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Fundamental genetics > Exam 3 > Flashcards

Exam 3 Flashcards

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Biology 101 flashcards
1
Q

Replica Plating

A
  • determines genotype for several colonies by seeing which bacteria grows when placed in a selective medium lacking a specific nutrient required
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2
Q

Prototroph

A
  • wild type bacteria that can synthesize all compounds needed for growth from simple ingredients
  • has all the nutrients required to grow

Ex: leu+, try+, F+

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3
Q

Auxotroph

A
  • mutant strain that LACKS one or more enzymes required for metabolizing nutrients
  • WILL grow on a supplemented media as long as new plate has the + on it

Ex: F-, leu- etc

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4
Q

Lederberg and Tatum Experiment

A
  • showed antibiotic resistance in bacteria
  • when they transferred the bacteria from the master plate to a new plate with penicillin on it the colonies that survived and grew were the ones with the resistance gene to penicillin
  • a filter plate placed between strain A and B did not allow the bacteria to pass through
  • THEREFORE genetic exchange requires direct contact between bacterial cells
    Ex: thi+ means can make it
    Thi- means can’t make it needs to be on plate
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5
Q

Conjugation

A
  • Direct (requires contact) transfer of DNA from one bacterium to another
  • bacterial conjugation transfers plasmids
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6
Q

Horizontal gene transfer

A
  • exchange genetic material between bacteria of the same generation (siblings)
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7
Q

Plasmids

A
  • extrachromosomal DNA replicates independently of chromosome

Ex: plasmid can contain an antibody resistance

F+/F-/Hfr/F’

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8
Q

F+/F-/Hfr/F’

A

Types of plasmids
F=fertility
F+=has the fertility plasmid (donates)
F-= doesn’t have the fertility plasmid (receives)
Hfr= high frequency of transferring F plasmid, part of bacterial chromosome is attached to the plasmid
- behave like F+, but sometimes does transfers both chromosome and plasmid and forms F’
- Hfr cross w F- can transfer both Hfr and F+
- so use to map bacterial chromosomes
F’= all plasmid not fully transferred

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9
Q

interrupted conjugation Mapping

A
  • interrupted mating experiment
  • Hfr used to map genes present
  • as transfers takes more time for different parts of the DNA genes
  • *GENES NEAR THE ORIGIN OF TRANSFER ARE TRANSFERRED FIRST aka are the HIGHEST on the graph**
  • the transfer times indicate the order and relative distances between genes as they are transferred in a linear fashion and can be used to construct a genetic DNA map
  • **for any two genes transferred from donor to recipient, all genes residing in between them have been transferred bc in a linear fashion
  • two genes that are very close may appear to transfer at the same time
  • it is necessary that all Hfr cells be absent from the population of cells recovered for genotyping**
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10
Q

Merozygote

A
  • bacteria that is now a partial diploid (some chromosome is transferred with the F gene)

the F factor and several adjacent genes are EXCISED from the chromosome of a Hfr cell and transferred to an F- strain

  • F factor is EXCISED from Hfr, some genomic genes are on the episome (plasmid that are able to integrate into the chromosome)
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11
Q

In an interrupted mating experiment, the purpose of plating cells on a Selective medium is to

A
  • ensure that only recombinant genotypes are recovered
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12
Q

Transduction

A
  • bacterial DNA is transferred to recipient by viruses (phages)
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13
Q

Transformation

A
  • donor cell dies and releases DNA fragments, recipient takes up these strands from the median, or environment
  • a cell must be competent to be able to take up these foreign cells
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14
Q

Mechanism of transformation

A

1) foreign DNA strands is released from death of donor cell and are floating around environment (lysed or hydrolyzed)
2) recipient bacteria cell will take up this DNA from the environment
3) foreign DNA and chromosomal DNA RECOMBINE OR HETERODUPLEX together,
4) some bacterial DNA adds to bacterial chromosome, whatever doesn’t remains strands and some bacterial chromosome becomes strands
- new DNA can give bacterial cell new functions
Ex: if it took in an antibody

  • homologous recombination occurs in regions where the two types of DNA are similar and then the donor cell DNA becomes part of the recipient cell chromosome
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15
Q

Gene mapping for transformations

A
  • genes that are closer together have higher %of being co-transformed
  • rate of contransformation and distance between genes are inversely proportional
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16
Q

Nucleosides

A
  • sugar+base
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17
Q

Nucleotides

A
  • makes up DNA
  • when a nucleoside goes through hydrolysis it becomes this
  • consist of a sugar, a phosphate and a nitrogen containing base
  • sugar-phosphate backbone
    Phosphate gives DNA a negative charge, hydrophilic and polar
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18
Q

Nitrogenous bases

A
  • purine and pyrimidine

Purine- “ Pure as silver(Ag)”
-> adenine and guanine

Pyrimidine- “pyramids have a sharp edge and sharp edges CUT”
-> cytosine, thymine(in DNA), uracil(in RNA)

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19
Q

DNA structure

A
  • DNA DOUBLE HELIX is constructed from two strands of DNA (B-DNA most common)
  • each strand is made up of polynucleotides consisting of nucleotides (sugar, phosphate and bases)
  • each strand has a SUGAR-PHOSPHATE BACKBONE and NITROGENOUS BASES that form HYDROGEN BONDS holding the two strands together
  • DNA strands run in opposite direction=ANTIPARALLEL-> 5’-3’ and 3’-5’
  • 3’ end has a HYDROXYL(OH) group on the DEOXYRIBOSE SUGAR
  • 5’ end has a PHOSPHATE GROUP
  • every turn of the helix means 10 bp
  • major and minor grooves for the binding
    sites of proteins and base pair recognition
  • sugar=deoxyribose
  • Phosphodiester linkage links nucleotides together assigns 5 and 3
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20
Q

Chargaffs Rule

A

A and T bond

G and C bond

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21
Q

Watson Crick DNA structure

A
  • found DNA structure using Wilkins and Franklins X-ray diffraction/crystallography photographs

1) DNA consists of 2 polynucleotide chains that run in an antiparallel fashion
- one strand goes 5’->3’ from top to bottom and the other goes 5’->3’ from bottom to top
2) backbone is made up of sugar and phosphates on the exterior part of the DNA (phosphodiester linkage)
- phosphate contains negative charges that are stabilized by aqueous environment that contains polar molecules
3) interior is made up of nitrogenous bases that are nonpolar and perpendicular to common axis
- 10 nitrogenous bases in every turn of DNA molecule

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22
Q

What holds the nitrogenous bases together

A
  • one purine always bonds to a pyrimidine
  • adenine to thymine and guanine to cytosine

1) HYDROGEN BONDS between the bases
2) vanderwaal forces
3) hydrophobic effect
- backbone is polar and nitrogenous bases are non polar

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23
Q

Frederick Griffith

A
  • first to find that there can be bacterial transformation
  • used streptococcus bacteria causes pneumonia
  • R (rough) can’t cause death, but live S (smooth) causes death when injected in mice
  • heats and kills S injects in mice they live aka only live bacteria affects mice
  • combines live R and dead S and the mice dies, bacterial transformation occurred and the bacteria inside the mouse became the live virulent (deadly) S strain
24
Q

Hershey and Chase Experiment

A
  • built the case for nucleic acids
  • labeled proteins with S35 and nucleic acids(DNA) with P32 radioactivity to see if they were passed on to progeny
  • the component associated with bacteria at the end of the experiment Must be the GENETIC material
  • if the blender was not used both preparations of bacteria would include ghosts and viral DNA, so both would be radioactive one with P32 and one with S35
    RESULT
  • used T2 bacteriophage
  • T2 proteins remained outside the host cell during infection and WERE NOT transmitted to progeny
  • T2 nucleic acids entered the cells and WERE transmitted to progeny
  • therefore since DNA not protein was associated with bacteria it must be the genetic material
25
Q

Avery, Macleod and McCarty

A

-expanded and isolated griffiths transformation principle

  • took the dead S strains and mixed with tubes that destroyed one thing (ex: lipid, RNA, proteins, polysaccharides, DNA and none were destroyed) then mixed with the R strain
  • in each one the R strain was transformed to the virulent S strain BUT in the strain that DNA was killed no live S strain was recovered
  • CONCLUSION: DNA IS HEREDITARY, THE TRANSFORMING SUBSTANCE IS DNA
26
Q

Ribose vs Deoxyribose structure

A

Ribose contains 3 OH groups

Deoxyribose contains 2 OH groups

27
Q

Purines Vs Pyrimidines in structure

A

Purines have 2 nitrogen containing rings

Pyrimidines have 1 nitrogen containing rings

28
Q

Central dogma

A

DNA-> RNA -> proteins(amino acids)
- flow of genetic info in living cells is from DNA to RNA to proteins

  • DNA can also participate in DNA replication and RNA can do RNA replication with the helping of protein
  • transcription is DNA to RNA
  • translation from RNA to proteins
  • reverse transcription is when RNA makes template for DNA
29
Q

DNA vs RNA structure

A
  • only difference is in the type of pentode sugar and in one base

DNA:

  • phosphate group
  • AGCT
  • deoxyribose (the presence in DNA as a sugar makes more stable and less susceptible to hydrolysis)

RNA:

  • phosphate group
  • AGCU
  • ribose (more susceptible to breakdown, to ensure correct proteins are maintained in cell)
30
Q

1) Conservative Replication
2) Dispersive Replication
3) Semiconservative replication

A

1) DNA molecule copies itself and makes a second new DNA molecule
2) DNA cuts itself at various parts and then makes copies and the old and new replicated DNA are attached to produce 2 DNA molecules

3) 2 DNA strands separate and each serves as a template to create a completely new identical DNA molecule strand to attach, thus producing 2 DNA molecules
- one old single strand and one new single strand form a double helix
- if goes through two cycles all four strands will serve as templates resulting in 4 duplexes or double stranded DNA, 2 will have only new strand other 2 will have one old one new

31
Q

Messelson and Stahl

A
  • proved Semiconservative replication to be true
  • all cells get an old DNA strand and one new strand
  • grew cells in a heavy nitrogen medium, sat on bottom of centrifuge
  • replicated it with light nitrogen and noticed it was half as heavy (half contained light nitrogen half contained heavy nitrogen)
  • replicated again and saw that either the DNA had two light strands or one light (new) one heavy(old)
32
Q

DNA replication in prokaryotes

A

1) initiation 2) elongation 3) termination

A. Initiation
- begins at a single origin of replication on a circular DNA cell
B. Unwinding
- helicase separates the 2 strands of DNA and creates a replication fork/bubble
- as being pulled apart it creates a strain on the DNA that is still in the helix
- Topoisomerase relieves the strain on the DNA strands around replication bubble

C. Priming

  • single strand binding proteins bind and stabilize the single stranded DNA
  • primase synthesizes an RNA primer at 5’ ends
  • polymerase cannot initiate DNA sequence without a 3 -OH group

D. Elongation
- DNA polymerase 3 synthesizes a new strand by polymerizing nucleophile using parent strand as a template

33
Q

DNA replication steps

A

1) the HELICASE binds to the origin of replication of double stranded DNA to help unwind it and moves to the fork of replication
2) it moves in the 3->5 direction and creates a replication fork made up of two single strands of DNA where DNA polymerase can replicate DNA, location of helicase
3) SSb or SINGLE STRAND BINDING PROTEINS bind to the exposed regions of the single stranded DNA to keep it from reassociating with the other strand
4) DNA GYRASE binds to the part of the DNA strand that is still double stranded to decrease stress that occurs for the unwinding
5) once unwinded into 2 single strands the DNA PRIMASE ENZYME synthesizes RNA PRIMERS to set a template of the sequence of nucleotides for polymerase to synthesize the daughter strands
6) DNA POLYMERASE 3 catalyzes the linkage of phosphodiester bonds that hold nucleotides together
- can ONLY synthesize DAUGHTER DNA in the 5’->3’ direction(add to 3 end), parent is 3’->5’ direction
7) the LEADING STRAND is synthesized continuously in the 5->3 direction towards the replication fork
8) the LAGGING STRAND can not be made by DNA polymerase 3 because it would go in the 3->5
- SO many RNA primers are formed and then DNA polymerase synthesizes away from the replication fork to create the 5->3 direction
- polymerase forms lagging strand in a piece wise fashion, short strands of DNA are synthesized in a discontinuous manner these pieces are known as OKAZAKI FRAGMENTS
9) once the fragments are formed DNA POLYMERASE I removes RNA primers and DNA LIGASE connects Okazaki fragments new and old to eachother by creating phosphodiester linkages and the DAUGHTER STRAND OF DNA is formed
- an RNA primer is found at the 5’ end of all newly synthesized DNA, required for DNA synthesis

34
Q

DNA helicase

A
  • breaks the hydrogen bonds between parietal DNA strands and unwinds the double helix to single strands
35
Q

Single stranded binding proteins

A
  • bind to the single strands of DNA preventing them from reassociating and allowing synthesis to occur on both strands
36
Q

DNA polymerase 3

A
  • synthesizes the new strands of DNA, but it requires 3’ hydroxyl (OH) group to add nucleotides
  • can only add nucleotides to the 3’ (OH) end of a new DNA strand
  • parental DNA strands go from 3->5 in opposite directions
  • so at a single replication fork one strand is synthesized away from the replication fork and the other is synthesized towards the replication fork
37
Q

Primase

A
  • creates short RNA primers, allowing for DNA polymerase 3 to initiate DNA synthesis on both template strands
38
Q

DNA polymerase 1

A
  • removes the RNA primers and replaces them with DNA Nucleotides
39
Q

DNA Ligase

A
  • on lagging strand joins catalyzes fornation of bonds between DNA fragments (Okazaki fragments) by forming phosphodiester bonds between them to fill in the nicks left there from removing the RNA primers
40
Q

Topoisomerase

A
  • Relieves the strain/supercooling on the DNA strands around replication bubble caused by the unwinding DNA
  • if did not work DNA would become tangled downstream of the replication fork
  • replication may start but would stop once it reaches the tangled DNA
41
Q

Heterochromatin vs Euchromatin

A
  • located inside DNA nucleus

Heterochromatin

  • HIGHLY CONDENSED CHROMATIN that are NOT ACTIVE
  • outer side of nuclear envelope

Euchromatin

  • inside middle of nuclear envelope
  • LIGHTLY PACKED CHROMATIN
  • ACTIVE
  • **contains OPEN/EXPOSED GENES that allow the polymerase to bind and begin the process of TRANSCRIPTION (DNA becomes mRNA to synthesize proteins that genes on DNA code for)
42
Q

Histones

A
  • proteins that provide structure and rule to DNA
  • DNA tightly WRAPS around the histone very tightly so it can fit in the nucleus
  • DNA+protein= chromatin
  • once histones are fully wrapped it is referred to as a NUCLEOSOME, made up of two copies of Each type of histone and DNA (8 total subunits+DNA)
  • histone subunits arrange themselves into 2 tetramers then they will combine with themselves and form an octamere
  • tetramere is composed of 2 units of each histones H2A and H2B and the other is composed of two units of histones H3 and H4
    4 Types of histones:
    1) H2A
    2) H2B
    3) H3
    4) H4
  • Histone tail is important for signaling process and control of gene regulation in eukaryotes
43
Q

Scaffold proteins

A
  • non histone proteins that play a role in the folding and packing of DNA
44
Q

Acetylation and Methylation (histone modifications)

A
  • chromatin remodeling: structure must change to allow access to DNA
  • controls expression of genes in eukaryotic cells

ACETYLATION(H4)- (HATS)

  • neutralizes the positive charge of histones that allow them to bond to a negatively charged DNA
  • decreases their positive charge and weakens the histone to DNA interaction
  • adds acetyl group to positively charged amino acid group present on the side chain of the amino acid lysine effectively neutralizing the charge
  • the histone will open/relax and the DNA will unwrap to allow access to the gene (Euchromatin)

METHYLATION(H3)- if have 2 X chromosomes in DNA want to deactivate one so will add a methyl group to H3 portion of gene and one chromosome will be inactive

45
Q

Chromatin

A
  • DNA wrapped(coil/condensed) around proteins (histones)
  • 4 subunits x 2 (tetrameres) = 8 (octamere)
  • chromosome is the pure coiled structure only seen during cell division
46
Q

Chromatin remodeling

A
  • when want to express gene we can remodel chromatin complex and unwind DNA from histone, remove histone from area and get access to the area
47
Q

Chromosome Banding

A
  • stains precise areas of chromosomes to see location, height, length, any issues
  • often based on G=C, A=T composition
  • unique to each chromosome

1) G banding
- most common

2) Q banding
- first banding technique, used UV light

3) R banding

48
Q

Condensing DNA into chromsomes Steps

A

1) HISTONES FORM NUCLEOSOME
- DNA (2nm in diameter) becomes smaller and wraps around histones, 8 histones form 2 tetrameres then become a octamere and form a nucleosome(11nm)

2) SOLENOID (30nm in diameter)
- nucleosome wraps in a helical fashion to form a coil called a solenoid

3) CHROMATIN/LOOPED DOMAINS (300nm in diameter)
- when supersolenoid are stacked on top of eachother they form a single fiber called chromatin

4) CHROMATID (700 nm diameter)
- chromatin coils until begins to form 2 chromatid strands

5) METAPHASE CHROMOSOME (1400nm diameter)
- chromatids coil together and become a chromosome ready for division

49
Q

H1 Histone

A
  • interacts with linker DNA and helps STABILIZE THE SOLENOID
50
Q

DNA vs RNA

A
  1. RNA is usually single stranded
    - can acquire shape
  2. RNA has a ribose sugar (3 OH)
  3. Ribonucleotides contain the pyrimidine base uracil instead of thymine
  4. RNA can be catalytic (ribozymes)

Structural differences
A=DNA B=RNA

  1. type of sugar
    A. Deoxyribose (2 OH) B. Ribose (3 OH)
  2. Presence of 2’ OH group
    A. No B. Yes
  3. Bases
    A. A, G, C, T B. A, G, C, U
  4. Double or single stranded
    A. Double B. Single
  5. Secondary structure
    A. Double helix B. Many, can change shape
  6. Stability
    A. Stable B. Easily degraded
51
Q

Genetic Code Rules

A
  • during translation from RNA to proteins translates sequence of nucleotides to corresponding sequence of amino acids on polypeptide chain

1) A 3 nucleotide sequence called a CODON is used to encode each amino acid
2) Code does not overlap, read via TRIPLETS
3) Code is read CONTINUOUSLY, sequence from beg to end
4) Code is DEGENERATIVE
- different codons that code for the same amino acid are called SYNONYMS
- minimizes effects of mutation

  • if reaches stop sequence tells the RNA to stop translation and it is the end of its protein synthesis process
  • read genetic code 1st left, 2nd up, 3rd right
52
Q

Messenger RNA (mRNA)

A
  • 5% of the cellular RNA
  • serves as a template for protein synthesis in a process called translation
  • in eukaryotes, distinct mRNA molecule for each gene
  • in prokaryotes mRNA molecules code for several genes
53
Q

Transfer (tRNA)

A
  • makes up 15% of cellular RNA
  • distinct stem loop structure and functions in carrying activated amino acids to the ribosome (the mRNA complex during protein synthesis)
  • each amino acid has at least one tRNA made specifically to carry it to the ribosome
54
Q

Ribosomal RNA (rRNA)

A
  • 80% of cellular RNA
  • major constituent of ribosomes (cell machinery that are responsible for synthesizing proteins)
  • rRNA gives the ribosome is structure and also plays a catalytic role in the formation of peptide bonds in polypeptide chain
  • in prokaryotes 3 types of rRNA make up ribosomes 235, 165 and 55 rRNA
55
Q

Codon

A
  • a 3 nucleotide sequence used to encode each amino acid
  • 3’ is the end of the codon
  • the possible codons are determined by repeating the copolymer use overlapping code to see combinations possible
    Ex: UG copolymer-> UGUGUGUG
  • will only ever get UGU or GUG

ANTICODON (tRNA)
- 5’ is the end of the anticodon

56
Q

Telomeres

A
  • maintain the end of chromosomes
  • serve as structural cap to block unraveling
  • replication of ends (not in somatic cells)
  • found in eukaryotes not prokaryotes
  • telomerase is an enzyme involved in the replication of the ends of eukaryotic chromosomes
57
Q

Nirrenberg and Mathhaei

A
  • figured out the sequence of input RNAs
  • short synthetic mRNAs of known sequence were added to a mixture of ribosomes and tRNA
  • tRNA BOUND TO THE RIBOSOME will STICK to the filter while the free one did not (triplet binding assay)

Fundamental genetics (15 decks)

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