DNA

Question 1

Griffith (1928)

Answer 1

• discovered transformation
• killed pathogenic bacteria with heat and then mixed the cell remains with living bacteria of non pathogenic strain
• some living cells became pathogenic and this trait was inherited by descendants of the transformed bacteria
• concluded that living R bacteria had been transformed into pathogenic S bacteria by unknown, heritable substance from dead S cells

Question 2

Avery (1944)

Answer 2

• purified various types of molecules from heat-killed bacteria and tried to transform live nonpathogenic bacteria from each type
• discovered that transformation only occurred when DNA was activated; determined transforming agent was DNA

Question 3

transformation

Answer 3

change in genotype and phenotype due to assimilation of external DNA

Question 4

Chargaff (1947)

Answer 4

• discovered DNA composition varies from one species to another (molecular diversity)
• the number of adenines approximately equaled the number of thymines; the number of guanines approximately equals the number of cytosines

Question 5

Chargaff’s rules

Answer 5

• for any given species, the number of A and T bases are equal and the number of C and G bases are equal
• the base composition varies between species

Question 6

Hershey & Chase (1952)

Answer 6

• concluded that injected DNA of viruses into the host cell during infection functions as genetic material of phage T2 that makes the host cells produce new viral DNA and proteins, which form new viruses
• used radioactive isotopes to label bacteriophages’ proteins and DNA to determine that phage DNA, not proteins, enter host cell

Question 7

Rosalind Franklin (1953)

Answer 7

-used x-ray crystallography to confirm double-stranded helical structure of DNA

Question 8

Watson & Crick (1953)

Answer 8

• build double-helix model of DNA with nitrogenous bases facing interior and antiparallel sugar-phosphate backbone on exterior
• base pairing of adenine with thymine and cytosine with guanine through hydrogen bonds

Question 9

Meselson & Stahl (1958)

Answer 9

• cultured E. coli for several generations in a 15N medium (heavy isotope) were transferred to 14N medium (lighter isotope) for further replication
• DNA sample centrifuged after first replication was hybrid DNA (supporting semi-conservative and dispersive model)
• DNA sample centrifuged after second replication produced both hybrid and light DNA, supporting semi-conservative model of DNA replication

Question 10

DNA structure: nucleotide

Answer 10

• sugar (deoxyribose)
• phosphate
• nitrogen base (A,T,G,C)

Question 11

DNA structure: directionality

Answer 11

• based on carbon in deoxyribose (5’ is the phosphate end and 3’ is attached to the OH)
• strands are antiparallel

Question 12

DNA structure: double helix

Answer 12

– Sugar and phosphate connected by covalent bonds to create the backbone
– Nitrogen bases connected by hydrogen bonds in the middle
– Van der Waal interactions help hold the stacked nitrogen bases together

Question 13

DNA structure: base pairing

Answer 13

• Purines contain two rings
–Adenine & Guanine
• Pyrimidines contain one ring
– Cytosine & Thymine
• To keep a consistent size in the diameter of the double helix purines must bond with pyrimidine 
–Adenine with Thymine
– Cytosine with Guanine
• These bases are held together by hydrogen bonds

Question 14

DNA replication: semiconservative replication

Answer 14

the basic model outlined by Watson & Crick, later shown to be correct by Meselson & Stahl

Question 15

DNA replication: origins of replication

Answer 15

• Replication occurs at multiple specific sites
• Replication will occur from both directions from this point
• Each side will have a
replication fork where the new strands are actively being synthesized

Question 16

DNA Polymerase

Answer 16

adds nucleoside triphosphate:
– The 3 phosphates provide energy to add the nucleotide to the growing polymer
– Two phosphate groups will be removed

Question 17

DNA Replication: order of enzymes

Answer 17

1. helicase
2. primase
3. DNA polymerase III
4. DNA polymerase I
5. DNA ligase

Question 18

DNA Replication: helicase

Answer 18

unwind the double helix at the replication fork & separates the two parental strands by breaking the hydrogen bond between nitrogen bases
-no directionality

Question 19

topoisomerase

Answer 19

alleviates extra strain on the DNA strands around the replication fork caused by unwinding DNA by helicase

Question 20

Single-strand binding proteins

Answer 20

bind to the separated DNA strands & stabilize them as replication occurs

Question 21

DNA Replication: Primase

Answer 21

creates a primer made of 5-10 RNA nucleotides that are complementary to the DNA strand
– This step is necessary because DNA polymerase cannot initiate a new strand, but only add nucleotides to an existing one
– moves from 5’ to 3’ direction, adding RNA nucleotides to the 3’ side of the new antiparallel and complementary strand

Question 22

DNA Replication: DNA polymerase III (aka: DNA pol III)

Answer 22

add new DNA nucleotides to the 3’ side of the newly synthesized strand until it reaches the next primer
– This will follow normal base pairing rules
– Only one primer is needed for the leading strand, but multiples are needed for the lagging strands creating Okazaki fragments

Question 23

DNA Replication: DNA polymerase I (aka: DNA pol I)

Answer 23

replace the RNA primers with the correct DNA nucleotides
– Only adds to the 3’ end
– It cannot join the last DNA nucleotide in the replacement to the existing DNA strand

Question 24

DNA Replication: DNA ligase

Answer 24

join the sugar-phosphate
backbone linking together gaps between the replaced primers & newly created DNA strand
– Links together the Okazaki fragments in the
lagging strand

Question 25

Antiparallel Elongation

Answer 25

DNA polymerase can only move in the 5’ to 3’ direction adding new nucleotides to the 3’ side of the molecule

Question 26

leading strand

Answer 26

DNA polymerase moves in the same direction of | helicase

Question 27

lagging strand

Answer 27

DNA polymerase moves in the opposite direction of helicase - requires multiple DNA polymerase enzymes resulting in the creation of Okazaki fragments

Question 28

Proofreading & Repairing | DNA

Answer 28

• Errors occur one in 100,000 bases • Luckily, DNA polymerase is also a proofreader and makes corrections as it goes dropping the number to one in 10 billion – The incorrect nucleotide is removed and the correct one is bond into place • Mismatches can sneak by polymerase & DNA can be damaged – Special enzymes do mismatch repair

Question 29

DNA Errors

Answer 29

• DNA is damaged frequently – Mutagens (ex. Reactive chemical, radioactivity, xrays, uv light, etc) constantly harm DNA – Spontaneous chemical changes within the cell also cause changes • Therefore, DNA is constantly monitored & repaired – 130 different repair enzymes have been found in humans – Corrections are typically done by nucleotide excision repair

Question 30

Nucleotide | Excision Repair

Answer 30

1. A mistake is present within a DNA strand 2. Nuclease (an enzyme)will remove the damaged section 3. DNA polymerase replaces the missing nucleotides 4. DNA ligase seals the ends together

Question 31

Telomeres

Answer 31

• Linear chromosomes are constantly being shortened on the 5’ end bc DNA pol I removes, but cannot replace RNA primer bc there is no 3' end from DNA nucleotide sequence – This may help prevent cancer by limiting cell divisions • As protection chromosomes have telomeres (repetitious sections of nucleotides that do not code for genes) that are removed rather than genes – linked to aging – Shortening is solved in germ cells by telomerase