Exam 2

Question 1

Repetitive DNA

Answer 1

1. Multi-copy genes: 2. Minisatellites and microsatellites: 3. Sines and lines:

Question 2

Multi-copy genes:

Answer 2

like rRNA, tRNA and histones genes these are considered both gene sequences and repetitive sequences.

Question 3

Minisatellites and microsatellites:

Answer 3

these are stretches of dinucleotide, trinucleotide or tetranucleotide repeats that occur sporadically along the length of all chromosomes. 5’- CACACACACA- 3’ long stretches (50 – 60 b)→heterochromatin regions → used in forensics.

Question 4

Sines and lines:

Answer 4

these are short or long stretches of DNA which reappear in many locations on the chromosomes; they appear to be viral in nature and may trace back to ancient retrotransposons.

Question 5

Position Effects:

Answer 5

• Genes that are moved from one chromosomal region to another (via inversion or translocation) or genes that are introduced by a virus, may suffer from “position effects”. This can happen if: 1. A gene (from a transcriptionally active region) is moved or inserted into a region of heterochromatin → silent the gene. OR 2. A gene from a transcriptionally repressed region is moved into a region of euchromatin. → overexpression (cancer)

Question 6

Heterochromatin

Answer 6

• Refers to chromosomal regions of highly repetitive DNA
• A-T rich
• Gene poor
• Transcriptionally inactive
• Remains condensed during all phases of cell cycle
• Retains Giemsa stain during karyotyping
• Dark bands
• Genes that are not needed that much
• Telomers, centromeres.

Question 7

Euchromatin

Answer 7

• Refers to chromosomal; regions which are not highly repetitive
• G-C rich
• Gene dense
• Transcriptionally active
• Condenses during M-phase; relaxed during interphase
• Does not retain Giemsa stain during karyotyping
• Light bands
• Genes that are needed a lot (housekeeping)
• Actin, tubuline for cell anatomy, ribosomes.
• 4 times more genes than heterochromatin

Question 8

Paternity and forensic tests

Answer 8

Repetitive sequences

Question 9

% of human genome that is repetitive DNA

Answer 9

about 50%

Question 10

Name for highly repetitive DNA

Answer 10

Satellites Ex. Centromeric and telomeric regions

Question 11

Telomeres:

Answer 11

• has 100s of repeating bp.
• G-rich-stranded overhang. It forms almost a 3 strand at the G-rich region
• Telomeric sequences usually consist of repeated units of a series of adenine or thymine nucleotides followed by G (guanine) nucleotides. 5’-TTAGGGG-3’ → toward end of chromosome.

Question 12

Chromatin structure

Answer 12

1) DNA double helix: 2nm
2) Histones (H2a,H 2b, H3, H4) (2 of each formthe octamer): Positively charged proteins (high % arginine and lysine).
3) Histone 1: It is not part of the core particle → binds 20-22 bp of DNA where the DNA joins and leaves the octamer
4) Nucleosome: DNA wrapped about 2 times around an octamer of 8 histone proteins
5) Chromatosome: 1 Nucleosome + 1H1 (histone)
6) Solenoids: Folding of nucleosomes 30nm
7) Long fiber loops: 300 nm (not coiled)
8) Wide fiber loops: 250 nm (coiled)
9) one chromatid fully condensed: 700 nm
10) Chromosome: thigh coiling of the 250 nm fiber produces the chromatid or a chromosome. 1,400 nm.

Question 13

3 Types of secondary structures that DNA can assume

Answer 13

1. B form
2. A form
3. Z form

Question 14

B-DNA structure form

Answer 14

• It is the biological relevant form of DNA, the structure described by Watson and Crick.
• Alpha helix, Right handed clockwise, spiral.
• Exist if plenty of water is present around the molecule

Question 15

A-DNA structure form

Answer 15

• Exists under dehydration conditions.
• Also a right-handed helix, but more compact.

Question 16

Z-DNA structure form

Answer 16

• LEFT-handed helix
• First observed from synthetic DNA containing only C-G base pairs (100bp machine would make it).
• Not clear whether Z-DNA exists in the cell, but if it does, it could be important for interaction with other molecules.

Question 17

Diameter of double helix of DNA

Answer 17

2 nm = 20 Å

Question 18

Separation between base pairs (bp)

Answer 18

0.34 nm = 3.4 Å

Question 19

1. Number of bp for every 360°
2. Distance for every turn of 360°

Answer 19

1. 10 bp
2. 3.4 nm = 34 Å

Question 20

Nucleotides:

Answer 20

1. Purines (Pure As Gold): Adenine and Guanine
2. Pyrimidine: Cytosine, Thymine and Uracil.

Question 21

Purines

Answer 21

• Adenine
• Guanine

Question 22

Pyrimidines

Answer 22

1. Citosine,
2. Thymine
3. Uracil

Question 23

DNA structure (# of bonds between nucleotides)

Answer 23

1. Adenine = Thymine (double bond)
2. Guanine = Cytosine (triple bond)

Question 24

Mischer (1868)

Answer 24

• Determined that the nucleus contained a novel substance that was lightly acidic and high in phosphorus. This material consisted of DNA and protein. Due to its occurrence in the cells’ nuclei, he termed the novel substance “nuclein”—a term still preserved in today’s name deoxyribonucleic acid.
• From nuclei of leukocytes found in pus from bandages.
• He investigated the chemistry of nucleic acids, but never determined their purpose or function.
• At that time, the consensus was that cells were largely made of proteins.

Question 25

Kossel Late 1800s

Answer 25

• (novel prize 1910)
• Discovered that DNA contains 4 nitrogen bases: adenine, cytosine, guanine, thymine, and uracil.

Question 26

Levene (around 1910)

Answer 26

• DNA consists of a larger number of linked repeating units called nucleotides; each nucleotide contains: Sugar + phosphate + base.
• Wrong hypothesis: DNA consists of a series of four-nucleotide units, each unit containing all 4 bases in a fixed sequence (tetranucleotide theory) → DNA is not variable enough to be the genetic material.

Question 27

Fred Griffith (1928)

Answer 27

• The pneumococcus bacterium occurs naturally in two forms with distinctively different characteristics. The virulent (S-strain) form has a smooth polysaccharide capsule that is essential for infection. The nonvirulent (R-strain) lacks the polysaccharide capsule, giving it a rough appearance. Mice injected with S-strain of the pneumococcus bacteria die from pneumonic infection within a few days, while mice injected with the R-strain bacteria continue to live. Injection with heat-killed S-strain bacteria also results in the mice surviving.
• Griffith was surprised to find in his experiments that mice injected with a mixture of heat-killed S-strain and live but non-virulent R-strain produced lethal results. In fact, Griffith discovered living forms of the S-strain bacteria in the infected mice !
• He hypothesize that the R-strain bacteria had somehow been transformed by the heat-killed S-strain bacteria. Some “transforming principle”, transferred from the heat-killed S-strain, had enabled the R-strain to synthesize a smooth polysaccharide coat and become virulent.

Question 28

Avery, McLeod, and McCarty
(1944)

Answer 28

• Discovered that “R” bacteria could become deadly simply when combined with inert lethal “S” form in a test tube—the mice were unnecessary to the equation.
• They, like Griffith, attempted to transform the R strain into the S strain by incubating living R and heat-killed S. They pre-treated the heat-killed bacteria (S strain) by making 3 samples one with RNase (destroys RNA), the second with Protease (destroys proteins) and the third one with DNase (destroys DNA).
• Then they mixed the 3 samples with R strain and got the following
• 1st R strain + RNas = Type R and type S
• 2nd R strain + Protease = Type R and type S
• 3rd R strain + DNase = Type R only
• They concluded that DNA was the transforming susbtance –> Genetic information resides in DNA

Question 29

Erwin Chargaff (1950)

Answer 29

• Found that a peculiar regularity in the ratios of nucleotide bases.
• In the DNA of each species he studies, the number of adenines approximately equaled the number of thymine, and the number of guanines approximately equaled the number of cytosine.
• In human DNA, for example, the four bases are present in these percentages: A=30.9% and T=29.4%; G=19.9% and C=19.8%. The A=T and G=C equalities, later known as Chargaff’s rules, helped Watson and Crick to discover the structure of DNA.

Question 30

Hershey and Chase (1952)

Answer 30

• In their experiments, they used two radioactive markers to label the proteins and the DNA of the phages. The proteins were labeled with 35S (a radioactive form of sulfur) and the DNA was labeled with 32P, a radioactive form of phosphorus. This allowed the researchers to easily differentiate between a sample that contained protein (thus would have 35S present) and a sample that only contained DNA (thus would have 32P present).
• During a phage infection, it was hypothesized that some part of the phage was injected into the bacterium and it was this injected material that conveyed the genetic material necessary to produce new phages. Hershey and Chase determined that the phage injected only the DNA into the bacterium and concluded that DNA must be the genetic material in phages.
• This was done by a series of two experiments in which different sets of non-radioactive bacteria were incubated with phages that had either their protein or their DNA labeled (as described above).
• They allowed the phages to infect the bacteria for a short time, they agitated the incubations to dislodge any loose parts of the phages.
• The bacteria cells were then pelleted (sedimentar) in a centrifuge and the location of the radioactivity. They found that the radioactive DNA was always found with the bacteria cells and that the radioactive protein was always in the supernatant. This suggested that the DNA was injected into the bacteria but the protein coat was not. Thus all of the information needed to produce new viruses was contained in the DNA and not the protein.

Question 31

Franklin and Wilkins

Answer 31

Used X-ray diffraction to study DNA.

Question 32

Watson, and Crick (1953)

Answer 32

Proposed the model for the 3-D structure of DNA by using Rosalind Franklin’s X-ray diffraction pictures of DNA.

Question 33

Three proposed models of DNA replication:

Answer 33

1. Conservative replication
2. Dispersive replication
3. Semiconservative replication:

Question 34

CONSERVATIVE DNA REPLICATION

Answer 34

In this model the two parental DNA strands are back together after replication has occurred. That is, one daughter molecule contains both parental DNA strands, and the other daughter molecule contains DNA strands of all newly-synthesized material.

Question 35

SEMICONSERVATIVE DNA REPLICATION

Answer 35

In this model the two parental DNA strands separate and each of those strands then serves as a template for the synthesis of a new DNA strand. The result is two DNA double helices, both of which consist of one parental and one new strand.

Question 36

DISPERSIVE DNA REPLICATION

Answer 36

In this model the parental double helix is broken into double-stranded DNA segments that, as for the Conservative Model, act as templates for the synthesis of new double helix molecules. The segments then reassemble into complete DNA double helices, each with parental and progeny DNA segments interspersed.

Question 37

Matthew Meselsohn and Frank Stahl

Answer 37

• In 1958, Matthew Meselsohn and Frank Stahl published evidence of a semi-conservative model of DNA replication in bacteria

Question 38

type of DNA Replication in eukaryotes (hint: 3 different models were proposed)

Answer 38

Semi-Conservative from multiple origens.

Question 39

• There is a “_____________” and “_____________” strand associated with each replication fork

Answer 39

• leading
• lagging

Question 40

Leading strand is also called ______________

Lagging strand is also called ______________

Answer 40

• Continuous DNA synthesis
• Discontinuous DNA synthesis

Question 41

Synthesis of DNA in Bacteria (4 requuirements)

Answer 41

1. Template
2. dNTPs (deoxynucleoside triphosphates)
3. Polymerase (Pol)
4. Primer (which is synthesized by primase)

Question 42

Five types of bacterial Pol:

Answer 42

Pol I, Pol II, Pol III, Pol IV, Pol V

Question 43

DNA Polymerase I

• Function

1. 5’ —> 3’ polymerization
2. 3’ —> 5’ exonuclease (proof-reading back)
3. 5’ —> 3’ exonuclease (replaces nucleotides ahead)

Answer 43

• Removes and replaces primers

1. yes
2. yes
3. yes

Question 44

• DNA Polymerase III
• Function

1. 5’ —> 3’ polymerization
2. 3’ —> 5’ exonuclease (proof-reading back)
3. 5’ —> 3’ exonuclease (replaces nucleotides ahead)

Answer 44

• Elongates DNA

1. yes
2. yes
3. no

Question 45

DNA Pol II, IV, and V functions

Answer 45

DNA Pol II, IV, and V have functions in DNA repair.

Question 46

of base pairs (bp) per second in DNA replication

Answer 46

1,000 bp/second

Question 47

Components required for replication in bacterial cells

Answer 47

1. Initiator protein: binds to origen and separates strands of DNA to initiate replication
2. DNA helicase: unwinds DNA at replication fork
3. SSBP (single-strand-binding proteins): attach to single stranded DNA and prevent secondary structures from forming
4. DNA gyrase: moves ahead of the replication fork, making and resealing breakes in the double-stranded helical DNA to release the torque that builds up as a result of unwinding at the replication fork
5. DNA primase: synthesizes a short RNA primer to provide a 3’-OH groups for the attachment of DNA nucleotides
6. DNA polymerase III: elongates a new nucleotide strand from 3’-OH group provided by the primer
7. DNA polymerase I: removes RNA primers and preplaces them with DNA
8. DNA ligase: joins Okazaki fragments by sealing breaks in the sugar-phosphate backbone of newly synthesized DNA.

Question 48

The “cohesive model” actually shows that

Answer 48

DNA Pol III is a dimer and extends on the leading and lagging strand simultaneously.

Question 49

Eukaryotes genomes and speed of replication vs. bacteria

Answer 49

• Eukaryotes have much bigger genomes and slower polymerases than bacteria, but could replicate their genomes faster (if all origins were activated at the same time).

Question 50

The Problem with Eukaryotic Telomeres

Answer 50

1. Chromosome ends shorten with each round of replication, because the primer removed from the end of the new chain cannot be filled back in.
2. For many somatic cells, shortening of telomeres is part of a normal biological clock for aging.
3. Leonard Hayflick discovered that most cells divide a certain number of times, then enter senescence. Apoptosis usually follows.
4. The number of cell divisions a normal cell can perform is called the “Hayflick limit.” It is usually between 30-50.

Question 51

How is Cellular “immortality” provided by telomerase?

Answer 51

1. Telomerase is a eukaryotic enzyme that can extend telomeres using a built-in RNA template.
2. Most somatic cells do not produce telomerase; thus, they eventually stop dividing and die.
3. Some normal cells (germ cells, white blood cells) and cancer cells contain active telomerase, which extends cell life.
   Telomerase is a target for some cancer drugs.

Question 52

1. Why do you think bacteria are never found that have null alleles for any of these proteins (DNA replications proteins)?
2. What kinds of mutations might be tolerated?

Answer 52

1. the reason is because null alleles are a mutant copy of a gene that completely lacks that gene’s normal function. Thus replication wouldn’t occur. –> no cell daugthers would be produced.
2. hypoactive proteins —> proteins still works but slow.

Question 53

DNA ligase

Answer 53

• After the last nucleotide of the RNA primmer has been repleaced, a nick remains in the sugar-phosphate backbone of the strand.
• DNA ligase seals this **nick **with a phosphodiester bond between the 5’-P groups of the initial nucleotide added by DNA polymerase III and the 3’-OH group of the final nucleotide added by DNA polymerase I.

Question 54

True or false: Histones are acidic and therefore negatively charged.

Answer 54

False. Histones are basic and therefore positively charged. The tails of histone proteins probably interact with the negatively changed phosphate groups of DNA.

Question 55

Polar amino acids

Answer 55

Serine

Threonine

Cysteine

Tyrosine

Asparagine

Glutamine

Question 56

Non-polar amino acids

Answer 56

Glycine

Alanine

Valine

Leucine

Isoleucine

Proline

Methionine

Phenylalanine

Tryptophan

Question 57

Amyotrophic lateral sclerosis (ALS)

Answer 57

1. Amyotrophic lateral sclerosis (ALS) is a devastating neurodegenerative disease that causes muscle weakness, disability and eventually death. ALS is inherited in most cases in an autosomal dominant pattern. Many patients would consider assisted suicide.
2. The disease severity, treatment ineffectiveness and increasing dependence on caregivers give rise to thoughts to terminate life among patients, especially those at an advance stage.
3. In ALS1, mutation in the Cu-Zn Superoxide Dismutase 1 gene (SOD1) on chromosome 21q22.1 correlates with the ALS phenotype. SOD1 is an enzyme that catalyzes the destruction of O2- free radicals, thus conferring protection to cells from the harmful effects of superoxide free radicals. The mutations in the SOD1 gene are thought to be pathogenic through a gain in toxic property (e.g. by misfolding and aggregation) rather than a loss of function of the protein.
4. Disease duration or rate of disease progression does correlate with some mutations. Particularly the A4V** mutation (Deng, 1998) causes about 50% of ALS1 in North America families being consistently associated with a rapid course from symptoms onset until death. This is a missense mutation where a single base pair substitution results in the translation of a different amino acid at that position. **The amino acid alanine is replaced by valine in the genetic code, introducing an incorrect amino acid into the protein sequence.
5. Another mutation is I113T (substitution of the amino-acid Isoleucine 113to the amino-acid threonine)**, which also occurs at the SOD1 dimer interface. These mutations exhibited reduced superoxide scavenging ability due to_misfolding of the homo-dimer at the interface._**
6. Currently there are two hypothesis that describe the mechanism by which mutations in SOD1 causes familial ALS.
7. The oligomerization hypothesis, in which normally soluble, functional proteins misfold to form some toxic multimeric species. These aggregated proteins can be also bind to other essential proteins inside the neurons making them unavailable to perform their function.
8. The second hypothesis is the oxidative damage where mutant CuZnSOD proteins are mutated at the metal-binding regions. This mutant CuZnSOD catalyzes reactions with hydrogen peroxide that damages important substrates for cell survival.
9. In summary oligomerization in CuZnSOD proteins seem to be a very important explanation for the toxicity of these proteins in the neurons in patients with ALS. The upper motor neurons and the lower motor neurons degenerate or eventually die, and stop sending messages to muscles. Unable to function, the muscles gradually weaken and atrophy.

Question 58

Totipotent stem cells

Answer 58

have unlimited capabilities and can become any type of cell

Question 59

Pluripotent stem cells

Answer 59

have the option to become many types of cells - BUT N0T ALL

Question 60

Multipotent stem cells

Answer 60

can become a limited specific type of cell (hematopoietic)