* 16

Question 1

transformation

Answer 1

A change in genotype and phenotype due to the assimilation of external DNA by a cell. When the external DNA is from a member of a different species, this results in horizontal gene transfer.

Question 2

phage T2 experiment

Answer 2

• 1952; Alfred Hershey and Martha Chase
• one of the many phages that infect E. coli
• phages grown w/ radioactive sulfur (S-35) were incorporated into phage protein
• phages grown w/ radioactive phosphorus (P-32) were incorporated into phage DNA

Question 3

Chargaff’s rules

Answer 3

1) base composition varies btwn species

2) within a species, the number of A,T bases are equal; number of G,C bases are equal

Question 4

double helix numbers

Answer 4

• Franklin’s X-ray data indicated that the helix makes 1 full turn every 3.4 nm along its length
• w/ the bases stacked just 0.34 nm apart, there are 10 layers of base pairs, in each full turn of the helix
• uniform width throughout (2 nm)
• purines (A,G) 2x as wide as pyrimidines (C,T)

Question 5

pairing

Answer 5

• A forms 2 H bonds w/ T

- G forms 3 H bonds w/ C

Question 6

3 models of DNA replication

Answer 6

• semiconservative: the replicated double helix consists of one old strand, derived from the parental molecule, and one newly made strand.
• conservative: the 2 parental strands come back together after the process.
• dispersive: all 4 strands of DNA following replication have a mixture of old and new DNA.

Question 7

replication fork

Answer 7

• Y-shaped region where the parental strands of DNA are being unwound
• HELICASES: enzymes that untwist the double helix at the replication forks, separating the 2 parental strands and making them available as template strands
• SINGLE-STRAND BINDING PROTEINS: bind to the unpaired DNA strands, keeping them from re-pairing.
• TOPOISOMERASE: the untwisting of the double helix causes tighter twisting and strain ahead of the replication fork; topoisomerase helps relieve this strain by breaking, swiveling, rejoining DNA strands

Question 8

initiation of replication

Answer 8

• the enzymes that synthesize DNA can’t initiate the synthesis of a polynucleotide; they can only add nucleotides to the end of an already existing chain that is base-paired w/ the template strand
• the initial nucleotide chain that is produced during DNA synthesis is actually a short stretch of RNA; this RNA chain is called a PRIMER, synthesized by the enzyme PRIMASE
• primer is 5 - 10 nucleotides long

Question 9

DNA synthesis rate; number

Answer 9

• 11 polymerases discovered in eukaryotes so far
• E. coli: several polymerases, but pol I and pol III play the major roles
• bacteria: 500 nucleotides per second
• humans: 50 per sec

Question 10

DNA synthesis

Answer 10

• enzymes called DNA polymerases add nucleotides to a preexisting chain
• each nucleotide added to a growing strand comes from a nucleoside triphosphate (DNA pol III)
• as each monomer joins the growing end of a strand, 2 hposphate groups are lost as a molecule of pyrophosphate
• subsequent hydrolysis of the pyrophosphate to 2 molecules of INORGANIC PHOSPHATE is a coupled exergonic rxn that helps drive the polymerization rxn

Question 11

leading strand

Answer 11

• DNA pol III remains in the replication fork on that template strand and continuously adds nucleotides to the new complementary strand as the fork progresses
• only 1 primer required

Question 12

synthesis of lagging strand

Answer 12

1. primase joins RNA nucleotides into a primer
2. DNA pol III adds DNA nucleotides to the primer, forming Okaz frag 1
3. after reaching the next RNA primer “ahead” of it, DNA pol III detaches
4. DNA pol I replaces the RNA (primer) w/ DNA
5. DNA ligase forms a bond btwn the newest DNA and the DNA of the “ahead” fragment

Question 13

DNA ligase

Answer 13

joins the final nucleotide of the replacement DNA segment (that replaces the RNA of the primer) to the first DNA nucleotide of the adjacent Okaz frag
- joins the sugar-phosphate backbones of all Okaz frag into a continuous DNA strand

Question 14

DNA replication complex

Answer 14

• the various proteins that participate in DNA replication form a single large complex
• the complex may not move along the DNA; rather, the DNA may move thru the complex
• eukaryotes: multiple copies of the complex may be anchored to the nuclear matrix

Question 15

roles of primase

Answer 15

by interacting w/ other proteins at the fork, primase apparently acts as a molecular break,

• slowing progress of the replication fork
• coordinating the placement of primers and the rates of replication on the leading and lagging strands.

Question 16

proofreading

Answer 16

DNA polymerases proofread each nucleotide against its template as soon as it is added to the growing strand. Upon finding an incorrectly paired nucleotide, the polymerase removes the nucleotide and then resumes synthesis.

Question 17

mismatch repair

Answer 17

• when mismatched nucleotides evade proofreading by a DNA polymerase, other enzymes remove and replace incorrectly paired nucleotides that have resulted from replication errors.
• a hereditary defect in one of these repair enzymes is associated w/ a form of colon cancer.

Question 18

UV light

Answer 18

• one resulting type of damage is the covalent linking of THYMINE bases that are adjacent on a DNA strand
• such THYMINE DIMERS cause the DNA to buckle and interfere w/ DNA replication.

Question 19

one DNA repair system

Answer 19

• a segment of the strand containing the damage is excised (cut out) by a DNA-cutting enzyme – a NUCLEASE – and the resulting gap is then filled in w/ nucleotides, using the undamaged strand as a template
• the enzymes involved in filling the gap are a DNA polymerase and DNA ligase
• one such DNA repair system is called NUCLEOTIDE EXCISION REPAIR

Question 20

UV light: disorder

Answer 20

• xeroderma pigmentosum
• in most causes, caused by an inherited defect in a nucleotide excision repair enzyme
• individuals are hypersensitive to sunlight; mutations in their skin cells caused by UV light are left uncorrected, resulting in skin cancer

Question 21

shortening of the ends of DNA molecules

Answer 21

• linear molecules
• lagging strand: after the end primer is removed, it can’t be replaced w/ DNA b/c there’s no 3’ end available for DNA polymerase to add nucleotides

Question 22

telomeres

Answer 22

• eukaryotic chromosomal DNA molecules have these special nucleotide sequences at their ends
• no genes; just multiple repetitions of one short nucleotide sequence
• in each human telomere: the 6-nucleotide sequence TTAGGG is repeated btwn 100 and 1,000 times
• specific proteins associated w/ telomeric DNA prevent the staggered ends of the daughter molecule from activating the cell’s systems for monitoring DNA damage. (staggered ends, which often result from double-strand breaks, can trigger signal transduction pathways leading to cell cycle arrest / cell death)

Question 23

telomerase

Answer 23

• enzyme that catalyzes the lengthening of telomeres in eukaryotic germ cells, thus restoring their original length and compensating for the shortening that occurs during DNA replication
• NOT active in most human somatic cells, but its activity in germ cells results in telomeres of max length in the zygote

Question 24

telomere shortening

Answer 24

• normal shortening may protect organisms from cancer by limiting the number of divisions that somatic cells can undergo
• cells from large tumors often have unusually short telomeres `
• further shortening would presumably lead to self-destruction of the tumor cells; however, researchers have found telomerase activity in cancerous somatic cells, suggesting that telomerase’s activity to stabilize telomere length may allow these cancer cells to persist

Question 25

levels of chromatin packing

Answer 25

(eukaryotic)
1. DNA, the double helix
2. histones
3. nucleosomes “beads on a string” (10-nm fiber)
4. 30-nm fiber
5. looped domains (300-nm fiber)
6. metaphase chromosomes

Question 26

histones

Answer 26

• each histone is small; contains only ~100 amino acids
• total mass of histone in chromatin approximately equals the mass of DNA
• > 0.2 of a histone’s amino acids are positively charged (LYSINE/ARGININE) and therefore bind tightly to the negatively charged DNA
• 4 types are most common: H2A, H2B, H3, H4

Question 27

nucleosome

Answer 27

• beads on a string (nucleosomes are the beads); 10 nm in diameter
• basic unit of DNA packing; the ‘string’ btwn ‘beads’ is called linker DNA
• a nucleosome consists of DNA wound twice around a protein core composed of 2 molecules each of the 4 main histone types
• the amino end (N-terminus) of each histone (HISTONE TAIL) extends outward from the nucleosome

Question 28

30-nm fiber

Answer 28

• results from interactions btwn the histone tails of one nucleosome and the linker DNA and nucleosomes on the other side. these interactions cause the extended 10-nm fiber to coil/fold, forming a chromatin fiber roughly 30 nm in thickness
• a 5th histone, H1, is involved
• quite prevalent in the interphase nucleus

Question 29

looped domains

Answer 29

• the 30-nm fiber forms loops called LOOPED DOMAINS attached to a chromosome scaffold made of proteins, thus making up a 300-nm fiber
• the scaffold is rich in one type of TOPOISOMERASE; H1 also present

Question 30

interphase chromatin

Answer 30

although it lacks an obvious scaffold, its LOOPED DOMAINS appear to be attached to the nuclear lamina, on the inside of the nuclear envelope, and perhaps also to fibers of the nuclear matrix

Question 31

heterochromatin

Answer 31

• even during interphase, the centromeres and telomeres of chromosomes, as well as other chromosomal regions, exist in a highly condensed state similar to that seen in a metaphase chromosome
• visible as irregular clumps w/ a LM
• largely inaccessible for transcription b/c of its compaction

Question 32

euchromatin

Answer 32

The less condensed form of eukaryotic chromatin that is available for transcription.