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Final Exam Flashcards

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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

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4
Q

complexity of an organism

A

— not necessarily able to predict relative genome size based on
— genome size is not the number of genes (seen in disproportional numbers)

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5
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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6
Q

all genes (both eukaryotic and prokaryotic) must have:

A
  1. regulatory region (info on where and when a gene will be transcribed during development [upstream])
  2. coding region (info for the structure of the expressed)
  3. transcription termination sequence (stop signal for where transcription should end [downstream])
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7
Q

prokaryotic vs eukaryotic genes

A

prokaryotic
- less variation of genes
- smaller genes (less bps)
- less compact genome

eukaryotic
- more variation of genes
- larger genes (exons and introns)
- more compact genes (histones)

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8
Q
A
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9
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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10
Q

histones

A

DNA and associated proteins in eukaryotes only

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11
Q

complexity of organism tends to…

A

increase with genome size
– not necessarily
bc genome size is not proportional to the number of genes

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12
Q

why is the mRNA length of eukaryotic 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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13
Q

P and E genes must contain:

A

1.) coding region (exon - info for protein being expressed)
2.) regulatory region (where and when a gene will be transcribed during development [upsteam])
3.) transcription termination (stop signal for where transcription should end [downstream])

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14
Q

gene organization eukaryotic vs prokaryotic

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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15
Q

Griffith Experiment

A

MC: some transforming factor is responsible for transformation of R into S cells

  • S-living = dead
  • R-living = alive
  • S-dead = alive
  • R-living & S-dead = dead + live S-cells
  • some transforming factor transformed the live R-cells into S-cells using dead S-cell material
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16
Q

Avery, McCarty, MacLeod Experiment

A

MC: DNA is the active component in transformation

  • tested by destroying a single part of transforming substance 1-by-1 and doing the experiment
  • no transformation occurred when DNA was destroyed and then introduced to R cells
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17
Q

Hershey-Chase Experiment

A

MC: DNA is the genetic material, not proteins

  • 32P DNA
  • 35S Protein
  • used phages (DNA+proteins)
  • testing by radiolabeling proteins and DNA and then infecting bacteria
  • proteins = no radioactivity entered the cell so supernatant showed radioactive 35S
  • DNA = radioactivity did enter the cell so pellet showed radioactive 32P

** “Blender experiment”

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18
Q

Watson/Crick

A

first model of DNA

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19
Q

Roseland Franklin

A

helical structure of DNA

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20
Q

Chargaff

A

how bases must pair together

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21
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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22
Q

Nitrogenous bases

A

*on 1’ carbon

A + G = Purine (2 rings)
C + T = pyrimindines (1 ring)

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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23
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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24
Q

phosphate group

A

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

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25
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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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 Polymerase III has 3 requirements for synthesizing new DNA

A

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

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28
Q

how does OH add new dNTPs onto strand?

A

3’ OH adds new dNTPS with a primer

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29
Q

bidirectionality of replication

A
  • each replication bubble has 2 replication forks… one traveling in each direction AWAY from the origin of replication (ori)
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30
Q

P vs E replication bubbles

A

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

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31
Q

topoisomerases

A

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

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32
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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33
Q

telomerase

A
  • ribonucleoprotein that extends telomeres by adding more repeats (repetitive DNA)
  • contains RNA that serves as a template for extending the overhang
  • 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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34
Q

PCR

A

polymerase chain reaction
(DNA replication in a test tube)

1) denaturation - breaking of H-bonds for dsDNA to ssDNA (95)

2) annealing primers - DNA primers bind to ss templates (55)

3) extension - Taq synthesizes the DNA (72)

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35
Q

where are DNA primers located in PCR?

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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36
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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37
Q

Low error rate of replication

A
  • many repair mechanisms
  • mistakes are 1 in a million
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38
Q

mismatch repair

A

fix errors in DNA replication that arent corrected by DNA polymerase
— occurs after DNA replication

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39
Q

MutS and MutL

A

recognize mismatches

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40
Q

MutH

A

nicks DNA (cuts backbone)

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41
Q

exonuclease

A

remove nucleotides around nick (including mismatch)

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42
Q

damaged nucleotides can be repaired by:

A

1) base excision repair
2) nucleotide excision repair
– occurs when BER doesn’t work

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43
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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44
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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45
Q

glycosylases

A

proteins that remove improper bases directly off the nucleotide backbone

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46
Q

AP endonuclease

A

nick DNA at AP site

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47
Q

UvrA and UvrB

A

recognize DNA distortions

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48
Q

UvrB and UvrC

A

nicks DNA (cuts backbone)

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49
Q

ds breaks repair mechanisms

A

1) homology-directed repair (HDR)
– more accurate than NHEJ bc it uses homologous DNA from sister chromatid

2) non-homologous end-joining (NHEJ)
– variable length indels
– broken ends are joined together
— less accurate than HDR
— can cause mutations

  • types of gene editing
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50
Q

mutation

A
  • PERMANENT alteration in DNA sequence; not repaired
  • new sources of ALLELES that are acted upon by evolution
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51
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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52
Q

indel

A

bases added or removed

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

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53
Q

substitutions can cause…

A

1) missense - changing from one amino acid to another
2) nonsense - STOP codon (truncation)
3) silent - null mutation with no change of amino acids

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54
Q

frameshift

A
  • causes by indels
  • many codons are affected
  • often causes truncation and disorders
  • can be fixed if shift is divisible by 3
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55
Q

loss of function

A
  • less function
  • recessive (must have homozygous genotype)
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56
Q

null

A

complete LOF

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57
Q

gain of function

A

new or more function
- dominant (homozygous/heterozygous genotype)

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58
Q

germline mutations

A
  • only type of mutation to be passed to offspring via gametes and gonads
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59
Q

spontaneous vs induced mutations

A

S: natural processes or random chance cause mutation
I: caused by mutagen

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60
Q

tautomers and tautomeric shifts

A
  • alternative, temporary configurations of bases
  • rare forms have different base pairing properties than typical forms
    (purine pairs with the wrong pyrimidine and vise versa)

spontaneous mutation example
- not a mutation, BUT can cause mutations
- only a temporary shift but can cause base pairing problems

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61
Q

positive vs negative controls

A

P: ensures assay gives (+) results when it should
- protects against false (-)

N: ensures that easy gives (-) results when it should
- protects against false (+)

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62
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

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63
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

n = 23 in humans

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64
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)

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65
Q

ploidy number

A

the number of homologous sets of chromosomes

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66
Q

homologous chromosomes

A
  • have the same genes BUT could have different alleles
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67
Q

c (DNA content)

A
  • the amount of DNA in a haploid gamete
  • c = amount of DNA in a haploid cell before DNA replication
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68
Q

DNA replication and cDNA content

A
  • doubles the amount of DNA in the cell

c + (ploidy x # of chromatid per chromosome)

2n x 2 + c = 2c

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69
Q

2 stages that product variation in meiosis

A

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

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70
Q

crossing over

A

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

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71
Q

independent assortment

A

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

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72
Q

aneuploidy

A

extra or missing chromosomes leading to an unbalanced chromosome complement

2n+1 = trisomy
2n-1 = monosomy

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73
Q

nondisjunction during meiosis

A

Anaphase I: failure of homologs to seperate

Anaphase II: failure of sister chromatids to segregate into different daughter cells

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74
Q

nondisjunction in MI VS MII

A

MI:
- heterozygous duplicate
- homologs don’t seperate
- all aneuploid gametes (2 n+1; 2 n-1)

MII:
- homozygous duplicate
- sister chromatids don’t segregate
- 1/2 aneuploid gametes ( 1 n+1, 1 n-1, 2 2n)

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75
Q

genetic variation arise because…

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

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76
Q

wild type VS dominant allele

A

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

*Alleles themselves are not dominant/recessive rather they have dominant/recessive relationships

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77
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)

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78
Q

why causes a 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)
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79
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
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80
Q

LOF vs GOF mutant allele types

A

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

GOF:
1. hypermorphic - more
2. neomorphic - new
3. antimorphic (dom-neg) - new, but antagonizes WT

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81
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

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82
Q

insertion right before the transcribed region will cause?

A

reduced transcription

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83
Q

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

A

reduced translation

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84
Q

replication 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
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85
Q

replication 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
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86
Q

leading strand

A

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

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

  • arrow points to 5’ end
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87
Q

lagging strand

A

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

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

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88
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
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89
Q

Ames test

A

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

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

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90
Q

cell cycle

A

G1: new cell birth
S: synthesize DNA and cells duplicate genetic material (chromosomes double to produce sister chromatids)
G2: grow more; synthesize proteins for mitosis
mitosis: 2 identical d. cells form

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91
Q

duplication

A

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

occur on homologous chromosomes

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92
Q

translocation

A

2 breaks 1 in each of 2 homologous chromosomes

– fragments switch places and attach on the nonhomo chromo

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93
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

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94
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.

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95
Q

what DNA repair mechanism would be disrupted in bacteria if they were unable to add methyl tags to DNA after replication?

A

mismatch repair

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96
Q

chromosomes are separated from one another into daughter cells during what stage of meiosis?

A

anaphase I

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97
Q

how are new strands built of entire chromosomes?

A

both built 5 to 3 with a mix of leading and lagging

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98
Q

2n = 22…chromosome and chromatid count at metaphase I?

A

44 chromatids
22 chromosomes

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99
Q

if base ratios are off…

A

may be PCR primer

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100
Q

During Prophase I, chromosomes will come together to form a connected structure called a…

A

Tetrad” or “Bivalent” or Chiasmata

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101
Q

Mendel’s Law of Segregation

A

2 alleles of each gene separate/segregate during gamete formation, and then unite at random (1 from each parent) at fertilization

  • MI separates homologs; then MII separates sisters. Each gamete ends up with 1 copy of each allele
  • Homologous chromosomes align in metaphase I and segregate into separate daughter cells
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102
Q

Mendel’s Law of Independent Assortment

A

During gamete formation, different pairs of alleles segregate independently of each other

50% chance of receiving alleles from mother vs father

Homologous chromosomes align in MetaphaseI with independent orientation; the orientation of 1 tetrad does not influence the orientation of another

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103
Q

IA on the same vs different chromosomes

A
  • alleles on different chromosomes = always independently assort
  • alleles on the same chromosome may/may not independently assort
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104
Q

purpose of test cross

A

cross recessive genotype with mystery genotype

  • all dominant –> homozygote dom
  • half dominant –> heterozygote dom

** figure out the genotype of an individual

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105
Q

true-breeding/pure-breeding

A

homozygous individuals whose line produces the same phenotype when selfed 100% of the time

**can assume genotype is homozygous

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106
Q

how to figure out which is the dominant individual?

A

look at heterozygous
-cross 2 pure-breeding individuals to get all heterozygous F1 generation and analyze the phenotypes

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107
Q

monohybrid self cross

A

heterozygotes of 1 gene crossed with each other

Ex/ Aa x Aa

1:2:1 genotypic ratio
3:1 phenotypic ratio

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108
Q

/ VS ;

A

/ - alleles on different homologs of the same chromosome

; - alleles on different chromosomes

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109
Q

dihybrid test cross

A

2 genes controlling 2 traits
-heterozygotes crossed with recessive homozygotes
Ex/ A/a;B/b x a/a;b/b

genotypic: 1:1:1:1
phenotypic: 1:1:1:1

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110
Q

dihybrid self cross

A

selfing of dihybrid

genotypic: 9:3:3:1
phenotypic: 9:3:3:1

9 - both dom
3 - 1 dom; 1 rec
3 - 1 rec; 1 dom
1 - both rec

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111
Q

product VS addition rule

A

product - AND
- the probability that 2 or more independent events occurring together is the product of the probabilities that each will occur by itself

addition - OR
- the probabilities of 2 mutually-exclusive events occurring is the sum of their individual probabilities

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112
Q

a scientific hypothesis makes — predictions and is —-.

A

testable and is falsifiable.

*null hypothesis must make a testable prediction.
ex/ IA will occur.

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113
Q

null hypothesis

A

there is no significant difference between the observed and expected frequencies
- must be very certain that you can reject the null hypothesis (5% error; 95% confidence)

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114
Q

Chi-Square Tests

A

determine p-value using a formula

total = (observed - expected)^2 /expected

compare values in chart

expected values:

  • look at total and use ratio expected based on the type of cross
  • ex/ monohybrid self cross = 3:1
    out of 400
    300 and 100 are expected values
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115
Q

P-value

A

represents the probability that the null hypothesis is TRUE

p > 0.05 - fail to reject the null
p < 0.05 - reject the null

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116
Q

degrees of freedom

A

the number of groups of observed/expected minus 1

n -1

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117
Q

genes controlled by single genes

A

display characteristic inheritance patterns
*though most traits are not controlled by a single gene

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118
Q

autosomal recessive disorders

A
  1. males and females equally affected
  2. unaffected individuals can have affected children via heterozygous carriers
  3. can skip generations
  4. rare
  5. becomes more common with inbreeding (homozygous recessive)
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119
Q

stipulation of rareness when discussing disease

A

when discussing diseases, you can assume these traits are rare so people entering the pedigree do NOT carry the disease allele
- unless you have info to suggest otherwise

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120
Q

autosomal dominant disorders

A
  1. males and females are equally affected
  2. affected individuals always have an affected parent (no heterozygous carriers bc they are affected too)
  3. does not skip generations
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121
Q

X-linked inheritance rules

A
  1. males inherit Y from their father and MUST inherit X from their mother
  2. females inherit one X from father and one X from mother
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122
Q

X-linked recessive disorders

A
  1. males more frequently affected
  2. never transmitted from fathers to daughters
  3. All sons of affected mothers will also be affected by the trait
  4. can “skip generations” via female carriers
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123
Q

X-linked dominant disorders

A
  1. females more frequently affected
  2. ALL of the daughters and NONE of the sons of affected fathers have the trait
  3. does not “skip generations”
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124
Q

penetrance

A
  • the percentage of individuals with a particular genotype that demonstate the expected phenotype
  • complete penetrance = 100%
  • incomplete penetrance = 1-99%
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125
Q

penetrance calculations over or underestimate?

A

overestimate
- there could be other nonpenetrant individuals that we are not certain about
- the demoninator is larger so the overall fraction/percentage will be smaller than originally estimated

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126
Q

variable expressivity

A

for individuals with the same genotype, there is a range of phenotype severity/expression

** the degree with which a genotype is expressed as a phenotype (how much phenotype is shown)

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127
Q

dominance types

A

heterozygous phenotype =
1) complete - same as homozygous dominant
2) incomplete - intermediate between the 2 homozygotes
3) codominance - mix of 2 homozygotes (shows both homo)

  • same genotypic but different phenotypic rations
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128
Q

monomorphic VS polymorphic traits

A

Monomorphic: have single “wild type” allele

Polymorphic: multiple common allele variants (no single WT allele)

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129
Q

trait classifications are NOT static

A
  • classifications can change
  • a mutant allele can become a common varient over time or a common varient can be lost from the population
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130
Q

Blood Types

A

AB: express 2 variants of the A and B enzymes [polymorphic]
- IAIB

O: null version/non-functional enzymes
- ii

B: enzyme adds B sugars
- IBIB
- IBi

A: enzyme adds A sugars
- IAIA
- IAii

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131
Q

with regard to blood type, IA is

A

codominant to IB and dominant to i

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132
Q

pleiotrophic allele

A

single alleles affects multiple properties/parts of an organism

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133
Q

lethality and the pelger

A

recessive for pelger is lethal phenotype

lethal means that they have severe defects or are never born

changes the phenotypic ratios

  • nuclear morphology: pelger allele is dominant
  • lethality phenotype: pelger allele is recessive
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134
Q

epistasis

A

the effect of one gene masks the effect of another

  • often occurs when 2 genes encode members of the same biochemical pathway
  • indicated by fewer phenotypic classes than expected
  • gene-gene interaction
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135
Q

bombay phenotype

A

recessive epistasis
hh genotype
- overrides the blood type for an O blood type
- doesn’t matter what parental blood types are
- 2 recessive h alleles mask the IA alleles phenotypic effect

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136
Q

duplicate/redundant genes

A

collapse all phenotypic categories with 1+ dominant allele for either gene

15:1 ratio

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137
Q

dihybrid cross ratio

A

9:3:3:1

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138
Q

complementary ratio

A

9:7

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139
Q

duplicate genes ratio

A

15:1

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140
Q

recessive epistasis ratio

A

9:3:4

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141
Q

dominant epistasis ratio

A

12:3:1

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142
Q

complementation / complementation test

A
  • when 2 individuals with the same mutant phenotype but different homozygous recessive genotypes produce offsprng with the wild-type phenotype when crossed

only works for recessive mutants
- always cross homo recessive mutants that are mutant for only one gene

-no complementaion if mutant trait appears (same gene)
- complementation if mutatnt trait does not appear (mutations in different genes)

  • failure to complement with itself bc if has mutant and crosses with itself, the mutant phenotype will remain.
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143
Q

MC1R1/MC1R1

A

can produce different doses of eumelanin vs phenomelanin depending on the available receptor variants

  • melanocytes are cells found in skin that produce pigmented melanosomes that give skin its color
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144
Q

dihybrids

A

individual that is heterozygous at 2 different genes

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145
Q

parental types VS recombinant type

A

P: phenotypes that reflect a previously existing parental combination of alleles that is retained during gamete formation

R: phenotypes reflecting a new combo of alleles that occurred during gamete formation via crossing over during prophase

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146
Q

hemizygote

A

genotype for genes present in only one copy in an otherwise 2n organism
ex/ x-linked genes in a male

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147
Q

advantageous vs disadvantageous alleles

A

an allele that is advantageous in one environment may be disadvantageous in another

  • UV damages folate for neural birth defects
    — alleles that increase pigmentation would be advantageous
  • UV is required for vitamin D production and without causes rickets in bone
    — alleles that increase pigmentation would be disadvantageous in a sunny environment
    — alleles that decrease pigmentation would be advantageous in a less sunny envt.
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148
Q

crossing over between 2 genes does not always happen…

A

so >50% of gametes will be parental. Fewer offspring will be recombinant.

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149
Q

cis VS trans dihybrids

A

cis - dominant alleles are on the same homolog [AB/ab]

trans - dominant alleles are on different homologs [Ab/aB]

  • cis and trans dihybrids differ in…what allele combinations are parental vs recombinant

(parental vs recombinant combos swap with cis and trans switch)

150
Q

distance between genes on a chromosome

A
  • the father apart genes are on a chromosome (mu), the more likely that they are able to be affected by a crossover event (recombination is more common)
151
Q

calculation recombination frequency:

A
  1. do a test cross of dihybrid (heterozygous with recessive genotype)
  2. determine which offspring are recombinant
  3. add up recombinants and divide by the total offspring
152
Q

recombinant frequency

A
  • RF = 0% (complete linkage; only parental genotypes)
  • RF = <50% (linkage; parental genotypes are more common)
  • RF = 50% (unlinked; parental and recombinant are equally likely; independent assortment)
  • RF will never exceed 50%
  • 1 m.u./cM = 1% RF
153
Q

getting probability of recombinance with map unit

A

6 mu = 6% will be recombinant

  • do a test cross and determine the parental and recomb genotypes
  • divide % by 2 and assign % to each recomb.
  • remaining % out of 100% divided by 2 will be the parental types

ex/

3% each recomb
47% each parental

1/2 of each recombinant will be of each recombinant type.

154
Q

recessive lethal allele

A

an allele that prevents survival of homozygotes
- although heterozygotes carrying the allele survive

  • decreases the denominator for a larger fraction/probability
155
Q

incomplete-/co-dominance changes only…

A

phenotypic ratios NOT genotypic ratios

156
Q

temperature sensitive alleles

A

function depends on the environmental temperature

(permissive conditions allows for allele; restrictive does not)

157
Q

hypostatic gene

A

gene (and its genotypic effects) that is being masked by the epistatic allele

158
Q

dominant vs recessive epistasis

A

R: the effects of recessive alley at one gene hid the effects of alleles at another another gene

D: the effects of a dominant allele at one gene hide the effects of alleles at another gene

159
Q

some people with disease have stump while some people have a 6th finger…

A

variable expressivity

160
Q

A trihybrid plant is self-fertilized. What is the probability that an offspring will have the dominant phenotype at 3 loci?

A

27/64

161
Q

Ducks that are homozygous for the crested allele do not survive birth. Mating 2 crested ducks results in what ratios of offspring?

A

2/3 crested
1/3 non-crested

162
Q

With regard to blood type, the h allele is dominant to the IA allele. (T or F)

A

False

163
Q

Dihybrids with mutations show a completely WT phenotype. (genes on different chromosomes)When the dihybrids are selfed, most of the offspring are fully WT meaning that…

A

The genes independently assort…cannot be linked because they are on different chromosomes

164
Q

genetic information flow (central dogma)

A

DNA (nucleic acids) –> RNA (amino acids) –> protein

165
Q

what contains al the info needs to transcribe DNA into mRNA

A

a gene!

166
Q

translation of start VS stop codons

A
  • start codon is translated and found within the transcribed region
  • stop codon is not translated and is found outside the amino acid sequence (not encoded within the protein)
167
Q

+1 nucleotide

A

(indicated often with an arrow)
- indicates the transcription start site and first nucleotide to be transcribed

*not necessarily before the AUG start codon

168
Q

coding vs template strand

A

coding 5’ to 3’ (codons)
template 3’ to 5’ (antiparallel)

see the codons within the coding strand (look for ATG)

169
Q

different genes can have different…

A

directions of transcription

this is bc different strands of dsDNA are used as the template for different genes (might be coding for one gene but template for another)

170
Q

cis and trans elements

A
  • recruit RNA polymerase to a gene

CIS: regions of DNA that are required for gene expression/regulation
*part of the same molecule as the genes they regulate

TRANS: diffusible molecules (proteins) that bind cis elements
- separate molecules from the genes they regulate!

171
Q

sigma subunit

A

key trans-factor that helps RNAP associate with promoters

in prokaryotes

many subunits functions together as 1 enzyme

172
Q

consensus sequences

A

a sequence of DNA having similar structure and function in different organisms.
(same sequence in the same location)

-10 and -35 consensus sequences are found in nearly all bacterial promoters

173
Q

differences in eukaryotic regulatory elements vs prokaryotic

A

1) no sigma subunit
2) different consensus sequences (TATA box most common)
3) additional cis-reg. elements besides promotor (enhancer/silencer) common. Enhancer often required!

174
Q

termination mechanism of transcription

A

hairpin/stem loop
- complementary base pairing with itself

  • strong C-G bonds hold structure together causing the stalling of RNAP
  • When stalling occurs, the weak AU bonds cannot hold mRNA and DNA complex together
175
Q

splicesome

A

snRNA and proteins

  • functions in recognizing introns and removing them
  • recognizes specific sequences in the DNA that will determine the sites of splicing
  • via BPing, the RNA component of splicesome recognizes the splice site sequences
  • found in nucleus
176
Q

w/o CAP and tail, RNA….

A

would be degraded

177
Q

alternative splicing may generate…

A

2 ore more types of mRNA from the same transcript

178
Q

Translation steps

A

1) Initiation
- complex of ribosome, first charged tRNA enters with mRNA

2) Elongation
- peptide bonds formed as charged tRNAs bring appropriate amino acids to site

3) Termination
- stop codons signal release factors and complex dissociates at the A site

179
Q

Shine-Delgarno sequence

A
  • ribosome binding site
  • positions ribosomes by start codon
  • found only in prokaryotes
180
Q

ribosomal peptidyl transferase

A

transfers the peptide in the P site to the amino acid in the A site

181
Q

dehydration synthesis reaction

A
  • catalyzed by ribosomes in order to form a peptide (covalent) bond between 2 amino acids
  • loss of a water molecule
182
Q

translation builds peptides in what direction?

A

N to C
amino acid to carboxylic acid

183
Q

open reading frame (ORFs)

A

indicate regions that could potentially encode for a protein
- sequence of codons within the same reading frame starting with 5’-AUG and ending with STOP-3’

** the longer the ORF, the more likely its a true protein-coding ORF

184
Q

mRNAs contain both…

A

translated and untranslated regions

185
Q

locating proteins via identification of ORFs

A
  • use coding strands that look like mRNA

1) locate potential start codons (ATGs in any direction if coding strand is unknown)
2) identify the ORFs by finding inframe STOP codons
3) Determine polypeptide length
4) Determine directionality
5) Determine mRNA sequence
6) Translate!

186
Q

tRNAs

A

translate codons into amino acids
- speak both languages bc can bind to codons that are both complementary and antiparallel

-function is to base pair with the codon on a strand of mRNA during translation.

***ensures that the correct amino acid will be added to the growing polypeptide chain.

187
Q

codon count

A

64 total
61 encode for amino acids

188
Q

aminoacyl-tRNA synthetase

A

enzyme that carries out the charging of the tRNA with its specific amino acid

  • attaches an a.acid to its tRNA
  • highly specific for a given amino acid and for a given tRNA

** product = charged tRNA

189
Q

wobble pairing

A
  • allows for a single anticodon of tRNA to interact with more than one mRNA codon
  • 61 codons but not 61 tRNAs
190
Q

Why doesn’t bacterium transcribe all of its genes all the time?

A
  • metabolize organisms
  • bacteria only produce the proteins needed for lactose metabolism when lactose is present
  • saves energy and resources only to transcribe/translate when a protein is needed
191
Q

add (+) regulation with…

A

activator

192
Q

remove (-) regulation by…

A

removing repressor

193
Q

inducible vs repressible operons

A

I: auto OFF
- transcription is turned ON

R: auto ON
- transcription is turn OFF

194
Q

constitutive

A

always on
- expression even when there is no lactose around and the operon should be turned off

195
Q

2 ways to induce a gene

A

1) add activator
2) remove repressor

196
Q

prokaryotes tend to regulate gene expression at the level of…

A

transcription

197
Q

polycistronic RNAs

A

mRNAs that code for more than one protein under the control of a single promoter. (share a promotor)

(prokaryotic only)

198
Q

operons

A

group of prok. genes that share a promoter and get regulated and transcribed as one unit

(energy/resource saver)

199
Q

5 components of the lac operon

A

Regulatory:
1) promoter (P)
2) operator (O)

Ensures Lac Y/Z expression:
3) LacI

Genes encoding metabolism proteins:
4) LacZ
5) LacY

200
Q

promoter (p)

A

DNA site where RNA polymerase initially binds

201
Q

operator (o)

A

the site where the repressor binds

202
Q

LacI

A

gene that encodes the repressor protein

**gene itself is NOT part of the lac operon
- has its own promoter and its regulated seperately
- still relevant bc protein it encodes regulated the lac operon

203
Q

LacZ

A

gene that encodes B-galactosidase, an enzyme that breaks down lactose into monosaccharides

204
Q

LacY

A

gene that encodes permease, an enzyme that makes it easy for lactose to enter the cell

205
Q

with lactose VS without lactose

A

with:
allolactose binds to repressor protein and changes shape so that it cannot bind to the operator
- transcription DOES occur

without:
repressor protein binds to operator and RNA poly cannot pass
- transcription DOES NOT occur

206
Q

2 different mutations that could prevent the lac repressor from binding to the operator…for constitutive expression

A

Lac I-
Lac Oc

207
Q

LacI- mutation

A

changes the shape of the Lac repressor DNA binding domain
- prevents lac repressor from being made

208
Q

Lac Oc mutation

A

changes DNA sequence (operator) that Lac repressor recognizes
- prevents lac repressor from binding

  • constitutive expression
209
Q

Lac Is mutation

A

prevents binding of allolactose
- prevents transcription (default OFF)

*** uninduciable transcription

super repressor

210
Q

Lac Z- mutation

A

produces nonfunctional B-gal protein
- cannot metabolize lactose

  • does not prevent transcription
211
Q

Lac Y- mutation

A

produces a nonfunctional permease protein
- stop the metabolism of lactose as it will not be able to enter the cell without permease.

212
Q

Lac I- and I+ relationship

A

Lac I- is recessive to Lac I+
- heterozygote is inducible, not constitutive
(Lac I+/I- => merozygote)

213
Q

regulatory elements

A
  • controls expression of genes
  • directly influences whether RNAP transcribes a gene
214
Q

Lac Oc causes constiutive expression only when…

A

alleles are in cis not trans (on same DNA molecule)

DOMINANT (O+ will falil to rescue the constiutive phenotype)

215
Q

effector cAMP induces the lac operon by…

A

adding (+) regulation

216
Q

E. coli prefers to metabolize…

A

glucose for carbon energy source

  • therefore, beneficial to transcribe lac operon only when glucose is absent
217
Q

CRP

A

trans-regulatory factor that binds the promoter and helps to recruit RNAP

  • w/o CRP, RNAP is inefficient for finding and binding to promoter (trans. levels LOW)
218
Q

cAMP

A

effector that causes allosteric changes to CRP to allow it to bind to the promoter

cAMP levels are HIGH when glucose levels are LOW

219
Q

bacterial growth in lactose and glucose

A
  1. metabolizes glucose for low level transcription
  2. runs out of glucose, changes gene expression, and plateaus
  3. metabolizes lactose instead for higher levels of transcription
220
Q

transcription levels of differing glucose and lactose

A

G + L = low transcription
No G + L = max transcription
G + no L = no transcription
No G + No L = no transcription

221
Q

catabolic vs anabolic pathway operons

A

catabolic (break down): inducible [auto OFF]

anabolic (build up): recessive [auto ON]

222
Q

trp operon

A

controls the expression of tryptophan synthesis of genes (anabolic)

contains:
- trp promoter
- trp operator
- repressor protein binding to operator (trpR)
- structural genes for tryptophan biosynthesis (trpE/B/C/D/A)

223
Q

when tryptophan binds to trp repressor protein bound to operator…

A

transcription does not occur (operon is turned off)

  • repressible (auto ON)
  • adds (-) regulation)
  • anabolic operon repressed in the presence of their metabolic end products (turned off when end product formed)
224
Q

7 Types of Gene Expression

A

Regulation of:
1. chromatin remodeling
2. transcription
——– post-translational
3. splicing and processing
4. transport (out of cell)
5. degradation of mRNA
6. translational
7. protein modification

225
Q

all cells in an organism have…

A
  • the same DNA
  • same genes and cis elements
  • DIFFERENT trans elements

**Therefore, the same gene in different cell types will have the exact same DNA (promoter and cis regulatory elements)

226
Q

each cell in a different environment…

A

has different TFs that may or may not bind to cis reg. elements of different genes

**Therefore, a gene will be off unless cis-elements are accessible to bind to TF

227
Q

TFs cannot interact with…

A

cis-elements that are packed into inaccessible chromatin

228
Q

euchromatin vs heterochromatin

A

Euchromatin:
loosely packed
open
accessible
transcription ON

Heterochromatin:
condensed
closed
inaccessible
transcription OFF

229
Q

chromatin accessibility can be regulated in 3 ways:

A
  1. histone modifications
    - acetylation increases accessibility (opens DNA)
  2. DNA methylation
    - decreases accessibility (closes DNA)
  3. Nucleosome sliding/reorganization
230
Q

interaction of cis and trans factors in a cell determines…

A

if/how a gene is transcribed

1) basal factors
2) TFs binding to enhancers/silencers
3) Tfs+cofactors

231
Q

basal factors

A

bind to promoters to recruit RNA poly to the gene

LOW LEVEL TRANSCRIPTION if only factor bound to DNA

232
Q

specific TFs bind to…

A

enhancers or silencers

increases rate of transcription in addition to basal factors

233
Q

Physical interaction with the poly and TFs + their co-factors will…

A

increase or decrease the rate of transcription

234
Q

if u change orientation or location of promoter VS enhancer….

A

Promoter: gene is NOT transcribed

Enhancer: cis/trans factors can usually still complex and function normally

235
Q

enhancers are always…

A

on the same chromosome (in cis) to genes they regulate
- nearby but can be far away (few hundred/thousand bp away on coding gene)

236
Q

silencers

A

type of enhancer that decreases transcription by binding to repressor proteins (trans-acting regulatory element)

237
Q

context-specific cis elements

A

context-specific if they interact with TFs present/active only in those specific contexts

(cell type, envt.)

238
Q

commonly responsbile for context-specific regulation of euk. genes?

A

enhancers and silencers

239
Q

Hox genes

A

encode TFs that specify the formation of specific parts of the body

Ex/ front leg hox gene produces… front leg hox protein (TF) that binds….to cis regulatory elements on gene

  • If body part in wrong spot…the body part will still be found in correct location
  • promoter = where
  • coding region = what
240
Q

transgenic construct that shows which cells express the eyeless protein from embryo to adult stage

A

design: link eyeless cis elements to reporter gene (GFP)
methods: add this hybrid transgene to normal early embryo
results: look for GFP expression as flyes develop
- indicates where eyeless is being expressed and functioning!

241
Q

GFP

A

green fluorescent protein
- reporter gene we can see in living tissues

242
Q

transgene

A

a gene that has been transferred naturally, or by any of a number of genetic engineering techniques, from one organism to another.

can change phenotypes !!

243
Q

invitation of transcription

A
  • RNA poly binds to dsDNA at promoter
  • RNAP unwinds dsDNA to expose unpaired bases on template strand
244
Q

elongation of transcription

A
  • sigma subunit is released and RNAP looses its affinity for promoter and gains affiinity to DNA
  • mRNA extended 5’ to 3’ antiparallel to template strand
245
Q

termination of transcription

A

terminator RNA sequences signal the end of transcription
- forms hairpin loops
- releases both RNA polymerase and mRNA chain from DNA

246
Q

hairpin loops

A

ssRNA folds back on itself and comp. base pairs with C/G’s

247
Q

primary transcript

A

single strand of RNA resulting from transcription

Pro: primary transcript is the mRNA used for synthesis
Euk: primary transcript undergoes RNA processing before protein synthesis

248
Q

5’ and 3’ untranslated regions (UTRs)

A

sequences located just after the methylated cap and before poly-A-tial
- encoded by exons and don’t include codons

249
Q

RNA splicing

A

deletes introns in euk. pre-mRNA and joins together adjacent exons to form mature mRNA

250
Q

RNAP I

A

transcribe genes that encode major RNA components of ribosomes (rRNAs)

251
Q

RNAP II

A

transcribes genes that encode proteins

252
Q

RNAP III

A

transcribes genes that encode tRNAs/other small noncoding RNA molecules

253
Q

enhancer

A

cis-acting regulatory element that regulates from nearby promoters
— function by acting as binding sites for TFs and are responsible for spatiotemporal specificity of transcription

254
Q

activator

A

TF that binds to specific DNA seq. with enhancer elements (CpG islands) and increases the level of transcription of a nearby promoter

255
Q

co-activator

A

protein that binds to a transcriptional activator and plays a role in increasing transcription levels

256
Q

If promoter is mutated…

A

no other genes will function

257
Q

A super repressor can only repress if…

A

it is bound to the operator

258
Q

Oc is ________ to I

A

epistastic
– makes the effects of I

259
Q

Can cell grow if lactose is its only carbon source?

A

no —– requires beta-galactosidase

260
Q

Introns are found….

A

within eukaryotic genes

261
Q

2 genes are expressed in the roots of a flower plant. Neither gene is expressed in the petals…meaning that?

A
  • not regulated as part of the same operon
  • must not be expressed in the same area
  • must not be expressed at the same time
  • do not have differnet enhancers in diffrent cell types
262
Q

Shine Dalgarno mutation

A

no translation (normal transcription)

263
Q

Is P+ Oc Z- Y+ / I- P+ O+ Z+ Y-

will it grow if lactose is the only carbon source available?

A

NO

264
Q
  • condition leads to altered hormone T levels
  • large deletion in thyroid-specific silencer that sits near the gene

What type of mutation?

A

GOF
hypermorphic

265
Q

Order of “things” that RNAP would encounter along a prokaryotic gene?

A

Promoter
Transcription start site
Shine-Dalgarno sequence
start codon
stop codon
3’ UTR
transcription termination site

266
Q

mutation in lac operon repressor that prevents it from binding to its effector?

A

uninducible

267
Q

mutation in trp operon repressor that prevents it from binding to its effector?

A

constitutive

268
Q

splicing can remove an exon from the primary transcript

A

TRUE

269
Q

Production of hGH/c-hGH in the 1950s

A
  • GH deficiency was treated by injected by cadaver-derived human GH
  • pool pituitary glands have GH
  • GH extracted with Wilhemi preparation
  • c-hGH is injected into hGH deficient humans
  • requires a lot of cadavers bc method only produces TINY amounts of hGH
    — prevented clinical testing for additional treatment uses
    — many people were denied due to scarcity
  • some were contaminated with prions giving people CJ disease
270
Q

organismal clones

A

exact genetic copies of entire organism

271
Q

cellular clones

A

groups of genetically-identical cells

272
Q

molecular/DNA clones

A

identical molecules (e.g. DNA)

273
Q

hgH protein biotechnology

A

1) isolate the DNA of the hGH gene
2) clone the gene – make many copies of the hGH gene in vivo
3) Make bacteria transcribe and translate the gene to make a ton of hGH

274
Q

Restriction digest

A
  • chops up one genome into millions of restriction fragments
  • still don’t know where the gene is and may only have a few copies total in the sample
275
Q

PCR

A
  • allows us to make many copies of the exact sequence we want
  • replicate target DNA into may copies (2^n)
    (n = number of cycles)
276
Q

what does PCR gel look like?

A
  • large dark band show many copies of the amplified DNA region
  • no band for the template (genomic) DNA on gel because there is not enough present to be seen
277
Q

defining primers of a specific region

A
  • Primer is the exact same sequence as the 5’ to 3’ next to the region of interest (on both sides)
  • The PCR product will extend from the 5’ end of one primer to the 5’ end of the other
278
Q

Although PCR allowed for isolation of the gene of interest….we are still not ready to put GH gene into bacteria

A
  • When bacteria replicates, linear DNA will be lost
  • need cloning vector that will be replicated when cell replicates DNA during mitosis
279
Q

3 regions of a plasmid

A
  1. origin of replication
    – so plasmid can be replicated
  2. multiple cloning site (MCS)
    – many different restriction enzyme sites so things can be inserted into plasmid (often LacZ gene)
  3. antibiotic resistace gene
    – any cells that uptake this plasmid will be resistant to a particular antibiotic
280
Q

Restriction enzyme sites are added to the…

A

5’ end of primers for isolation of the region of interest

281
Q

Isolation of GH gene by PCR and engineering of EcoRi sites on either side…

A
  • EcoRi addition creates identical sets of sticky ends on both the plasmid and GH gene
  • sticky ends by EcoRi allows for recombination and GH gene to be added to the plasmid after annealing of primers and ligase
282
Q

recombinant plasmid DNA

A
  • transformed into bacteria which replicate it with its endogenous DNA replication machinery
  • the plasmid replicates with the bacteria and we now have a lot of recombinant molecules inside the bacteria if transformation worked
  • even if it did…there may be some untransformed bacteria mixed in!!!
283
Q

antibiotic selection

A

is used to identify bacteria that contain the plasmid

colonies = transformants (have plasmid)
no colonies = non-transformants (no plasmid)

284
Q

problem with antibiotic selection

A
  • know the Amp-resistant colonies have a plasmid but we do NOT know if they have a plasmid that contains the GH gene
285
Q

what is the significance of having a MCS located in a lacZ coding gene?

A

B-galactosidase activity can be visualized using X- gal (blue colored reporter gene)

*If B-gal is functional, X-gal is converted into a blue chemical

  • MCS doesn’t disrupt LacZ alleles and functional B-gal is produced (blue color)
  • If gene is inserted at MCS, it does disrupt LacZ allele so NO B-gal is produced (white color)
286
Q

The strain of E.coli must be LacZ-…

A

so functional lacZ is from the plasmid

287
Q

Steps of creating recombinant DNA

A

1) genomic DNA

PCR

2) hGH gene and restriction sites + cloning vector (plasmid)

digest

3) complementary sticky ends pair

ligate

4) final vector

5) transform into bacteria

6) look for white colonies

288
Q

Gel with antibiotic selection and white-blue screening

A

1) pick colonies
2) grow each colony in individual culture
3) Isolate plasmid DNA
4) Cut with EcoRi
5) Run gel

Blue: one band of DNA
White: one DNA plasmid band and one band for GH gene
— GH gene travels farther bc it is smaller than plasmid DNA

288
Q

Gel with antibiotic resistance (not blue/white assay)

A

1) Pick colonies
2) Grow each colony in individual culture
3) Isolate plasmid DNA
4) Run PCR with primers specific to GH gene

one large band representing the PCR product

289
Q

regulatory element allows for the bacteria to express the hGH gene

A

promoter

  • prokaryotic if end goal is to make protein in bacteria
    (bacterial RNA polymerase won’t recognize eukaryotic promoter)
  • same for eukaryotic promoter
  • prokaryotic and eukaryotic ==> recombinant DNA
290
Q

Why is cDNA library used instead of genomic DNA?

A
  • cDNA does not contain introns and will have a shorter length of bp
  • no extra material
290
Q

Why is mRNA way longer than it is supposed to be in transformed bacteria?

A
  • prokaryotes don’t splice (so introns are still present)
  • they do not have the proper machinery to splice a eukaryotic gene
291
Q

GMO

A
  • Genetically Modified Organisms
  • recombinant organism
  • any organism that contains DNA that has been recombined from multiple sources

What do people not support GMOS?
- can pass transgenic DNA to you
- unanticipated ecological effects if released into the environment
- Organic foods are more nutritious
- There is not a consensus among scientists (though 88% believe they are safe)

292
Q

GMO corn

A
  • corn borers are small caterpillars that eat and destroy corn
  • farmers often spray corn with Bacillus thuringiensis toxin (Bt)
  • toxin considered an organic pesticide when purified directly from bacteria that normally produce it
  • Bt toxin unfortunately degrades in the sun and other concerns about affects on pests/bugs
  • INSTEAD…
    – Agrobacterium tumefaciens can insert pieces of its Ti plasmid into plant genomes
    – Recombinant DNA allows plants to have new trait eliminating the need for pesticide

*** typically pathogenic but Ti plasmid is altered to remove harmful part and still include gene of interest

293
Q

steps for creating GMO corn

A
  1. isolate DNA from the Bacillus thuringiensis
  2. Amplify (PCR) the Bt toxin gene
    3/4. Digest (RE) vector/plasmid and digest gene of interest
  3. Create recomb. DNA molecule
  4. Transform bacteria
  5. Select for transformed Agrobacteria
  6. Transform corn plant
  7. Select for transformed corn plants
294
Q

if we want to express gene in eukaryote…

A

eukaryotic promoter must be upstream the gene of interest
- RNA polymerase does not recognize a prokaryotic promoter

295
Q

functions of a vector

A
  1. a cloning vector to be replicated and selected for in bacteria (agrobacterium)
  2. serve as an expression vector once in plants
  3. carry info for “selection” of transformed plants
296
Q

antibiotic resistance genes will have ___ promoter

A

prokaryotic

297
Q

CRISPR-Cas9

A

bacterial defense system against viruses.
- Bacteria integrate bits of viral genetic material into their genomes.
- encode for RNAs that can bind to viral genomes by complementary base pairing, allowing the bacteria to detect and destroy viral DNA

**cut DNA at precise location using guide RNA and Cas9

  • SPECIFIC & PROGRAMMABLE
298
Q

Cas9

A

enzyme that cuts DNA to form single-stranded breaks

299
Q

cutting DNA with CRISPR

A
  • guide RNA directs Cas9 to cut at specific locations based on base pairing
  • sgRNA functions the same as 2 separate RNAs
300
Q

specific and programmable

A
  • Cas9 will only cut a particular DNA sequence
  • DNA sequence can be determined by the guide sequence
301
Q

why wouldnt restriction enzymes be particularly useful for targets genome editing?

A
  • restriction enzymes have specific DNA sequences and are NOT programmable
    Ex/ EcoRi will always cut at a specific point (GAATTC)

*cannot be used for targeted gene sequencing

302
Q

PAM site

A

A short DNA sequence, the protospacer-adjacent motif (PAM), is frequently used to mark proper target site

NGG
(N= any nucleotide)

  • found in the non-complementary strand

*N is adjacent to 3’ end of guide on the opposite strand

303
Q

2 conditions for Cas9 to recognize and cut DNA:

A
  1. sgRNA
  2. PAM sequence
  • sgRNA does not contain the PAM
  • sgRNA found right before PAM
  • sgRNA contains the same sequence (Us instead of Ts) as the bottom strand (where PAM is found)
304
Q

2 ways to repair ds breaks

A

1) non-homologous end joining (NHEJ)
- repair double-stranded break by doing 2 ends directly together
* could ADD or REMOVE nucleotides to the sequence first before ending joining

2) homology directed repair (HDR)

break is repaired using homologous chromosome as template to ensure proper sequence

  • homologous template “fills in the blank” of what was lost
305
Q

NHEJ and CRISPR-Cas9

A
  • random change (indel) will be introduced at the targeted site
  • if break is repaired correctly, it will be recognized by sgRNA and cut again!
    (used to create knockouts/loss-of-function mutations)
    *random mutation is likely a loss-of-function mutation

-Two cells edited with the same guide could end up with a different repair.
(null, hypomorphic, silent)

306
Q

HDR and CRISPR

A

-relies on homologous chromosomes/sister chromatids
– UNLESS scientists provide a repair template with homology arms (regions of homology to the L/R of break site); the cell can be used instead

**used to generate gene knock-ins

307
Q

transgenic mouse model

A

1) mouse embryonic stem (ES) cells are genetically modified
2) edited stem cells are injected into a blastocyst
(cells are pluripotent and totipotent
3) mosaic pups are born
(cells are mix of original blastocyst cells and edited stem cells)
4)mosaic pup are crossed to WT

308
Q

pluripotent and totipotent

A
  • cells can develop into any kind if cell type given the proper cues

P: adult
T: embryo and adult (all stages)

309
Q

mosaicism

A

occurs when a person has two or more genetically different sets of cells in his or her body.

310
Q

mosaic mouse phenotypic ratios

A
  • 0% homozygous for edited DNA
  • 25% homo for transgene
  • 50% hetero for transgene
  • 25% homo for WT
  • always heterozygous for edited gene bc only female or male is affected most often
311
Q

how to get mice homozygous for edited genes?

A
  • selfing of edited gene heterozygotes

50/50: heterozygotes/homozygous dominant

312
Q

problems with transgene mice

A
  • germline may be derived from the blastocyst rather than the ES cells injected
    — if no pups inherit the genome edit when crossing WT and mosaic
313
Q

CRISPR-based gene therapy for treatment of SCA and b-thalassemia

A
  • BCL11A is a TF that promotes the switch from expressing y-globin(fetal) to b-globin(adult)

-Casgevy introduces mutations in an ENHANCER that normally turns BCL11A on after birth

  • gene that produces BCL11A has its own promoters and enhancers that are controlled by other TFs!
  • Casgevy mutates an enhancers of the BCL11A gene so that it is not made and affects how the globin genes are regulated
  • preventing BCL11A expression means that y-globin can be expressed
314
Q

transcription factors

A
  • regulate gene expression
  • TFs themselves are encoded by the genes
  • genes that produce TF have its own promoters and enhancers that are regulated by other TFs
315
Q

advantages to the hemoglobin SC engineering strategy

A

1) causing a gene knockOUT is easier than a gene knockIN –> it doesn’t not require HDR with a template (you just break the gene)

2) Casgevy works for multiple kinds of b-globin disorders regardless of what specifically is wrong with a particular patient’s B-globin gene

3) using a patient’s own (edited) marrow cells makes rejection of a transplant unlikely

316
Q

sanger sequencing elements

A
  • DNA synthesis for Sanger sequencing includes ddNTPs, so synthesis will stop whenever one is added even though only 1 primer is used
  • dNTPS
  • ddNTPS
  • primer
317
Q

reading sanger sequencing gels

A
  • top to bottom
  • use separate lanes OR one lane with dyes
318
Q

modifications to Sanger sequencing

A
  • capillary gel electrophoresis (used laser through DNA sample and detector)
  • automated detection (sequence analysis and reconstruction on computer)
319
Q

current sequencing methods generate…

A

reads that are 100s or 1000s of base pairs long

human genome is billions of bp long

320
Q

read

A

the sequence determined by a single sequencing reaction

321
Q

contig

A

a set of DNA segments/sequences that overlap in a way that provides contiguous (touching) representation of a genomic region

322
Q

assembling contigs requires…

A

identification of overlapping reads

(reads are like sentence fragments that need to be assembled into contigs like sentences)

323
Q

the human genomes was first sequenced in …via…

A

2000s
- competition and collaboration between the government (Human Genome Project) and the private sector (Celera Genomics)

324
Q

reference genomes are not…

A
  • the genome of a single individual
  • synonymous with wild type or normal
  • 100% perfectly assembled or complete (it may improved over time!)
325
Q

can u identify a causative mutation by checking all the bases that differ from the reference genome?

A

NO there will be millions of differences to check because the genome is so BIG

  • extracting useful into from genome data remains a challenge
  • most of the human genome is shared yet we still have a tremendous amount of genetic variation
326
Q

can u identify a causative mutation (a specific disease) in a genome sequence?

A
  • MAYBE…we can check what differences we can find in the regions of the genome that have been annotated as the disease gene or one of its known regulatory elements
326
Q

SNPs

A
  • common differences between genomes (single nucleotide variant will be considered a SNP if it becomes a frequency of at least 1%)
  • every SNP begins as a substitution mutation (rare variant that becomes COMMON)
  • may or may not be the cause of a mutation
  • can only change protein if in coding region
327
Q

linked SNPs

A

no effect on protein production or function (just bc a SNP is in an exon does NOT mean it must be causative)

328
Q

causative SNPs

A
  • in the regulatory region = changes the amount of protein produced
  • in coding region/protein = changes amino acid sequence
329
Q

ideal marker SNP

A

one that is completely linked to whatever causes the trait of interest (recombination frequency is ~0%)

330
Q

associations between easily genotyped SNPs

A
  • take advantage of known associations between easily genotyped SNPs and whatever linked variant is responsible for a phenotype even if we do NOT know what the causative nearby variant !!!

*may not know what allele is responsible for the trait, but linked SNPs can sometimes still help us make predictions about phenotype

(haplotype presence can show commonality)

331
Q

haplotype

A
  • sets of SNPs that do not independently assort (inherited as one unit
  • many SNPS are completely linked to all other nearby SNPS
332
Q

recombination hot spots

A
  • SNPs not located in recomb. hot spots are almost never separated from each other by recombinations
  • not all regions of the genome are equally likely to undergo recombination
333
Q

How to use SNPs to determine certain genetic outcomes

A
  1. Locate SNPs in a region of DNA
    - (examine DNA in and around gene)
  2. Create haplotype group (create groups with certain SNPs and determine people that fall into certain groups)
  3. Test haplo response to certain condition/treatment
334
Q

PGD

A
  • pre-implantation genetic diagnosis tests embryos for genetics traits before implantation in uterus
  1. mothers eggs are collected
  2. each is fertilized with sperm
  3. fertilized eggs are placed on Petri dish to grow
  4. embryos divide for 3 days
  5. blastomere is removed from each embryo
  6. blastomere is tested to see if its embryo contains the defective gene carried by one or both parents
  7. defective gene discarded or donated to reserach
  8. good genes are implanted or frozen for later use
335
Q

eugenics

A

immoral and pseudoscientific theory that claims it is possible o perfect people and groups through genetics and the scientific laws of inheritance
- use incorrect and prejudiced understanding of Charles Darwin and Gregor Mendel to support the idea of “racial improvement”

336
Q

DNA fingerprinting

A
  • relies on highly variable SSR loci.
  • 13 different loci are tested
  • each person has 2 alleles for each locus and there are many differently sized alleles for each locus the population

*** fingerprinting looks at the sizes of PCR products made from all 13 loci

337
Q

DNA profile gel electrophoresis

A
  • size standard added to every lane
  • other lanes are fluorescently labeled PCR products
  • compare the fragments to see if relatives DNA fingerprints match relatives
  • babies have STR loci from each parent
338
Q

genomic VS cDNA library

A

genomic: represent all regions of DNA equally and show what the intact genome looks like in the region of each clone

cDNA: reveal which parts of the genome are transcribed in specific tissues and how those transcripts are processed into mRNAs

339
Q

knockout

A

homozygous for an amorphic allele of a gene induced by gene targeting

340
Q

SSRs, indels, copy number variants VS SNPs

A

SSRS, indels, CNVs DO change the number of base pairs in a sequence

SNPs do NOT [PCR/gel cannot differentiate between fragments]

341
Q

2 primers are used in a PCR reaction…the final product will extend from…

A

the 5’ end of one primer to the 5’ end of the other primer

342
Q
A
343
Q

if the HDR pathway is nonfunctional for CRISPR….

A

NHEJ can be used to make a random insertion or deletion in a specific location

344
Q

choosing a restriction enzyme

A
  • directly upstream of the region we want to utilize
  • sticky ends > blunt ends
345
Q

If you digest with a RE…

A

the size of the band on the gel will be larger than the original sequence due to the addition of RE

346
Q

why might you utilize a plasmid for transformation into corn?

A

industrial production of Bt toxin in bacteria for use by organic farmers

347
Q

When isolating the Bt gene for this application, it was first necessary to generate the Bt gene cDNA.

A

false
-can use genomic DNA bc prokaryotes do NOT have introns

348
Q

CRISPR and phenotypic ratios

A
  • doesn’t change phenotypic ratios for crosses and offspring
349
Q

DNA fingerprinting looks at alleles that vary in size based on…

A

SSRs

350
Q

cell theory

A

all cells arise from preexisting cells

351
Q

meiosis VS mitosis responsibilities for cell reproduction

A

Meiosis: generating heritable genetic variation
Mitosis: growth by cell division

352
Q

All cells in our body have the same DNA EXCEPT…

A

somatic mutations that arise in development

353
Q

somatic mutations

A
  • generate clones of genetically distinct cells
  • most cancers have more than 1 somatic mutation
  • all mutated cells will contain the first mutation
354
Q

cancer vs normal cells

A
  • cancer cells continue to divide when normal cells stop
  • genetic changes lead to inappropriate growth or survival
355
Q

somatic vs germline mutations

A

S: mutation after the embryonic stage somewhere during development
G: mutation occurs before sperm and egg fertilize

356
Q

ways that cancer cells can grow

A
  • loss of apoptosis cell death
  • loss of contact inhibition that stops growth of cells when cells of certain type come in contact with one another
357
Q

cancer cells can invade other parts of the body
examples:

A
  • angiogenesis (blood vessel development)
  • metastasis (changing skin cells)
  • growth in areas of the body away from the primary initiation of cancer
358
Q

proto-oncogenes

A
  • WT function is to promote growth (at appropriate times and locations)
  • GOF mutation in a proto-oncogene can promote cancer
359
Q

tumor suppressors

A
  • WT function is to restrict growth OR maintain genome stability
  • LOF mutation in a tumor suppressor can promote cancer
360
Q

Ras

A
  • proto-oncogene
  • promotes cell proliferation when active
  • normally, activation is regulated by cell signaling
  • mutation causes Ras to be stuck in active form regardless of signaling with growth factor
361
Q

Bcr/Abl

A
  • proto-oncogene
    -Bcr and Abl genes are on separate chromosomes
  • Abl promotes cell proliferation but does so in a way that is tightly regulated
  • A reciprocal translocation generates a fusion protein that retains the pro-proliferation abilities of Abl, but it cannot be regulated properly

“Philidelphia chromosome”

362
Q

Rb

A

*tumor-suppressor
- Rb normally functions to restrict the expression of cell cycle genes

363
Q

p53

A

*tumor-suppressor
- prevents cells from dividing when there is DNA damage
- makes sure DNA is repaired so replication errors or damage don’t become permanent mutations
- anti-growth gene

  • does not directly promote growth, but increases the likelihood of more mutations occurring within tumor suppressors or proto-oncogenes
364
Q

retinoblastomas

A
  • cancerous tumors associated with LOF mutations in the Rb tumor suppressor
  • recessive LOF
  • WT are haplosufficient (but are at high risk for cancer bc already have one allele disabled in every cell
  • autosomal dominant inheritance pattern with high penetrance (~90%)
365
Q

spontaneous VS familial retinoblastoma

A

S: requires 2 mutations (very unlikely)
F: rb1 allele is inherited for heterozygosy and another mutation occur

*converting a SINGLE retina cell from WT phenotype to Rb-/- phenotype can initiate a tumor (random mutation)

366
Q

bilateral vs unilateral retinoblastomas

A
  • nearly all familial
  • spontaneous RB in 2 eyes is highly unlikely
  • loss of heterozygosy is more statistically likely to happen in 1 or more retinal cells

unilateral - tumor in 1 eye
bilateral - tumor in 2 eyes

367
Q

radiation treatment

A
  • can mutagenize already-destabilized cancer cells
  • precise spatial targeting gives high mutagenic dose to tumor , with much less exposure to healthy tissues
  • genomic instability drives cancer progression but also can make cancer cells more vulnerable to catastrophe
  • use cancer-causing radiation to stop radiation
368
Q

targeted chemotherapy

A
  • knocking which mutations cause a person’s cancer can enable targeted chemotherapy
  • often target DNA replication that happens a lot in fast-dividing cells
    *therefore affects rapidly growing cells like hair and stomach cells

genetics (5 decks)

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