∙ The process through which the genetic information of a cell is copied through a series of enzymatic reactions to pass identical DNA to daughter
5’ to 3’ Polarity
∙ refers to the carbon numbers of the sugars ribose and
deoxyribose which contain hydroxyl groups to which
incoming nucleotides are
polymerization, with addition specifically to the 3' hydroxyl group, resulting in increases in length at the 3' end, in
polymerization in the 5' to 3' direction
∙ The configuration of a DNA molecule that is made up of two strands held together by complimentary base pairing. This results in one strand
reading 5’3’ one direction and 5’3’ in the opposite
∙ An enzyme that lays down complimentary RNA
nucleotides to the template strand of DNA known as
∙ Necessary before DNA Polymerase III can lay down new DNA
∙ The enzyme that “unzips” the original DNA strand to start replication and create We also discuss several other topics like What is an example of personal distress?
replication forks and two
Single strand binding protein ∙ These proteins bind to the unpaired, separated DNA
strands and keep them from repairing.
∙ The enzyme that stays ahead of the replication fork
relieving the strain on the yet to be separated DNA because the untwisting of the double helix causes strain.
DNA Polymerase III
∙ The enzyme that begins If you want to learn more check out What is weber's law?
laying down new nucleotides complimentary to the template strand on both the leading and lagging strand after the RNA
primer has been established o alphasubunit: Enzymatic polymerization of DNA in
the 5’3’ direction
Exonuclease that proofreads in a 3’5’ direction
o betasubunit: holds the whole complex onto the
o tausubunit: Tethers the two alpha subunits of DNA
polymerase III to each other DNA Polymerase I
∙ Replaces the RNA primer with DNA after RNase
degrades the RNA primer.
∙ Polymerizes in the 5’3’ direction
∙ An enzyme that connects the fragments of lagging strand DNA If you want to learn more check out What did hominids evolve from?
∙ Lagging strand segments that contain the RNA primer and nucleotides laid down by
DNA Polymerase III
∙ DNA synthesized
continuously 5’3’ following the replication fork (helicase) Lagging Strand
∙ DNA synthesized in segments away from the replication fork also in the 5’3’ direction
∙ An enzyme that works by cleaving nucleotides one at a time from the end (exo) of a polynucleotide chain.
Endonuclease If you want to learn more check out How do plants control viral diseases?
∙ An enzyme that works by cleaving nucleotides one at a time from the
∙ Rare forms of nucleotides that can “confuse” DNA
polymerase when the enzyme selects nucleotides to pair up ∙ Nucleotides have a floating proton that can give the
molecule a different structure
causing an incorrect pairing that could lead to point
mutations due to DNA’s
semiconservative replication properties
∙ Repetitive DNA added to the 3’ ends of chromosomes early in development
∙ End replication problem occurs because lagging
strands would be staggered leaving them uncopied
∙ Telomeres prevent the staggered ends from
activating the cells systems for monitoring damaged
DNA. Don't forget about the age old question of What is the definition of consumer culture?
∙ Also acts as a buffer that provides protection against the organism’s genes
shortening. They do get
shorter after every round of replication eventually leading to somatic cell apoptosis or senescence= no more cell
∙ The enzyme that adds
telomeres to DNA strands.
∙ Set of genes that are related and coordinate functionally for a protein
∙ Includes the switch segment of DNA called an operator, the promoter sequence and the genes they control.
∙ Describing a type of
messenger RNA that can
encode more than one
polypeptide separately within the same RNA molecule.
∙ An RNA molecule that encodes for only one protein Promoter
∙ The DNA sequence where RNA polymerase attaches and initiates transcription.
∙ An enzyme that pries the two strands of DNA apart and
joins together RNA
nucleotides complementary to the DNA template strand.
∙ Assemble a polynucleotide in
the 5’3’ direction.
∙ TATA binding protein that recognizes the promoter
sequence and recruits general transcription factors to bind and assemble
∙ A helicase enzyme that unwinds parent DNA and
polymerase II at the C
∙ Phosphates become binding sites for “hitchhiker” proteins such as Capping Enzymes and Cleavage factors
∙ Also used as a promoter clearance
∙ A nucleotide sequence containing TATA, about 25 nucleotides upstream from the transcriptional start point.
∙ The strand of DNA that is used to make a new strand of RNA during transcription by matching complimentary
nucleotides to it.
∙ The other strand of DNA not being used to create RNA
∙ The carboxyl end of an amino acid where more amino acids are attached to create the
growing polypeptide chain
General Transcription Factor ∙ class of protein transcription factors that bind to specific sites (promoter) on DNA to activate transcription of
genetic information from
DNA to mRNA.
Poly A Signal
∙ Polyadenylation signal sequence in DNA (TTATTT) that specifies a signal
sequence in mRNA
∙ Once this stretch of
nucleotides appears, it is
immediately bound by
cleavage factors to release the mRNA to head to ribosomes
Poly A Tail
∙ At the 3’ end, Poly A
polymerase adds about 200 more adenine nucleotide
∙ Used to facilitate the export of the mRNA from the nucleus, used to protect the mRNA
from degradation by
exonucleases, and they help ribosomes identify and attach to the 5’ end of the mRNA.
∙ A modified form of guanine nucleotide added to the 5’ end after transcription
∙ Done for same reasons as PolyA tail. (Export of
nucleus, prevent degradation, point of attachment for
∙ Regions of mRNA that are expressed by being translated into AAs.
∙ Regions of mRNA that are noncoding segments that are usually between coding
regions and are cut out by
large complex of proteins and snRNA called splicesomes.
∙ A complex of snRNA and proteins that bind to introns to splice them out of mRNA
∙ The formal name of the complex made up of snRNA and binding proteins that
cleave out introns from
∙ Molecule of nucleotides that is transcribed from DNA and translated into AAs via
∙ Transfer RNA used to transfer amino acids from the
cytoplasmic pool to a growing polypeptide in a ribosome.
∙ RNA molecules that, together with proteins, make up
ribosomes; the most abundant type of RNA
∙ Small nuclear RNA that, together with proteins, make up spliceosomes
∙ Micro RNA that regulates the expression of genes by
interacting directly with DNA or w mRNA copy
∙ Capable of binding to
complementary sequences in mRNA molecules.
∙ Allows complex to bind to any mRNA molecule with at least 7 or 8 nucleotides of
∙ miRNA and protein complex either degrades the target
mRNA or blocks its
∙ RNA molecules that function as enzymes
∙ found in the ribosome where they join amino acids together to form protein chains
∙ Parts of the mRNA that will not be translated into protein, but they have other functions, such as ribosome binding.
∙ Codon AUG that signals the proteinsynthesizing
machinery to begin translating the mRNA at that location.
∙ Termination codon that signals the end of a
Aminoacyl tRNA Synthetase (AARS)
∙ Correct matching up of tRNA and amino acid is carrier out by these enzymes.
∙ Active site of each AARS fits only a specific combo of AA and tRNA making there be 20 different synthases for each AA.
∙ Catalyzes the covalent attachment of the AA to
tRNA driven by hydrolysis of ATP
∙ Small subunit
o Binds to both mRNA and a specific initiator tRNA,
which carriers the AA
methionine. Binds the
mRNA at a specific RNA
sequence, just upstream
from the start codon, AUG
o Binds to the 5’ cap of the mRNA and then moves, or scans, downstream along
the mRNA until it reaches
the start codon
∙ Large subunit
o Completes the initiation complex.
o Hydrolysis of GTP
provides the energy for the assembly. Initiator tRNA
is in the P site; the A site is
available to the tRNA with the next AA
∙ Proteins that are required to bring all the translation
∙ Several proteins that are required to add AAs one by one to the previous amino
acid at the Cterminus of the growing polypeptide chain.
∙ Set of genes that are related and coordinate functionally for lactose
o Lac Z: encodes beta
cleaves and begin
metabolism of lactose
o Lac Y: encodes lactose permease which is a
o Lac A: encodes
transacetylase which is still
∙ Positioned within the
promoter, or between
promoter and enzymecoding genes.
∙ Controls the access of RNA polymerase to the genes.
∙ A protein that binds to the operator and blocks
attachment of RNA
polymerase to the promoter preventing transcription of the genes.
∙ Small molecule that
cooperates with a repressor protein to switch an operon off
∙ Small molecule that
inactivates the repressor if the operon is always active and requires a repressor to turn it off.
Catabolite Activator Protein (CAP)
∙ A protein (activator) that binds to DNA and stimulates transcription of a gene.
∙ “senses” the transport of another sugar and binds to site near promoter and interacts with RNA polymerase,
stabilizes it on a weak
o Promoters have specific sequences of nucleotides
and if sequence doesn’t
match ideal sequence of
promoter, it will not recruit
RNA polymerase well
∙ A protein that accumulates when glucose is scarce that allosterically regulates by
binding to CAP protein.
∙ Distal control element which maybe thousands of
nucleotides upstream or
downstream of a gene or even within an intron.
∙ Activator proteins bind to the enhancer area which has three binding sites. DNA bends via bending proteins that bring
activators closer to the
transcription factors, mediator proteins, and RNA
polymerase II are nearby.
∙ Activators bind to certain mediator proteins and general tx factors helping them form an active transcription
initiation complex on the
Regulatory Transcription Factor
∙ Activate or inhibit gene expression by recruiting
chromatin modifying enzymes to genes.
∙ Only a dozen or so nucleotide sequences appear in control elements for different genes. Each enhancer is composed of about ten control elements
each of which can bind only one or two specific tx factors. ∙ The particular combination of control elements in an
enhancer associated with a
agene, rather than the
presence of a single unique control element that is
important in regulating
Histone Acetyltransferase (HAT) ∙ Modify chromatin so its more open and RNA
factors can bind easier and
transcribe a gene.
Histone Deacetylase (HDAC) ∙ Remove functional group from histone tail causing a
highly condensed chromatin state.
∙ histonemodifying enzymes that catalyze the transfer of methyl groups to histone
∙ gene is brought into a closed chromatin state so it may not be expressed such as silencing pancreatic cell from
expressing the sequence for hemoglobin production.
∙ also referred to as closed chromatin or gene poor
because it is inaccessible to the machinery of the cell
responsible for transcribing the genetic information coded in the DNA
∙ true chromatin is loosely packed making its DNA
accessible to machinery so the genes present in euchromatin can be transcribed.
Maternal Effect Gene
∙ these genes specify polarity of
a developing cell into anterior, posterior, dorsal, and ventral. ∙ Expressed by mother during oogenesis (egg development) ∙ Bicoid and Nanos mRNA are localized at opposite poles of the oocyte.
∙ Established by the maternal effect gene
∙ Determine the position and polarity of each segment of a developing organism created by morphogen concentrations that encode for transcription factors
Homeotic (HOX) Gene
∙ Homeobox genes encode for
tx factors with conserved DNA binding domains
∙ Similar sequence amongst many organism and they
control pattern formation in the late embryo, larva, and adult
∙ Gene that is expressed and mRNA is transported from nurse cell to oocyte and
specifically localized to the region in contact with the nurse cells to become the anterior of the organism
∙ Gene that is expressed and mRNA is transported from
nurse cell to oocyte and localized to opposite end to become posterior of organism
1. Where does the energy for the synthesis of nucleic acid molecules come from? How does this necessarily constrain the direction of synthesis of new nucleic acid strands?
∙ The linkage of nucleic acid molecules is a dehydration synthesis reaction. Adjacent nucleotides are joined by a phosphodiester linkage consisting of a phosphate group that links the sugars of two nucleotides. The bond results in a repeating patter of sugarphosphate units called the backbone. The two free ends of the polymer are different. One has a phosphate attached to a 5’ carbon and tother end has a hydroxyl group on the 3’ carbon.
∙ This restricts a nucleic acid polymer to only be able to add more sugarphosphate molecules to the 3’ end of the previous sugarphosphate structure because during hydrolysis, a water molecule is added and the Oxygen molecule leaves the hydroxyl on the 3’ carbon the oxygen on the phosphate group can bind to the carbon
2. How did the MeselsenStahl experiment provide support for the semiconservative model of DNA replication? How would the results have been different if the mode of DNA replication were conservative?
∙ Their experiment eliminated the conservative model for DNA because the first replication produced a band of hybrid DNA. The second replication produced both light and hybrid DNA, a result that refuted the dispersive model and supported the semiconservative model. This proves that unlike a conservative model, daughter cells will contain one old strand from the parent molecule and a newly made strand. A conservative model would have the two parental strands somehow come back together after the process and be seen in its entirety in the daughter cells.
3. What are the specific functions of E. coli DNA Polymerases I and III in DNA Replication? ∙ DNA Polymerase III adds DNA nucleotides to an RNA primer using the parental strand as a template to add complimentary nucleotides.
∙ DNA Polymerase I replaces the RNA nucleotides of the primer after RNase degrades the primer. 4. Define the functions of the alpha, epsilon, beta and tau subunits of E. coli DNA Polymerase III during replication. ∙ Alpha: the enzymatic polymerization of DNA in the 5’3’ direction
∙ Epsilon: exonuclease 3’5’ that acts as a proofreader to ensure nucleotides are complimentarily paired ∙ Beta: The subunit that clamps polymerase to the DNA strands
∙ Tau: Tethers two parental strands and the polymerase to one another.
5. What is an exonuclease? How do 5’ to 3’ and 3’ to 5’ exonucleases differ from one another functionally? How does an endonuclease differ from an exonuclease?
∙ An exonuclease is an enzyme that degrades nucleotides from the one end of the molecule which is different than an endonuclease which degrades nucleotide molecules from with in the molecule.
∙ The difference is on which strand the exonuclease is acting on. DNA polymerase III, for example, proofreads the template strand in the 3’5’ direction but lays new nucleotides in the 5’3’ direction. RNase acts as an exonuclease, deleting RNA primer in the 5’3’ direction on the leading/lagging strands.
6. Define the roles of each of the following in DNA replication: Helicase, DNA Polymerase I, DNA Polymerase III, Primase, Single Strand Binding Protein, Topoisomerase, DNA Ligase, RNAse.
∙ Helicase: unwinds parental double helix at the replication forks
∙ DNA Polymerase I: replaces the RNA nucleotides of the RNA primer with DNA nucleotides.
∙ DNA Polymerase III: adds new DNA nucleotides to the RNA primer that must first be laid down ∙ Single Strand Binding Proteins: Binds to and stabilizes singlestranded DNA until it is used as a template to prevent the just unwound helix from coming back together
∙ Topoisomerase: Relieves overwinding strain ahead of replication forks by breaking, swiveling, and rejoining DNA strands
∙ DNA Ligase: Joins okazaki fragments of lagging strand once the RNA primer is replaced with DNA nucleotides. On the leading strand, it joins the 3’ end of DNA that replaces primer to the rest o the leading strand of DNA ∙ RNase: an exonuclease that degrades the RNA primer before DNA polymerase I can replace the nucleotides 7. What’s the difference between the leading strand and the lagging strand? Why is it necessarily true that every replication bubble will have two leading strands and two lagging strands?
∙ The difference between the leading strand and the lagging strand is that the leading strand is continuously formed once the primer is laid down, DNA polymerase III moves towards the replication fork adding nucleotides. The lagging strand must wait as the double helix is unzipped because nucleotides can only be added on the 3’ end causing the DNA polymerase III to add nucleotides segment by segment in the opposite direction of the replication fork.
∙ There are two of each strand because replication occurs in a bidirectional format from the origin of replication creating two replication forks.
8. Although both catalyze new covalent bond synthesis in DNA, DNA Polymerase requires no ATP for energy, whereas DNA ligase does require the energy of ATP. Explain why this is so.
∙ This is because hydrolysis can drive the connection of new nucleotides to a growing strand by DNA polymerase but to connect a sugarphosphate backbone of a 3’ carbon to a 5’ carbon the the strand “in front” of it requires ATP to create the new bond.
9. Define the end replication problem in eukaryotes, and explain how eukaryotic cells deal with the problem. ∙ Replication machinery cannot complete the 5’ ends of daughter DNA strands because a DNA polymerase can add nucleotides only to a 3’ end of a preexisting polynucleotide. Once a primer is removed from the last fragment there is no 3’ end for new nucleotides to be added so repeated rounds of replication produce shorter and shorter DNA molecules with staggered ends. Eukaryotic chromosomal DNA molecules have special sequences called telomeres at the end that do not contain genes but multiple repetitions of one short nucleotide sequence.
∙ Telomeres prevent the cell from activating mechanisms to handle damaged DNA such as cell cycle arrest or apoptosis. Also the telomeric DNA acts as a buffer so the DNA does not shorten too fast. An enzyme called telomerase catalyzes the lengthening of telomeres in eukaryotic germ cells, thus restoring their original length and compensating for the shortening that occurs during DNA replication.
10. What is the Hayflick limit? How is it related to telomeres, cellular aging and senescence? ∙ The Hayflick limit is the number of times a normal human cell population will divide until cell division stops ∙ This is related to telomeres, cellular ageing, and senescence in that telomeric DNA tends to be shorter in dividing
somatic cells of older individuals and in cultured cells that have divided many times. It has been proposed that shortening of telomeres is somehow connected to the ageing process of certain tissues and even to aging of the organism as a whole. Once a telomere is too short the cell will be signaled for apoptosis or senescence which is no more cell division leading to aging tissues/aging organism.
11. What is meant by the terms deamination and depurination? How does the cell “deal with” each of these problems? ∙ Deamination is the removal of an amine group from a molecule. Enzymes that catalyze this reaction are called deaminases. In DNA, deamination of cytosine is corrected by the removal of uracil. This site is then recognized by endonucleases that break a phosphodiester bond in the DNA, permitting the repair of the resulting lesion by replacement with another cytosine. A DNA Polymerase may perform this replacement by its 5'>3' exonuclease activity, followed by a fillin reaction by its polymerase activity. DNA ligase then forms a phosphodiester bond to seal the resulting nicked duplex product, which now includes a new, correct cytosine. This is done is similar fashion for all nucleotides
∙ Depurination occurs when a purine (A or G) is released from a sequence. The cell can repair this with an endonuclease that cuts a segment of DNA and then DNA polymerase rewrites the sequence to insert the missing base pair. 12. What is an intercalating agent? How do intercalating agents lead to mutation and DNA damage? ∙ Such as ethidium bromide and proflavine, are molecules that may insert between bases in DNA, causing frameshift mutation during replication. Frameshift mutation can lead to cancer cell growth or cause the cell to create proteins in incorrect order once transcribed to RNA and translated by ribosomes
13. How does exposure to ultraviolet radiation lead to DNA damage and potentially to mutations? ∙ UV is a physical mutagen known as mutagenic radiation which can cause disruptive thymine dimers in DNA. This new structure no longer hydrogenbonds to adenine, and creates a noticeable kink that will signal a base mismatch. If left to its own devices, DNA polymerase will “idle” at this site. The other side can be replicated, leaving a gap at the dimer. This would be lethal in the next round of replication, because it would produce a doublestranded break.
14. Define the four essential steps that excision repair pathways in both prokaryotic and eukaryotic cells have in common. ∙ Teams of enzymes detect and repair damaged DNA, such as a thyamine dimer (caused by UV radiation) which can distort a DNA molecule. A nuclease enzyme cuts the damaged DNA strand at two points, and the damaged section is removed. Repair, synthesis by a DNA polymerase fills in the missing nucleotides. DNA ligase seals the free end of the new DNA to the old DNA, making the strand complete.
15. Why is cancer the logical consequence of damage to DNA repair mechanisms like the UV Repair pathway? ∙ Skin cancer can be caused by an inherited defect in a nucleotide excision repair enzyme because this can cause mutations to go unnoticed by the cell. The formation of thyamine dimers like those in UV damage caused DNA to buckle and interfere with DNA replication and if these go uncorrected, skin cancer can result.
16. How do some chemotherapeutic strategies exploit defective DNA repair mechanisms in cancer cells? ∙ Some chemotherapeutic strategies target the DNA repair and replication mechanisms in cancerous cells to prevent them from repairing DNA that has mutations and continuing to replicate it once repaired. Many cancerous cells do not have a replication limit and they will continue to divide and move mutated DNA to new cells spreading cancer. 17. Compare the enzyme responsible for transcription to the enzyme responsible for initiating each new nucleic acid strand in replication and the enzyme that does the bulk of DNA replication. How are the enzymatic activities similar? How are they distinct?
∙ RNA Polymerase II and DNA Polymerase III are similar in that they both read template DNA in the 5’3’ direction. ∙ The enzymes are different in that RNA Pol can initiate new nucleic acids whereas DNA Pol must have an RNA primer laid down first. RNA Pol also requires general transcription factors to initiate the transcription of mRNA. 18. What is the relationship between the sequence of the transcribed RNA molecule and the template and nontemplate strand of the gene?
∙ The mRNA molecule that is transcribed comes from an RNA polymerase II enzyme reading the template strand and producing a complimentary strand. The nontemplate strand is inside the RNA polymerase enzyme, however, it is not read to create an mRNA strand.
19. During transcription, where does the energy for the synthesis of the RNA molecule come from? How does this necessarily constrain the direction of synthesis of the transcribed RNA? How is this related to the synthesis of DNA during replication?
∙ This is the same as the construction of a new DNA molecule during replication in that nucleotides are added to the 3’ end of the RNA molecule via hydrolysis reactions. Only sugarphosphate backbone molecules can be attached at the 3’ carbon of the last molecule.
20. What is meant by coding versus noncoding RNA? What classes of RNA fall into each category and what are their major functions?
∙ Coding RNA refers to triplet nucleotide sequences that will code for a protein. Noncoding RNA consist of nucleotide sequences that will not be translated into proteins but have other important functions.
∙ mRNA: (coding) strand that is read by ribosome known as “messenger” RNA. Molecule of nucleotides that is transcribed from DNA and translated into AAs via ribosomes.
∙ tRNA (noncoding) Transfer RNA used to transfer amino acids from the cytoplasmic pool to a growing polypeptide in a ribosome.
∙ rRNA (noncoding) RNA molecules that, together with proteins, make up ribosomes; the most abundant type of RNA ∙ snRNA (noncoding) Small nuclear RNA that, together with proteins, make up spliceosomes ∙ miRNA (noncoding) Micro RNA that regulates the expression of genes by interacting directly with DNA or w mRNA
copy. Capable of binding to complementary sequences in mRNA molecules. Allows complex to bind to any mRNA molecule with at least 7 or 8 nucleotides of complementary sequence. miRNA and protein complex either degrades the target mRNA or blocks its translation
21. What is meant by the term ribozyme? What classes of RNA are categorized as such and what are their major functions?
∙ Ribozymes are RNA molecules that function as enzymes. Intron RNA functions as a ribozyme and catalyzes its own excision. (rRNA) RNA is singlestranded, a region of an RNA molecule may basepair, in an antiparallel arrangement, with a complementary region elsewhere in the same molecule; this gives the molecule a particular 3D structure. Specific structure is essential to the catalytic function of ribozymes, just as it is for enzymatic proteins. ∙ Some of the bases in RNA contain functional groups that can participate in catalysis.
∙ Acid molecules add specificity to its catalytic activity
22. What distinguishes eukaryotic RNA Polymerase II from Pol I, III, IV and V?
∙ RNA polymerase I synthesizes a prerRNA which matures and will form the major RNA sections of the ribosome. ∙ RNA polymerase II synthesizes precursors premRNA. It controls transcription for the most part and a range of
transcription factors are required for its binding to promoters.
∙ RNA polymerase III synthesizes tRNAs, rRNA and small RNAs found in the nucleus and cytosol. ∙ RNA polymerase IV synthesizes siRNA in plants.
∙ RNA polymerase V synthesizes RNAs involved in siRNAdirected heterochromatin formation in plants. 23. Why are general transcription factors required for transcription by Pol II? What are the major functions of the two general transcription factors we discussedTFIID(TBP) and TFIIH?
∙ Transcription factors mediate the binding of RNA polymerase and the initiation of transcription. Only after tx factors are attached to the promoter does RNA polymerase II bind to it.
∙ TFIID/TBP: TATA binding protein that recognizes the promoter sequence and recruits general transcription factors to bind and assemble
∙ TFIIH: A helicase enzyme that unwinds parent DNA and phosphorylates RNA polymerase II at the Cterminal domain. Phosphates become binding sites for “hitchhiker” proteins such as Capping Enzymes and Cleavage factors. Also used as a promoter clearance
24. In general terms, how do the functions of general transcription factors like TFIID and regulatory transcription factors differ?
∙ General tx factors are essential for transcription of all proteincoding genes. A few general tx factors bind to a DNA sequence such as the TATA box with in the promoter but most bind to proteins, including other tx factors and RNA pol II.
∙ Regulatory tx factors determine the rate of gene expression. Tx factors bind to activation domains that bind other regulatory proteins and tx machinery, facilitating a series of proteinprotein interactions that result in enhanced transcription off a given gene. Some tx factors act as repressors and can inhibit gene expression in several ways (directly binding to control element of DNA, blocking activator binding or interfering with the activator itself so it can’t bind to the DNA)
25. What is a promoter? What is a TATA box and why is it called that?
∙ A promoter is the DNA sequence where RNA Pol II binds to begin transcription. The TATA box is a crucial sequence of DNA that is part of the promoter sequence where RNA polymerase binds. It is called this because the sequence is just that TATA.
26. What is the polyA signal sequence? What role does this sequence play in terminating transcription of a eukaryotic gene? What is its relationship to the polyA tail of eukaryotic mRNAs?
∙ The polyA signal sequence refers to the sequence TTATTT on DNA that is seen at the 5’ end of the template strand that is transcribed in the 3’5’ direction. When transcribed into mRNA it reads AAUAAA and is the sequence recognized by cleavage facts to end the transcription to create the premRNA strand. The polyA tail is added to the 3’ end of the newly transcribed mRNA by polyA polymerase to increase efficiency of translation and to slow degradation 27. What is the 5’ cap on eukaryotic mRNA? What is its functional significance?
∙ The 5’ cap on mRNA is a methylated Guanine nucleotide that is added immediately after tx begins. The cap protects the mRNA from exonucleases and is the initiation of translation for the ribosome to bind.
28. What are exons and introns? What molecular machine is responsible for recognizing these sequences and controlling their processing in eukaryotic mRNAs?
∙ Exons are regions of mRNA that are expressed by being translated into AAs and Introns are regions of mRNA that are noncoding segments that are usually between coding regions and are cut out by large complex of proteins and snRNA called spliceosomes.
∙ Transcription factors that bind to a DNA sequence, other transcription factors, and RNA polymerase II facilitate the correct positioning of the DNAenzymetx factor complex on the promoter sequence of DNA and initiate transcription (RNA synthesis). The interactions of enhancers also affect the rate of gene expression by binding to activators or repressor proteins that in turn interact with the transcription initiation complex. It is the combinatorial code of all these things that regulate transcription.
29. What advantage does alternative splicing confer to eukaryotic cells?
∙ This is one of several opportunities for regulating gene expression. Different mRNA molecules are produced form the same primary transcript depending on which RNA segments are treated as exons and which as introns. Regulatory proteins specific to a cell type control intronexon choice by binding to regulatory sequences within the primary transcript.
∙ This expands the repertoire of a eukaryotic genome. The extent of alternative splicing greatly multiplies the number of possible human proteins, which may be better correlated with the complexity of the organism.
30. What is the relationship between the exon and a “functional domain” on a protein? How is this related to the concept of protein evolution over time?
∙ Different exons code for the different domains of a protein including the functional domains of the active sites and cell
membrane binding sites. The presence of introns in a gene may facilitate the evolution of new and potentially beneficial proteins as a result of a process known as exon shuffling. Introns increase the probability of crossing over between the exons of alleles of a gene simply by providing more terrain for crossover without interrupting coding sequences.
∙ This might result in new combos of exons and proteins with altered structure and function. The occasional mixing and matching of exons between different genes could occur and this may also lead to new proteins with novel combinations of functions.
31. What are the four fundamental properties of the genetic code?
∙ Genetic code is written in triplets: 3 nucleotides in mRNA known as codons specify 1 amino acid. ∙ Codons are redundant so that each amino acid (except Met and Trp) encoded by more than 1 codon. ∙ Codons are unambiguous; each codon specifies only 1 amino acid
∙ “Universal” coding for all organism that use essentially the same codes for amino acids
32. What are the two “business ends” of a tRNA molecule? What is their functional significance in translation? ∙ The two business ends include the Amino acid attachment site which uses the AARS to attach an amino acid to the tRNA and the Anticodon end which has the complimentary nucleotides to the mRNA being translated. 33. How is it that the process of linking tRNAs to their cognate amino acids is a relatively high fidelity process, given the similarity in structure between some amino acids, for example leucine and isoleucine?
∙ The active site of each type of AARS fits only a specific combination of amino acid and tRNA. There are 20 different synthetases, one for each amino acid. Each synthetase is able to bind to all the different tRNAs that code for its particular amino acid. The synthetase catalyzes the covalent attachment of the amino acid to its tRNA in a process driven by the hydrolysis of ATP.
34. Describe the structure of the eukaryotic (80S) ribosome, what role is played by the large and small subunits of this molecular complex?
∙ Small subunit Binds to both mRNA and a specific initiator tRNA, which carriers the AA methionine. Binds the mRNA at a specific RNA sequence, just upstream from the start codon, AUG. It binds to the 5’ cap of the mRNA and then moves, or scans, downstream along the mRNA until it reaches the start codon
∙ Large subunit Completes the initiation complex. Hydrolysis of GTP provides the energy for the assembly. Initiator tRNA is in the P site; the A site is available to the tRNA with the next AA
35. What do the “E”, “P” and “A” stand for in describing the three tRNA binding sites within the large ribosomal subunit? Describe the cycling of tRNAs through these three sites during the process of translation. What is happening to the mRNA molecule as this is occurring?
∙ The P site (peptidyltRNA binding site) holds the tRNA carrying the growing polypeptide chain while the A site (aminoacyltRNA binding site) holds the tRNA carrying the next amino acid to be added onto the chain. Discharged tRNAs leave the ribosome from the E site (exit site). The positions of the ribosome hold the tRNA and mRNA in a close proximity and positions the new amino acid so that it can be added to the carboxyl end of the growing polypeptide. It then catalyzes the formation of the peptide bond.
36. What is the general role of initiation factors in translation? What roles are played by elongation factors? What role is played by release factors?
∙ Initiation factors are required to bring all the components of translation together. The cell also expends energy obtained by hydrolysis of a GTP molecule to form the initiation complex.
∙ Elongation factors are required by the addition of amino acids to the growing polypeptide chain. The mRNA is moved through the ribosome in one direction (same as saying ribosome moves 5’3’ down RNA). Codon recognition requires hydrolysis of one molecule of GTP, which increases the accuracy and efficiency. One more GTP is hydrolyzed to provide energy for the translocation step.
∙ Release factors bind directly to the stop codon in the A site. The release factor causes the addition of a water molecule instead of an amino acid to the polypeptide chain. This reaction breaks the bond between the completed polypeptide and the tRNA in the P site releasing the chain through the exit tunnel of the ribosomes large subunit.
37. How does a given mRNA direct protein synthesis by either free ribosomes in the cytoplasm, or bound ribosomes at the rough ER?
∙ Polypeptide synthesis begins on a free ribosome in the cytosol. A signalrecognition particle binds to the signal peptide which targets the protein to the ER. The SRP binds to a receptor protein complex that forms a pore and has a signalcleaving enzyme. The SRP leaves and polypeptide synthesis resumes, with simultaneous translocation across the membrane. The signal cleaving enzyme cuts off the signal peptide. The rest of the completed polypeptide leaves the ribosome and folds into tits final conformation.
38. What is meant by the terms silent, missense, nonsense and frameshift mutation? How can a base change in a protein coding gene that causes a single amino acid substitution have either no effect, minimal effect or devastating effect on the protein function?
∙ Point mutations are changes in a single nucleotide causing the following mutations:
o Silent mutation is a change in a nucleotide pair that may transform one codon into another that is translated into the same amino acid and there is no observable effect on the phenotype.
o Missense mutations are substitutions that change one amino acid to another. This mutation may have little effect on the protein because the new amino acid has similar properties as the one it replaced or it may be in a region of the protein where the exact sequence of amino acids is not essential to the proteins function. This may be detrimental if the new
amino acid is substantially different than the one it replaced or if the region in the protein complex is sensitive to changes.
o Nonsense mutations occur when a point mutation to a nucleotide changes an amino acid to a stop codon and it causes translation to be terminated prematurely; the resulting polypeptide will be shorter than the polypeptide encoded by the normal gene. These usually lead to nonfunctional proteins.
∙ Insertion and deletions are additions or losses of nucleotide pairs in a gene. These have devastating effects on the resulting proteins and the organism as a whole.
o Frameshift mutations are the alterations of the reading frame of the genetic message, the triplet grouping of nucleotides on the mRNA that is read during translation. All nucleotides downstream of the deletion or insertion will be improperly grouped into codons and will result in extensive missense usually ending sooner or later in nonsense and premature termination.
39. How could a mutation within an intron lead to disruption of protein function? What percentage of human diseases are thought to be the result of this type of mutation?
∙ A mutation in an intron may disrupt protein function because a mutation in an intron would cause a different site for RNA splicing. This leads to different outcomes of exons which ultimately serve as the coding regions of mRNA. 40. What is the molecular explanation for the diauxic growth curve of a culture of E. coli grown in the presence of glucose and an alternate sugar like lactose?
∙ The biphasic bacterial growth in the presence of glucose and alternate sugars like lactose is due to the preferred sugar by E. coli. A shift then occurs in the expression in genes to metabolize the next sugar. Bacteria utilize the glucose first because more energy from the monosaccharide than having to isomerize and split the disaccharide lactose.
41. What are the functions of the three structural genes of the lac operon of E. coli? Which, if any, of the three genes are indispensable for normal lactose metabolism?
∙ Lac Z: encodes the lactose betagalactosidase which cleaves and begins metabolism of lactose. ∙ Lac Y: encodes for lactose permease which is a membrane transport protein
∙ Lac A: (dispensible) encodes galactosidase transacetylase enzyme which is not understood completely
42. Distinguish between the promoter, operator and CAP binding sites of the lac operon. Where are they located relative to the structural genes and to one another? Which of these are considered “regulatory elements”? What are they regulated by? Why is the I gene expressed independently of the lac operon structural genes?
∙ Promoter directs RNA polymerase to bind to gene and begin transcription located upstream next to operator and upstream from the +1 location.
∙ Operator is the binding site for the regulatory protein that negatively regulates transcription known as a repressor. It is located next to +1 and the lac Z gene.
∙ CAP Site is where the regulatory protein CAP binds which is located next to and upstream from the promoter. CAP protein interacts with RNA polymerase II to stabilize promoter because the promoter is “weak” without it. ∙ The CAP site is considered regulatory because when cAMP binds to the CAP protein it is because glucose is not present. This occurs because glucose will block adenylyl cyclase from catalyzing cAMP.
∙ The I gene is expressed independently because it encodes for a repressor regulatory protein. This repressor will bind to the operator site when no lactose is present because there is no allolactose to allosterically inhibit the repressor from binding.
43. Why is the lactose operon referred to as an inducible operon? What is the molecular mechanism of induction? How does this mechanism effectively allow the E. coli genome to “sense” when lactose is present?
∙ The lac operon is referred to as inducible because it is usually off but can be stimulated when a specific small molecule interacts with a regulatory protein. The regulatory protein is the, usually inactive, repressor protein that gets activated in the presence of allolactose (lactose).
44. What is the molecular mechanism that leads to higher levels of lac operon transcription when glucose is absent? How does this mechanism effectively allow the E. coli genome to “sense” when glucose is present? ∙ When glucose is not present cAMP concentrations can increase because its catalysis is not inhibited by the presence of glucose. The cAMP allosterically activates CAP which can bind to the CAP site on the operon. When CAP binds it increases the transcription rate of RNA polymerase II. It acts as a sensor for the presence of glucose because if glucose
where present, concentrations of cAMP would be lower and CAP would be less likely to be activated. 45. Generally speaking, what “types” of eukaryotic genes are regulated and what types would you expect to be expressed constitutively over the life of the cell? Why?
∙ Constitutive genes are not regulated and may code for proteins that are necessary for normal cell function such as cell membrane proteins, organelles, etc. Facultative genes are regulated as needed such as genes signaling rapid cell replication as seen during embryonic development.
46. What functional distinction can be made between general transcription factors like TFIID and gene specific transcriptional activators and repressors?
∙ General transcription factors are used by both eukaryotes and prokaryotes; regulatory transcription factors are used by eukaryotes. General tx factors like TFIID are required for transcription to first occur, whereas, gene specific tx factors and regulatory tx factors alter the expression/repression of genes.
47. What is meant by the term “enhance”? How is it that such elements can carry out their function in a distance, location and orientation independent fashion?
∙ Enhancers may be thousands of nucleotides upstream or downstream of agene or even within an intron. Activator proteins bind to the enhancer groups in the DNA. A DNA ending protein brings the activatorenhancer complexes closer to the promoter. General tx factors, mediator proteins, and RNA pol II are now closer. The activators bind to certain mediator proteins and general tx factors helping them form a transcription initiation complex. 48. What is meant by the “combinatorial code” model of transcriptional regulation?
∙ The correct combination of regulatory transcription factors must bind to enhancer elements to achieve transcription. Each combo of control elements will be able to activate tx only when the appropriate activator proteins are present, which may occur at a precise time during development or in a particular cell type.
49. How is coordinate control of functionally related genes accomplished in eukaryotes? How is this distinct from coordinate regulation of operons in prokaryotic organisms?
∙ In prokaryotes, genes are often clustered into an operon, which is regulated by a single promoter and transcribed into a single mRNA molecule. This is different then eukaryotes, in that, coordinate gene expression depends on the association of a specific combination of control elements with every gene of a dispersed group. Activator proteins in the nucleus that recognize the control elements bind to them, promoting simultaneous transcription of the genes, no matter where they are in the genome.
∙ Coordinate control of dispersed genes in a eukaryotic cell often occurs in a response to chemical signals from outside the cell. Steroid hormones, for example, enter a cell and bind to a specific intracellular receptor protein forming a complex that serves a tx activator.
50. What are the two general mechanisms by which eukaryotic transcriptional activators and repressors influence the rate of transcription?
∙ DNA Methylation occurs usually to the cytosine nucleotide. The gene is brought into closed chromatin state which is often not expressed. Protein chromatin remodeler is recruited to alter the expression of genes (silencing). Removal of the extra methyl groups can turn on some of these genes. Once methylated, genes usually stay that way through successive cell divisions in a given individual.
∙ Histone Acetylation appears to promote transcription by opening up the chromatin structure. The Nterminus of each histone molecule in a nucleosome protrudes outward from the nucleosome. These tails are susceptible to alterations by the addition/subtraction of acetyl groups.
51. Define the term chromatin. What does it mean when we say that chromatin has been “remodeled”? ∙ Chromatin is the complex of DNA and proteins that makes up eukaryotic chromosomes. When the cell is not dividing, chromatin exists in its dispersed form, as a mass of very long, thin fibers.
∙ Chromatin remodelers such as ncRNAs change the spacing and level of coiling to create modification. Chromatin that is more open can express genes more easily. Chromatin that is more tightly packed are considered gene poor. 52. What is meant by the terms heterochromatin and euchromatin? In what chromosomal regions is heterochromatin characteristically found?
∙ Heterochromatin is a more condensed form of chromatin structure. This is usually found in the centromere or telomere regions of DNA.
∙ Euchromatin refers to a more open chromatin structure that can express genes more easily.
53. What functional distinction can be made between DNA methylation and histone modification, in terms of the type of regulation each confers to a gene?
∙ DNA methylation occurs on certain bases like cytosine of DNA instead of histone tails. Unlike histone modification, methylation causes genes to usually stay that way through successive cell divisions. Histone modification occurs via enzymes that add or remove methyl, acetyl and phosphate groups to the exposed tails of histones. This type of regulation is reversible. Adding acetyl promotes transcription by opening up the chromatin structure while adding methyl usually condenses structure.
54. What is the molecular mechanism by which a transcription factor induces remodeling of chromatin in the region of a specific gene?
∙ Regulatory tx factors recruit chromatin modifying enzymes to genes. Nterminal domains tails get modified via covalent functional group binding or removal. The addition/removal of phosphate, methyl, or acetyl groups changes chromatin structure. Acetylation is associated with more open chromatin (less condensed means more accessible to transcription machinery). More transcription as a result creates more gene expression.
55. What are HATs and HDACs? What role do they play in regulating gene expression?
∙ HATs (Histone Acetyltransferase) modify chromatin so its more open and RNA polymerase/transcription factors can bind easier and transcribe genes.
∙ HDACs (Histone Deactlyase) remove functional groups from histone tail causing a highly condensed chromatin state. ∙ Regulatory tx factors that bind to enhancers can recruit HATs and HDACs to gene region.
56. Generally speaking, what types of genes might be subject to irreversible versus reversible gene regulation over the life of the cell? Describe one molecular mechanism by which cells commonly irreversibly silence genes. ∙ Genes that are related to embryo development would most likely be irreversibly regulated to be turned off such as rapid replication or genes that are specialized.. Genes that can be reversibly regulated include those regulate hormone production etc.
57. What are maternal effect genes? What developmental role is played by the maternal effect genes bicoid and nanos in the immediate postfertilization period in drosophila?
∙ these genes specify polarity of a developing cell into anterior, posterior, dorsal, and ventral. Expressed by mother during oogenesis (egg development). Bicoid and Nanos mRNA are localized at opposite poles of the oocyte and they establish the posterior and anterior structure cells in drosophila.
58. What is the function of maternal effect genes like bicoid and nanos? Why is it accurate to say that their expression in the earliest stages of embryonic development determines and ultimately controls events that happen much later, for example, correct limb formation?
∙ Maternal effect genes in nurse cell nucleus diffuse across the organisms’ egg and establish at the anterior and posterior regions. High concentrations of Bicoid create the nuclei of cells that will populate the anterior region and ultimately create the structures such as head. Nanos will create the cells farther from the nurse cells to make posterior structrures such as back legs and tail.
59. What would happen to an embryo that had a loss of function mutation in the bicoid gene? Could an embryo ever be affected by a loss of function mutation in the bicoid gene? If so, how? If not, why not?
∙ If there was a loss of function mutation to the Bicoid it would cause the development of two anterior tails on the embryo and as it began to develop improperly it would eventually die.
60. Why is it accurate to say that in drosophila the establishment of the basic body axes (anterior/posterior, dorsal/ventral) depends on regulated gene expression in both the mother and the embryo?
∙ Cytoplasmic determinants that are localized in the unfertilized egg provide positional information for the placement of anteriorposterior and dorsalventral exes even before fertilization. These determinants are coded by genes in the mother called maternal effect genes. Egg polarity genes which are the Bicoid and Nanos set up the posterioranterior axes of the embryo.
61. What do segmentation genes encode? What is their developmental relationship to the maternal effect genes? What functional role do segmentation genes play in drosophila development?
∙ Maternal effect genes establish downstream segmentation genes which determine the next phase of embryo development for drosophila. They encode for the segmentation and determine position and polarity of each segment to a more specific structure of the embryo and will eventually, downstream, influence homeotic genes.
62. What do homeotic genes encode? What is their developmental relationship to segmentation genes? What functional role do homeotic genes play in drosophila development?
∙ Homeotic genes encode for specific body structures and are influenced by segmentation genes. Every segment is associated with turning on effector genes (homeotic) and encode a family of tx factors that specify body parts. ∙ Homeobox (HOX genes) encode tx factors with condensed DNA binding domains and gene duplication events are the reason HOX gene family has expanded so much.
63. How do the effects of a homeotic gene mutation differ from the effects of a mutation in the maternal effect or segmentation genes?
∙ Mutation of homeotic genes can cause abnormalities in the structures once they are established but may not be lethal. Mutations in maternal effect or segmentation genes are referred to as embryonic lethals because as the embryo develops, it will not be able to survive past a certain point because necessary events will not occur properly.
64. Researchers in the field of molecular genetics and the regulation of development sometimes refer to drosophila as “little people with wings”. In what sense(s) are flies at all similar to people?
∙ This is a comparison of genetic material that can relate to many eukaryotic organisms. Both humans and flies (along with many other organisms) contain similar machinery to construction of an embryo with maternal effect, segmentation, and homeotic genes. Studying these across organisms shows that the genes are similar and their influence are similar. The difference between humans and flies is that humans have 4 sets of HOX genes which is why mutations in some of these do not show extreme abnormalities like legs growing where our antennae should be.