BIOMG 3300: Principles of Biochemistry -Unit 6

Question 0

Given the structure of the four common bases, identify each as a purine or pyrimidine and give it’s common name. ( you will no be asked to draw it’s structure.

Answer 0

-Purines Contain have two cyclic rings, Pyrimidines have one. -Guanine is the only purine with a double bonded oxygen -Thymine is the only pyrimidine with a methyl group attached -Uracil is used in RNA, it is distinguishable from the other pyrimidines via the presence of two double bonded oxygens -Purines are Adenine,Guanine. (Remember that pure gold is AU) -Pyrimidines are Cytosine and Thymine (Remember Pyramids in CT)

Question 1

What is a Nucleoside? Given the base, you must be able to draw the structure.

Answer 1

A Nucleoside consists of a Sugar ( Deoxyribose, or Ribose) and a Nucleic Base. There is no bound phosphate. -Remember : S in Nucleoside.. stop drawing the molecule at the phosphate.

Question 2

What is a Nucleotide? Given the base, you must be able to draw the structure.

Answer 2

A nucleotide consists of the sugar, nucleotide base, and at least one phosphate. - A name denoted with 5’ d- or triphosphate indicates a nucleotide with multiple phosphates.

Question 3

Use Fig. 8-1 (p. 281) to discuss the numbering convention for the pentose ring. Which atom in the sugar is the base attached to? What kind of bond joins the sugar to the base (p. 282)? Using Fig. 8-4 (p. 283), identify the atom that is found at the 2’ position in deoxyribonucleotides? - in ribonucleotides? What functional group is at the 3’ position in ribo- and deoxy-ribonucleotides? To which atom in the sugar is the phosphate bound?

Answer 3

-The carbons are numbered in a clockwise fashion starting from the carbon next to the oxygen in the pentose ring. -The atom in the sugar that the base is attached to is the 1’ carbon. -N-B-Glycosyl bond -In Deoxyribonucleotides, the 2’ consists of just a hydrogen ( hence de-oxygenated). In ribonucleotides, the 2’ consists of an Oxygen and hydrogen ( hydroxyl group) -The 3’ group of both nucleotides consist of a hydroxyl group. -The phosphate is bound to the 5’ carbon of the pentose ring

Question 4

Use Fig. 8-4 (p. 283) to discuss the nucleotides found in DNA. Compare and contrast these nucleotides to those found in RNA.

Answer 4

The DNA Nucleotides have a hydrogen at the 2’ carbon. They also consist of the four common bases Adenine, Cytosine, Guanine, and Thymine. The RNA Nucleotides possess a hydroxyl group at the 2’ carbon. Instead of a nucleotide possessing thymine like in DNA, these nucleotides have a Uracil based nucleotide denoted as uradine.

Question 5

Draw a section of DNA (use the letters A, C, G, and T to represent the structure of the base). Point out the following (Fig. 8-7, p. 285): a. phosphodiester linkage b. sugar-phosphate backbone c. 5’ and 3’ ends d. overall charge

Answer 5

Question 6

Distinguish between an oligonucleotide and a polynucleotide (p. 286).

Answer 6

• A short nucleic acid is referred to as an Oligionucleotide. Generally referring to polymers containing less than 50 or fewer nucleotides. -A longer Nucleic acid is called a polynucleotide

Question 7

Use Fig. 8-11 (p. 287) to discuss the hydrogen bonds that stabilize the DNA double helix.

Answer 7

-The functional groups of purines and pyrimidines are ring nitrogens, carbonyl carbons, and exocylcic amino groups. Hydrogen bonds involving the amino and carbonyl groups are the most important interaction between two complementary strands of nucleic acid. -Adenine and Thymine form two hydrogen bonds. -Cytosine and Guanine for three hydrogen bonds.

Question 8

How was DNA shown to be the genetic material (p. 288)? Why was the base composition (Chargaff’s rules) important in suggesting a model for the structure of DNA (p. 288)? What other information was crucial in the structure determination (p. 288)?

Answer 8

-Avery, MacLeod, and McCarty found that DNA extracted from a virulent bacterium could transform a non- virulent strain into a virulent strain of bacterium. - Later Alfred Hershey performed experiments involving radioactively labeled proteins and nucleic acids, and their transmission from bacteriophages to bacteria cells. - Chargoff Concluded that different species varied in base composition. Regardless the relative composition of complementary bases remained the same. A+G=C+T -Rosalinda Franklin’s X-Ray Diffraction Patterns provided evidence of DNA’s Helical Structure.

Question 9

Use Fig. 8-13 (p. 289) to discuss the structure of double-stranded DNA. Point out the major and minor grooves, the sugar-phosphate backbones, and the stacked base pairs.

Answer 9

• The DNA structure consists of two helical DNA Chains wound around the same axis to form a right handed double helix. The hydrophilic backbones consisting of deoxyribose and phosphate are on the outside of the double helix facing the water. The offset pairing of the two strands creates a major and minor groove. The stacked base pairs enable a greater increase in stability

Question 10

Use Fig. 8-14 (p. 289) to discuss the strand polarity in double- stranded DNA. What word is used to describe the polarity of the two strands? What base pairs are found in DNA? Discuss the key difference between these two base pairs. What is the position of the sugar-phosphate backbone relative to the bases?

Answer 10

-The two DNA strands run in opposite directions. ( whether the 3’ and 5’ phosphodiester bonds should run in the same direction) -The word used is Anti-parallel. - The base pairs found in DNA : A=T,G=C - Adenine and Thymine form twi hydrogen bonds, while Cytosine and Guanine form three hydrogen bonds. - The sugar phosphate bonds run perpendicular to the base pairing in the double helical model.

Question 11

What two forces stabilize the DNA Double helix? p.289

Answer 11

• The hydrogen bonds between complementary bases - The Base stacking interactions make the major contribution to the stabilization to the stability of the helix.

Question 12

Double-stranded DNA can be reversibly denatured and renatured (pp. 297 - 298). a. Discuss the types of bonds broken during denaturation (p. 297). b. Use Fig. 8-27 (p. 298) to discuss the following: 1) What is the melting point (tm)? 2) What is the relationship between tm and the base composition? c. Define what is meant by “annealing” (p. 297).

Answer 12

a.- Hydrogen bonds between paired bases and of base stacking causes unwinding of the the double helix to form two separated or partially separated single strands. No covalent bonds in DNA are Broken . b1.- tm is the melting point at which half the DNA is present as separated single strands. b2.- The higher the concentration of Guanine and cytosine, the higher the temperature of tm. this is because cytosine and Guanine form three hydrogen bonds between the bases, thus raising the melting point. c.- The term Annealing refers to the process in which DNA reforms the double helix.

Question 13

Define and distinguish between DNA replication, transcription, and translation (p. 977).

Answer 13

DNA replication - the copying of parental DNA to form daughter DNA molecules with identical nucleotide sequences. Transcription - the process by which parts of the genetic message encoded in DNA are copied precisely into RNA Translation - the genetic message encoded in messenger RNA is translated onto the ribosomes into a polypeptide with a particular sequence of amino acids.

Question 14

What is a gene (p. 980)? How are genes named (p. 1010)? Later you will learn about sequences that control expression of genes. These and introns are part of the modern definition of a gene.

Answer 14

-A gene is all the DNA that encodes the primary sequence of some final gene product ( such as polypeptide or RNA) - bacterial genes are generally named using three lowercase, italicized letters, often reflecting their apparent function

Question 15

What distinguishes a plasmid from a chromosome (p. 981)?

Answer 15

Many bacteria contain one or more small circular DNA molecules that are free in the cytosol. Chromosomes are the repositories of genetic information.

Question 16

Is all eukaryotic DNA stored in the nucleus (p. 983)? Explain.

Answer 16

No, there is some DNA in the mitochondria and there is some DNA in the chloroplasts of plant cells.

Question 17

Eukaryotic chromosomes are complex. Define the following (pp. 984 - 985): a. intron b. exon c. satellite DNA d. centromere e. telomere

Answer 17

a.Intron- Non translated DNA segments in genes b.Exon-The coding segments of genes c.Satellite DNA- highly repetitive short sequences of DNA. These repetitive on coding sequences have are associated with telomeres and centromeres. c. Centromere-is a sequence of DNA that functions during cell division as an attachment point for proteins that link the chromosome to the mitotic spindle. d. Telomere- are sequences at the ends of eukaryotic chromosomes that help stabilize the chromosome.

Question 18

DNA Supercoiling (pp. 985 - 993): a. What is meant by “linking number” (p. 988; do not worry about the equations)? What is a “topoisomer” (p. 989)? b. Discuss how over-winding or underwinding the DNA double helix leads to the formation of positive or negative supercoils respectively (Figs. 24-13 and 24-16; pp. 987, 989). c. List two reasons why it is advantageous for cells to maintain their DNA in the underwound state (pp. 987 - 988). d. What is a topoisomerase (pp. 989 - 990)? 1) Differentiate between type I and type II topoisomerases in E. coli (p. 990). 2) Describe how E. coli topoisomerase I removes negative supercoils, thereby increasing the linking number (Fig. 24-20, p. 991).

Answer 18

DNA is coiled in in the form of a double helix. The coiling about itself is termed supercoiling. a. Linking Number- the Number of times closed cirular DNA intersect each other b. Overwinding of DNA leads to positive supercoiling or the formation of a right handed helix. Underwinding of DNA leads to negative supercoiling or the formation of a left handed helix. c. Cells keep DNA in the underwound state to facilitate it’s compaction via coiling. Underwound DNA is also easier to access the genetic information. d1.type I topoisomerase break 1 strand of DNA, pass the unbroken strand through the break and rejoin the ends, they increase the linking number by 1. Type II break both strands of DNA, they increase the linking number in increments of 2. d2. Tyr residue cleaves the phosphodiester bondds.

Question 19

Use Figs. 24-25 (p. 995) and 24-26 (p. 996) to discuss the packaging of DNA in eukaryotes into nucleosomes. What is unique about the amino acid composition of histone proteins that is crucial to their DNA-binding function (p. 995)?

Answer 19

DNA is bound tightly to beads of protein in regularly spaced intervals. The beads are complexes of histones and DNA. The bead plus the DNA that leads to the next bead to form a nucleosome. This height units of

Question 20

Use Figs. 24-29 (p. 1000), 24-30 (p. 1000), and 24-31 (p. 1001) to discuss higher order organization of eukaryotic DNA.

Answer 20

First there is double stranded helical DNA. Then H1 stimulates the formation of the 30 nm fiber. Further coiling and looping occurs into the form of rosettes which eventually coil into thicker structures and eventually the mitotic chromosome.

Question 21

Explain what is meant by semi-conservative replication (p. 1011).

Answer 21

Each DNA strand serves as a template for the synthesis of a new strand producing two new DNA molecules, each with one new strand and one old strand. This is semiconservative replication.

Question 22

Define what is meant by the term origin of replication (p. 1012).

Answer 22

Replication loops always initiate at a specific point ( Inman showed this using denatured A-T base pairs in bacteriophages and comparing how far along DNA replication had occurred.)

Question 23

What is a replication fork (p. 1012)? Use Fig. 25-3 (p. 1012) to discuss bi-directional replication.

Answer 23

Inside the loop of Ecoli, One or both ends of the loop are dynamic points, where parent DNA is being unwound and the separated strands are quickly replicated.

Question 24

Using Fig. 25-4 (p. 1013), discuss leading and lagging strand synthesis. What is the direction of DNA synthesis? In what direction is the template strand read? Which strand is replicated continuously, and which discontinuously? What are Okazaki fragments?

Answer 24

-A new strand of DNA is always synthesized in a 5'-\>3' direction. ( the 3' OH is the point of Elongation) -Because the to DNA Strands are antiparallel, the strand serving as the template is read from it's 3' end toward it's 5' end. - The leading strand is replicated continuously. It is the strand in which the 5'-\> 3' synthesis proceeds in the same direction as the replication fork movement. -The lagging strand is replicated discontinuously, it is the strand where 5'-\>3' synthesis occurs opposite to the replication fork - The lagging strand is synthesized in short fragments called Okazaki fragments.

Question 25

Distinguish between an exonuclease and an endonuclease (p. 1013).

Answer 25

-Both enzymes degrade DNA ( These enzymes are known as nucleases) -Exonucleases: Degrade nucleic acids from one end of the molecule. Many operate in the 5' -\>3' or the 3'-\>5' direction, removing nucleotides only from the 5' or 3' end, respectively, of one strand of a double-stranded DNA or single stranded DNA. -Endonucleases: can begin to degrade at specific internal sites in a nucleic acid strand or molecule, reducing it to smaller and smaller fragments

Question 26

DNA Polymerases a. What are the two "central requirements" of DNA polymerase for DNA polymerization (pp. 1013 - 1014)? b. Use Fig. 25-5 (p. 1014) to discuss the reaction catalyzed by DNA polymerases. 1) Which type of nucleotide (dNMP, dNDP, or dNTP) are used as substrates? 2) What process drives the reaction toward the formation of products (p. 1013)? Discuss in terms of Le Chatelier's principle. c. Define processivity (p. 1014).

Answer 26

a All DNA polymerases require a template.Polymerases also require a primer ( a strand segment with a free 3' OH group o which a nucleotide can be added) b.The catalytic mechanism involves two Mg^(2+), coordinated to the phosphate groups of incoming nucleotide triphosphate. the 3' hydroxyl group of the primer will act as aaanucleophile. The Mg(2+) ion depicted at the top facilitates the attack of the 3' hydroxyl group of the primer on the alpha phosphate of the nucleotide triphosphate. The other Mg(2+) facilitates the displacement of pyrophosphate. b1. dNTP b2. Non-covalent base stacking and base pairing interactions provide extra stabilization to lengthened DNA. The formation of products is helped by energy generated in the hydrolysis oh the pyrophosphate product by the enzyme pyrophosphates. c. average number of nucleotides before a polymerase dissociates.

Question 27

Use Fig. 25-6 (p. 1015) to discuss how the geometry of the active site of DNA polymerase contributes to the fidelity of DNA replication.

Answer 27

The standard A-T and C-G couplings have similar geometries, and an active site that can fit one will fit the other. This will exclude incorrectly paired bases from binding with the polymerase.

Question 28

Use Fig. 25-7 (p. 1016) to discuss proofreading by DNA polymerase. Which exonuclease activity is responsible for DNA proofreading (p. 1015)?

Answer 28

A mismatched base impedes translocation of DNA polymerase I to the next site. The DNA bound to the enzyme slide backward into the exonuclease site, and the enzyme corrects the mistake with it's 3'-5' exonuclease activity. The enzyme then resumes it's activity in the 5'-3' direction.

Question 29

Use Table 25-1 (p. 1016) to compare DNA pol I and DNA pol III. Pay particular attention to the following: 1) 3' to 5' exonuclease activity (proofreading) 2) 5' to 3' exonuclease activity 3) Rates of polymerization 4) Processivity

Answer 29

Question 30

Use Fig. 25-9 (p. 1018) to explain why DNA pol III is more processive than DNA pol I.

Answer 30

The two Beta subunits of polymerase III form a circular clam that surrounds the DNAThe Clamp slides along the DnA molecule, increasing the processivity of the polymerase II.

Question 31

Initiation of Replication Use Fig. 25-10 (p. 1019) to discuss the sequence features at the origin of replication. (Note: Do not concern yourself with actual sequences, just the number and different types of repeated sequences present!)

Answer 31

* R-sites:9 bp sequence that serve as binding sites for the key intitiator protein DnaA * DNA Unwiding Element - A region rich in A-T base pairs (Abbreviated as DUE) * There are also three additiona DnaA binding sites (I sites), ( They have different sequences.) * Binding sites for IHF (integration host factor) and FIS (factor for inversion stimulation) are present. * Another protein HU (Histone binding protein), is also present but does not have a specific binding site

Question 32

Use Fig. 25-11 (p. 1020) to discuss the events that initiate DNA replication 1)What is the function of the R and I sites? What happens at these sites that leads to unwinding of the double helix at the DUE (p. 1020)? 2)What is the role of the helicase that enters next (p. 1020)?

Answer 32

1. Eight DNaA pretin molecules with bound ATP bind to the R and I sites in OriC (Ecoli Origin) 2. The DNA wraps around this complex ina right handed helical structure 3. This induces a strain in the DUE (A-T Rich region) region. The net effect is a positive supercoil. 4. The DUE region becomes denatured. 5. Several binding proteins help with DNA Bending 6. DnaC protein than loads DnaB on the seperated strands in the denatured region. The ATP on DnaC is hydrolyzed leaving DnaB. 7. DnaB Helicase continue in the 5'-3' direction unwinding DNA as it travels.. creating two replication forks.

Question 33

Elongation (pp. 1021-1023) a.Using Fig. 25-12 (p. 1021) a s a guide, discuss the following: 1)As the fork enlarges, what enzyme continues to be involved in unwinding the double helix? 2) What enzyme counteracts the positive supercoiling caused by the helicase, by introducing negative super coils? 3)What protein binds to the resulting single-stranded DNA? Why is this important? 4)What process is required before DNA synthesis can occur on the single-stranded template? Explain. What enzyme carries this out? Why is RNA used as a primer instead of DNA? 5) What enzyme now carries out the bulk of DNA replication?

Answer 33

1. Parent DNA is first unwound by DNA helicase 2. DNA topoisomerase relieves topological stress 1. Leading Strand 1. DnaG primase creates short RNA primer at the replication origin. (Dna G is linked to DnaB) 2. Nucleotides are added by DNA polymerase III tehetered tot the opposite DNA strand. 2. Lagging Strand 1. The primase binds to helicase creating a new primer 2. beta clamp of polymerase 3 associates, polymerase 3.

Question 34

Using Fig. 25-13 (p. 1022) as a guide, discuss how a dimer of DNA polymerase at the replication fork can copy both parental strands simultaneously. Discuss the role of the beta subunit.

Answer 34

The beta subunit helps to increase processivity.

Question 35

Name the enzyme that removes the RNA primer (p. 1023). Which activity from table 25-1 (p. 1016) is used for RNA primer removal?

Answer 35

DNA Polymerase I removes the Primer and replaces the RNA with DNA

Question 36

Name the enzyme that then catalyzes the formation of phosphodiester bonds between the newly synthesized DNA fragments (p. 1023).

Answer 36

DNA ligase seals the nicks between the newly synthesized DNA fragments

Question 37

DNA replication is bi-directional. Use the diagram below to explain bi-directional DNA replication from a single origin of replication (represented by the dots in the middle of the single stranded region). Use squiggled lines to represent RNA primers and straight lines for DNA. Use arrow heads pointing from 5' to 3' to denote the direction of polynucleotide synthesis.

Answer 37

Question 38

Telomeres (pp. 1089-1092) a. What are telomeres (pp. 1089-1090)? b. Discuss why linear chromosomes can not be entirely copied by lagging strand synthesis (p. 1090). b2. Name the enzyme that adds telomeric ends (p. 1090). What type of enzyme is it (p. 1091)? c. Use Fig. 26-38 (p. 1091) to discuss telomere addition. What serves as the primer for telomere addition? What serves as the template? d. In humans, what types of cells (germline or somatic) have telomerase (p. 1092)? e. Discuss the role of telomeres in cell senescence (p. 1092).

Answer 38

a. repeating tandem sequences of DNA Base pairs at the ends of chromosomes. b. because DNA polymerase requires a primer b2. DNA telomerase, a cellular reverse transcriptase ( it uses RNA to create DNA) c. TG primer, or repeating sequences of T and G. The enzyme telemorase is composed of an RNA sequence that is used as a template d. Germlines e. Telemere shortning is believed to be responsible for cellular aging and cellular death. As dna is degrading when the telemeres are depeleted

Question 39

Base-excision Repair (pp. 299-302, 1030-1031) a. Use Fig. 8-30a (p. 300) to discuss the spontaneous cytosine deamination reaction. b. Use Fig. 25-24 (p. 1032) to discuss base -excision repair, naming each enzyme and discussing the function of each.

Answer 39

a. roughly 100 events b. 1. glycosalase breask the glycosidic linkage between the sugar and the phsoshate bond 2. ap endonuclease cleaves the phosphodiester bond at the 5' site 3. dna polymerase comes in and at nucleotides 4. ligase comes in and seals the nicks of the phosphodiester bonds.

Question 40

List three key processes that contribute to the high fidelity of DNA replication (p. 1027, summary).

Answer 40

1. replication is semiconservative 2. It is carried out in three phases 3. the process is directional in bacteria