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Microbiology > Chapter 9 + 10 > Flashcards

Chapter 9 + 10 Flashcards

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Biology 101 flashcards
1
Q

Genetics

A

The study of inheritance of biological characteristics by living things (heredity)
-The study of genes

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2
Q

Genetic Material

A

DNA (deoxyribonucleic acid), a Polymer of Nucleotides

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3
Q

Gene

A

A sequence of nucleotides that codes for a product (a polypeptide)

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4
Q

Levels of Genetic Study

A

Organism level–> Cell Level–> Chromosome level–> Molecular level

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5
Q

DNA: Deoxyribonucleic acid

A

The genetic material

  • DNA Molecule: 2 strands of nucleotides bonded to each other with hydrogen bonds
  • DNA Nucleotide: Subunit or Monomer; a nucleotide strand consists of nucleotides bonded together with covalent bonds
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6
Q

DNA Molecule Consists of?

A

2 strands of nucleotides bonded to each other with hydrogen bonds

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7
Q

DNA Nucleotide consists of?

A

Nucleotides bonded together with covalent bonds

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8
Q

DNA Bases

A

A=T

G=C

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9
Q

RNA Bases

A

A=U

G=C

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10
Q

Properties of DNA Molecule

A
  • Polymer of nucleotides
  • Double-stranded; Helix
  • Antiparallel
  • Complementary Base Pairing (A=T, G=C)
  • Semiconservative Replication (open DNA molecule, save strand from original and make a new one)
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11
Q

Genome

A

The sum total of genetic material carried within a cell

  • Eukaryote- plasmid, nucleus, nucleolus, chloroplast, , mitochondrion, extrachromosomal DNA, chromosomes
  • Prokaryote- chromosome + plasmids
  • Viruses- DNA + RNA
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12
Q

Genotype

A

Chemical composition of an organisms DNA
ex:
-homozygous
-heterozygous

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13
Q

Phenotype

A

How genes are expressed (hair color, skin type, personality, color of colony, shape + arrangement)

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14
Q

Flow of Information

A

DNA–>DNA
>Cell To Cell:
-Horizontal Gene Transfer (transformation, transduction, conjugation)
-Vertical Gene Transfer (mitosis and binary fission) (prokaryotes do; done before mitosis or binary fission)

DNA–>RNA–>Polypeptides
-Within a cell

DNA–(Replication)–>DNA–(Transcription)–>RNA–(Translation)—>Protein
DNA–>DNA–>RNA–>Protein

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15
Q

Semiconservative Replication

A

A. The parent molecule has 2 complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C.
B. The first step in replication is separation of the two DNA strands.
C. Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand.
D. The nucleotides are connected to form the sugar-phosphate backbones of the new strands. Each “daughter” DNA molecule consists of one parental strand and one new strand.
-(Has a parental helix, replication fork and replicas)

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16
Q

Eukaryotic Replication

A 
Location: Nucleus
Enzymes involved:
-Helicase
-Primase
-DNA polymerase III
-DNA polymerase I
-Ligase
-Gyrase
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17
Q

Eukaryotic Replication Enzyme: Helicase

A

unzipping the DNA helix

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18
Q

Eukaryotic Replication Enzyme: Primase

A

Synthesizing an RNA primer

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19
Q

Eukaryotic Replication Enzyme: DNA Polymerase III

A

adding bases to the new DNA chain; proofreading the chain for mistakes

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20
Q

Eukaryotic Replication Enzyme: DNA polymerase I

A

removing RNA primer, closing gaps, repairing mismatches

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21
Q

Eukaryotic Replication Enzyme: Ligase

A

final binding of nicks in DNA during synthesis and repair

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22
Q

Eukaryotic Replication Enzyme: Gyrase

A

supercoiling

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23
Q

Eukaryotic Replication: Leading Strand

A

-Unwind the helix (helicase)
-Stabilize open (SSBP)
-Primase (RNA polymerase) starts attaching to RNA nucelotides in 5’ to 3’ direction
-DNA polymerase takes over and attaches new nucleotides to already existing 3’ end of primer in 5’ to 3’ direction
>Template: 3’ to 5’
>New strand: 5’ to 3’

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24
Q

Eukaryotic Replication: Lagging Strand

A

-Unwind the helix (helicase)
-Stabilize (SSBP)
-Primase (RNA polymerase) starts attaching RNA nucleotides in 5’ to 3 direction
-DNA polymerase takes over and attaches new nucleotides to already existing 3’ end of primer
-DNA polymerase copies a short fragment 5’ to 3’
-DNA ligase joins the short fragments together
>Template: 5’ to 3’
>New Strand: 3’ to 5’ overall

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25
Q

Prokaryotic Replication

A

Location: Cytoplasm

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26
Q

DNA Transcription

A

Location:
>Prokaryotic Cell: Cytoplasm
>Eukaryotic Cell: Nucleus
-Prokaryotic Cell: in a cell lacking a nucleus, mRNA produced by transcription is immediately translated without additional processing
-Eukaryotic Cell: The nucleus provides a separate compartment for transcription. The original RNA transcript, called pre-mRNA, is processed in various ways before leaving the nucleus as mRNA

27
Q

DNA Transcription in Prokaryotic Cell

A

Location: Cytoplasm

-In a cell lacking a nucleus, mRNA produced by transcription is immediately translated without additional processing

28
Q

DNA Transcription in Eukaryotic Cell

A

Location: Nucleus
-The nucleus provides a separate compartment for transcription. The original RNA transcript, called pre-mRNA is processed in various ways before leaving the nucleus as mRNA

29
Q

DNA Transcription: RNA Synthesis 3 Steps

A
  1. Initiation
  2. Elongation
  3. Termination
30
Q

DNA Transcription Example

A
  1. A gene composed of exons and introns is transcribed to RNA by RNA polymerase
  2. Processing involves ribozymes and proteins in the nucleus to remove the intron-derived RNA and splice together the exon-derived RNA into mRNA
  3. After further modification, the mature mRNA travels to the cytoplasm, where it directs protein synthesis
31
Q

Triplet

A

3 DNA bases

  • a Triplet= 1 codon or
  • 3 DNA bases= 1 codon
32
Q

DNA-Protein Relationship

A 
3 bases (triplet) codes for 1 codon;
1 codon codes for 1 amino acid
33
Q

Genetic Code

A
  • all polypeptides start with AUG (methionine); initiates translation
  • stop codes: UAA, UAG, UGA (stop translation)
34
Q

all polypeptides start with what in the genetic code

A

AUG (methionine); initiates translation

35
Q

Stop codes in the genetic code

A

-UAA
-UAG
-UGA
(stops translation)

36
Q

Translation Location

A

Location:

Both Prokaryotic and Eukaryotic Cells: Cytoplasm

37
Q

Translation Participants

A

-mRNA
-rRNA
-tRNA
-Amino Acids
-ATP
>Components needed to begin translation come together

38
Q

Translation Process

A
  • Initiation
  • Elongation
  • Termination
    1. Entrance of tRNAs 1 and 2; tRNA 1 is in P site, tRNA 2 is in A site
    2. Formation of peptide bond between 1 and 2
    3. Discharge of tRNA 1 at E site
    4. First translocation: tRNA 2 shifts over into p site; tRNA 3 enters ribosome at A site
    5. Formation of peptide bond between tRNA 2 and 3
    6. Discharge of tRNA 2; second translocation; tRNA 3 enters P site, tRNA 4 enters ribosome in A site
    7. Formation of peptide bond between tRNA 3 and 4
    8. Steps 3 to 5 are repeated until stop codon is reached and protein synthesis is terminated
39
Q

Polyribosome

A

complex of ribosomes strung along a single strand of mRNA that translates the genetic information coded in the mRNA during protein synthesis
-complex of an mRNA molecule and two or ore ribosomes that act to translate mRNA instructions into polypeptides

40
Q

Regulation of Gene Expression

A
  • Induction- turns on transcription

- Repression- Turns off transcription

41
Q

Regulation of Gene Expression: Induction

A

turns on transcription

42
Q

Regulation of Gene Expression: Repression

A

turns off transcription

43
Q

Lactose Operon in bacteria

A

inducible genes are controlled by substrate

44
Q

Lactose Operon in bacteria ON

A

upon entering the cell, the substrate (lactose) becomes a genetic inducer by attaching to the repressor, which loses its grip and falls away. The RNA polymerase is now free to bind to the promotor and initiate transcription, and the enzymes produced by translation of the mRNA perform the necessary reactions on their lactose substrate

45
Q

Lactose Operon in bacteria OFF

A

in the absence of lactose, a repressor protein (the product of a regulatory gene located elsewhere on the bacterial chromosome) attaches to the operator of the operon. This effectively locks the operator and prevents any transcription of structural genes downstream (to its right). Suppression of transcription (and consequently of translation) prevents the unnecessary synthesis of enzymes for processing lactose

46
Q

Repressible Operon

A

Genetic control through excess nutrient

47
Q

Repressible Operon ON

A

a repressible operon remains on when its nutrient products (here, arginine) are in great demand by the cell. The repressor cannot bind to the operator at low nutrient levels

48
Q

Repressible Operon OFF

A

the operon is repressed when (1) arginine builds up and, serving as a corepressor, activates the repressor. (2) The repressor complex affixes to the operator and blocks the RNA a polymerase and further transcription of genes for arginine synthesis

49
Q

Mutations

A
  • changes in the genetic material of a cell
  • point mutations are changes in just one base pair of a gene
  • the change in a single nucleotide in the DNA’s template strand may lead to production of an abnormal protein
50
Q

Example of Mutation

A
  1. During DNA replication, a thymine is incorporated opposite guanine by mistake
  2. In the next round of replication, adenine pairs with the new thymine, yielding a A-T pair in place of the original G-C pair
  3. When mRNA is transcribed from the DNA containing this substitution, a codon is produced that, during translation, codes for a different amino acid, tyrosine instead of cysteine
51
Q

Wild-Type hemoglobin DNA / Mutant hemoglobin DNA

A
  1. In the DNA, the mutant template strand has an A where the wild-type template has a T
  2. The mutant mRNA has a U instead of an A in one codon
  3. The mutant (sickle-cell) hemoglobin has a valine (Val) instead of glutamic acid (Glu)
52
Q

Types of Mutations

A
  1. Spontaneous- random, by chance, for no reason

2. Induced- something that causes a mutation

53
Q

DNA Repair Mechanism

A
  • Endonuclease- cuts the DNA
  • DNA polymerase- fills the gap
  • DNA ligase- joins it together
54
Q

DNA Repair Mechanism in action

A
  1. Exposure to ultraviolet light causes adjacent thymines to become cross-linked, forming a thymine dimer and disrupting their normal base pairing
  2. An ENDONUCLEASE cuts the DNA, and an exonuclease removes the damaged DNA
  3. DNA POLYMERASE fills the gap by synthesizing new DNA, using the intact strand as a template
  4. DNA LIGASE seals the remaining gap by joining the old and new DNA
55
Q

Ames Test for Mutagenicity

A

-In the control set up, bacteria are plated on a histidine free medium containing liver enzymes but lacking the test agent
-The experimental plate is prepared the same way except that it contains the test agent
a.Control Plate; minimal medium, lacking histidine and test chemical
b. Test Plate; minimal medium, with test chemical and no histidine
>Incubation (12 hrs) any colonies that form have back-mutated to his(+)
After Incubation:
a. Control plate; his (+) colonies arising from spontaneous back-mutation
b. Experimental plate; his (+) colonies in presence of the chemical
c. The degree of mutagenicity of the chemical agent can be calculated by comparing the number of colonies growing on the control plate with the number on the test plate. Chemicals that induce an increased incidence of back-mutation are considered carcinogens.

56
Q

Genetic Recombination

A
  • Transfer of genetic material between two different sources
  • exchange of genes between DNA molecules to form a new combination of genes on a molecule
  • Required: Donor + Recipient
57
Q

Recombinant DNA

A

the newly created molecule containing genes from two different sources
-Recipient with incorporated donor DNA

58
Q

Eukaryotic Recombination

A
  • meiosis

- fertilization

59
Q

Prokaryotic Recombination

A

-Natural
>rare event between closely related species
>three processes; conjugation, transduction, transformation)
-Artificial
>manipulation of genes to cause transfer of genetic material between closely related species as well as between completely unrelated species

60
Q

Natural Recombination in Prokaryotes: 3 processes

A
  • Transformation
  • Transduction
  • Conjugation
61
Q

Recombination: Transformation

A
  • Fred Griffith
  • 1928
  • Subject: Streptococcus pneumoniae
  • Transfer of naked DNA in solution
  • Factors involved: Free donor DNA (fragment or plasmid), live, competent recipient cell
  • Indirect
  • Examples of Genes Transferred: Polysaccharide capsule; unlimited with cloning techniques; metabolic enzymes
62
Q

Recombination: Conjugation

A

-Transfer of DNA via a Plasmid
-Donor= F+ (contains a fertility factor aka a plasmid)
-Recipient= F- (does not contain a fertility factor)
-Factors involved:
>Donor cell with pilus
>fertility plasmid in donor
>both donor and recipient alive
>bridge forms between cells to transfer DNA
-Direct
-Examples of genes transferred: Drug resistance; resistance to metals; toxin production; enzymes; adherence molecules; degradation of toxic substances; uptake of iron

63
Q

Recombination: Transduction

A

-Transfer of DNA via a Bacterial Virus (phage)
-Factors involved:
>donor is lysed bacterial cell
>defective bacteriophage is carrier of donor DNA
>live recipient cell of same species as donor
-Indirect
-Examples of genes transferred: Exotoxins; enzymes for sugar fermentation; drug resistance

Microbiology (20 decks)

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