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Genetics > Exam 2 > Flashcards

Exam 2 Flashcards

1
Q

Repetitive DNA

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

Multi-copy genes:

A

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

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3
Q

Minisatellites and microsatellites:

A

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.

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4
Q

Sines and lines:

A

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.

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5
Q

Position Effects:

A

• 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)

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6
Q

Heterochromatin

A
  • 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.
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7
Q

Euchromatin

A
  • 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
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8
Q

Paternity and forensic tests

A

Repetitive sequences

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9
Q

% of human genome that is repetitive DNA

A

about 50%

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10
Q

Name for highly repetitive DNA

A

Satellites Ex. Centromeric and telomeric regions

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11
Q

Telomeres:

A
  • 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.
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12
Q

Chromatin structure

A

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.

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13
Q

3 Types of secondary structures that DNA can assume

A
  1. B form
  2. A form
  3. Z form
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14
Q

B-DNA structure form

A
  • 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
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15
Q

A-DNA structure form

A
  • Exists under dehydration conditions.
  • Also a right-handed helix, but more compact.
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16
Q

Z-DNA structure form

A
  • 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.
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17
Q

Diameter of double helix of DNA

A

2 nm = 20 Å

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18
Q

Separation between base pairs (bp)

A

0.34 nm = 3.4 Å

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19
Q
  1. Number of bp for every 360°
  2. Distance for every turn of 360°
A
  1. 10 bp
  2. 3.4 nm = 34 Å
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20
Q

Nucleotides:

A
  1. Purines (Pure As Gold): Adenine and Guanine
  2. Pyrimidine: Cytosine, Thymine and Uracil.
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21
Q

Purines

A
  • Adenine
  • Guanine
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22
Q

Pyrimidines

A
  1. Citosine,
  2. Thymine
  3. Uracil
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23
Q

DNA structure (# of bonds between nucleotides)

A
  1. Adenine = Thymine (double bond)
  2. Guanine = Cytosine (triple bond)
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24
Q

Mischer (1868)

A
  • 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.
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25
Q

Kossel Late 1800s

A
  • (novel prize 1910)
  • Discovered that DNA contains 4 nitrogen bases: adenine, cytosine, guanine, thymine, and uracil.
26
Q

Levene (around 1910)

A
  • 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.
27
Q

Fred Griffith (1928)

A
  • 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.
28
Q

Avery, McLeod, and McCarty
(1944)

A
  • 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
29
Q

Erwin Chargaff (1950)

A
  • 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.
30
Q

Hershey and Chase (1952)

A
  • 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.
31
Q

Franklin and Wilkins

A

Used X-ray diffraction to study DNA.

32
Q

Watson, and Crick (1953)

A

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

33
Q

Three proposed models of DNA replication:

A
  1. Conservative replication
  2. Dispersive replication
  3. Semiconservative replication:
34
Q

CONSERVATIVE DNA REPLICATION

A

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.

35
Q

SEMICONSERVATIVE DNA REPLICATION

A

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.

36
Q

DISPERSIVE DNA REPLICATION

A

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.

37
Q

Matthew Meselsohn and Frank Stahl

A

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

38
Q

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

A

Semi-Conservative from multiple origens.

39
Q

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

A
  • leading
  • lagging
40
Q

Leading strand is also called ______________

Lagging strand is also called ______________

A
  • Continuous DNA synthesis
  • Discontinuous DNA synthesis
41
Q

Synthesis of DNA in Bacteria (4 requuirements)

A
  1. Template
  2. dNTPs (deoxynucleoside triphosphates)
  3. Polymerase (Pol)
  4. Primer (which is synthesized by primase)
42
Q

Five types of bacterial Pol:

A

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

43
Q

DNA Polymerase I

  • Function
  1. 5’ —> 3’ polymerization
  2. 3’ —> 5’ exonuclease (proof-reading back)
  3. 5’ —> 3’ exonuclease (replaces nucleotides ahead)
A
  • Removes and replaces primers
  1. yes
  2. yes
  3. yes
44
Q
  • DNA Polymerase III
  • Function
  1. 5’ —> 3’ polymerization
  2. 3’ —> 5’ exonuclease (proof-reading back)
  3. 5’ —> 3’ exonuclease (replaces nucleotides ahead)
A
  • Elongates DNA
  1. yes
  2. yes
  3. no
45
Q

DNA Pol II, IV, and V functions

A

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

46
Q

of base pairs (bp) per second in DNA replication

A

1,000 bp/second

47
Q

Components required for replication in bacterial cells

A
  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.
48
Q

The “cohesive model” actually shows that

A

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

49
Q

Eukaryotes genomes and speed of replication vs. bacteria

A

• 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).

50
Q

The Problem with Eukaryotic Telomeres

A
  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.
51
Q

How is Cellular “immortality” provided by telomerase?

A
  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.
52
Q
  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?
A
  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.
53
Q

DNA ligase

A
  • 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.
54
Q

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

A

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

55
Q

Polar amino acids

A

Serine

Threonine

Cysteine

Tyrosine

Asparagine

Glutamine

56
Q

Non-polar amino acids

A

Glycine

Alanine

Valine

Leucine

Isoleucine

Proline

Methionine

Phenylalanine

Tryptophan

57
Q

Amyotrophic lateral sclerosis (ALS)

A
  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.
58
Q

Totipotent stem cells

A

have unlimited capabilities and can become any type of cell

59
Q

Pluripotent stem cells

A

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

60
Q

Multipotent stem cells

A

can become a limited specific type of cell (hematopoietic)

Genetics (15 decks)

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