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Exam 1 Flashcards

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Biology 101 flashcards
1
Q

genome

A

entire set of genetic info in a given organism

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2
Q

circular vs linear chromosomes

A

circular:
- found in pro/euk
- found in pro cytoplasm
- found in euk mitochondria/chloroplasts
- loosely packed

linear:
- found in euk
- found in euk nucleus
- tightly packed (compact around histone proteins)

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3
Q

histones

A

DNA and its associated proteins in eukaryotes only

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4
Q

complexity of an organism tends to increase with…

A

genome size
– not necessarily able to predict relative genome size based on the complexity of the organism

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5
Q

genome size is NOT proportional to

A

the number of genes
**there must be other factors at play for size differences

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6
Q

why is the mRNA length of euk. genes more variable as compared to prokaryotes?

A

1) introns account for mRNA and gene length changes in eukaryotic genes (mRNA length will be smaller)
2) differences in genes that these proteins are encoding for

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7
Q

All genes (pro. and euk.) contain 3 things:

A

1.) coding region (exon)
2.) regulatory region
3.) transcription termination sequence

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8
Q

coding region

A

EXON
- contains the information for the structure of the expressed protein

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9
Q

regulatory region

A

info on WHERE and WHEN a gene will be transcribed during development; usually upstream of the coding region

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10
Q

transcription termination sequence

A

STOP signal for where transcription should end; downstream of coding region

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11
Q

gene organization of euk VS pro

A

prokaryotes:
- less variation of genes
- smaller genes (less bps)
- less genes (less compact genes)

eukaryotes:
- more variation of genes
- larger genes (more bps and introns)
- more genes (more compact genome)
- more space between genomes (other function genes. not coding genes)

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12
Q

Griffith experiment

A
  • S-living = dead
  • R-living = alive
  • S-dead = alive
  • R-living + S-dead = dead (living S were present when rough transformed to smooth by transfer of transforming factor)
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13
Q

Avery, McCarty, MacLeod Experiment

A
  • tested for transformation by destroying components of transforming substance 1-by-1
    *determined that the active component (transforming factor) in S.pneumonae is DNA

DNAase –> DNA destroyed –> introduced into R cells –> R cells
(no transformation to S cells like in protease, RNase, centrifuging)

*physical/chemical analysis indicated a predominance of DNA

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14
Q

Hershey and Chase experiment

A

“blender experiment”
- 32P = DNA
- 35S = proteins of phages

  • phages were infected with bacteria and blended to separate phage particles from bacteria; then centrifuged

1) 35S showed radioactivity in supernatant
- no radioactivity in the cells
- *proteins are NOT genetic material

2) 32S showed radioactivity in pellet
- radioactivity enters cells
- *DNA is genetic material

**Phages transform DNA, not protein, into their host
(protein is not genetic material)

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15
Q

phage

A

protein + DNA

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16
Q

Watson and Crick

A

1st physical model of DNA

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17
Q

Chargaff

A

how bases must pair (ratio of purine to pyrimidine)

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18
Q

Roseland Franklin

A

helical properties of DNA (width, rotation)

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19
Q

DNA

A
  • polymer of repeating nucleotide monomeric units

Monomers are made up of:
1) Nitrogenous bases
2) pentose sugar
3) phosphate group

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20
Q

nitrogenous bases

A

*on 1’ carbon

A + G = Purine
C + T = pyrimindines

A + T = 2 bonds
C + G = 3 bonds

  • equal ratios of purines to pyrimidines (50/50)
  • purine and pyrimidine pairing maintains constant width
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21
Q

purines vs pyrimidine ring #

A

purine = 2 rings
pyrimidine = 1 ring

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22
Q

pentose sugar

A

*pentose with oxygen and hydroxyl group
- the other nucleotides part attach to sugar backbone

  • deoxyribose in DNA; ribose in RNA
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23
Q

phosphate group

A

*on 5’ carbon of pentose sugar
- has a (-) charge at physiological pH

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24
Q

polar

A
  • DNA has an overall direction with distinct ends
  • 5’ and 3’ ends
  • polarity of the monomer is preserved in the polymer
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25
Q

DNA is antiparallel

A

parallel in 2 different directions

5’ –> 3’
3 <– 5’

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26
Q

DNA structure summary

A

5’ phosphate group
3’ hydroxyl
1’ nitrogenous base
*all attached to sugar pentose

  • phosphate attaches to 3’ carbona and replaces OH-
  • 3 bonds between C & G
  • 2 bonds between A & T
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27
Q

DNA growth

A
  • DNA grows when a 5’ triphosphate reactions with 3’ OH of another nucleotide
  • cleaves the high-energy phosphate bond – makes this an energetically favorable reaction
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28
Q

DNA Polymerase III has 3 requirements for synthesizing new DNA

A

1) dNTPS
2) 3’ OH group
3) DNA template

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29
Q

dNTPs

A

deoxyribonucleotide triphosphate monomers
- buildings blocks for new DNA that bring their own energy to the reaction

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30
Q

how does OH add new dNTPs onto strand?

A

3’ OH adds new dNTPS with a primer

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31
Q

DNA template

A

determines which complementary dNTP gets added to the new strand

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32
Q

DNA synthesis direction

A

5’ –> 3’ direction of strand being built

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33
Q

bidirectionality

A
  • DNA replication!!
  • each replication bubble has 2 replication forks… one traveling in each direction AWAY from the origin of replication (ori)
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34
Q

AZT nucleotide mimic

A
  • can be added to chain, but no addiational additions can occur bc it is missing a 3’ OH
  • first drug for HIV therapy, but there are better modern therapies
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35
Q

pro VS euk replication

A

pro: replicate bidirectionality from 1 ori
euk: replicate from many oris so many replication bubbles per chromosomes

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36
Q

topoisomerases

A

relieve tension and disentangle the 2 daughter DNA chromosomes when DNA replication is complete

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37
Q

Linear DNA and end of replication problem

A
  • end of replication problem when final primers are removed –> unfinished section of the lagging strand
  • linear chromosomes shorten after every cell direction
  • telomerase counteracts this by extending the telomeres
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38
Q

telomerase

A
  • ribonucleoprotein that extends telomeres by adding more repeats (repetitive DNA)
  • contains RNA that serves as a template for extending the overhang
  • extends telomeres to prevent linear chromosomes from shortening after every cell division

*primer attaches after initial extension of the overhang so that the other strand can be lengthed too
1) lengthen overhang
2) go back and extend short section

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39
Q

reverse transcription

A

RNA –> DNA

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40
Q

PCR

A

polymerase chain reaction
(DNA replication in a test tube)

1) denaturation
2) annealing primers
3) extensions

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41
Q

denaturation

A

HOT - 95 C
- dsDNA is separated into single strands
- H-bonds break

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42
Q

annealing primers

A

COLD - 55 C
- primers bind to single-stranded templates (DNA primers)

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43
Q

extension

A

HOT - 72 C
- Taq synthesizes DNA
- thermophilic bacteria that can withstand heat of denaturation)

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44
Q

where are DNA primers located?

A

on 3’ ends of DNA strands
- synthesized in the opposite direction
(5’ –> 3’ at 3’ end of DNA along DNA strand toward the 5’ end of the template )

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45
Q

in vivo vs pcr

A

1) helicase VS heat for strand separation
2) DNA polymerase VS Taq for elongation
3) RNA primers VS DNA primers
4) Ligase needed for OFs of lagging strand VS no lagging
5) both need nucleotides for elongation; use DNA at a template for new strand synthesized from 5” –> 3’

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46
Q

mismatch repair

A

fix errors in DNA replication that arent corrected by DNA polymerase

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47
Q

MutS and MutL

A

recognize mismatches

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48
Q

MutH

A

nicks DNA (cuts backbone)

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49
Q

Exonucleases

A

remove nucleotides around nick (including mismatch)

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50
Q

methylation in prokaryotes

A

the strand that is methylated is recognized for errors
- otherwise it may not be recognizable as to whether the parent or daughter strand has the error

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51
Q

damaged nucleotides can be repaired by:

A

1) base excision repair
2) nucleotide excision repair

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52
Q

when does NER occur?

A

nucleotide excision repair only occurs when base excision repair doesn’t work

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53
Q

base excision repair steps

A

1) deaminated DNA with uracil (C–> U)
2) glycosylase removes uracil, leaving an AP (apurinic site)
3) AP endonuclease cutes the backbone to make a nick at the AP site
4) DNA exonucleases remove multiple nucleotides near the nick, creating a gap
5) DNA polymerase synthesizes new DNA to fill in the gap
6) DNA ligase seals the nick

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54
Q

nucleotide excision repair steps

A

1) exposure to UV light
2) thymine dimer forms
3) UvrB and C endonucleases nick strand containing dimer
4) Damaged fragment is released from DNA
5) DNA poly fills in the gape with new DNA
6) DNA ligase seals the repaired strand

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55
Q

glycosylases

A

proteins that remove improper bases directly off the nucleotide backbone

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56
Q

AP endonuclease

A

nick DNA at AP site

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57
Q

UvrA and UvrB

A

recognize DNA distortions

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58
Q

UvrB and UvrC

A

nicks DNA (cuts backbone)

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59
Q

ds breaks repair mechanisms (2)

A

1) homology-directed repair (HDR)
2) non-homologous end-joining (NHEJ)

*gene editing

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60
Q

homology-directed repair (HDR)

A

more accurate than NHEJ bc it uses homologous DNA from sister chromatid
*more precise

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61
Q

non-homologous end-joining (NHEJ)

A

variable length indels
broken ends are joined together
- less accurate than HDR
- can cause mutations

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62
Q

mutations

A
  • PERMANENT alteration in DNA sequence; not repaired
  • new sources of ALLELES that are acted upon by evolution
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63
Q

4 categories of point mutations

A
  1. molecular nature
  2. phenotypic effect
  3. location
  4. cause
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64
Q

substitution

A

changing one nucleotide to another
1) transition: Pu-Pu or Py-Py
2) transversion: Pu-Py or Py-Pu

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65
Q

indels (insertion and deletion)

A

bases added or removed

*structural change to molecular nature (large scale change to chromosome organization
- large deletions
- inversions
- translocations

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66
Q

substitutions can cause..

A

1) missense
2) nonsense
3) silent

mutations

67
Q

missense

A

changing from one amino acid to another

68
Q

nonsense

A

changing from one amino acid to STOP; truncates protein (loss of function)

UAG UAA UGA

69
Q

silent

A

no change to protein amino acid sequence

70
Q

indels cause…

A

frameshifts

71
Q

frameshifts

A

many codons are affected; often has a disorder and early truncation

*avoided/fixed if shift is divisible by 3

72
Q

loss of function

A

less function (recessive –> must have phenotype in a WT/mutant heterozygote)

73
Q

null

A

complete loss of function

74
Q

gain of function

A

new or more function

75
Q

only — mutations are passed to offspring

A

germline mutations (can arise in any type of cell though)

76
Q

spontaneous mutation

A

natural processes or random chance cause mutation

77
Q

induced mutation

A

caused by mutagen

78
Q

tautomers

A

alternate, temporary configurations of bases

  • rare forms have different base pairing properties than typical forms
    (pu pairs to wrong py and vise versa)
79
Q

tautomeric shift

A

spontaneous mutation example
- not a mutation, BUT can cause mutations

(only a temporary shift but can cause base pairing problems down the line that cause permanent change for mutation)

80
Q

positive control

A

ensures that the assay gives positive results when it should
* protects against false negatives

81
Q

negative control

A

ensures that the assay gives negative results when it should
* protects against false positives

82
Q

mitosis VS meiosis

A

Mitosis:
- makes somatic cells (asexual)
- 2 identical cells as product
- no variation
- 1 equal division
- somatic cells
- developmental problems, cancer if goes wrong

Meiosis:
- makes gametes
- 4 non-identical gametes
- variation present
- 2 divisions (not equal)
- cells for gonads (gametes for egg and sperm)
- nonviable gametes or chromosome imbalances if goes wrong

83
Q

n (haploid number)

A
  • number of chromosomes in a haploid gamete
  • how many unique chromosomes in one haploid “set”
  • each unique chromosomes has different genes than the others
84
Q

human chromosome count

A

23 unique types of chromosomes (1-22 + X/Y)
- diploid (2n)
- somatic cells have 2 homologous sets of 23 chromosomes (1 from egg and 1 from sperm)

85
Q

ploidy

A

how many homologous “sets” of chromosomes a cell has

86
Q

homologous chromosomes

A

have the same genes BUT could have different alleles

87
Q

2n

A

the number of chromosomes in a diploid cell
2n = the total # of chromosomes in a diploid cell

88
Q

c (DNA content)

A

the amount of DNA in a haploid gamete
- c = amount of DNA in a haploid cell before DNA replication

89
Q

DNA replication doubles the amount of DNA in a cell…

A

cells DNA content goes from 2c –> 4c
- if 2n cell before S phase has a DNA content of 2c

90
Q

2 stages that produce variation in meiosis

A

1) prophase I = crossing over of alleles
2) metaphase I = independent orientation leading to independent assortment

91
Q

crossing over

A

generates variation by generating new combinations of alleles on a particular chromosome

92
Q

alignment of one set of homologs in meiosis –

A

independent of all other sets

93
Q

independent assortment

A

generates variation by generating new combinations of chromosome homologs in a particular gamete

94
Q

euploidy

A

having only complete sets of chromosomes
(1n, 2n, 4n..)

95
Q

aneuploidy

A

the occurrence of one or more extra or missing chromosomes leading to an unbalanced chromosome complement

96
Q

2n with trisomy

A

2n +1

97
Q

2n with monosomy

A

2n -1

98
Q

nondisjunction

A

failure of homologs (anaphase 1) or sister chromatids (anaphase 2 or mitotic anaphase ) to segregate into different daughter cells

99
Q

Nondisjunction in MI vs MII

A

MI:
- heterozygous duplicate (Aa); homologs don’t separate
* all aneuploid gametes/offspring
* n+1 gametes contain 2 homologs
MII:
- homozygous duplicate (AA); sister chromatids don’t separate
* 1/2 aneuploid gametes and offspring
* n+1 gametes contain 2 sisters

100
Q

genetic variation arises bc…

A

1) mutation generates new alleles
2) in sexually-reproducing organisms, meiosis generates gametes with new haploid combinations of alleles by crossing over or independent assortment
3) fertilization brings together new 2n allele combos

101
Q

crossing over vs independent assortment

A

new combos on the same chromosome (CO) vs across different chromosomes (IA)

102
Q

allele interactions (dominance relationships) emerge in…

A

diploids
(phenotype could reflect one or both alleles of allele 1 or 2 compared to haploids who reflect allele 1 phenotype)

103
Q

WT vs dominant allele

A

WT appears as most common trait
dominant allele appears in heterozygote phenotype

104
Q

alleles themselves are not dominant or recessive..

A

dominance relationships describes how 2 alleles interact in a heterozygote

“The mutant allele is dominant to the WT allele”

105
Q

haplo(in)sufficiency

A

whether a single WT allele can produce a WT phenotype
- determined by examining the phenotype of a hemizygote (an organism containing only 1 allele for a gene)

106
Q

hemizygote

A

an organism containing only 1 allele for a gene

107
Q

why hemizygote

A
  • aneuploidy
  • heterozygous for a chromosomal deletion (deficiency)
  • sex chromosomes (found on X, but not on Y)
  • 2n with monosomy by nondisjunction
  • deletion on homolog (1 copy only of effected genes)
108
Q

threshold of WT phenotype for haplo(in)sufficiency

A
  • threshold of WT phenotype when WT allele is haploinsufficient will be HIGHER than the threshold when WT allele is haplosufficient
109
Q

loss of function mutant allele types

A
  1. amorphic (null) - less (none)
  2. hypomorphic (leaky) -less (some)

as compared to WT allele

110
Q

gain of function mutant allele types

A
  1. hypermorphic - more
  2. neomorphic - new
  3. antimorphic (dom-neg) - new, but antagonizes WT
111
Q

loss/gain of function mutations can reduce/increase…

A

DNA –> RNA –> protein

LOF
- reduces transcription (less RNA and protein)
- reduces translation (less protein)
- reduces protein activity

GOF
- increase transcription (more RNA and protein)
- increase translation (more protein)
- increase protein activity OR cause new activity

112
Q

complementary base pairing ensures…

A

semiconservative replication

113
Q

transformation

A

bacteria transfer genes from one strain to another
- DNA from a donor is added to the bacterial growth medium and is taken up from the medium by recipient or transformat
(DNA is the active agent of transformation)

114
Q

semiconservative replication

A

1 strand original and 1 strain new DNA for each 2 daughter cells (1st gen)
- 2nd gen is 2 new strands and 2 mixed strands (identical to 1st gen)

115
Q

conservative

A

1 helix is original and 1 helix is new DNA (1st gen)
-2nd gen is 3 new strands and 1 original strand

116
Q

dispersive

A

2 helixes are both mixed with new and original DNA blocks (1st gen)
- 2nd gen is 4 of the same mixed strands

117
Q

insertion right before the transcribed region will cause?

A

reduced transcription

118
Q

insertion right before the protein-coding region and within the transcribed region will cause?

A

reduced translation

119
Q

semiconservative replication conserves…

A

chromosome number and identity during mitotic cell division
- bc the 2 sister chromatids are identical in base sequence to each other and to the original parental chromosome

120
Q

DNA polymerase

A

enzyme that forms a new DNA strand during replication by adding nucleotides reverse complementary to a template (1-by-1), to the 3’ end of a growing strand

121
Q

DNA replication requirements for DNA polymerase action

A
  1. 4 DNTPS
  2. ssDNA template (dsDNA is unwound by DNA poly)
  3. primer with 3’ hydroxyl (primer is a short ssDNA or RNA that base pairs as part of template strand)
122
Q

initiation

A
  • proteins bind to inititator protein
  • initator protein attracts helicase which unwinds DNA
  • DNA unwinds into replication bubble with 2 Y shaped area called replication forks
  • SSBPs stabalize DNA and keep them seperated
  • DNA poly III adds nucleotides to 3’ end of preexisiting DNA strand
  • RNA primer initates DNA synthesis with primase
123
Q

elongation

A
  • DNA poly III catalyzes polymerization
  • DNA grows 5 to 3
  • DNA poly moves 3 to 5
  • DNA poly moves in the same direction as the fork to synthesize the leading strand
  • the new DNA strand is the lagging stand and replicated 5 to 5 away from Y-fork in Okazaki fragments
  • DNA poly I replaces the RNA primer of OFs with DNA
  • Ligase bonds fragments
124
Q

leading strand

A

replicated continuously 5’ to 3’ toward the unwinding y-fork

125
Q

lagging strand

A

the new DNA strand is the lagging stand and replicated 5 to 5 away from Y-fork in Okazaki fragments

126
Q

DNA topisomerases

A

relax supercoils by nicking DNA strands and cleaving the sugar-phosphate backbone between 2 adjoining nucleotides

  • supercoiling is when chromosomes accommodates strain of distortion by twisting back upon itself
127
Q

bacteria chromosomes

A
  • circular
  • 1 ori
  • 1 termination region
  • replication is bidirectional
128
Q

how is integrity of genetic info preserved?

A

redundancy
precision of replication machinery
enzymes repair chemical damage to DNA

129
Q

HDR definition

A

removal of a small region from the DNA strand that contains the altered nucleotide and then using the other strand as a template to resynthesizes the region removed

130
Q

mismatch repair occurs

A

AFTER DNA replication

131
Q

depurination

A

DNA alternation in which a purine base (G+A) is hydrolyzed from the deoxy-phosphate backbone

132
Q

deamination

A

removal of an amino (NH2) group from normal DNA

133
Q

radiation

A

cosmic rays and x-rays break sugar-phosphate background

134
Q

thymine dimers

A

covalent linkage between adjacent thymine residues in DNA that cause mutation

135
Q

base tautomerization

A

interconversion of bases between 2 similar forms or tautomers (each base has 2 tautomers)

  • wrong base is incorporated and point mutation occurs if It is not fixed
136
Q

Ames test

A

screen for chemicals that cause mutations in bacterial cells
- Many His- – His+ will grow without histidine

137
Q

gametes

A

specialized cells (eggs and sperm) that carry genes between generations (each contain half the material for making new progeny)

138
Q

chromosomes

A

self-replicating DNA/protein complexes in the nucleus that contain genes

139
Q

mitosis

A

division that produces 2 d. cells that are the same number and type of chromosomes as the original parent cell

140
Q

meiosis

A

process of 2 consecutive cell divisions in the progenitors of gametes

1) homologs separate into 2 different daughter cells
2) chromatids of each chromosome segregate into 2 different daughter cells

results in 4 unique haploid daughters containing half of the # of chromosomes found

141
Q

sister chromatid

A

2 identical copies of chromosome that exist after DNA replication

  • held together by cohesins
142
Q

centromere

A

specific place at which sister chromatids are attached to each other

143
Q

homologous chromosomes

A

homologs, pair of chromosomes containing the same linear sequence of genes (1 from each parent)

144
Q

nonhomologous chromosomes

A

carrying different sets of unrelated genetic info

145
Q

cell cycle steps

A

G1
Synthesis (S)
G2
mitosis

146
Q

G1

A

new cell birth (growth)
- chromosomes are not duplicating or dividing

147
Q

Synthesis (S)

A

cell duplicate its genetic material by synthesizing DNA
- each chromosome doubles to produce sister chromatids

148
Q

G2

A

cells grows more
synthesizes protein for mitosis

149
Q

mitosis

A

sister chromatids separate and 2 daughter nuclei form

150
Q

crossing over

A

exchange of sections of DNA and then the rejoining of chromosomes

PROPHASE 1

151
Q

recombination

A

offspring derive of combo of alleles different from that of either parent

152
Q

haploinsufficiency requirements

A

property of a gene in 2n’s where 2 WT functional gene copies are required

153
Q

deletion

A

loss of nucleotide pairs from DNA

154
Q

inversion

A

180 degree rotation of a segment of a chromosomes relative to the rest

155
Q

duplication

A

chromosomal rearrangement where the number of DNA copies increases (paired with duplications often)

occur on homologous chromosomes

156
Q

translocation

A

2 breaks 1 in each of 2 homologous chromosomes

– fragments switch places and attach on the nonhomo chromo

157
Q

tandem vs nonrandom duplications

A

repeated copies lie adjacent to each other VS duplications are not adjacent to one another

158
Q

unequal crossing over

A

recombination following misalignment of homologs where 1 homo chromo ends up with a duplication and the other sustains a deletion

159
Q

lagging strand

A

synthesized discontinuously toward the 5’prime end of template strand
(closest to the 3’ of template)

160
Q

leading strand

A

synthesized continuously toward the 5’ end of the template strand from (5’->3’)

  • arrow points to 5’ end
161
Q

pulse vs pulse chase experiments

A

pulse:
pulse of radioactive dnTPS are used up for synthesis quickly and then cells are killed
– dna is extracted and denatured and separated by size
–**many small pieces bc ligase doesn’t have time to mend pieces

pulse chase:
- allows some time for DNA synthesis to occur (chase)
– less pieces and larger sizes bc ligase had time to mend pieces together

162
Q

ames test

A

his- –> his+ (colonies found in disk bc His is necessary for bacteria to grow) with the addition of chemical indicates that it has some mutational properties that allows the His- to mutate

look for high number of revertants (suggests that mutation occurred)

163
Q

base vs excision repair

A

Base excision repair is a pathway that repairs replicating DNA throughout the cell cycle. Nucleotide excision repair is a pathway that repairs constantly damaging DNA due to UV rays, radiation and mutagens.

genetics (5 decks)

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