Chapter 27 covers the biochemistry of the transmission of genetic information

The text looks at the detailed biochemical mechanisms underlying the process of genetic information flow.

The techniques used to analyze, construct, and clone DNA were presented in Chapter 6.
Chapter 27 will be studied if you review these chapters.

The transmission of genetic information from parent to child is covered in Chapter 27.

The problems that a cell must overcome to duplicate its duplex DNA are presented in the chapter.
The text expands upon earlier coverage with a more thorough description of the A, B, and Z forms and the underlying chemical determinants of these structural variations.
The text explains how the sequence dependent variation of the structure of DNA provides a basis for its sequence specific recognition.
The roles the template, primer, and metal ion play in their activities are explained in the previous description.
There is a basis for the fidelity of the polymerases.
Next, the helicases that relax DNA are described.
The text gives a description of the topoisomerases that modify their linking numbers.
It describes the fork and initiation of the replication.
There are mechanisms and roles of DNA ligases.

Chapter 27 contains information about DNA polymerase III in replication.

The nature and functions of the telomeres and telomerase are described and the special problems of replication arising from the amount of eukaryotic DNA in a cell are introduced.

To give the new sequence of nucleotides in DNA upon which evolution can act, and also to provide the recombination between two different DNA molecules.

A mechanism for repairing damaged DNA and regulating gene expression can be found in the breakage and joining of fragments of DNA.
Section 5.6.2 of Chapter 5 contains information on plasmids and bacteriophages.
The recombinases that form and resolve it are described.
There is a description of the nature, causes, and consequences.
The chapter ends with examples of pathological deficiencies of DNA repair in humans, the relationship of repair impairments and mutation to carcinogenesis, and a test system for detecting potential carcinogens through their mutagenic action onbacteria.

You should be able to complete the objectives once you have mastered this chapter.

In creating a duplicate copy of a double-strand DNA molecule, there are problems facing a cell.

The challenges of DNA replication can be solved by the use of enzymes and DNA-binding proteins.

The evolutionary relationships among DNA polymerases are appreciated.

List the proteins that interact with the DNA in this region and give their functions and reactions.

Understand the complexity of the machinery.

Evaluate the reactions and identify the genes that carry out the replication.

The mechanisms for repairing damaged DNA are outlined.

The features or characteristics in the right column should be compared with the type of DNA helix in the left column.

The properties or functions in the right column should be compared with the DNA polymerase in the left column.

There is a problem with the replicative machinery of a human cell due to the duplication of the ends of linear, duplex DNA.

Recombinases make and break bonds in a reaction.

In the right column, match the type of mutation with the appropriate one in the left column.

B-DNA has local variations from the average that were observed in DNA fibers.

The hydrogen-bonded base pairs are often twisted and tilted out of the plane that is parallel to the helix axis.

It is possible to recognize specific sequences in DNA without disrupting the helix.

A-T base pairs have two symmetrically related acceptor atoms in the minor grooves and thus have less information than G-C base pairs.
The recognition of DNA may be aided by the interaction of the methyl groups on T. A given local DNA sequence gives rise to a particular shape.
The sequence per se might not be as important as the shape that the sequence specifically recognizes.

The structures of A-DNA are similar to those of double-strand RNA, RNA-DNA hybrid, and some short sequences of double-strand DNA embedded in B-DNA.

Knowledge about the structure of A-DNA contributes to an understanding of other similar helices.

The B-DNA to Z-DNA transition would require the complete unraveling of the right-handed helix to form the left-handed one.

A segment of B-DNA can be converted into Z-DNA by negative supercoiling.

A, b, c, d, e.

The two strands of the helix are quired so that they can be a template for DNA polymerases.

Energy is required to drive helicase action because DNA is so stable.

There are no discontinuities in either strand of the helix.

These are not substrates for DNA ligase because they lack ends.

The answer is not correct because the topoisomer need not be bound by a topoisomerase.

Although all topoisomerases break and reseal bonds, an ex ternal energy source is not always required.

The relaxing of a negatively supercoiled DNA molecule by relaxing it with topoisomerase I is exergonic and requires no energy input.
The specific mechanisms of given topoisomerases determine whether they are coupled.

The lagging strand is only synthesised discontinuously.

The answer is incorrect because the primer is not a template.

Answer (b) is incorrect due to the fact that DNA polymerase III is a highly pro cessive enzyme that synthesises extensively before leaving its template.

A torus is formed by the b2 subunit with the duplex DNA.

The b2 ring is used to hold the replication machinery on the DNA.

The answer is incorrect because the enzyme joins a 3,,-hydroxyl to a 5,,-phos phate.

Answer (d) is correct; although not completely described in the text, the discontinuities in the DNA need to be sealed.

ShortRNA chains are used as a primer to start the replication of the leading strand at the origin of the replication and to start the Okazaki fragments.

Adding a nucleotide to an NTP can be done with relatively low accuracy.

The information in the complimentary strand and the exonucleases in a nick translation reaction allow DNA polymerase I to replace theRNA with the DNA in the beginning of a new chain.

Answer 5 is not a correct match because each fragment is synthesised continuously.

They prevent the gyrase from doing its job by blocking its action.

Human cells don't have anidase like gyrase, so they are unaffected by antibiotics.

Because the two strands of the parental DNA duplex are antiparallel, the removal of the primer from the end of the parental template would leave an overhanging 3.

Ordinary DNA polymerases can't initiate DNA chains.
The portion that could not be copied would shorten the DNA.
Humans have a segment of repeating G-rich DNA at their ends that can be circumvented.
The telomerase, which carries its own RNA template, can extend the uncompleted end at each round of replication.
The repeating telomere sequence is renewed with the help of the RNA template.

As a result of the initial reactions of recombination, the four strands of two interacting duplex DNA molecule become joined to form one molecule.

When two separate duplexes are reformed, the Holliday junction is resolved.

The topoisomerases have reaction mechanisms similar to those of the recombinases.

The bond linking the enzyme to the DNA has a high free energy of hydrolysis and can be re-synthesised.
Recombinases do not need an external source of energy to form a bond.

Transitions are the substitution of a purine for a pyrimidine.

A purine is substitute for a pyrimidine.

The rare enol tautomer of G could base-pair with a T in the template to allow its incor poration into a growing DNA strand.

The resulting daughter DNA duplex would have a G-T base pair.
The T would put an A into its daughter strand during the next round of replication.
The substitution of an A-T base pair for the original G-C base pair was the final result.

Most of the misincorporated nucleotides that do not form a base pair with the template are removed by the nuclease.

There is a second chance to incorporate the correct nucleotide.
There are systems that can detect and repair a mismatch between base pairs.

Damage to one strand of the DNA can be repaired by using the undamaged strand as a template to replace the removed incorrect ones.

A, b, c, d, e are all related to the removal of the 12-nucleotide long oligonucleotide.

It is possible for C to form U in DNA.

U would pair with A changing what was a C-G base pair into an U-A base pair.
The repair machinery of a cell that used U normally in its DNA wouldn't be able to distinguish the U in an A-U base pair arising from a C deamination.
The uracils formed by deamination can be repaired with the help of the methyl group on T.

It is possible that the inability to repair mutagenic lesions in DNA will lead to their accumu lation.

Cancer may be caused by malfunctioning genes regulating cellular proliferation.

These strains can be converted from auxotrophs, which are unable to grow in the absence of histidine, to Prototrophs by changing the DNA in them.

The revertants can grow on media without histidine and are detected with high sensitivity.

These strains are used as an inexpensive initial test of the cancer potential of a compound.

Because animals sometimes convert innocuous compounds to carcinogens, it is possible to mimic what would happen to a chemical in the body with a human extract.
The test's capacity to detect potential human carcinogens has been expanded.

Potential hydrogen-bond donors and acceptors are found in the major and minor grooves of B-DNA.

A region of B-DNA is G-C-rich.

The two a-helical recognition units that bind to DNA are separated by a distance of 34 A.

Thephosphoryl group has been added as an energy source.

Topoisomerase I acts on it.

It is acted on by the two genes.

A template for the synthesis of a strand by DNA polymerase can be found in a single-strand circular DNA with the base composition 30% A, 20% T, 15% C, and 35% G.

The copy-choice model was an attractive mechanism for genetic recombination.

Recombination between two parental DNA duplexes occurs when DNA polymerase jumps from one parental duplex to the other so as to produce a daughter DNA duplex that is derived from the templates of two different DNA duplexes.
The copychoice model is not always used.

Chapter 27 was found under the conditions in which DNA is blocked.

Relate genetic recombination to exon shuffling.

There are more copies of genes B and C than there are of genes D and A.

If a bacterium is found to replicate its genetic material at a very low rate, then that is a good sign.

It is found to have normal levels of activity of the genes.
It makes normal amounts of the wild-types.

A product that base pairs with ade nine can be produced with Hydroxylamine.

Multinucleate cells containing the nucleus of both genotypes can be formed in the presence of Sendai virus.

When fibroblasts from two patients were fused, they showed no deficiency in DNA repair.

Physical studies show that the b2 subunit binding is more tightly to circular than to linear DNA.

At the C-5 position, the Eukaryotic DNA can be highly methylated.

The degree of methylation is related to the expression of genes.
It is known that C-5-methylated cytosines can cause changes in the expression of genes.

Acyclovir is used to reduce the pain and promote the healing of skin wounds caused by chicken pox.

This is due to the fact that all of the herpesviruses have a thymidine kinase gene.

One hydrogen bond donor and two hy drogen bond acceptors are in the major grooves of the G-C and A-T base pairs of B-DNA.

An A-T pair has only two acceptors, whereas a G-C pair has one donor and two acceptors.

Both (b) and (c) have the same sequence of purines and pyrimidines.

Sequences that are rich in GC dinucleotides form Z-DNA more easily than those that are rich in AT dinucleotides.
The latter can be driven into the Z conformation if they are flanked by GC-rich sequences.

34 A is the distance between the major grooves.

The grooves are wide enough to accommodate the recognition helix.

RV can recognize a palindromic DNA sequence.

RV doesn't use helices to contact the DNA.

In the overall reaction,AMP and pyrophosphate are involved.

The only thing that would be labeled was AMP.
The polynucleotide chain does not produce thephosphate involved in the formation of the bond.

The linking number of DNA is increased by 1 each cycle.

The increase was done at the expense of the negative supercoil.

The number of negative supercoils is increased by 2, the linking number is decreased, and the reaction in which both DNA strands are broken.

They only have 1% of the activity of their wild-type counterpart.

They are able to replicate their DNA at normal rates because of the fact that DNA polymerase III is their main source of replication.
The exonuclease, which is used to process the Okazaki fragments, is likely to be replaced by the RNaseH.

If copy-choice were a correct model, no phage should be produced in the absence of new DNA synthesis.

According to that model, there should be no bromouracil and 32P in the duplexes.
Different labels from different DNA molecule could occur in the same molecule.

Exons often contain genes.

The order of exons in a gene can be changed by genetic recombination.
The rearranged genes could give rise to new genes with new domain orders.

From the information given, the order cannot be established.

The low rate of DNA replication can be attributed to a decrease in the activity of primase.

The prior synthesis ofRNA primers is required for the synthesis of DNA.

Slow replication could be caused by decreased rates of dNTP synthesis.

5-bromouracil could be used to induce the mutation in (c).

The sequence AAG, which codes for leucine, could be changed to the sequence AAA, which codes for phenylalanine.
The treatment with 5bromouracil did not cause the other mutations.
The text has a genetic code in it.
The sequence AAG on the informational strand of DNA corresponds to the sequence CUU on the RNA.

Unless otherwise specified, the 5, D3 direction is where the nucleotide sequences are written.

The change of C-G pairs to T-A pairs is caused by Hydroxylamine.

The action of hydroxylamine cannot cause the muta tion in (a).
TTC could change to TTT in (b).
ATG could be converted to ATA.
In (d), the ATC could change to GTC.

The supply of dTTP is sufficient to support the synthesis of DNA at normal rates.

The rate of division of the tumor cells will be affected by the suppression of DNA synthesis.
This type of treatment takes advantage of the fact that tumors divide more quickly than normal cells.
The drug's dosage is adjusted so that it affects more quickly dividing cells.
The division of some cells that are rapidly dividing may be retarded as well.

The fibroblasts from the two patients show complementation, so it is likely that the two patients have different genetic variations.

Several genes are likely to be involved in the repair of damaged DNA.
One patient could have produced a nuclease that excises damaged DNA but could have been deficient in a ligase.
The other patient could have been deficient in nuclease activity.
There are at least nine different complementation groups.

C-5 can deaminate just as cytosine can.

When C-5 deaminates, it forms thymidine.
The product of deamination will not be removed from the DNA because it is not an appropriate base.

There is a free :OH group on which further nucleotides can be linked.

The cells will not have the susceptible enzyme.

The deoxyribonucleoside triphosphates are used in the DNA polymerase I.

The leaving group is joined to the 5,phosphate by DNA-adenylate.
Topoisomerase I uses a DNA-tyrosyl intermediate that is linked to the OH.

FAD, CoA, and NADP are plausible alternatives.

All three of these molecules have a PP linkage.

Positive supercoiling resists the transfer of genetic material.

The melting temperature of DNA goes from negative supercoiled to positive supercoiled.

Positive supercoiling is an adaptation to high temperature.

Skeletal muscle sarcomere shortening is 15-fold faster than the movement of the polymerase at a fork.

The sarcomere shortening is the same speed as the fork movement.

Positive supercoils are caused by the unwind of DNA at the fork.

The action of topoisomerase II overcomes this effect by introducing negative supercoils.
The DNA would become too wound ahead of the fork without topoisomerase II.

Cell di vision requires the synthesis of the ends of new linear chromosomes.

Because cancer cells are dividing rapidly, it is likely that the telomerase gene must be activated for a cell to become a cancer.

The activity will be similar to the replacement of a primer with a new one.

The combined activities of exonuclease and 5,D3 are used.
The free 3,,-OH will be extended by the endonuclease and then taken from the internal 5,,-phosphate to make room for the newly synthesized DNA.
The result is a "nick translation" event in which an unlabeled portion of one DNA strand is replaced with a radioactive stretch of DNA.

Tracks with a low grain density at one end and a high grain density at the other end would be seen.

The middle of a track would have a low density if it were not for replication.

The last base of each codon could be U.

Potentially harmful side reactions are avoided.

If light could not be activated by a pyrimidine dimer, the enzyme might be damaged.
The absorption band is similar to the activity of hexokinase.

Supercoiled DNA can be relaxed by catalyzing the cleavage of a phosphodiester bond in a DNA strand.

The group is attached to thephosphoryl group at the site of scission.

After it has acted on its genetic material, it is required to release DNA gyrase.

Negative supercoiling only requires the binding of ATP.

The supercoiled DNA has a compact shape.

There is a larger circle of gyration in relaxed DNA and it moves more slowly through the gel matrix.

The bands that are next to each other are different by a superhelical turn.

The experimental compound should be classified as either non-mutagenic or slightly mutagenic.

The experimental compound's metabolic product is mutagenic.