three levels of DNA compaction within the nucleus of the cell? What is the difference between euchromatin and heterochromatin?

1.DNA wrapping **Wrapped around histones to form nucleosome **Shortens length of DNA molecule 7-fold 2.30-nm fiber **Asymmetric, 3D zigzag of nucleosomes **Stable nucleosome units connected by bendable linker regions **Shortens length another 7-fold 3.Radial loop domains **Interaction between 30-nm fibers and nuclear matrix **Loops are 25,000-200,000 bp and are anchored to the nuclear matrix *Level of compaction is not uniform **Heterochromatin **Euchromatin *Heterochromatin - highly condensed chromatin (radial loops are compacted), Euchromatin - less condensed regions in which the 30-nm fibers forms radial loop domains. Cell divison *Metaphase chromosomes become even more compacted into heterochromatin. **Euchromatin is further compacted **Width of metaphase chromosome - 1,400nm **Much shorter than same chromosome during interphase.

7. What are three kinds of cell surface receptors?

1.Enzyme-linked receptors- *all living species *Signal molecule binds to receptor->then conformational change in extracellular domain of recptor-> conformational change in intercellular domain->activation of enzyme (by posphorlation of amino acid)->attachment of negatively charged phosphate to protein will alter its conformation *enzyme that adds P= potein kinase **named for amino acid to which the P is added=serine/threonine kinase 2.G-protein coupled receptors (GPCR)- *eukaryotes *G protein=able to bind GTP (guanosine triphophate) and GDP *ligand binds receptor (GPCR)-> confmational change in recptor enables it to bind to G protien->G protein (lipid anchored protein) releases GDP and binds GTP instead-> GTP causes conformational change in G Protein-> alpha subunit binds GTP or Beta and Gamma subunits-> celluar response *7 transmembrane helices **1. ligand binds receptor **2.activated recptor binds to G protein **3. Releases GDP and binds GTP instead **4.GTP causes G protein to dissociate **5.alpha subunit and beta/gamma dimer interact with other proteins in asingaling pathway. *Regeneration G protein after cell response: alpha times GTP->GTP->GDP times Pi-> alph + beta/gamma subunits reassociate-.inactive G protein complex 3.Ligand-gated ion channels- *plant and animal cells (role in signal transduction) *ligand binding opens ion channels (ex Ca2+) *ions to flow through the mmebrane, changing the ion concentrion in the cell. * in animals, these transmit synaptic signals between neurons and muscles or between two neurons **Ligand-gated ion channels allow the influx of Ca2+ into the cytosol, which can play a role in signal transduction.

26. What are the three stages of translation?

1.Initiation: qmRNA, first tRNA and ribosomal subunits assemble 2.Elongation: Synthesis in 5' to 3' direction from start codon to stop codon 3. 3.Termination Complex disassembles at stop codon releasing the completed polypeptide

6. What are the five levels of DNA structure?

1.Nucleotides - the building blocks of DNA and RNA 2.Strand - a linear polymer strand of DNA or RNA (nucleotides linked covalently together). 3.Double helix - the two strands of DNA 4.Chromosomes - DNA associated with an array of different proteins into a complex structure 5.Genome - the complete complement of genetic material in an organism

2. What are two reasons with examples of why cells need signals?

1.To respond to a changing environment *Cellular responses to the environment can be critical for survival = cellular response. EX: Glucose signals yeast cells to increase glucose transporters and enzymes - glucose binds to a receptor, causing cellular response. **Allows efficient uptake and use of glucose 2. To communicate with each other *Cell-to-cell communication EX: Phototropism in plants **Signaling molecule (auxin) produced in cells in the light **Auxins diffuse cell-to-cell **Auxins cause cells to elongate (dark side) **Except elongation is inhibited by light ***Cells in shadows growbending the plant toward the light

If two heterozygote individuals (Tt x Tt) mate, what is the probability that their first two children will both be affected with the recessive disorder (Child 1 and Child 2)? 1 1/2 1/4 1/16

1/16 Explanation In this question, the probability of one child being affected is 1/4, so the question asks what is the probablity that Child 1 AND Child 2 will be affected, so you apply the product rule: 1/4 x 1/4 = 1/16.

A woman has six sons. The chance that her next child will be a daughter is: 1/2 1 0 5/6 1/6

1/2

If two heterozygote individuals (Tt x Tt) mate, what is the probability that their first child will be an affected boy OR an affected girl? 1/2 1/4 1/8 1

1/2 Explanation Since this problem is asking you to calculate the probability of two mutually exclusive events (see "or" in question), you will apply the sum rule to the mutually exclusive events, but first you need to calculate the probablity of EACH event.The probability of being an affected male is: 1/2 (being a male) x 1/4 (being affected) = 1/8.The probability of being an unaffected girl is: 1/2 (being a girl) x 3/4 (being unaffected) = 3/8The SUM of 1/8 + 3/8 = 4/8 or 1/2.

How many replcation forks are there at an orgin of replication?

2

If non-disjunction occurred with Chromosome 1 during Meiosis I, in a diploid organism with three pairs of homologous chromosomes, how many chromosomes would each daughter cell have after Meiosis I is complete? 3 and 5 4 and 8 2 and 4 1 and 4

2 and 4

8. What is the second stage of transcription and what does RNA polymerase do? What is the template strand and what is the coding strand of DNA? What is the relationship of each to the mRNA? In what polarity is the transcript synthesized? To what end are nucleotides added? What is substituted for thymine?

2.Elongation *RNA polymerase synthesizes RNA transcript *Template strand - used as template for RNA synthesis. *Coding strand - the opposite DNA strand. *The transcript is synthesized 5' to 3' (add to 3' end) *Uracil substituted for thymine

There are _____ aminoacyl-tRNA synthetases in a cell. 64 12 20 3

20

RNA polymerase reads sequence in which direction along the DNA

3' to 5' along the template strand

RNA polymerase reads sequence in which direction along the DNA? 3' to 5' along the template strand 3' to 5' along the coding strand 5' to 3' along the template strand 5' to 3' along the coding strand 5' to 3' along both strands of DNA

3' to 5' along the template strand

In bacteria, how does the small ribosomal subunit know where to associate with the mRNA? a. by recongnizing the ribsomal binding sequence at the 5' end of the mRNA. b. by binding to a release factor in the large ribosomal subunit c. by recognizing the ribsomal binding sequence at the 3' end of the mRNA d. by recognizing the guanosine cap at the end of the mRNA e. all of the above

A

7. What is the building block of nucleic acids? What are the three components of this building block? What distinguishes a purine from a pyrimidine base? Which are purines and which are pyrimidines?

A nucleotide has three components: a phosphate group, a pentose (five-carbon) sugar, and a nitrogen-containing base (Figure 11.4). The nucleotides in DNA and RNA contain different sugars. Deoxyribose is found in DNA, and ribose is found in RNA. Five different bases are found in nucleotides, although any given nucleotide contains only one base. The five bases are subdivided into two categories, the purine and the pyrimidine, due to differences in their structures. The purine bases, adenine (A) and guanine (G), have a double-ring structure, whereas the pyrimidine bases, thymine (T), cytosine (C), and uracil (U), have a single-ring structure. Adenine, guanine, and cytosine are found in both DNA and RNA. Thymine is found only in DNA, whereas uracil is found only in RNA.

What defines the open reading frame? The first C base in the transcript An early stop codon. AUG - the first codon in a transcript

AUG - the first codon in a transcript

2. What proportion of genes encode ncRNAs versus protein?

About 50% of "genes" in the human genome do not code for protein These genes encode small ncRNAs and long ncRNAs

32. What happens to the mRNA and ribosome after the peptidyl transferase reaction takes place? Why is this important? What direction does the ribosome move? What occupies the P site, A site and E site after this step takes place?

After the peptidyl transfer reaction is complete, the third step involves the movement or translocation of the ribosome toward the 3′ end of the mRNA by exactly one codon. This shifts the tRNAs in the P and A sites to the E and P sites, respectively. The uncharged tRNA exits the E site. Notice that the next codon in the mRNA is now exposed at the unoccupied A site. At this point, a charged tRNA can enter the A site, and the same series of steps will add the next amino acid to the polypeptide. A single ribosome in the act of translating an mRNA. In living cells, it is common for multiple ribosomes to be gliding along the same mRNA and synthesizing polypeptides. The complex of a single mRNA and multiple ribosomes is called a polysome (see the chapter opening micrograph).

What is the inducer of the Lac Operon? Lac A Allolactose Lac I ATP

Allolactose

How does SRP RNA function during co-translational sorting at the ER membrane? Decoy Alters SRP function so it can hydrolyze GTP Ribozyme Blocker of mRNA

Alters SRP function so it can hydrolyze GTP

11. Why are some receptors inside the cell (as opposed to spanning across the plasma membrane)? What are two examples of hormones that bind to intracellular receptors? What does the estrogen-receptor complex do?

Although most receptors for signaling molecules are located in the plasma membrane, some are found inside the cell. In these cases, an extracellular signaling molecule must diffuse through the plasma membrane to gain access to its receptor. *Estrogen and testosterone **Hormone passes through cell membrane moves to the nucleus where it binds estrogen receptor *Estrogen•receptor dimerized complex binds DNA and regulates gene transcription *Intracellular recptors: (ex: steroid homone receptors) Estrogens/androgens secreted in bloodstream by endocrine glands-> estrogen (hydrophoic) passes into cell through plasma membrane-> binds to recptor inside cell (cytosol receptor or nuclear receptor(estrogen))->estrogen:recptor complec in nulceus=transcription facotor->gene expression=cellular response

onsider a diploid species where n=5. If an individual of this species was found to have 11 chromosomes, it would be categorized as: Polyploid Aneuploid Tetraploid Monosomic Triploid

Aneuploid

1. In the CRISPR-Cas defense system in bacteria, where does the spacer DNA come from that will ultimately give rise to CRISPR RNA? a. DNA of the bacteria b. DNA from a virus that invades the bacterial cell in the past c. RNA from the lac operon d. DNA from another bacterial cell

Answer: (b). The whole point of the defense provided for the bacteria by the CRISPR-Cas system, is to enable the bacterial cell to protect itself from a virus that has previously invaded that cell. The beauty of this system, is that the cell literally collects pieces of the viral DNA from previous infections and keeps these to make CRISPR RNA when infected by that same virus in the future. The CRISPR RNA targets the invading virus as being one that had previously infected the cell and enables the cell to cut up the viral DNA before it is replicated. Learning the process by which the CRISPR-Cas system is used to defend the bacterial cell from future infections by the same viruses is key to understanding this process as a future gene editing technology. If you can understand how the bacterial cell uses this system naturally, you will be able to follow the various applications that are being proposed and tested for using this system to do gene editing in higher organisms. If you are interested to learn more about gene editing using CRISPR-Cas 9, please watch the video on the Video Page for Chapter 13. For the exam, I am asking you to understand the process of Adaptation, Expression and Interference in the CRISPR-Cas system in bacteria.

1. What type of ncRNA is associated with neurological and cardiovascular diseases? a. rRNA b. tRNA c. miRNA d. RNA in SRPs e. RNA in telomerase

Answer: (c). Probably one of the most profound advancements in our understanding of RNA and disease has been the relatively recent discovery of the role miRNAs play in diseases that are complex. Be sure to remember this important connection as well as the fact that miRNAs play an important role in the NORMAL development of the nervous system and other complex organ systems in the body.

10. If this is a sex-linked recessive pedigree, what is the probability of individual 18 being neither affected nor a carrier? a. 0% b. 100% c. 25% d. 50%

Answer: Choice a, because as stated above half of their daughters will be affected and half of their daughters will be carriers.

7. If this is a sex-linked recessive pedigree, what is the probability of individual 22 being affected? a. 0% b. 100% c. 25% d. 50%

Answer: Choice b because #22 has an affected mother who must be homozygous for this X-linked trait. Therefore, she has only a mutant carrying X chromosome to pass on to all of her sons.

6. Consider the pedigree of a recessive trait. What is the mode of inheritance? a. It can only be autosomal b. It can only be sex-linked c. It can be either autosomal or sex-linked d. Mitochondrial inheritance

Answer: Choice b, #4 is an affected male from a carrier mother (#1) but his father (#2) is not a carrier, therefore, the only way #4 could be affected is if this were an X-linked trait. His mother, #1, was a carrier and passed her X-chromosome carrying the mutation to her son, #4 and he is affected. Let's test this hypothesis by looking at #16. His mother, #3, is also a carrier, but #16 has an affected daughter, #19. This could only happen if his mate (#10) was a carrier. So the evidence supports a sex-linked recessive trait. If it were autosomal recessive, then none of the offspring of #1 could be affected. If it were mitochondrial inheritance, then all of the offspring of females would be affected and none of the offspring of the males in the pedigree would be affected.

12. If this is a sex-linked recessive pedigree, if individual 20 is not affected, what is the probability individual 21 will be a carrier? a. 0% b. 100% c. 25% d. 50%

Answer: Choice b, because females get their X-chromosomes from their mother. If the father, #20, is not affected, the affected mother, #21, still can only give an X-chromosome with the mutation to her offspring because she is homozygous for this mutation (both of her X-chromosomes have this mutation). Therefore, it is with 100% probability that all of her daughters will be carriers because they can only receive an X-chromosome bearing the mutation from their mother.

Pedigree for Questions #5 through #12 These questions test your ability to recognize a particular pattern of inheritance from a pedigree and determine the probability of particular genotype from the pedigree. Note the pedigree here shows the carrier as well as the affected. This will NOT be the case on your exam. 5. Consider the pedigree. Is the trait dominant or recessive? a. The mode of inheritance cannot be determined. b. Dominant c. Recessive d. Codominant or incompletely dominant

Answer: Choice c, the fact that you find even one affected offspring (II-4) with unaffected parents, tells you that this trait is a recessive one.

9. If this is a sex-linked recessive pedigree, what is the probability of individual 18 being a carrier and not affected? a. 0% b. 100% c. 25% d. 50%

Answer: Choice d, because this individual has a hemizygous mutant father and a carrier mother. Choice d, because this individual has a hemizygous mutant father and a carrier mother. Half of their daughters will be affected and half of their daughters will be carriers. You know that individual 18 is a female because she is represented as a circle on the pedigree.You know that individual 18 is a female because she is represented as a circle on the pedigree.

8. If this is a sex-linked recessive pedigree, what is the probability of individual 18 being affected? a. 0% b. 100% c. 25% d. 50%

Answer: Choice d, because this individual has a hemizygous mutant father and a carrier mother. Half of their daughters will be affected and half of their daughters will be carriers. You know that individual 18 is a female because she is represented as a circle on the pedigree.

11. If this is a sex-linked recessive pedigree, if individuals 12 and 17 have a son, what is the probability of him being affected? a. 0% b. 100% c. 25% d. 50%

Answer: Choice d, since the mother (#12) is a carrier and the father (#17) is not, there is a 50% change that a son will inherit an X chromosome from his mother that carries the mutation.

Beadle and Tatum identified mutations in genes that encoded enzymes in the pathway that synthesized _____________ as the last product in the pathway. Ornithine Glucose Citrulline Arginine Homogentistic Acid

Arginine

23. What are some of the ways that a cellular response is reversed? What enzyme converts cAMP to AMP?

As mentioned, signaling that involves second messengers is typically of short duration. When the signaling molecule is no longer produced and its concentration falls, a larger percentage of the receptors are not bound by their ligands. When a ligand dissociates from a GPCR, the GPCR becomes deactivated. Intracellularly, the α subunit hydrolyzes its GTP to GDP, and the α subunit and β/γ dimer reassociate to form an inactive G protein . The amount of cAMP decreases due to the action of an enzyme called phosphodiesterase, which converts cAMP to AMP. As the cAMP level falls, the regulatory subunits of PKA release cAMP, and the regulatory and catalytic subunits reassociate, thereby inhibiting PKA. Finally, enzymes called protein phosphatases are responsible for removing phosphate groups from proteins, which reverses the effects of PKA

4. What experiment was responsible for identifying DNA and NOT protein as the genetic material? Who carried out this experiment and what did they do and observe?

Avery, McCarty and MacLeod, 1944 *Purified DNA from type S could transform type R (initial observation) *DNA might still contain traces of contamination that may be the transforming principle **Added DNase, RNase and proteases ***RNase and protease had no effect **When DNase was added, no transformation took place *Surprising conclusion: DNA is the genetic material *Still, many biologists were skeptical. **Reflected a belief that the genes of bacteria could not be similar in composition and function to those of more complex organisms. *They asked the question: "What substance is being transferred from the dead type S bacteria to the live type R bacteria?" Purified DNA from type S bacteria and mixed it with type R bacteria. After allowing time for DNA uptake, they added an antibody that aggregated any nontransformed type R bacteria, which were then removed by centrifugation. The remaining bacteria were incubated overnight on petri dishes. So they essentially carried out the transformation in culture, looking for a change in phenotype of the bacteria. As a control, if no DNA extract was added, no type S bacterial colonies were observed on the petri plates. When they mixed their S strain DNA extract with type R bacteria, some of the bacteria were converted to type S bacteria.

The DNA sequence of a one strand of a gene to be transcribed is: 5' - AGTCCGATGGGCTGA - 3' The sequence of the mRNA is: 5' - AGUCCGAUGGGCUGA - 3' The sequence of the DNA strand shown above is that of the: a. Template strand b. Coding strand (identical in sequence and polarity to the mRNA sequence) (except U for T).

B

What of the following is NOT one of the three discrete sites for tRNA binding and polypeptide synthesis in the ribsome? a. the aminoacyl site (A) b. the entrance site (E) c. the peptidyl site (P) d. the exit site (E)

B

Nucleosomes are composed of: Acidic proteins called Histones. Basic proteins called Histones Positively charged lipids Negatively charged carbohydrates

Basic proteins called Histones

3. Predict the phenotype of a promoter (lac P) mutant which has a mutation in the promoter for the lac operon. a. The lac genes would be expressed efficiently only in the absence of lactose. b. The lac genes would be expressed efficiently only in the presence of lactose. c. The lac genes would be expressed continuously. d. The lac genes would never be expressed efficiently.

Be sure to be able to predict whether AND explain why the lac operon is expressed or not under the following conditions: no lactose, no glucose; no lactose, glucose present; both lactose and glucose present; lactose present but no glucose. This question requires that you understand the function and position of each of the components of the lac operon, how they interact with one another and control gene expression. Do you know how an inducer (Allolactose) affects the lac operon? Answer: Choice (d). Since the Lac P region is the promoter that binds RNA polymerase, a mutation that would block the function of this region would prevent the structural genes (Lac A, B and Y) of the lac operon from ever being expressed, regardless of the lactose conditions.

How do miRNAs and siRNAS function in RNA interference in controlling the translation in the cytosol? Scaffold Alters protein stability Ribozyme Blocker of mRNA

Blocker of mRNA

How is a new DNA strand synthesized in nucleotide excision repair? By crossing-over between sister chromatids By DNA polymerase By DNA ligase By Uvr A

By DNA polymerase

Epinephrine is responsible for regulating the synthesis of glycogen and glucose. What effect would a mutation in the gene that encodes Glycogen synthase have on this pathway? a. Glycogen breakdown would be stimulated. b. The synthesis of Glucose-1-Phosphate would be stimulated. c. Glycogen synthesis would be blocked. d. Glycogen breakdown would be blocked.

C

Uncontrolled growth is a necessary step for the development of all cancers. Which of the following drugs, that target the MAP/ERK pathway (EGF receptor pathway), could inhibit uncontrolled cell growth? a. Drugs that bind and activate the EGF receptor b. Drugs that block the inactivation of the Ras G-protein c. Drugs that block the phosphorylation of Mek d. Drugs that inactivate Protein Kinase A (PKA).

C

24. How does caffeine affect the epinephrine signaling pathway?

Caffeine inhibits phosphodiesterase, which is the enzyme that converts cAMP to AMP. Phosphodiesterase functions to remove cAMP once a signaling molecule, such as epinephrine, is no longer present. When phosphodiesterase is inhibited by caffeine, cAMP persists for a longer period of time and prolongs the effects of signaling molecules like epinephrine. Therefore, even low levels of epinephrine have a greater effect. This is one of the reasons why drinks containing caffeine, including coffee and many energy drinks, provide a feeling of vitality and energy

What enzyme is responsible for carrying out apoptosis? Caspase Catalase Calcium Apoptase

Capase

37. What is the mechanism of five different antibiotics listed in Table 12.5 in your text with regard to translation?

Chloramphenicol: Blocks elongation by acting as a competitive inhibitor of the peptidyltransferase complex. Erythromycin: Binds to the 23S rRNA and blocks elongation by interfering with the translocation step. Puromycin: Binds to the A site and causes premature release of the polypeptide. This early termination of translation results in polypeptides that are shorter than normal and usually nonfunctional. Tetracycline: Blocks elongation by inhibiting the binding of aminoacyl tRNAs to the ribosome. Streptomycin: Interferes with normal pairing between aminoacyl tRNAs and codons. This causes misreading of the code and thereby produces abnormal proteins.

Which is one way a tumor suppressor gene could become mutated and lose function? Gene duplication Chromosome loss Insertion of active promoter infront of tumor suppressor gene

Chromosome loss

The cell cycle is controlled by the changing concentrations of different ________: Cdks Cyclins Chromosomes Autosomes

Cyclins

The Kd for the insulin receptor is 4.5 x 10-9M. At what [insulin] are 50% of the total receptors bound by insulin? a. 0.75 x 10-9M b. 1.5 x 10-9M c. 3.0 x 10-9M d. 4.5 x 10-9M

D Remember: Kd = [Receptor][Ligand] [Receptor bound to Ligand]

12. What is a new and exciting application of the CRISPR-Cas 9 system?

DNA editing

Which of the following is responsible for separating (melting) the two DNA strands in preparation of copying the template strand? DNA polymerase DNA helicase DNA topoisomerase

DNA helicase

24. Be familiar with the DNA polymerases in Bacteria versus humans. Which human DNA polymerase is most similar in function to DNA primase in E.coli?

DNA polymerase III has a subunit called the clamp protein that allows the enzyme to slide along the template strand without falling off, a characteristic called processivity. Each DNA polymerase differs in the rate and accuracy of DNA replication and/or the ability to prevent the formation of DNA gaps. Bacteria have 5 DNA polymerases while humans have 12. E. Coli - DNA polymerase I and III stall when they encounter DNA damage, while II, IV and V do not stall, but their rate of synthesis is not as rapid as DNA pol I and III, they will ensure that DNA replication is complete. Humans - general DNA polymerases (alpha, delta, gamma) may be unable to replicate over an abnormality in DNA structure (a lesion). Lesion-replicating polymerases are attracted to damaged DNA. These polymerases have special properties that enable them to synthesize a complementary strand over the lesion. (over different types of DNA damage).

Because more than one codon can specify the same amino acid, the genetic code is said to be

Degnerate

21. What provides the energy to connect nucleotides during DNA replication?

Deoxynucleoside triphosphates **Free nucleotides with three phosphate groups **Breaking covalent bond to release pyrophosphate (two phosphates) provides energy to connect nucleotides

The mode of DNA replication that predicts that EACH DNA STRAND will be a combination of segments from the old and new DNA is: Dispersive Conservative Semiconservative

Dispersive

Which of the following is NOT required for the initiation of translation? a. small subunit of the ribsome b. mRNA c. initiator tRNA d. hydrolysis of GTP e. elongation factor

E

Which of the following is NOT required for the initiation of translation? Small subunit of the ribosome mRNA Initiator tRNA Hydrolysis of GTP Elongation factors

Elongation factors

Which of the following is NOT an example of a secondary messenger? cAMP DAG Ca2+ ions Epinephrine All are examples of secondary messengers

Epinephrine

A gene that when mutant exhibits only a single trait or mutant phenotype is called Pleiotropic. True False

F

After Meiosis 1, cells are still diploid. True False

F

Controlled matings meant that a geneticist would allow plants or animals to mate at random. True False

F

Heterochromatin allows transcription of genes tightly packed into nucleosomes. True False

F

Mice that do not have a functional p53 gene will be born with many abnormalities. True False

F

Mutations can only occur in the coding sequence of a gene. True False

F

Polyploidy in plants is usually lethal, as in polyploid plants can never be bred or made. True False

F

The Mediator binds directly to the promoter in a eukaryotic gene. True False

F

A gene always encodes a polypeptide. True False

False

A high Kd means that more ligand is bound to its receptor than unbound to its receptor.. T/F

False

Homologous chromosomes are completely identical to one another because they are direct copies of each other. True False

False Explanation No, sister chromatids are exact copies of each other.

Who proposed the Central Dogma of Molecular Biology?

Francis Crick

13. Which base pair combinations have more H-bonds and how does this affect the melting temperature of DNA?

G-C base pairs have 3 hydrogen bonds, while A-T base pairs have two. Therefore, double-stranded DNA with a higher number of G-C base pairs will be more strongly bonded together, more stable, and will have a higher melting temperature.

When epinephrine binds to its receptor on muscle cells, its ____________ is activated, resulting in the activation of __________. Ligand-gated ion channel receptor; PKA G-protein coupled receptor; PKA Enzyme-linked receptor; ATP

G-protein coupled receptor; PKA

What protein(s), together with RNA polymerase II, bind to the core promoter of a gene to form the pre-initiation complex?

General transcription factors

When Hershey and Chase labelled T2 bacteriophage with 35-S and then allowed the bacteriophage infect bacterial cells, where did they find the 35-S label after the infection took place? Inside the bacterial cells In the bacteriophage on the outside of the bacterial cells (never entered bacterial cells). Both inside and outside the bacterial cells

In the bacteriophage on the outside of the bacterial cells (never entered bacterial cells).

Where is the TATA box located in a eukaryotic gene? In the upstream regulatory region In the core promoter In the coding region (part that contains exons and introns) At the termination site of the gene

In the core promoter

31. What is the process called when an amino acid is added to the growing polypeptide chain? Why is it called a "transfer" reaction, what is being transferred from what to what? What catalyzes this reaction?

In the second step, a peptide bond is formed between the amino acid at the A site and the growing polypeptide, thereby lengthening the polypeptide by one amino acid. As this occurs, the polypeptide is removed from the tRNA in the P site and transferred to the amino acid at the A site, an event termed a peptidyl transfer reaction. This reaction is catalyzed by a region of the 50S subunit known as the peptidyltransferase center, which is composed of several proteins and rRNA

Which of the following statements best describes DNA polymerase? It is an enzyme required to glue pieces of DNA fragments together. It is an enzyme required to produce a primer needed for DNA replication. It is an enzyme that catalyzes the addition of nucleotides to the 5' end of a growing DNA strand. It is an enzyme that catalyzes the addition of nucleotides to the 3' end of a growing DNA strand.

It is an enzyme that catalyzes the addition of nucleotides to the 3' end of a growing DNA strand.

2. In Drosophila flies, eye color is sex-linked. The allele for red eyes (XW) is dominant to the allele for white eyes (Xw) . A male fly with red eyes XWY and a carrier (heterozygote) female XWXw are mated. What is the probability that the female offspring will have white eyes? a. 0% b. 100% c. 25% d. 75% e. 50%

It is important for this problem to make a Punnet square. This is the only way to reliable find the answer to this problem. Predict the gametes each parent will make and complete the Punnet square to solve this sex-linked inheritance question. Answer: Choice a, because when you complete the Punnet square you will see that the females are 50% WW and 50% Ww (or carriers).

6. What is Kd and what does it mean? When the concentration of the ligand is equal to Kd, how much of the receptors are bound to ligands? What is meant by a low Kd and a high Kd with regard to the binding affinity of the ligand to its receptor?

Kd is called the dissociation constant between a ligand and its receptor. The Kd value is inversely related to the affinity between the ligand and receptor. *Kd = [Ligand][Receptor]/[Ligand.Receptor Complex] *Small Kd - equilibrium favors the complex, there is a high affinity of the receptor for the ligand. *High Kd - equilibrium favors the separation of the receptor and the ligand, there is a low affinity of the receptor for its ligand. *Kd = [Ligand], when half of the receptor has bound ligand. *If the concentration of a ligand is far below the Kd value, a cellular response is not likely because relatively few receptors form a complex with the signaling molecule.

Make sure you know that_________ phosphorylate amino acids in proteins while ______ remove phosphate groups from amino acids. Their actions oppose one another.

Kinase, phosphatase

The structural genes in the Lac Operon are all dedicated to the breakdown of: Glucose Glycogen Lactose Sucrose

Lactose

When a neurotransmitter released from one neuron binds to a ____ on a second neuron, that neuron depolarizes and fires. Nuclear receptor Enzyme-linked receptor G-protein coupled receptor Ligand-gated ion channel

Ligand-gates ion channel

24. How does the difference between prokaryotic and eukaryotic ribosomes benefit humans today in medicine (treatment of antibiotic infections in humans)?

Many diseases that affect humans and domesticated animals are caused by pathogenic bacteria. An antibiotic is any substance produced by a microorganism that inhibits the growth of other microorganisms, such as pathogenic bacteria. Most antibiotics are small organic molecules, with masses of less than 2,000 Da. In some cases, antibiotics exert their effect because they inhibit or interfere with bacterial translation. Because the components of translation differ somewhat between bacteria and eukaryotes, some antibiotics inhibit bacterial translation without affecting eukaryotic translation. Therefore, they can be used to treat bacterial infections in humans, pets, and livestock.

A chromosome with its centromere in the middle of the chromosome is called: Acrocentric Metacentric Telocentric

Metacentric

The stage of Mitosis when the pairs of sister chromatids are aligned along the metaphase plate is: Prophase Prometaphase Metaphase Telophase

Metaphase

n Griffith's experiment, when heat killed S bacteria was mixed with living R cells and then this mixture was injected into mice, what was observed? Mice died and had living R cells inside. Mice didn't die and no bacteria was inside. Mice died and had living S cells inside

Mice died and had living S cells inside

A point mutation that results in an amino acid substitution is a __________ mutation. Nonsense Silent Missense

Missense

9. How are the carbons numbered in the nucleotide? Which one is attached to the base and which to the sugar?

Numbering system: Prime (') symbol describes the locations of carbon in the sugar. The atoms in the ring structures of the bases are not given the prime designation. The sugar carbons are designated 1', etc, with the carbon atoms numbered in a clockwise direction starting with the carbon atom to the right of the ring oxygen that is attached to the nitrogenous base. The 5' carbon is outside the ring and is attached to the phosphate group. Deoxyribose lacks a single oxygen atom at the 2' carbon. *Sugar carbons are 1' to 5' *Base attached to 1' carbon on sugar *Phosphate attached to 5' carbon on sugar

Bacteria are grown in 15N (heavy) medium and then transferred to 14N (light) medium and are allowed to replicate for 2 generations. The DNA is subsequently isolated and centrifuged in a CsCl2 gradient to yield what type of band(s), as observed by Meselson and Stahl in 1958? One heavy band and one half-heavy band One light and one half-heavy band One heavy band only One half-heavy band only

One light and one half-heavy band

In humans, having dimples in the cheeks is a dominant trait. If a child has dimples but only one of her parents does, what are the genotypes of her parents? (D = dimples; d = no dimples) a. One parent must be DD, the other parent could be either dd or Dd b. One parent is dd, other parent must be DD c. One parent is Dd, other parent is DD d. One parent dd, other parent must be Dd e. One parent must be dd, the other parent could be either Dd or DD

One parent must be dd, the other parent could be either Dd or DD

1. What percentage of the genome is made up of protein coding genes?

Only 1.5% of the human genome encodes protein

Which of the following is true concerning a somatic cell mutation? a. Half of the gametes carry the mutation. b. A small fraction of the gametes carry the mutation. c. Only a small group of cells within the organism is affected by the mutation. d. All cells within the organism are affected by the mutation. e. All of the gametes carry the mutation.

Only a small group of cells within the organism is affected by the mutation.

The site in the ribosome where the tRNA that is attached to the growing peptide is located is the _________ site. A P E T

P

In order to produce a single strand of DNA, the nucleotides combine to form what type of bond?

Phosphodiester

Telomerase uses a _______________ sequence as a template for making the extended 3' end of the chromosome. RNA DNA Protein

RNA

What protien do transcription facotors directly affect?

RNA polymerase

What protein is involvd in synthesizing messenger RNA in eukaryotes?

RNA polymerase II

What happens when the cAMP-CAP complex binds the CAP site near the lac promoter?

RNA polymerase binds to the promoter and transcription rate increases.

16. The EGF signal transduction pathway involves a "relay" protein, a protein kinase cascade and activation of transcription factors. Be familiar with the proteins involved in these steps.

Relay Proteins: Grb binds to the activated receptor; then to Sos, Sos stimulates Ras to release GDP and bind GTP *Activated receptor binds first relay protein, Grb *Grb binds Sos *Sos binds inactive Ras with GDP bound.Stimulates GDP release and GTP binding. *Active Ras-GTP conformation *Activated receptor binds to Grb which results in a conformational change in Grb which causes it to bind to another relay protein, Sos. This activates Sos which causes it to bind to a third relay protein, Ras, Ras releases GDP and binds GTP. The GTP form of Ras is the active form. -or- EGF singla transduction pathway (2) *Protein Kinase Cascade starts with Ras-GTP activating Raf (first protein kinase in cascade) *Active Raf phosphorylates Mek to active form *Active Mek phosphorylates Erk nActive Erk enters nucleus phosphorylates transcription factors, Myc and Fos *The function of Grb, Sos, and Ras is to relay a cellular signal to additional proteins in the signal transduction pathway that form a protein kinase cascade. *Activated Ras binds to Raf, the first protein kinase in the cascade. Raf then phosphorlates Mek, which becomes active, and in turn, phosphorylates Erk. Raf, Mek, and Erk, the protein kinase cascade, are all examples of mitogen-activated protein kiinases (MAP-kinases). Mitogen = agent that causes cell to divide. EGF is a mitogen -or- EGF Cellular REsponse (3) *Activated Erk phosphorylates Myc and Fos *Activated Myc and Fos transcription factors bind specific genes for cell cycle activation and activate transcription (production of RNA) nmRNAs for cellular division are translated *Cellular division process is started *Growth factors such as EFG cause a rapid increase in the expression of many genes in mammals, perhaps as many as 100. Growth factor signaling pathways are often involved in cancer. Mutations that cause proteins in these pathways to become hyperactive result in cells that divide uncontrollably. Best way-- *Relay proteins in EGF Signal Transduction Pathway *Grb (1st relay protein) activated by phosphorylation of EGF intercellular subunit->Grg undergoes conformational change-> Grg binds to Sos (2nd relay protein)->Sos activated and binds to Ras (3rd relay protein)->Ras releases GDP and binds GTP=active Ras-> Active Ras binds to Raf (1st protein kinase in pathway)-. Raf phosphorylates Mek-> active Mek phosphorylates Erk->active Erk entere nucleus->phosphorylates transcription factors: Myc, Fas-> genes turned ON cell divides *Raf, Mek, and Erk= mitogen-activated protein kinases (MAP-kinases)

In the Meselson and Stahl experiment, two rounds of DNA replication were observed. After the first round of replication, but before the second round had taken place, Meselson and Stahl could be confident of which of the following conclusions about replication?

Replication is not conservative

How does ncRNA in the ribosome (ribosomal RNA) function with ribosomal protein to form the shape of the ribosome? Scaffold Guide sequence Ribozyme Blocker of mRNA

Scaffold

17. What is the type of replication proposed for double stranded DNA?

Semiconservative??

What effect do chromatin remodeling ezymes have on chromatin?

Shifts the position of the nucleosome on the DNA

What separate from each other in Anaphase II? Homologous chromosomes Sister chromatids

Sister chromatids

4. What are the three stages of cell signaling and what is happening in each. Can you identify these three stages in the EGF signaling pathway?

Stage 1: Receptor Activation- In the initial stage, a signaling molecule binds to a receptor of the target cell, causing a conformational change in the receptor that activates its function. In most cases, the activated receptor initiates a response by causing changes in a series of proteins that collectively forms a signal transduction pathway, as described next. Stage 2: Signal Transduction- During signal transduction, the initial signal is converted—or transduced—to a different signal inside the cell. This process is carried out by a group of proteins that form a signal transduction pathway. These proteins undergo a series of changes that may result in the production of an intracellular signaling molecule. However, some receptors are intracellular and do not activate a signal transduction pathway. As discussed later, certain types of intracellular receptors directly cause a cellular response. Stage 3: Cellular Response- Cells respond to signals in several different ways. Three common categories of proteins that are controlled by cell signaling: enzymes, structural proteins, and transcription factors. *Many signaling molecules exert their effects by altering the activity of one or more enzymes. For example, certain hormones provide a signal that the body needs energy. These hormones activate enzymes that are required for the breakdown of molecules such as carbohydrates. *Cells also respond to signals by altering the functions of structural proteins in the cell. For example, when animal cells move during embryonic development or when an amoeba moves toward food, signals play a role in the rearrangement of actin filaments, which are components of the cytoskeleton. The coordination of signaling and changes in the cytoskeleton enables a cell to move in the correct direction. *Signaling molecules may also affect the function of transcription factors—proteins that regulate the transcription of genes. Some transcription factors activate gene expression. For example, when cells are exposed to sex hormones, transcription factors activate genes that change the properties of cells, which can lead to changes in the sexual characteristics of entire organisms. Estrogens and androgens are responsible for the development of secondary sex characteristics in humans, including breast development in females and beard growth in males.

Epinephrine signaling causes glycogen to be broken down into glucose-1-Phosphate to be used in aerobic respiration. T/F

T

One result in Mendel's monohybrid cross was that the F1 offspring did not exhibit a blending of tall and short traits together. True False

T

Repressors and activators affect the level of transcription in the cell. True False

T

The amplification of the Myc gene will lead to the over replication of cells because Myc is a transcription factor that activates target genes in the EGF signaling pathway. True False

T

The major groove of the double stranded DNA is the space were proteins can associate with the bases in the DNA. T/F

T

The progression from one phase in the cell cycle to the next (e.g. G1 to S) is dependent on the increase in concentration of a cyclin-cdk complex. True False

T

Tumor suppressor genes are activated at checkpoints when DNA damage is detected. True False

T

9. What is the third stage of transcription? How does RNA polymerase know to terminate transcription? What does it cause to happen?

Termination: Eventually, RNA polymerase reaches a terminator, which causes it and the newly made RNA transcript to dissociate from the DNA. This event constitutes the termination of transcription.

What observation made by Thomas H. Morgan, suggested that the inheritance of the white eyed mutation was not autosomal? a. More females than males were white eyed in the F2 generation. b. The F2 generation from the parental crossing of a white eyed male to a red eyed female resulted in an equal number of male and female white eyed progeny. c. The F2 generation from the parental crossing of a white eyed male to a red eyed female resulted in white eyed flies that were male and not female.

The F2 generation from the parental crossing of a white eyed male to a red eyed female resulted in white eyed flies that were male and not female.

Folowing its initial activation, how does a G protein become inactivated?

The G-protein alpha subunit GTP is hydrolyzed to GDP allowing reassociation of the G-protien alpha subunit and beta-gamma dimer.

If the cell's control of replication is like the brakes and accelerator of a car. An oncogene is similar to: The accelerator stuck in the "on" position The brakes of the car not working The accelerator not working The brakes of the car "on" all the time

The accelerator stuck in the "on" position

What part of the histone protein is modified to control nucleosome packing? a. The carboxyl end of the histone can be methylated, phosphorylated or acetylated b. The amino end of the histone can be methylated, phosphorylated or acetylated c. The entire length of the histone is covered with basic acetyl groups d. Segments of the histone are cleaved and removed so the DNA can be packed tighter around the nucleosome core.

The amino end of the histone can be methylated, phosphorylated or acetylated

What part of the histone protein is modified to control nucleosome packing?

The amino end of the histone can be methylated, phosphorylated or acetylated.

What part of the histone protein is modified to control nucleosome packing? a. The carboxyl end of the histone can be methylated, phosphorylated or acetylated. b. The entire length of the histone is covered with basic acetyl groups. c. The amino end of the histone can be methylated, phosphorylated or acetylated. d. Segments of the histone are cleaved and removed so the DNA can be packed tighter around the nucleosome core.

The amino end of the histone can be methylated, phosphorylated or acetylated.

What feature of the structure of DNA did Chargoff's rule address? The diameter of the double helix The base pairing of T with A, and G with C The sugar phosphate backbone

The base pairing of T with A, and G with C

17. What is the functional relationship between the codons and amino acids?

The genetic code consists of 64 different codons (Table 12.1). The sequence of three bases in most codons specifies a particular amino acid.

17. What type of modifications of the EGF MAPK pathway could lead to uncontrolled growth of the cell?

The phosphorylated form of Erk enters the nucleus and phosphorylates transcription factors such as Myc and Fos. What is the cellular response? Once these transcription factors are phosphorylated, they stimulate the transcription of genes that encode proteins that promote cell division. After these proteins are made, the cell is stimulated to divide. Growth factors such as EGF cause a rapid increase in the expression of many genes in mammals, perhaps as many as 100. As discussed in Chapter 15, growth factor signaling pathways are often involved in cancer. Mutations that cause proteins in these pathways to become hyperactive result in cells that divide uncontrollably!

Which of the following is NOT a way that mRNA is processed in eukaryotes: Poly A tail is added to the 3' end of the mRNA The 5' end of the mRNA is capped with a modified guanosine The introns are spliced out of the pre-mRNA, leaving only the exons in the mature mRNA. The pre-mRNA is processed in the cytoplasm. All of these are ways that pre-mRNAs are processed in eukaryotes.

The pre-mRNA is processed in the cytoplasm.

Which of the following is most likely to occur when a tumor-suppressor gene is mutated? a. The tumor-suppressor gene may be overactive. b. The tumor-suppressor gene and resulting protein may lose its function and ability to suppress cell proliferation. c. The resulting tumor-suppressor protein would further suppress cell proliferation. d. The resulting tumor-suppressor protein would activate an oncogene. e. None of the choices are possible.

The tumor-suppressor gene and resulting protein may lose its function and ability to suppress cell proliferation.

9. Aneuploidy is the presence of an abnormal number of chromosomes in a cell. Monosomy is a "subtractive" form of aneuploidy due to the presence of only one chromosome from a pair while trisomy is an "additive" form where there are three instances of a particular chromosome, instead of the normal two. A strawberry plant is normally diploid, but a monosomic strawberry plant was observed with 9 chromosomes in its cells. Which of the following is NOT a correct possibility regarding the chromosomal material of a strawberry plant? a. A trisomic strawberry plant would have 11 chromosomes in its cells. b. A trisomic strawberry plant would have 15 chromosomes in its cells. c. Doubling the chromosomal number found in the normal haploid state would result in a strawberry plant with 10 chromosomes in its cells. d. During meiosis of the normal diploid, the chromosomal number is halved during the first meiotic division, yielding 5 chromosomes.

This question expects you to be able to use the information provided and first determine the normal diploid and then haploid number of chromosomes in this organism. With that information, you should be able to pick the correct answer. Answer: Choice b. To answer this question, first focus on the fact that the monosomic plant has 9 chromosomes. This tells you that the normal diploid number is 10, so the normal haploid number is 5. So, a trisomic plant would have one more chromosome in addition to the diploid number: 10 + 1 = 11 (choice a). Now choice c is what you expected, the diploid number is 10 and choice d is what you expected, that half of 10 = 5. So, the only choice that is NOT a correct possibility is choice b. Make sure that if the number of chromosomes in the monosomic plant is different, you could still answer the question. Also, be sure you know the difference between a trisomy (10 + 1) and a triploid (5 x 3).

4. In some chickens, feather color is controlled by codominance. When a black feathered chicken mates with a white feathered chicken, all of the offspring are covered in both black and white feathers. A farmer mates a black feathered chicken (BB) with a black-and-white feathered (BW) chicken. What are the predicted phenotypes of their offspring? a. All of the offspring will have black feathers. b. All of the offspring will have black-and-white feathers c. 75% of the offspring will be black feathered, and 25% of the offspring will be white feathered. d. 50% of the offspring will be black feathered and 50% of the offspring will be black-and-white feathered.

This question is testing your understanding of inheritance of codominant traits, where the phenotype associated with both alleles is expressed in the heterozygote. Answer: Choice d, because if you make a Punnet square for this mating (which I strongly suggest), you will find that 50% of the offspring are BB (black feathered) and 50% are BW which will be black and white feathered, due to the codominant inheritance pattern of these alleles.

8. A cell isolated from a tumor was studied and found to rapidly advance from G1 to S phase. This continued despite exposing the cell to UV irradiation. The irradiated cell continued to divide rapidly and passed through the G1 to S phase transition quickly. What aspect of cell cycle regulation could have been abnormal in this cell? a. Lack of production of cyclins b. Faulty G1 checkpoint c. Overexpression of p53 d. Lack of production of cyclin dependent kinases e. All of these choices could be responsible for the rapidly dividing tumor cell after UV irradiation.

This question is testing your understanding of the role of cyclins, checkpoints, p53 and cdk's in regulating the cell cycle. Answer: Choice (b), because the G1 checkpoint is where the cell can pause before advancing to S phase if DNA damage (such as UV irradiated induced damage) is detected. A lack of cyclins (choice a) and cdk's (choice d) would inhibit the cell advancing from G1 to S. In addition, the overexpression of p53 would immediately stop the cell from advancing in the cell cycle. So the only correct answer is (b).

6. Histone Acetyltransferases are enzymes responsible for modifying histone structure. These enzymes add acetyl groups to amino acids on the histone protein that contain amine groups on their side chains (arginine and lysine). By adding acetyl groups to these amino acid side chains, the structure of the entire histone protein complex will change, thereby altering the expression of different genes associated with these histones. If histone X were to be acetylated, the expression of gene Y would be increased. If gene Y codes for protein Z, what would happen to the levels of protein Z if the concentration of histone acetylase were to decrease? a. The concentration of protein Z would increase, as the expression of gene Y is increasing. b. The concentration of protein Z would decrease, as the expression of gene Y is decreasing. c. The concentration of protein Z would increase, as the expression of gene Y is decreasing. d. The concentration of protein Z would decrease, as the expression of gene Y is increasing.

This question requires an understanding of the effect histone modifying enzymes have on gene expression. Most methylation of histone amino acids results in repressing transcription, while acetylation of certain histone amino acids result in the activation of transcription. Answer: Choice (b). The production of protein Z is dependent upon gene Y. Acetylation of the histones associated with gene Y will promote transcription of gene Y and thereby increase the production of protein Z. If acetylation of histones associated with gene Y would decrease, then the production of protein Z would also decrease.

2. A non-conservative mutation is referred to as one that results in the exchange of a single, new amino acid that has different biochemical properties for the original one in the polypeptide. Which one of these could be a non-conservative mutation? a. Silent b. Missense c. Nonsense d. Frameshift

This question requires that you understand each of these four types of mutations, how they are different from one another, and what effect they have on the polypeptide product. Answer: Choice (b). The definition of a missense mutation is when one amino acid is switched for a different amino acid. A silent mutation results in no amino acid change, a nonsense mutation substitutes a stop codon for the codon calling for an amino acid, and a frameshift mutation inserts a base in the code, so that the reading frame for all the codons that follow has changed. This results in a complete change in many amino acids.

4. Which of the following will least likely affect the length of a protein product? a. Nonsense mutation b. Single-base deletion c. Missense mutation d. Frameshift mutation

This question requires that you understand the effect each of these types of mutations have on the resulting polypeptide. Answer: Choice (c) because a single amino acid is substituted for another in a missense mutation and this doesn't change the length of the protein. In choice a, the nonsense mutation will lead to a shortened protein (introduces an early stop codon, in choice b, a single-base deletion results in a frame-shift and that will often lead to an early stop codon, and choice d, as just mentioned, often leads to an early stop codon.

1. All the following mutations can result in a reduction of β-galactosidase synthesis (in the Lac Operon) except: a. A mutation in adenylate cyclase b. A mutation in catabolite activator protein (CAP). c. A mutation in the CAP site in the lac control region. d. A mutation in the repressor binding site in the operator.

This question requires that you understand the function and position of each of the components of the lac operon, how they interact with one another and control gene expression. Answer: Choice (d). A mutation in the Lac O region, binding site of the repressor, would not reduce the amount of B-gal because binding of the Lac repressor to Lac O normally blocks the synthesis of B-gal. All the other choices would prevent or reduce the synthesis of B-gal because the production and binding of the CAP protein (bound to cAMP made by adenyl cyclase) would be compromised by these mutations.

4. Predict the phenotype of an operator mutant (Oc) which prevents the binding of the repressor to lac O (the operator), where "c" means that the operator cannot bind to a repressor protein. a. The lac genes would be expressed efficiently only in the absence of lactose. b. The lac genes would be expressed efficiently only in the presence of lactose. c. The lac genes would be expressed continuously, sometimes at a basal level. d. The lac genes would never be expressed efficiently.

This question requires that you understand the function and position of each of the components of the lac operon, how they interact with one another and control gene expression. Mutations in the lac operon were critical in defining the role of each gene in the lac operon. It is important that you know which genes are regulatory (lac P = promoter, lac O = operator, binding site of lac repressor) and those that actually encode for proteins (lac Z = B-galactosidase that breaks lactose into glucose and galactose, lac Y = lactose permease that is a transporter that brings lactose into the bacterial cell, and lac A = B-galactoside transacetylase that modifies lactose. Answer: Choice (c). The whole purpose of the Lac repressor is to allow the lac operon to be repressed when it is not needed, when there is no lactose to be metabolized into glucose. Since a mutation in the operator (binding site of the Lac repressor) would prevent the shutting down of the lac operon in the absence of lactose, the lac genes would be expressed continuously. The lac operator is situated between the lac promoter and the structural genes of the lac operon to block the expression of the lac operon genes in the absence of lactose. If this repression doesn't take place, the lac structural genes are free to be expressed, even when there is no lactose to be metabolized into glucose.

5. A point mutation within which functional part of a DNA sequence would be most likely to ultimately result in the production of proteins which differ from the non-mutated form by only a single amino acid? a. Exon b. Promoter c. Intron d. Telomere e. Any choice could be correct.

This question requires that you understand the function of each of these parts of the gene and which encodes amino acid sequence. What if the mutation didn't change the amino acid sequence but just affected the level of the protein in the cell? Answer: Choice (a), which is the only choice given that contains DNA sequence that codes for amino acid sequence. All the other choices are either not found in the mature/processed mRNA (choice b and choice c) or are a part of the chromosome that is never transcribed (choice d).

2. Which of the following is true about the DNA-binding and activation domains of transcription activators: a. The DNA-binding domain is always at the N-terminus, and the activation domain is at the C-terminus. b. The activation domain of one protein can be fused to the DNA-binding domain of another protein to generate a functional activator protein. c. The DNA-binding domain is always at the C-terminus, and the activation domain is at the N-terminus. d. The distance between a DNA-binding domain and activation domain cannot be altered.

This question requires that you understand the role of DNA-binding domains and activation or repression domains in regulatory protein. An activator is comprised of two separate domains within the same protein: a DNA-binding domain that directs this protein to bind to a certain DNA sequence and an activation domain that activates the transcription of the gene that the protein has bound to. One controls guiding the protein to its site in the genome, and the other domain executes the actual function, to activate the expression of the gene. The same is true for a repressor protein, only it has a domain that executes the repression of the target gene's expression when it binds. Both of these proteins have to have DNA-binding domains, or they cannot affect the transcription of their target genes. These domains (DNA-binding and activation/repression are interchangeable). For example, if a protein normally represses gene A, but has had its repressor domain replaced with an activator domain, now the same protein will activate gene A. Answer: Choice (b). The distance between a DNA-binding domain and an activation domain can be variable and these domains are not required to be near either end of the protein. Hybrid proteins are made routinely by fusing the coding region of the DNA-binding domain of one gene with the activation domain of another gene. This creates a protein that will bind to the DNA corresponding to the DNA-binding domain, while activating the expression of the target gene.

3. Which of the following represents a frameshift mutation of the given template strand (with the codons indicated) here: 5'-AGC-CTT-AGC-3' a. 5'-AGC-CTT-AGG-3' b. 5'-AGC-GCT-TAG-C-3' c. 5'-TTT-AGC-CTT-AGC-3' d. 5'-TGC-CTT-AGC-3' e. 5'-CTT-AGC-3'

This question requires that you understand what a frameshift mutation is and how to recognize it in a changed nucleotide sequence. Answer: Choice (b). This question requires that you understand that a frameshift mutation is the introduction of a base (or deletion of a base) that changes the reading frame or codons of the remaining sequence. In choice a, the last base in the sequence is changed from a C to a G, so this is not a frameshift mutation. In choice c, three T's are added to the 5' end of the template, so this would not change the reading frame because it is an insertion of three bases. In choice d, the first base is changed from A to T, so like choice a, this is not a frameshift mutation. Finally, choice e deletes the first three bases of the sequence and this itself doesn't change the reading frame of the sequence.

6. During which phase of the cell cycle are DNA repair mechanisms least active? a. G1 b. S c. G2 d. M

This question requires that you understand what is taking place in each of the phases of the cell cycle and when DNA repair mechanisms take place. [There is more than one way to repair DNA.] Answer: Choice d, because during Mitosis, the chromosomes are condensed, align during metaphase and move to opposite daughter cells during anaphase. There is no time for DNA repair during M phase, but DNA repair must take place during the checkpoints in G1, G2 and by DNA polymerase during the S phase (when the DNA is being synthesized).

Practice LEAPing - Chp 17 Here are examples of the types of questions you will see on your take home, open book exam. These are not identical to the test questions, nor do they include all the examples of the types of questions you will see on the exam. 1. In humans, hair texture exhibits incomplete dominance. The gene for curly hair (H) is incompletely dominant to the gene for straight hair (h). Individuals that are heterozygous (Hh) have wavy hair. Two heterozygous parents have a child. What is the chance that the child will have wavy hair? a. 0% b. 25% c. 50% d. 75% e. 100%

This question tests your ability to apply Mendelian genetics to human genetics. You should be able to interpret genetic questions that include incomplete dominance, complete dominance and sex-linked inheritance. Answer: Choice c, because a punnet square of the mating Hh x Hh will produce the following offspring: 25% HH, 50% Hh and 25% hh. Therefore, since the wavy hair trait is associated with the Hh genotype, 50% of their offspring can be expected to have wavy hair.

4. A cell has 12 pairs of homologous chromosomes. How many chromosomes does each daughter cell have after the completion of Meiosis I? a. 6 b. 12 c. 24 d. 48

This question tests your ability to follow the number of chromosomes versus chromatids versus chromosome pairs during Meiosis. Answer: Choice (b) because Meiosis I is called the reductive division when the homologous pairs of chromosomes separate into different daughter cells. This splits each homologous pair into a chromosome in each daughter cell.

5. A cell has 12 pairs of homologous chromosomes. How many chromosomes does each daughter cell have after the completion of Meiosis II? a. 6 b. 12 c. 24 d. 48

This question tests your ability to follow the number of chromosomes versus chromatids versus chromosome pairs during Meiosis. Answer: Choice (b) because the second Meiotic division just separates sister chromatids from one another (as in Mitosis). After the completion of Meiosis II, each chromatid represents a chromosome in the final daughter cells.

2. A cell has 12 pairs of homologous chromosomes. How many chromosomes does it have in Metaphase of Mitosis? a. 6 b. 12 c. 24 d. 48

This question tests your ability to follow the number of chromosomes versus chromatids versus chromosome pairs during Mitosis. Answer: Choice (c) because you are told that there are 12 pairs of homologous chromosomes which makes 24 total chromosomes. DNA synthesis prior to Mitosis does not increase the number of chromosomes, but does make sister chromatids that are attached at the centromere. Could you answer this question and the next three if the number of homologous chromosomes was different?

3. A cell has 12 pairs of homologous chromosomes. How many chromatids does it have in Metaphase of Mitosis? a. 6 b. 12 c. 24 d. 48

This question tests your ability to follow the number of chromosomes versus chromatids versus chromosome pairs during Mitosis. Answer: Choice (d) because as described above the number of chromatids doubles after DNA synthesis but the number of chromosomes does not.

6. Consider a diploid species where n = 6, where n is the number of chromosomes in a haploid gamete. If an individual of this species was found to have 18 chromosomes, it would be categorized as a. Aneuploid b. Tetraploid c. Polyploid d. Monosomic e. Trisomy:

This question tests your ability to interpret an abnormal chromosome number with regard to the different types of chromosomal abnormalities listed. Make sure you can determine which chromosome number would represent each of the five cases listed. Answer: Choice (c) because there are three copies of the haploid number of chromosomes (n = 6 x 3 = 18). This could also be expressed as triploid but that was not one of the answer choices. Choice a would have been either 11 or 13 (a change by one chromosome number from diploid = 12). Choice b would have been 24 which is 4 x 6. Choice d would have been 11 which is the diploid number (12) - 1. And Choice e would have been 13 which is the diploid number (12 + 1), when you would have 3 copies of one chromosome.

13. The pedigree shown tracks the presence of attached earlobes through a family's generation. Having attached earlobes is an autosomal recessive trait. What is the genotype of individual II-3? a. XeY b. XEY c. ee d. EE e. Ee

This question tests your ability to recognize a pattern of inheritance and to determine the chances of offspring having a particular genotype/phenotype in a given pedigree. Answer: Choice e, since you are told that this is an autosomally recessive trait, individual II-3 is not affected but has a homozygous, affected mother (I-2). So this makes clear that II-3 is a heterozygote since his mother can only pass a recessive allele to all of her offspring.

3. Hemophilia is an X-linked recessive condition in which blood does not clot properly. Queen Victoria of England had one allele for hemophilia. Most of her male descendants had the disorder, but few females had it. Why did hemophilia occur more frequently in Queen Victoria's male descendants? a. Only one copy of the X chromosome is found in cells of males, but two copies are found in cells of females. b. Replication of the X chromosome occurs often in males but rarely in females. c. Males have hormones that enhance the expression of X-linked traits. d. Males mature more slowly than females, allowing recessive traits more time to appear.

This question tests your understanding of sex-linked (or X-linked) recessive disorders and their inheritance. You should review the sex-linked pedigrees we will work out in class. Answer: Choice a, because this is the true issue in X-linked inheritance. Since males have only one X chromosome and one Y chromosome, there is not a homologous pair of like sex chromosomes. Any recessive mutation on the X in the male will be expressed. Not so with females who have 2 X chromosomes. The other choices make no sense. There is no difference in the replication of the X chromosome between male and female (choice b). Males cannot enhance the expression of the X-linked traits (choice c) and males do not mature slower than females (choice d).

5. You have discovered a mutation in a gene that completely shuts down the initiation of transcription. You look at the sequence of the gene and find that the mutation in the: a. TATA box b. 5' splice site between the first exon and intron c. Termination of transcription sequence d. 5' cap site

This question tests your understanding of the different parts of the gene and the role certain parts of the gene play in regulating transcription. Answer: Choice (a). Initiation of transcription is focused in the promoter region of a gene. The only choices that are relevant to the initiation of transcription is the TATA box (choice a). The other choices relate to steps that occur after the initiation of transcription such as processing the mRNA (choice b), ending transcription (choice c) and processing the mRNA (choice d).

7. A cell that is dividing lacks the ability to make kinetochore microtubules while it is still able to synthesize and use astral and polar microtubules and microfilaments. During which stage of Mitosis will this defect first be visibly evident under the microscope? a. Prophase b. Metaphase c. Anaphase d. Telophase e. Cytokinesis

This question tests your understanding of the function of the three types of microtubules (astral, polar and kinetochore) as well as the function of microfilaments during Mitosis. Answer: Choice (b) because it is during Metaphase that the effect of kinetochore microtubles attaching to and lining the chromosomes up on the Metaphase plate can be clearly visualized. During Prophase the astral, polar and kinetochore microtubules are all extending and making first contact with their targets but the chromosomes are still randomly distributed. Anaphase would not take place but that would not be the first visible sign of a problem. Telophase is after Anaphase and Cytokinesis is dependent on microfilaments constricting the cell membrane and the positioning of the spindle fiber by the polar microtubules.

1. Which is an example of aneuploidy in a human? a. A male somatic cell has one X chromosome and one Y chromosome. b. A female gamete contains two X chromosomes. c. A female somatic cell contains 46 chromosomes. d. A male gamete has only one Y-chromosome.

This question tests your understanding of what aneuploidy means, which is an abnormal number of chromosomes, but not a multiple of the entire genome (= polyploidy). Answer: Choice (b) because a female gamete would only have one X chromosome. The other choices describe normal genomes with the correct number of chromosomes. Be sure to know how to interpret the different sex chromosome aneuploidies shown in the textbook and discussed in class and that these sex chromosome aneuploidies do not result in the death of the individual.

10. If a cell with 4 pairs of chromosomes, skips Meiosis II while making a total of two gametes, how many total chromosomes and chromatids will each gamete have at the end of this abnormal Meiosis? a. 8 chromosomes and 8 chromatids b. 8 chromosomes and 4 chromatids c. 4 chromosomes and 4 chromatids d. 4 chromosomes and 8 chromatids

This question will require you to think carefully about what happens to the chromosomes and chromatids during Meiosis. Answer: Choice d. During Meiosis I, called the reductive division, the homologous pairs of chromosomes align on the Metaphase I plate and separate into separate daughter cells. Then in Meiosis II, the chromosomes align on the Metaphase II plate and the chromatids separate to opposite cells. So, if you begin with 4 pairs of chromosomes, each pair will align on the Metaphase I plate and the pairs will separate, resulting in each daughter cell after Meiosis I having 4 chromosomes and 8 chromatids (no pairs, but each chromosome has 2 chromatids). Now if Meiosis II did take place (which is doesn't in this question), at the end of Meiosis II, you would still have four chromosomes but four chromatids in each daughter cell, because the chromatids separate in Meiosis II.

When part of a chromosome becomes attached to a non-homologous chromosome, this type of chromosomal mutation is called a: Duplication Deletion Inversion Translocaton

Translocaton

Thymine Dimers are caused by: Chemical mutagens UV irradiation Smoking Spontaneous errors in DNA replication

UV irradiation

Which of the following statements about the cell cycle is correct? a. In actively dividing cells, the S and G2 phases are collectively known as interphase. b.T he phases of the cell cycle are G1, S, and M phases. c. When the S phase of the cell cycle is finished, a cell has twice as many chromatids as the number of chromosomes in the G1 phase. d. During G2 phase, the cell grows and copies its chromosomes in preparation for cell division.

When the S phase of the cell cycle is finished, a cell has twice as many chromatids as the number of chromosomes in the G1 phase.

What does DNA polymerase require to start replicating a bare template strand at the replication fork?

a RNA primer

What is the kinetochore? a protein complex that attaches to the centromere the microtubule organizing center another name for the centromere another name for the mitotic spindle

a protein complex that attaches to the centromere

Mendel's second law, Independent Assortment, has its basis in which of the following events of meiosis I? crossing over separation of homologs at anaphase separation of cells at telophase alignment of tetrads at the equator synapsis of homologous chromosomes

alignment of tetrads at the equator

What ensyme catalyzes the attachment of amino acids to tRNA molecules?

aminoacyl-tRNA synthhetase

The production of second messengers in signal transduction offers at least two advantages, speed and ______

amplification

Consider a diploid species where n=5. If an individual of this species was found to have 11 chromosomes, it would be categorized as triploid. tetraploid monosomic. aneuploid. polyploid.

aneuploid.

Microtubules that connect the centromeres to the inner edge (cortex) of the cell and keep the spindle fibers correctly positioned are called: polar microtubules astral microtubules kinetochore microtubules

astral microtubules

In a diploid cell, the major way that Meiosis II differs from Mitosis is that: a. during Meiosis II, a pair of sister chromatids move to one pole. b. during Meiosis II, sister chromatids align at the metaphase plate as a tetrad, while in Mitosis this doesn't occur. c. at the end of Meiosis II, the cells are haploid, while at the end of Mitosis the cells are diploid. d. during Meiosis II, a synaptonemal complex is formed, while this doesn't occur in Mitosis.

at the end of Meiosis II, the cells are haploid, while at the end of Mitosis the cells are diploid.

The likely outcome from a mutation in the lacO site (lac Operator) of the lac operon would be

binding of a repressor protein could be hindered

The likely outcome from a mutation in the lacO site (lac Operator) of the lac operon would be a. the order in which the genes of the lac operon are transcribed could be altered. b. binding of the activator protein could be hindered. duplication could be affected. c. binding of RNA polymerase could be hindered. d. binding of a repressor protein could be hindered.

binding of a repressor protein could be hindered.

A cross between two pea plants results in 732 purple and 268 white plants. What is (are) the genotype(s) and phenotype(s) of the parents that produced these offspring? a. both parents are heterozygous purple b. one parent is homozygous purple, second parent is homozygous white c. one parent heterozygous purple, second parent homozygous white d. one parent is heterozygous purple, second parent is heterozygous white e. one parent is homozygous purple, second parent is heterozygous white

both parents are heterozygous purp

How is a new DNA strand made in nucleotide excision repair? by crossing over between sister chromatids by DNA polymerase by DNA ligase by DNA exonuclease

by DNA polymerase

What immediate effect does activating GPCR have when binding epinephrine?

causes G-protein to bind to GTP

Liver cells, mammary cells and skin cells all contain the same genome; however, their respective proteomes vary drastically. This observation is best explained by what phenomenon?

cell differentiation

Liver cells, mammary cells, and skin cells all contain the same genome; however, their respective proteomes vary drastically. This observation is best explained by what phenomenon? a. evolution b. cell division c. cell differentiation d. crossing over

cell differentiation

Which of the following statements best describes DNA polymerase? a. It is an enzyme required to glue pices of DNA fragments together. b.it is an enzyme required to produce a primer for DNA replication. c. it is an enzyme that catalyzes the addition of nucleotides to the 5' end of growing DNA strand. d. it is an enzyme that catalyzes the addition of nucleotides to the 3' end of growing DNA strand.

d

The movement of a hormone from the brain to its target receptors in the pancreas is an example of: Autocrine signaling Paracrine signaling Endocrine Signaling Direct intercellular signaling Contact-dependent signaling

endocrine signaling

When epidermal growth factor binds to its _____, the receptor phosphorylates itself, triggering a signal transduction pathway. Ligand-gated ion channel receptor Thermoreceptor G-protein coupled receptor Enzyme-linked receptor

enzyme-linked receptor

The best description of the role of the mediator protein complex in eukaryotes is it a. is composed of a single protein. b. facilitates interactions between RNA polymerase II and regulatory transcription factors. c. controls the rate at which RNA polymerase translates mRNA. d. completely enfolds RNA polymerase and the general transcription factors.

facilitates interactions between RNA polymerase II and regulatory transcription factors.

An organized unit of DNA sequences that enables a segment of DNA to be transcribed into RNA and ultimately results in the formation of a functional product is called a ______.

gene

How does the ncRNA in telomerase function while extending the length of the chromosome? scaffold guide sequence decoy blocker

guide sequence

The chemical, EMS (ethyl methanesulfonate) can cause ____________ mutations. induced spontaneous

induced

Which statement is TRUE regarding the effect of alternative splicing of pre-mRNAs? a. it increases the size of both the genome and the proteome b. it has no effect on the genome, while increasing the size of the proteome c. it has no effect on the proteome, while increasing the size of the genome d. it decreases the size of both the genome and the proteome e. it increases the size of both the genome while decreasing the proteome

it has no effect on the genome, while increasing the size of the proteome

When a neurotransmitter released from one neuron binds to a _______ on a second neuron, that neuron deploarizes and fires.

ligand-gated ion channel

When glucose is high, cAMP is _____________. high low doesn't change

low

Which of the following represents the order of increasingly higher levels of organization of chromatin?

nucleosome, 30-mn chromatin fiber, looped domain

Which is the most common type of DNA repair? base excision repair methyl-directed mismatch repair nucleotide excision repair direct repair indirect repair

nucleotide excision repair

Beginning with the simplest level of structure, whoch order of organization of gentic material is CORRECT?

nucleotide, DNA, gene, chromosome, genome

Bacteria grown in 15N (heavy) medium and then transferred to 14N (light) medium and are allowed to replicate for 2 generations. The DNA is subsequently isolated and centrifuged in a CsCL2 gradient to yield what type of gradient band(s)?

one light and one half-heavy band

How does a mutation in p21 Ras result in an oncogene? p21 Ras normally encodes a G-protein that represses cell division p21 Ras normally encodes a cyclin protein kinase that promotes cell division p21 Ras normally encodes a G-protein that promotes cell division p21 Ras normally encodes the Philadelphia chromosome

p21 Ras normally encodes a G-protein that promotes cell division

How does a mutation in p21 Ras result in an oncogene? p21 Ras normally encodes a G-protein that represses cell division. p21 Ras normally encodes a G-protein that promotes cell division.

p21 Ras normally encodes a G-protein that represses cell division.

When a cell secretes a growth factor that binds to receptors only on neighboring cells causing them to proliferate this is an example of ___.

paracrine signaling

The ensyme that is directly responsible for breaking down cAMP to inactive AMP is

phosphodiesterase

What hormone can induce apoptosis?

prednisolone

An activator is to an enhancer as a(n) _______ is to a silencer.

repressor

In bacteria, RNA polymerase requires a __________ to recognize and bind to the promoter. mRNA nucleus tRNA sigma factor

sigma factor

The ensyme that attaches DNA sequences at the ends of eukaryotic chromosomes in order to prevent chromosome shortening is called ________.

telomerase

How does caffeine affect the epinephrine signaling pathway?

the conversion of cAMP to AMP is blocked

What happens to EGF receptors when they each bind one molecule to EGF?

they dimerize and autophosphorylate

In cats, black fur color is caused by an X-linked allele; the other allele at this locus (gene) causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male? orange females; orange males tortoiseshell females; tortoiseshell males tortoiseshell females; black males black females; orange males orange females; black males

tortoiseshell females; black males

The event that occurs when alloclactose levels increase in E. coli (in the absence of glucose) is

transcription of genes within the lac operaon increases.

To determine if a plant that displays the dominant tall trait is homozygous (TT) or heterozygous (Tt), the tall plant is crossed to a: TT Tt tt

tt