There is only one strand of the DNA that can be used as a template.
The duplex DNA has a G + C composition.
The cell's genes are contained in theRNA of the cell.
Answer (f) is correct because it is necessary for the plate to function.
The promoter and terminator sequence are required to specify the precise start and stop points for the transcription.
The sequence will be written in the 3' di rection.
The a-phosphate of the ribonucleoside triphosphate is selected by base pair to the template strand of the duplex DNA.
The reaction is made by the use of the RNA polymerase.
A ribonucleoside monophosphate is added to the chain and it has grown in the 3' direction.
The strands are antiparallel, that is, they are assembled in the 3' 5' direction with respect to the polarity of the template strand of the DNA.
Answer (d) is incorrect because the interaction of the two genes takes place during translation.
The minimum that can serve as a codon is three contiguous nucleotides.
There are four different types of nucleotides.
Only 16 possible combinations can be made with a codon consisting of only two nucleotides.
It's not enough to specify all 20 of the amino acids.
A codon with three nucleotides allows 64 combinations, more than enough to specify the 20 amino acids.
There is a sequence of the polypeptide.
The fourth codon of the transcript is UAA, which is a translation codon, and the reading frame is set at the 5' end of the transcript.
The answer is correct because both AGU and AGC specify serine.
Answer (b) is correct because the codon on the transcript from AGU could be changed by the change of a single nucleotide in the DNA.
The anticodon ACU would base-pair with the codon AGU.
Mitochondria can use a genetic code that is different from the standard code because they have a distinct set of tRNAs that are matched to the genetic code used in their mRNAs.
The location and nature of the substitution of the amino acid for each of the identified genes could be deduced from the sequence of the wild-type and the mutant genes.
The result would be that the order of the genes on the map is the same as the order of the polypeptides produced by them.
There were two stretches of extra nucleotides between exons.
There are intron sequences in the DNA that don't have the same sequence in the mRNA.
The shuffling of exons allows them to interact in new ways while preserving the functional units.
exon shuffling takes place at the DNA level through breakage and rejoining of DNA notRNA.
There are 5243 base pairs in the genomes of mammals.
One can observe an increase in the absorbance of ultraviolet light when a solution containing intact DNA is heated.
A decrease in absorbance is observed when the solution is cooled slowly.
heating causes a similar effect if breaks are made in the sugar-phosphate backbones.
The reduction in absorbance is slower when the solution is cooled than it is when it is intact.
There are a number of factors that affect the behavior of a linear, double-strand DNA molecule.
Explain each of the observations considering this.
You have a double-strand linear DNA molecule, the appropriate primers, all the enzymes required for DNA replication, four 32P-labeled deoxyribonucleoside triphosphates, and the means to detect newly synthesized radioactive DNA.
The deoxyribonucleases do not hydrolyze base-paired DNA.
Formaldehyde forms hydroxymethyl derivatives.
If you have a solution with separated strands of DNA, what should it look like?
When double-strand DNA is placed in a solution containing tritiated water, hydrogens associated with the bases readily exchange with protons.
While many experiments were suggesting that the chromosomes are very long and continuous, it was found that the primer chain has to have a DNA template in order for it to function.
The mole percentage is G + C.
Early studies of the denaturation of double-strand DNA did not know if the strands separated from each other.
If you have two strands of double- strand DNA, one strand is labeled with 14N and the other is labeled with 15N.
The N-1 and N-7 of adenine, the N-7 of guanine, the N-3 of cytosine, and the O-4 of thymine are all protonsated under strongly acidic conditions.
Predict the effects of low pH on double-stranded DNA.
Some types of single- strandRNA can be used to form double- strand molecules.
There are many cells that can make deoxyuridine 5'-triphosphate.
The DNA of bacteriophage l is a linear double-strand molecule.
When two "cohesive" ends on the same molecule join, they can form a closed circular molecule.
When denatured and allowed to reassociate under certain conditions, single-strand circles can be formed.
Resolution is not enough to see the ends of the molecule.
A circle formed by adenoviruses.
There are volcanic sulfur springs with a pH 2 and tempera tures as high as 85oC.
Incubation with a circular DNA template at 100oC can extend a 20-nucleotide primer by more than 100 nucleotides.
The concentration of the enzyme-to-primer must be at least 3:1.
TdT can be used to extend a DNA primer by 5' 3' using deoxyri Bonucleoside triphosphates.
The primer must have a free 3'-OH end.
Theidase does not need a template or copy one.
The 2',3'-dideoxynucleoside can be used as a reagent.
To have a measurable effect on DNA synthesis, these analogs must be converted to dideoxynucleoside triphosphates.
A single dideoxyribonucleoside can effectively block subsequent chain extension when incorporated into a growing DNA chain.
In each chain-elongation reaction, a phosphodiester bond is formed and pyrophosphate is released.
Most cells have a potent pyrophosphorylase, which makes it easy tolysis of pyrophosphate.
One of the products of the chain-elongation reaction is partially responsible for the forward progress of the polymerization.
If the double-strand helix is allowed to form, isolated DNA polymerases can efficiently carry out chain extension.
In his studies of DNA in the late 1940s, Chargaff found that all organisms have the same number of bases.
Considering that the ability to form hydrogen bonds with adenine is the same as the ability to form single-strandedRNA from tobacco mosaic virus, do you expect similar constraints on base composition to be found in the following?
Some DNA endonucleases degrade double-strand DNA to yield mononucleotides and dinucleotides, but they don't degrade the duplex sequence to which other proteins are tightly bound.
The codon in the mRNA sequence for the enzyme is GUU, which is the same as the valine in the primary sequence.
If the codon to GCU is changed, the activity of the enzyme is unaffected, but if the codon to GAU is changed, the activity of the enzyme is completely inactivated.
Explain the observations.
To remove introns precisely, it's important to remove them between the first and second nucleotides of an exon.
The sequence at the normal junction in a pre-spliced mRNA between an intron and an exon is...
The viral genome is the most important factor in determining the amount of the proteins synthesized by a cell after it has been exposed to T2 bacteriophage.
Some of the genes overlap, that's one reason for the increased capacity.
The coding sequence for genes B and A is located in the same sequence.
The two proteins specified by these genes are completely different.
There is no nuclease capability in RNA polymerase.
Tell us why the two enzymes are different.
The genome of G4 is a small circle of DNA.
There is a small segment ofRNA.
An RNA molecule consisting of 350 nucleotides can be found in the transcript of the gene that codes for tyrosine tRNA.
A 41-base segment on the 5' side of the tRNA sequence and a 224-base segment on the 3' side of the tRNA sequence are removed by at least three ribonuclease enzymes.
The primary transcript's tRNA sequence is continuous, and no nucleotides are removed from that part of the transcript during processing.
This primary transcript has a sequence on one side of the rRNA sequence and another on the other side.
There is a transcript with all three rRNA sequences.
The codons UAA, UAG, and UGA are not read by the tRNA molecule.
These codons are found at the end of the coding sequence.
Premature termination of the protein chain can be caused by single-base mutations in certain codons.
A codon can change to UCG.
There is a change in the anticodon of the tRNA molecule that allows it to read a terminated codon.
Suppressive tRNAs suppress the effect of a chain-termination change.
The normal coding sequence has a UAG codon in it.
Each case will have a single base change.
In the laboratory, poly ribonucleotides were used to determine the genetic code.
Tell us how the cell uses this enzyme.
density gradient centrifugation can be used to separate the two strands of the same DNA.
The stabilbity of a particu lar a helix that is buried in the myoglobin is being studied.
Section 6.4 of the text describes site-specific mutagenesis, a technique used to replace leucine in the helix.
Explain your predictions.
There is a compound called cordycepin that can block the synthesis of RNA because it doesn't have the 3'-OH end needed for chain extension.
The structure of cordycepin is shown.
H O H (a) Cordycepin does not affect the growth ofbacteria, but it does affect the growth and division of mammals.
Consider the reactions that are required for cordycepin to be converted into a substrate for RNA polymerase and then propose a reason for its ineffectiveness inbacteria.
Raney nickel can be used to convert cysteinyl-tRNACys to alanyl-tRNACys.
This experiment tells you about the ability of the machinery to recognize alanyl-tRNACys.
2'- and 3'-monophosphates are included in the products of the cleavage of RNA.
The flow of genetic information is subject to regulation.
The production of macromolecules limits the amount of energy the cell can use.
It is possible to achieve the greatest economy in energy expenditure by a mature cell by regulating the steps in storage and transmission of genetic information.
David Baltimore was looking into the activity of the polymerase in the leukemia virus.
The virus causes leukemia in mice.
He created a mixture with either the four dNTPs or the four NTPs in a buffered solution after disrupting the virus particles.
Radiolabeled was one of the dNTPs.
After allowing time for a reaction to occur, the mixture was treated with strong acid to make nucleic acids and leave unreacted triphosphates in solution.
He was able to detect the formation of the product of a polymerase by measuring the precipitated radioactivity.
CHAPTER 5 product was not incorporated into it, dNTPs were incorporated into it, and the isolated radiolabeled product was destroyed by RNase.
Radioisotopes can be used to identify specific molecule involved in biochemi cal processes.
The medium is 4 or 35SO4.
The Bacteriophage T2 has a genome.
After infecting the cells with the two different virus preparations, they put the culture of the cells in a blender to remove any leftover virus.
They collected the stripped cells and compared the amount of radioisotope in them to the amount left in the supernatant.
An increase in the absorbance of light is observed when the intact double-strand circular DNA molecule is heated in solution.
The circles are so intertwined that they remain closely associated.
When the molecule is cooled, the interlocked strands move relative to each other until their bases are aligned.
The molecule absorbs less light than the pair of strands.
During denaturation, breaks in one or both strands of a double-strand DNA molecule allow the two strands to separate from one another.
In order to form a doublestrand molecule, the separate strands collide randomly until at least a small number of correct base pairs is formed.
A difference in the reduction of absorbance will be observed if a pair of interlocked circles are compared to a pair of separate strands in solution.
The longer the DNA molecule, the more thermal energy is required to disrupt it.
Experiments show that there is a relationship between the length of a molecule and the number of base pairs.
It is easier to separate the two strands when the concentration of NaCl decreases.
Once a short stretch of base pairs is formed, reassociation to form the longer double-strand molecule occurs quickly.
The higher the concentration of DNA, the quicker the solution will find and pair with each other.
The tendency of bases to stack contributes to the stability of the helix.
Base stacking allows the sugar-phosphate chain to be located on the outside of the helix, where it can besolvated, and it also reduces the contact of the relatively insoluble bases with water.
The helix may be destabilized by the presence of urea, which may allow bases to associate more with water.
The diameter of the helix is not likely to be a factor in determining a specific sequence.
The grooves of the intact helix contain hydrogen-bond donors and acceptors.
A hydrogen bond can be formed between a group of atoms in one of the grooves of the helix.
The edges of the bases can contribute to the specificity of the interaction between the side chains.
Although the system described could yield 32P-labeled daughter DNA molecule, CHEMICAL methods can't distinguish between strands of the same molecule in which one strand is labeled and the other is unlabeled.
In their experiments, Meselson and Stahl used a physical technique to separate the labeled molecules according to their content of 14N and 15N, which differ in their specific densities.
In solution, the oligodeoxyribonucleotide forms an interchain double-strand molecule with flush ends and a small single-strand loop.
A small double-strand linear molecule remnant containing seven base pairs is left after the deoxyribonuclease hydrolyzes the phophodiester bonds.
Formaldehyde could react with the exocyclic groups on the carbon of adenine, the C-2 of guanine, and the C-4 of cytosine to form hydroxymethyl derivatives.
Because these derivatives can't form hydrogen bonds with bases, single strands would reassociate to a lesser extent.
The ring nitrogen atoms in pyrimidines may be present at the actual sites of the reaction of formaldehyde with DNA.
The experiment suggests that the hydrogen bonds of base-paired regions of double strand DNA may become bubbles.
The tritiated water allows the exchange of protons.
A continuous, linear double-strand DNA molecule has only two 3'-OH groups available for the initiation of DNA synthesis by DNA polymerase; because each is located at opposite ends of the molecule, no template sequence is available.
In order to create a relatively simple mechanism for chromosomal replication, one could postulate that each of the breaks offered the 3'-OH group required for the initiation of the new DNA strand.
The strand opposite the break would be the location of the template required for replication.
It has been established that the DNA in the chromosomes is long and continuous.
The fact that there are no breaks in the molecule makes the mechanism of replication complexample.
See page 760 for details.
Most organisms live at temperatures that are considerably lower than 65oC.
The integrity of the double-strand DNA molecule is important for the transmission and expression of genetic information.
Determine the temperature at which the hydrogen bonds are disrupted and single strands are formed.
You can measure the extent of hyperchromicity by heating the double-strand DNA to different temperatures.
If you want to separate the 14N-labeled strands from the 15N-labeled strands, you have to use the density-gradient equilibrium sedimentation technique.
It would suggest that the strands separate during thermal denaturation if you are successful.
Normal hydrogen bonding is impossible because the atoms can no longer serve as hydrogen-bond acceptors.
Double-strand DNA is less stable at neutral pH values than it is at low pH ones.
Deprotonation of other ring atoms denatured DNA at high pH values.
A hybrid mole cule can be formed by the association of a molecule ofRNA with a molecule of DNA.
The formation of hydrogen bonds between bases will allow the creation of a double helix.
The structure and hydrogen-bonding properties of uracil are very similar to those of thymine and can be used as a base for DNA poly merase.
uracil and adenine pair with each other when incorporated into a double-strand DNA polymer.
The reasons uracil is not normally incorporated into DNA can be discussed on page 771 of the text.
Interchain joining to form multimers is more likely at higher concentrations.
The most reasonable model for the structure of the l phage DNA molecule is a dou ble-strand molecule with single-strand protrusions at the 5'-ends, as illustrated in the margin.
The 3'-ends have groups that allow them to be used as a primer for DNA synthesis.
The molecule with flush ends is created by the filling of the single-strand regions of the molecule.
The required circular molecule can no longer be formed with cohesive ends.
Molecules treated with exonuclease III may not be infective because they have a longer single-strand sequence at both ends, which means that the ends could no longer be joined.
When the exonuclease-treated DNA is treated with DNA polymerase I, the single-strand regions are filled in to reform a molecule that has protruding single strands that are approximately the same length as those in the native molecule.
The molecule becomes infective again.
I will once again produce a molecule with flush ends that is no longer infective after further treatment of the molecule.
There are a pair of inverted repeats on each strand.
Each strand would need a primer.
The association between primer and the enzyme may protect it from thermal denaturation.
There are many examples where the binding of the Substrate is stable.
You might expect a higher G + C composition from a thermophile.
It's not unusual for the base ratios of the DNA to be the same as those found in the hot springs where they grow.
The thermophiles must be protected from thermal denaturation with the help of the bacterial cells.
TdT does not use a template and cannot copy from one, so that the base composition of the newly synthesized single strand of DNA will depend on the relative concentrations of the deoxynucleoside triphosphate substrates.
The base composition of the chains will be similar to the template strand.
TdT does not need an exonuclease activity that removes bases from newly synthesized strands.
TdT can be used to introduce sequence variation.
The 2',3'-dideoxy analogs must be converted to nucleoside triphosphates in order to be used as a base for DNA polymerase.
In studies on the inhibition of DNA synthesis in living cells, cells with the neutral forms of the analogs are used instead of the negatively charged triphosphate forms.
Once inside the cell, nucleoside analogs are phosphorylated by cellular enzymes that normally function to "salvage" nucleosides.
The error-correcting exonuclease activity of some DNA polymerases is blocked by the lack of a 3'-OH group.
The reaction is driven by noncovalent forces.
There are hydrogen bonds between A and T bases in the antiparallel chains.
Stacking interactions between adjacent bases on the same strand may contribute to helix formation and stability.
The noncovalent forces are likely to account for the forward progress of the chain.
The base ratios will not conform to Chargaff's established rules because a single-strand polynucleotide might form hydrogen bonds between bases.
This is also true for single-stranded DNA.
The number of uracil and adenines would be the same as those of cytosine, and the number of guanine bases would be the same as well.
Adding the DNA endonuclease will degrade the DNA that is not protected.
The size of the protected fragments of DNA and the base sequence can be determined using methods discussed in Chapter 6 of the text.
When this process is performed, they will allow the polymerase molecule to move from the promoter site to the template where transcription begins, as well as beyond.
The promoter site was no longer protected from endonuclease degradation.
Alterations in the structure of the enzyme do not affect the activity of the enzyme.
The substitution of aspartate for valine is specified in the GAU codon.
The side-chain carboxyl group of aspartate has a negative charge at neutral pH.
The charged group could disrupt the native structure of the enzyme.
The exon coding sequence will be altered because of the loss of two bases if the spliceosome cleaves the initial transcript.
Instead of beginning with the codon GCU in the normal exon, the reading frame will begin with the codon UAA.
This codon is a signal for the end of the translation of the polypeptide specified by the messenger RNA.
The synthesis of viral-directed proteins can't begin until T2 messenger RNA is made.
The ribonucleoside triphosphates synthesized by the bacterial cell must be used for the Transcription of the T2 DNA.
If each of the coding sequence is read in a different frame, the weight of the molecule can be 201,000.
The location of the AUG initiation signal is critical to the establishment of the proper reading frame.
Both reading frames 1 and 2 are established when the first AUG codon is binding to the second AUG codon.
The two different genes specified by the two different genes will have different sequence of the same genes.
The genetic information that is passed on to progeny cells comes from the duplication of the DNA of the chromosomes.
Errors that occur in the copying of a DNA template will be transmitted to the duplicated chromosome and all the messenger RNA molecule that was transcribed from it.
Errors that occur due to the inclusion of mismatched nucleotides can be corrected by DNA polymerase.
Many copies of mRNA are made, but they have relatively brief lives in the cell, and very few are passed on to offspring cells.
It appears that the cell can tolerate errors in transcription if not many occur.
A primer nucleotide and a template are required for the synthesis of a DNA chain.
A primer can be served by either an oligodeoxyribonucleotide or an oligoribonucleotide.
When replication has extended around the circle and the 5' end of the primer is reached, the primer is hydrolyzed and the small gap is filled in with a DNA sequence.
In some organisms, the pre-tRNA transcript is removed in order to make tRNA.
Only one promoter is required for rRNA synthesis.
The removal of a few amino acids from the C-terminal end of many, but not all, does not affect their normal function.
The premature end of the chain would be suppressed by each tRNA molecule's placement of an amino acid at the corresponding site.
Along with their codons, the amino acids that could be found at the site are GAG, Trp, Leu, AAG, Ser, and Tyr.
When a cell contains a suppressor tRNA, it may not be possible to end it with a single stop codon.
Most cells are tolerant of suppression, even if the extension could be lethal.
One explanation is that the stop codon may be seen by other proteins that are involved in chain termination.
A full understanding of the reasons for toleration of suppression is needed.
The equilibrium for the reaction is in the direction of the degradation of the RNA.
It is likely that the concentrations of ribonucleoside diphosphates in the cell are not enough to drive net polynucleotide synthesis.
The polyribonucleotides it synthesises contain random sequences, which makes them useless for protein synthesis.
The nucleases that regulate the lifetimes of the genes are used by the cell.
The lifetimes of mRNAs inbacteria are short.
Only one of the two strands at a given location along the DNA was being transcribed into DNA if less than all the DNA could form a hybrid.
In rare cases, theRNA is made from both strands of the template DNA.
The results show that only one of the two strands of the SP8 virus is transcribed.
In most other organisms, different regions of each strand are used for different things, but in the case of the SP8 virus, only one strand is used.
The conversion of cordycepin to the triphosphate form is carried out by a number of kinases.
It's likely thatbacteria can't phosphorylate cordycepin efficiently, which makes them less susceptible to inhibition of RNA and DNA synthesis.
You will learn later that cordycepin might affect the synthesis of DNA.