Chapter 2- DNA and RNA: Composition and Structure

Question 1

Pneumococcus experiment

Answer 1

The first evidence that DNA is the genetic material was obtained in 1920 through an experiment involving two strains (R and S forms) of pneumococcus (a bacteria that causes pneumonia). The R strain forms rough colonies, and the S strain forms smooth colonies. S strains are more virulent and cause pneumonia, while R strains aren’t virulent. Treatment of the R-strain (non virulent) with DNA extracted from S-strain (virulent) resulted in their transformation into the S-form. This demonstrated that DNA was a transforming agent, as the traits were inherited from the S strain. It also showed that DNA was responsible for the transmission of information from one generation to the next.

Question 2

How much DNA does a human cell contain?

Answer 2

It contains enough information for the synthesis of about 25,000 proteins.

Question 3

Traditional vaccines

Answer 3

Traditional procedure for making vaccines include attenuated (inactive) intact cells or dead organisms that will have body produce antibodies, which will provide immunity. Examples include successful vaccines that provided protection against diseases such as polio, smallpox, whooping cough, typhoid and diphtheria. However, in addition to the growing problem of antibiotic resistance, it is difficult to make vaccines for some pathogens like HIV and malaria via traditional methods, this prompted the use of alternate methods to make vaccines. A malaria vaccine was recently approved using protein fragments from the falciparum parasite.

Question 4

How could DNA vaccines be made?

Answer 4

DNA vaccines would typically be a bacterial plasmid engineered to include the sequence of an antigenic protein from the pathogen. DNA can enter various cells and replicate followed by transcription and translation, acting similarly to viruses. These DNA contain a limited amount of genetic information and cannot become infectious. The mechanism of uptake and induction is still unclear. Promising results have been obtained against viruses and bacteria and may help fight HIV/AIDS, TB and malaria. Could be used for cancer in the future

Question 5

Antigenic

Answer 5

Refers to the mechanism by which an infectious agent changes the proteins or carbohydrates on its surface. It helps the pathogen to avoid the host immune response

Question 6

Pros of DNA and RNA vaccines (3)

Answer 6

1. Cost effective
2. Ability to be developed more quickly than traditional vaccines, which rely on actual inactivated viruses and can take years to develop.
3. Can be made more readily available because they rely on genetic code – not a live virus or bacteria. This also makes the vaccines cheaper- only requires a plasmid of the DNA sequence to produce a protein

Question 7

Protein vaccines

Answer 7

Protein vaccines have been used for decades- protect people from hepatitis, shingles, and other viral infections. The vaccine contains the protein or protein fragments from the bacteria or virus. Protein based vaccines are under currently investigation for COVID

Question 8

Advantages of mRNA vaccines

Answer 8

If you know what protein you want to produce in the body, it is easy to synthesize corresponding mRNA. Proteins can be difficult molecules to use because they are large molecules, in contrast to DNA and RNA, which are smaller. Proteins also have more complicated structures than DNA and RNA do, and proteins have some stability issues. Also mRNA vaccines can be safer, and mRNA vaccines can be more readily translated into proteins. mRNA cannot be integrated into host DNA

Question 9

Cons of DNA vaccines

Answer 9

With a DNA vaccine, there is always a risk it can cause a permanent change to the cell’s natural DNA sequence. However, there are ways to minimize risk. mRNA cannot get integrated into the DNA, and is readily translated into protein. mRNA vaccines are safer and mRNA can be more readily translated into proteins

Question 10

Hybridization probe

Answer 10

Probes are single stranded DNA or RNA sequences that are complementary to the specific sequences we are interested in. The probe is tagged for easy detection of the hybrid double helix. Based on the association of complementary polynucleotide strands with the probe, this method has been developed for the detection and quantitation of specific sequences of target nucleic acid.

Question 11

Use of hybridization probes (4)

Answer 11

1. Determines whether a certain sequence occurs on the DNA of a particular organism
2. Determines genetic or evolutionary relatedness between different organisms
3. Determines the number of genes transcribed in a particular mRNA
4. Determines the location of any given DNA sequence

Question 12

Hybridization Experiments

Answer 12

A mixture of denatured DNAs is treated with a DNA probe bearing a label. The probe can then hybridize with those DNAs with complementary sequences, and the unhybridized probes can be washed away. Probe labels can be radioactive to help with detection. Detection of the double-helical complexes allows for detection and quantitation of DNA that contains the sequence of interest.

Question 13

DNA Arrays

Answer 13

New methods are being developed to monitor gene expression and to analyze genes rapidly for mutations. The arrays consist of a number of gene-specific DNA probes immobilized at specific sites on a chip. Chips can then be treated with labeled target DNA or RNA derived from cells of an organism. Arrays can contain thousands of DNA probes. Hybridization of the targets with complementary probe sequences allows for immobilization of the label at specific sites on the chip.
Rapid screening of disease associated mutations is an advantage of this technique. They profile gene expression by determining the amount of mRNAs, which may further assist in detection of cancer and individualized treatments.

Question 14

Uses of DNA arrays (5)

Answer 14

1. To detect mutations leading to ataxia telangiectasia,
2. To detect recurrent respiratory infections
3. To detect dilated blood vessels in the skin and eyes
4. To detect mutations in the hereditary breast and ovarian cancer gene BRCA
5. To identify pathogens present in the sample.

Question 15

Conformations of Double helical DNA

Answer 15

A-, B-, and Z-DNA are different DNA conformations that are associated mainly with variation in the conformation of the nucleotide constituents of DNA.
Depending on the conditions and base sequence, the double helix can acquire various geometries

Question 16

Noncanonical DNA structures

Answer 16

Formed when DNA interacts with certain proteins. This structure of DNA is not straight, it bends and forms unusual structures such as cruciforms or triple-stranded arrangements as it interacts with proteins. These variations in DNA conformation are an important recurring theme in the process of molecular recognition of DNA by proteins and enzymes. Variations in DNA structure or conformation are favored by specific DNA sequence motifs such as inverted repeats, mirror repeats, and direct repeats.

Question 17

Bent DNA

Answer 17

Noncanonical DNA structure. DNA sequences with runs of 4 to 6 adenines separated by 10 base-pair produce bent conformations. This structure is important for DNA replication and transcription. Bending variations in DNA structure are associated with different DNA sequences or motifs. DNA damage may also lead to bending, which helps the cell to recognize it needs to initiate repair

Question 18

DNA bending purpose

Answer 18

A fundamental element in the interaction between DNA sequences and proteins that catalyzes replication and transcription. Bending also occurs because of photochemical damage or mispairing of bases and serves as a recognition signal for initiation of DNA repair.

Question 19

Cisplatin

Answer 19

Tetracoordinate platinum complex, which is used as a chemotherapy drug in ovarian, bone, testicular and lung cancers. It forms inter and intrastrand cross-links in double stranded DNA and the adduct (bond) itself represents 90% of DNA lesions or damaged sites of DNA. The bonds are formed from displacement of chloride ligands on platinum. Structural studies of intrastrand cross-linked DNA adducts show the double helix is bent

Question 20

How are cisplatin-DNA adducts recognized in the cell?

Answer 20

Bent structures of the cisplatin-DNA adduct recognized by (1) several DNA-binding proteins that helps in repair (nucleotide excision repair - NER proteins), and (2) nonhistone DNA binding proteins - high mobility group (HMG) proteins, which regulate transcription

Question 21

How does cisplatin cause cell death?

Answer 21

Transcription and apoptosis are affected by cisplatin-DNA adducts. Nucleotide excision repair proteins are recruited but excision repair is prone to produce DNA strand breaks and accumulation of these breaks will ultimately induce apoptosis as the DNA becomes too damaged to function.

Question 22

High mobility group (HMG) domains and cisplatin

Answer 22

These are nonhistone DNA binding proteins, and are considered high mobility as they exhibit high mobility on gel electrophoresis. These proteins also help to regulate transcription. In binding to DNA, the HMG domain prefers single-stranded or bent double-stranded structures. HMG-domain proteins may affect the antitumor properties of cisplatin by repairing cisplatin DNA adducts

Question 23

Cruciform DNA

Answer 23

Cross shaped DNA, which is another noncanonical DNA structure. Disruption of hydrogen bonds between the complementary strands and formation of intra- strand hydrogen bonds within the region of an inverted repeat produce a cruciform structure.
The loops generated by cruciform formation require the unpairing of 3 to 4 bases at the end of the “hairpin.” Bases base pair with other complementary bases to form a branch of the cross

Question 24

Function of cruciform DNA

Answer 24

Cruciform structures at origins of DNA replication in mammalian cells have been shown to recruit cruciform binding proteins that function during the initiation of DNA synthesis. Depending on the sequence, these structures may be only slightly destabilizing because residues in the loop can remain stacked at the end of the helix.

Question 25

Triple-stranded DNA

Answer 25

Noncanonical DNA structure- some polynucleotides such as poly(dA) and poly(dT) combine to form triple-stranded complexes rather than the expected double helices. Intramolecular triple helices can be formed by disruption of double-helical DNA with polypurine sequences in mirror repeats.

Question 26

Mirror repeats

Answer 26

A mirror repeat is a region such as AGGGGA that has the same base sequence when read in either direction from a central point. Refolding generates a triple-stranded region and a single-stranded loop in a structure called H-DNA. These repeats may contribute to the development of triple-stranded DNA

Question 27

Function of triple-stranded DNA

Answer 27

Many sequences in eukaryotic genomes have the potential to form triple-stranded DNA structures. These potential triple helical regions may constitute as much as 0.5% of some eukaryotic genomes and are especially common near sequences involved in gene regulation. Because of this, it has been proposed that H-DNA may play a role in the control of RNA synthesis and including possible roles in initiation and termination of replication and recombination. The ability to interfere with transcription have also led to efforts to use intermolecular triple helices to control RNA and protein synthesis.

Question 28

Hereditary Persistence of Fetal Hemoglobin (HPFH) pathophysiology

Answer 28

A group of conditions in which fetal hemoglobin synthesis is not terminated at birth but continues into adulthood. Under normal conditions, fetal hemoglobin synthesis would be terminated at birth. It is characterized by changes in RBC similar to those found in the genetic blood disorder β-thalassemia and by delayed production of adult hemoglobin. The condition results from failure to stop transcription of human γ-globin genes, leading to elevated levels of fetal hemoglobin.

Question 29

Fetal hemoglobin

Answer 29

Involved in transporting oxygen from the mother’s bloodstream to the organs and tissues of the fetus. Produced at 6 weeks of pregnancy, levels remain high after birth for 2-4 months until adult hemoglobin is produced. γ-globin genes are expressed in the fetal liver and make fetal hemoglobin

Question 30

Hereditary Persistence of Fetal Hemoglobin (HPFH) symptoms

Answer 30

Associated with mild clinical or hematologic abnormalities. Mild musculoskeletal pains may occur infrequently, but patients with HPFH are often asymptomatic.

Question 31

Hereditary Persistence of Fetal Hemoglobin (HPFH) genetics

Answer 31

The Condition results from failure to stop transcription of human γ-globin genes, leading to elevated levels of fetal hemoglobin. The formation of an intramolecular DNA triple-helical structure located about 200 bp upstream from the initiation site for transcription of the globin genes acts as a brake for their expression. Hemoglobin genes of patients contain mutations in one or more positions in this region, decreasing the stability of the triple helix and reducing its ability to inhibit the protein synthesis.

Question 32

Four-stranded DNA

Answer 32

Four-stranded DNA guanine nucleotides and highly G-rich polynucleotides form novel tetrameric structures called G-quartets that contain a planar array of guanines connected by hydrogen bonds. Polynucleotides can interact to form tetraplexes where G-quartets stack on each other to form a multilayered structure. The ends of eukaryotic chromosomes (telomeres) contain repetitive G-rich sequences and contain 800–2400 copies of the hexameric repeat sequence d(TTAGGG)n. Oligonucleotides with this sequence can form tetraplex structures.

Question 33

Four stranded DNA implications

Answer 33

Telomeres are attracting attention as targets for new anticancer drugs and G-tetraplexes have been implicated in recombination of immunoglobulin genes and in dimerization of double-stranded RNA of the HIV.

Question 34

Telomeres

Answer 34

The ends of linear eukaryotic chromosomes, which are critical for maintaining the stability of the genome.

Question 35

How is telomerase linked to cancer?

Answer 35

Telomerase activity present in most tumor cell lines may be responsible for their immortalization and when increased, correlates to poorer clinical prognosis. Two approaches are being examined for selective inhibition of telomerase.

Question 36

Telomerase as a Target for anticancer agents- 2 approaches

Answer 36

1. The first involves targeting of the RNA-containing portion of the enzyme. The RNA portion is used as a template to extend the repeat sequences of the telomere. Nucleic acids with chemically modified sugar phosphate backbones bind to telomerase RNA in immortal human cells, inhibit activity, and ultimately cause cell death or apoptosis. 2. Second approach involves drugs that bind to G-quadruplex DNA, such as large aromatic molecules like porphyrins and anthraquinones. The drugs selectively bind and stabilize G-quadruplexes DNA structure

Question 37

Slipped DNA (SMP-DNA)

Answer 37

DNA regions with direct repeat symmetry can form slipped, mis-paired DNA. Formation involves unwinding of the double helix, realignment, and subsequent pairing of one copy of the direct repeat with an adjacent copy on the other strand. Can form single stranded loops, which are a single strand of nucleotides. Although SMP-DNA has not yet been identified in vivo, genetic evidence suggests that it is involved in spontaneous frameshift mutations that result in base addition or deletion. Deletions and duplications of DNA segments that are longer than a single base can occur during DNA replication between direct repeats, causing slipped-looped structures.

Question 38

DNA Triple Repeats and Human Diseases

Answer 38

The fragile X-syndrome, myotonic dystrophy, X-linked spinal and Kennedy’s disease, Friedrich ataxia, and Huntington disease are the genetic diseases involving presence of repeated three-base pair DNA sequences. Associated with expansion of nucleotide triplet repeats.

Question 39

Kennedy’s disease

Answer 39

Caused by a CAG repeat in the first exon of the androgen receptor gene. It is a neurological disease leading to muscular atrophy, muscle weakness, and poor muscle coordination. Motor neurons are lost

Question 40

Friedrich ataxia

Answer 40

Caused by a GAA sequence found within an intron. Symptoms- difficulty walking, impaired speech, brain damage

Question 41

Myotonic dystrophy

Answer 41

Caused by a CTG sequence found in the 3’-untranslated region. Symptoms- muscle weakness

Question 42

Fragile X-syndrome

Answer 42

Characterized by expansion of GCC triplet of the FMR-1 gene, expanding from 30 to >2000 copies. Develops when normal expression of the FMR-1 gene is halted/stops. Causes intellectual disabilities

Question 43

How do DNA triple repeats cause disease?

Answer 43

Overall, the expansion of the triplet interferes with normal functioning of the related protein. Loss of protein function but sometimes gain of deleterious function may occur, and diseases can increase in severity with each successive generation. Triplet expansion may result from slipped mispairing during DNA synthesis. Because of massive amplification that characterizes these diseases, repeated or multiple slippages would have to be involved. One possible mechanism - slippage of nascent DNA during lagging strand synthesis, which may be aided by formation of a stable hairpin structure by a slipped loop.

Question 44

Topoisomerases

Answer 44

Regulate the formation of superhelices. These enzymes catalyze the concerted breakage and rejoining of DNA strands, producing a DNA that is more or less superhelical than the original. The precise regulation of the cellular level of DNA super-helicity is important to facilitate protein interactions with DNA.

Question 45

Mechanism of action of topoisomerase I

Answer 45

If DNA is overwound, TP1 nicks one strand of DNA. The intact strand passes through the break and the ends of broken strands are ligated

Question 46

Topoisomerase as Treatment

Answer 46

Topoisomerases are important targets of antimicrobial and antineoplastic agents. Trap complex between topoisomerase and DNA resulting in degradation of DNA, introduction of mutations, or inhibition of translation and replication - leading to apoptosis. Anti-cancer durg Camptothecin and its derivatives act on topoisomerase 1, while other drugs target topoisomerase 2

Question 47

Drugs targeting topoisomerase 1

Answer 47

Anti-cancer drug Camptothecin and its derivatives act on topoisomerase 1. The activity of Camptothecin and its derivatives may be improved by increased levels of topoisomerase I in tumor cells (colon cancer).

Question 48

Which drugs act on topoisomerase 2

Answer 48

1. Amsacrine and etoposide 2. Anthracycline (Adriamycin and doxorubicin), synthetic intercalators, ellipticines, and podophyllotoxins 3. Ciprofloxacin

Question 49

Amsacrine and etoposide mechanism

Answer 49

Act on topoisomerase II. These drugs kill cells by stabilizing covalent topoisomerase II-DNA cleavage complexes, which accumulates leading to permanent DNA breaks.

Question 50

Which cancers are treating by targeting topoisomerase 2?

Answer 50

Lymphoid and nonlymphoid leukemias, high-grade non-Hodgkin lymphomas, and Hodgkin disease are treated mostly with combinations of topoisomerase II inhibitors with or without additional cytotoxic agents. Leads to DNA breaks and apoptosis

Question 51

Ciprofloxacin

Answer 51

Topoisomerase 2 inhibitor that gained considerable attention for its use in the treatment of inhalation anthrax. It is representative of the quinolones, a class of powerful broad-spectrum antibacterial activities. Cell death occurs due to accumulation of lethal double stranded breaks in the bacterial DNA. Ciprofloxacin shows good activity against Gram-positive and Gram-negative bacteria. Newer drugs show even broader activity, for example, Gemifloxacin is useful for treating respiratory infections including those by multidrug resistant Staphylococcus pneumoniae, mycoplasma and legionella.

Question 52

Epigenetics

Answer 52

Changes in phenotype that do not involve changes in genotype. This occurs when the environment interacts with genes. Environmental factors (food, chemicals, smoking, drugs, stress) impact the way your genes are expressed. Your lifestyle determines which genes get “switched on” or “switched off”

Question 53

Genotype

Answer 53

An individual's DNA sequence

Question 54

Phenotype

Answer 54

How genes are expressed

Question 55

Polymorphism

Answer 55

A change in DNA sequence

Question 56

Chromatin

Answer 56

A complex of DNA and protein (histones). Their function is packaging long DNA molecules into more compact structures, preventing the strands from becoming tangled. Plays important roles in preventing DNA damage, regulating gene expression, and DNA replication.

Question 57

Histones

Answer 57

Proteins that are critical in the packing of DNA into the cell and into chromatin and chromosomes. They interact with DNA to form a periodic beads on a string structure called a polynucleosome. Also very important for regulation of gene expression. Because of their unusually high content of the basic amino acids - lysine and arginine, histones are highly cationic and interact with the polyanionic phosphate backbone of DNA. Histones are proteins of relatively low molecular weight, exhibit known interspecies homologies in their amino acid sequences, and are noncovalently attached to DNA in stoichiometric amounts.

Question 58

Histone deacetylases

Answer 58

Enzymes that removes the acetyl group from lysine/arginine amino acid of the histone proteins on DNA, making the DNA less accessible to transcription factors. Overexpressed in colon, prostate, breast and cervical cancers.

Question 59

Epigenetic Treatments of Cancer

Answer 59

Cancer progression can involve epigenetic alterations that may affect cell-cycle progression, repair of DNA damage and apoptosis. This can involve histone modification, as histone deacetylases are overexpressed in colon, prostate, breast, and cervical cancers. Trichostatin A and suberoylanilide hydroxamic acid (SAHA) bind to a zinc ion in the enzyme active site and block deacetylation of histone lysines and arginines. These molecules inhibit tumor cell growth and block cell-cycle progression. SAHA is currently approved for the treatment of cutaneous T cell lymphoma.

Question 60

Dietary component to the epigenetic control of cancer

Answer 60

Short chain fatty acids such as butyrate are generated by bacterial flora of the large intestine and show a weak ability to regulate epithelial cell mitosis, differentiation, and apoptosis. Studies have shown that butyrate increases the levels of histone acetylation, presumably from inhibition of histone deacetylate. These data suggest that the nutritional state of an individual can affect the histone code and its role in cancer.

Question 61

Why is DNA well suited to its function?

Answer 61

Because of its chemical stability and because it can encode a lot of information using only 4 types of nucleotides

Question 62

Central dogma of molecular biology

Answer 62

DNA stores information that determines the sequence of RNA, which in turn determines the structure of the protein. Much of the structure and biochemistry of cells is due to the proteins they produce, but the properties of the proteins is determined by the DNA sequence.

Question 63

Transcription

Answer 63

How genetic information is transmitted from DNA to RNA.

Question 64

Ribozyme

Answer 64

When RNA acts as a catalyst and constitutes the active component of ribosomes

Question 65

Translation

Answer 65

The RNA sequence is translated into a protein sequence at the ribosome.

Question 66

Reverse transcription

Answer 66

When RNA acts as a template for DNA synthesis, reversing the normal flow of information

Question 67

Nucleic acids

Answer 67

Linear polymers of nucleotide units. Every nucleotide is made of a phosphate ester, a pentose sugar, and a nitrogenous base. In DNA, the sugar is 2-deoxy-D-ribose. In RNA, it's D-ribose.

Question 68

Beta-N-glycosidic bond

Answer 68

The bond that attaches nucleotide bases to the 1-position of the sugar

Question 69

Purines

Answer 69

Guanine and adenine. They both occur in DNA and RNA and are attached to the sugar at N-9

Question 70

Pyrimidine

Answer 70

Cytosine, uracil (RNA), and thymine (DNA). They are linked to the sugar at the N-1 position.

Question 71

Nucleoside

Answer 71

A base that is glycosylated with either pentose sugar. Nucleosides with ribose are ribonucleosides and nucleosides with deoxyribose are deoxyribonucleosides

Question 72

4 ribonucleosides in RNA

Answer 72

1. Adenosine (A) 2. Guanosine (G) 3. Cytidine (C) 4. Uridine (U)

Question 73

4 deoxyribonucleosides in DNA

Answer 73

1. Deoxyadenosine (dA) 2. Deoxyguanosine (dG) 3. Deoxycytidine (dC) 4. Deoxythymidine (dT)

Question 74

Nucleotides

Answer 74

Phosphate esters of nucleosides. Any of the sugar hydroxyl groups can be phosphorylated.

Question 75

Nucleoside monophosphates

Answer 75

Nucleotides that contain a phosphate monoester

Question 76

Physical properties of nucleosides and nucleotides (6)

Answer 76

1. Nucleosides and nucleotides are soluble in water over a wide range of pHs 2. Nucleobases and nucleosides are neutral molecules at physiological pH 3. Under strongly basic conditions, the slow hydrolysis of phosphate esters can occur 4. Molecules that contain purine or pyrimidine bases strongly absorb UV light. Purines can absorb light more strongly than pyrimidines 5. N-glycosidic bonds of nucleosides and nucleotides are stable under basic conditions, but less stable under acidic conditions 6. At elevated temperatures, the bond between the sugar and purine bases is more likely to break in an acidic environment. Pyrimidines are more resistant to acids

Question 77

Sugar puckering

Answer 77

5 membered rings are highly flexible, and in pentose rings, one or two atoms of the ring can twist out of place. In cyclopentane rings, there are several envelope and half chair conformations that rapidly interconvert. Because of the asymmetric substitution pattern of the pentose ring in nucleosides, two conformations are preferred, which is referred to as sugar puckering.

Question 78

Endo face

Answer 78

The endo face of the pentose is termed "above" and this is where C5 and the base project from the ring. If C2 is displaced above the pentose ring, the conformation would be called C2-endo. C3-endo is another common pucker, where C3 is displaced above the ring. C2-endo and C3-endo conformations are in rapid equilibrium

Question 79

Which conformations do RNA and DNA sugars prefer?

Answer 79

An electronegative substituent at the 2' position of the pentose favors the C3 endo conformation. Therefore, ribonucleosides in RNA prefer the C3 conformation as they have a 2' hydroxyl group. DNA prefers the C2 conformation because the 2' deoxynucleoside contains a hydrogen in place of a hydroxyl group

Question 80

Glycosidic conformations of purines

Answer 80

In purines, the syn and anti conformations readily interconvert, with the anti conformation being more stable in most cases

Question 81

Glycosidic conformations of pyrimidines

Answer 81

In pyrimidines, steric clashes between the sugar and the oxygen of the base strongly disfavor the syn conformation.

Question 82

Anti and syn conformations

Answer 82

The anti conformer has the smaller H-6 (pyrimidine) or H-8 (purine) atom above the sugar ring. The syn conformer has the larger O-2 (pyrimidine) or N-3 (purine) in that position. The syn conformation is stabilized in guanosine 5-phosphates because of favorable interactions between the 2-NH2 group and the phosphate oxygens

Question 83

Phosphodiester bonds

Answer 83

Links strands of nucleotides to form nucleic acids. The bond links the 5' hydroxyl group of one the sugar molecule of one residue to the 3' hydroxyl group of the next sugar molecule

Question 84

Oligonucleotides

Answer 84

Strands of nucleic acids that contain 50 or fewer nucleotides. Strands longer than that are called polynucleotides

Question 85

Directionality of phosphodiester bonds

Answer 85

Linear nucleotide strands have one end that terminates in a 5'-OH (the 5'-terminus) and one end that terminates in a 3'-OH (the 3'-terminus).

Question 86

What is responsible for polynucleotide conformations?

Answer 86

The bases are largely responsible for the conformations of polynucleotides.

Question 87

Base stacking

Answer 87

The edges of nucleotide bases contain nitrogen and oxygen atoms that can interact with other polar groups or surrounding water molecules. However, the faces of the rings cannot participate in these interactions and tend to avoid contact with water. Instead, they interact with one another to produce a stacked conformation (base stacking). This reduces the hydrophobic surface area that must be solvated by polar water molecules, which is more energetically favorable. van der Waals forces also help to facilitate base stacking

Question 88

Stability of the polynucleotide backbone

Answer 88

The half life for the spontaneous hydrolysis of phosphodiester linkages in DNA is millions of years, which makes DNA very stable and suitable for the long term storage of genetic information. RNA is much more prone to hydrolysis

Question 89

Nucleases

Answer 89

Catalyze the breaking of the phosphodiester bonds in the DNA backbone. Nucleases are characterized by the type of polynucleotides they hydrolyze and the specific bonds they break. Exonucleases cleave the last nucleotide residue at either the 5' or 3' terminus of a polynucleotide. Endonucleases cleave phosphodiester bonds located in the interior of the polynucleotides. They do not require a free terminus. DNase 1 and DNase 2 are examples of endonucleases that hydrolyze DNA with little selectivity. Nucleases also exhibit specificity with respect to the overall structure of polynucleotides. Some are specific to DNA or RNA, and some are specific to whether the polynucleotide is single or double stranded

Question 90

Restriction endonucleases

Answer 90

Recognize and cleave very specific double stranded DNA sequences by cutting each strand. Most endonucleases recognize sequences 4-6 nucleotides long.

Question 91

How was it determined that DNA formed a double helix? (3)

Answer 91

1. It was found that DNA did not contain equal amounts of the 4 nucleosides, there were variable amounts in different organisms. However, the abundance of deoxyadenosine equaled that of deoxythymidine, and the abundance of deoxyguanosine always equaled that of deoxycytidine 2. X-ray diffraction data showed that DNA contained double-helical structures, and symmetry suggested that the two polynucleotide strands were antiparallel 3. It was suggested that the nucleobases were in the keto and amino tautomeric forms rather than enol and imino forms

Question 92

Double helix

Answer 92

The structure of DNA, as represented in Watson and Crick's model, is a double-stranded, antiparallel, right-handed helix. The sugar-phosphate backbones of the DNA strands make up the outside of the helix, while the nitrogenous bases are found on the inside and form hydrogen-bonded pairs that hold the DNA strands together.

Question 93

Anti parallel

Answer 93

Double-stranded DNA is an antiparallel molecule, meaning that it's composed of two strands that run alongside each other but point in opposite directions. In a double-stranded DNA molecule, the 5' end (phosphate-bearing end) of one strand aligns with the 3' end (hydroxyl-bearing end) of its partner, and vice versa. This produces a stable association between strands

Question 94

Tautomers

Answer 94

Isomers of a molecule that differ only in the position of a hydrogen atom. Each of the 4 nucleobases has 2 or more possible tautomeric forms that are in equilibrium

Question 95

Chargaff's rules

Answer 95

Due to the hydrogen bond specificities of DNA, the double stranded DNA must contain ratios of nucleosides that agree with experimental observations (dA=dT and dG=dC). The geometry of each of these base pairs result in similar distances between the first carbons on the sugar and glycosidic bond orientations. This is a structural isomorphism, which means that any of the 4 possible base pairs can be placed into the double helix without significantly changing the structure of the backbone. Therefore, this could lead to possible replacement of one base pair by any other, providing a key to understanding how mutations occur

Question 96

Complementarity

Answer 96

The relationship between bases on opposing stranded within the double helix. Bases are considered complementary because every nucleobase of one strand is matched by shape and hydrogen bonding to a complementary base on the other strand. Only thymine can appropriately fill the space across from adenine

Question 97

Grooves

Answer 97

Interwinding of the two antiparallel strands produces a structure that has 2 distinct helical grooves between the sugar-phosphate backbones. There is a major groove that is much wider than the minor groove. This disparity arises from the geometry of the base pairs. The glycosidic bonds between the bases and the backbone pentose are not arranged directly opposite to each other, but are displaced toward the minor groove. The nucleotide sequence of DNA can be determined without dissociating the double helix by looking inside these grooves. Each base always displays the same atoms into each of the grooves of the double helix

Question 98

Factors that stabilize double-helical DNA (3)

Answer 98

1. Stacking interactions- these interactions are the most important factor in stability 2. Networks of cooperative hydrogen bonds 3. Electrostatic forces

Question 99

Base stacking in the double helix

Answer 99

The separation between the hydrophobic core of the stacked nucleobases and the hydrophilic exterior of the charged sugar-phosphate group is pronounced in the double helix. It helps the bases to avoid contact with water, as base stacking essentially removes nucleobases from the aqueous environment. These interactions help to stabilize the double helix

Question 100

Hydrogen bonds in the double helix

Answer 100

Hydrogen bonds are relatively weak and are not the "glue" that maintains the double helix structure. However, hydrogen bonds are highly directional and help discriminate between correct and incorrect base pairs. Because of the directionality, hydrogen bonds tend to orient the nucleobases to favor stacking

Question 101

Electrostatic forces in DNA

Answer 101

Phosphodiester groups are ionized at physiological pH, so the double helix carries 2 negative charges with each base pair. The interstrand electrostatic repulsion between negatively charged phosphates is destabilizing and tends to separate the complementary strands

Question 102

When must the DNA double helix be disrupted?

Answer 102

During almost every important biological process- DNA replication, transcription, repair, and recombination. The forces that hold the strands together are adequate for providing stability but weak enough to allow easy strand separation.

Question 103

Denaturation/melting

Answer 103

When the temperature of a solution is increased, a few base pairs will be initially disrupted at low temperatures, creating one or more open-stranded bubbles. These bubbles form initially in sections that contain higher proportions of adenine-thymine pairs because of the pair's lower stacking energies. As the temperature increases further, the size of the bubbles increases, and the thermal motion of the polynucleotides eventually overcomes the forces that stabilize the double helix. At even higher temperatures, the strands fully separate and acquire a random-coil conformation. Stacking interactions gradually decrease as the helix is disrupted

Question 104

What physical changes occur in a DNA-containing solution during denaturation?

Answer 104

There is an increase in buoyant density, a reduction in viscosity, a change in ability to rotate polarized light, and changes in absorbance of UV light. The change in absorbance of UV light can be used to follow the process of denaturation experimentally. The absorbance by individual bases is reduced by electronic interactions between stacked bases, which is called hyperchromicity. As denaturation occurs and stacking interactions decrease, the absorbance of UV light can increase

Question 105

At which pH can DNA become denatured?

Answer 105

DNA becomes denatured at a pH greater than 11.3 as the N-H groups on the bases become deprotonated, preventing them from participating in hydrogen bonding. Alkaline denaturation is often used to prevent damage to the DNA that can occur at a high temperature or low pH.

Question 106

Renaturation/annealing

Answer 106

If DNA strands have been separated by denaturation, they can reform a double helix if appropriately treated. If denaturation is not complete and a few nucleobases remain hydrogen bonded between the 2 strands, the helix-to-coil transition is rapidly reversible. Annealing is possible if the strands have been completely separated, but this depends on the DNA strands meeting in a manner that can lead to reformation of the original structure. This is a slow and concentration dependent process. Once correct nucleobases begin to pair by chance, the double helix is rapidly reformed over the entire DNA molecule

Question 107

Cooperative base pairing

Answer 107

After the formation of the initial base pair, stacking and base pairing are not independent events, but are influenced by the neighboring pairs. Therefore, this process can be referred to as cooperative

Question 108

A-DNA

Answer 108

Found under conditions of low humidity and high salt concentration. Consecutive guanines on one strand favor A-DNA conformations. The A-DNA structure is shorter and thicker than B-DNA, and has about 11 base pairs per turn

Question 109

B-DNA

Answer 109

Appears under conditions of high humidity and low salt concentration, and is the basis of the Watson-Crick structure- the average structure of DNA in living organisms is believed to be close to B-DNA. The phosphate groups in B-DNA are more accessible to water molecules than in A-DNA. In AT-rich sequences, an ordered array of water molecules occupies the narrow minor groove of B-DNA. In B-DNA, considerable local variation of individual nucleotides may occur. Such variations may be important in regulation of gene expression, since they can influence the extent of DNA binding with various types of regulatory proteins

Question 110

Z-DNA

Answer 110

Incorporates a left handed helix in contrast to the typical right handed helix. It is generally seen in sequences of alternating purines and pyrimidines. The phosphodiester backbone assumes a zigzag arrangement compared to the smooth conformation that characterizes A-DNA and B-DNA. The Z-DNA structure is longer and much thinner than that of B-DNA and has 12 base pairs per turn. The base pairs are displaced so far into the major groove that a distinct channel no longer exists. This places the stacked nucleobases on the outer part of Z-DNA instead of their conventional positions in the interior of the double helix. Z-DNA may have a role in gene regulation

Question 111

Structure of B-DNA

Answer 111

The base pairs are nearly perpendicular to the helical axis, which passes through the base pairs. The helix is long and thin, and has approximately 10 base pairs per helical turn.

Question 112

Contour length

Answer 112

Length of the DNA assuming a B-form double helix. The contour length of genomic DNA is usually much larger than the size of the cell that contains it.

Question 113

Conformation of DNA complexes

Answer 113

Depending on the source, DNA complexes can be circular or linear. DNA in most higher organisms is linear and forms chromosomes. Some organisms have DNAs that are either linear or circular at different points in their life cycles. With E. coli, DNA is initially linear, but during infection, circularization occurs.

Question 114

Chromosomes

Answer 114

Humans have 46 chromosomes in diploid cells, which are linear double-helical DNA complexed with proteins.

Question 115

DNA ligase

Answer 115

In DNA replication, ligase’s job is to join together fragments of newly synthesized DNA to form a seamless strand. The ligases used in DNA cloning do basically the same thing. If two pieces of DNA have matching ends, DNA ligase can join them together to make an unbroken molecule. Using ATP as an energy source, ligase catalyzes a reaction in which the phosphate group sticking off the 5’ end of one DNA strand is linked to the hydroxyl group sticking off the 3’ end of the other. This reaction produces an intact sugar-phosphate backbone.

Question 116

Relaxed DNA

Answer 116

Circular double stranded DNA is considered relaxed, it is formed by ligating the free termini of a linear DNA. The relaxed DNA has greatly reduced activity in replication, translation, and recombination.

Question 117

Superhelical DNA

Answer 117

Superhelical DNA is biologically active. It is a topologically strained isomer created by underwinding or overwinding the double helix. Any segment of double stranded DNA that is immobilized at both ends can be superhelical. The DNA of eukaryotic cells can acquire a superhelical form because its anchoring by nuclear proteins creates numerous closed topological domains.

Question 118

Topological domains

Answer 118

Defined as a DNA segment contained in a manner than restrains rotation of the double helix

Question 119

Function of supercoiling

Answer 119

Promotes packaging of DNA within the confines of the cell by facilitating formation of compact structures. Superhelices can also have a tendency to generate regions with disrupted hydrogen bonding (bubbles) which may be instrumental in facilitating the process of localized DNA strand separation during DNA repair, synthesis, and recombination

Question 120

Topoisomerase 2

Answer 120

Dimeric proteins that bind to a double helical DNA and cleave both strands. Passage of another double helical DNA segment through the break removes or adds two supercoils. All eukaryotic and many prokaryotic topoisomerases 2 only relax supercoiled DNA. ATP hydrolysis is required for the turnover of the enzyme, but not for the actual relaxation reaction

Question 121

Junk DNA

Answer 121

Only 2-4% of DNA in a mammalian cell may suffice for all its genes. Some of the remaining DNA, like in centromeres and telomeres, have a well-defined function. The majority of noncoding DNA has no defined function and is called junk DNA. However, junk DNA may have an important role in regulation of gene expression during development. The amount of junk DNA actually correlates better to the complexity of the organism than the number of genes

Question 122

5 classes of histones

Answer 122

H1, H2A, H2B, H3, and H4. The amino acid sequences of histones are very highly conserved between species. For example, histone H4 only differs by 2 amino acids between cows and peas. H3 is also very highly conserved, although H2A and H2B are less highly conserved. H1 is larger, more basic, and is the most tissue-specific and species-specific histone.

Question 123

Nucleosome

Answer 123

The elementary units of a polynucleosome, formed when histones interact with DNA to form a beads on a string structure. Each nucleosome is disk shaped and consists of a DNA segment and a histone cluster, composed of two molecules each of H2A, H2B, H3, and H4 histones. The histones are in contact with the minor groove of DNA and leave the major groove available for interaction with proteins that regulate gene expression and other DNA functions

Question 124

Stoichiometry

Answer 124

The determination of the proportions in which elements or compounds react with one another. The rules followed in the determination of stoichiometric relationships are based on the laws of conservation of mass and energy and the law of combining weights or volumes

Question 125

Base pair complementarity in DNA is a consequence of

Answer 125

The size, shape, and chemical composition of the bases

Question 126

The stability of the DNA double-helix is due to

Answer 126

Hydrogen bonds between complementary bases, hydrophobic interactions and favorable van der Waals radii.

Question 127

Histone remodeling

Answer 127

Packing changes at various stages of the cell cycle, which is controlled in part by covalent modification of the core histones (called histone remodeling). Reversible acetylation on the amino groups of lysine and phosphorylation of serine and threonine residues are involved in regulating the activity of nucleosomal DNA. These modifications reduce the numbers of positive charges on histones, causing them to bind less tightly to DNA. The resulting decondensation of chromatin produces loosely packed euchromatin, which is transcriptionally active

Question 128

Heterochromatin

Answer 128

Methylation of basic amino acid side chains (lysine and arginine) promotes packing more tightly. When accompanied by decreased acetylation, it characterizes a highly condensed, inactive form of chromatin. This is called heterochromatin. Throughout the cell cycle, permanently repressed genes and other untranscribed regions like telomeres, centromeres, and "junk" DNA remain condensed

Question 129

Histone code

Answer 129

The pattern of covalent modifications that provides a highly tunable means of regulating gene expression. This code is epigenetic and therefore does not cause changes in the sequence of DNA. However, it is heritable, due to coordination between histone modification and methylation of DNA at the 5-position of deoxycytidine.