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CU: Genetics > Exam 2: Chapters 5-9 > Flashcards

Exam 2: Chapters 5-9 Flashcards

Linkage, recombination, DNA structure, replication

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1
Q

What is special about the female X chromosomes?

A

they silence an X chromosome
- inactivate during embryonic development “Barr body”
- Barr body: so condensed, inaccessible for making proteins
- ie calico cat: orange or black on X chromosomes, silencing not affect homozygous genes, heterozygous leads to calico (some orange silenced, some black silenced)

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2
Q

What is linkage? What does it affect?

A
  • Genes within 50 map units of each other have a decreased likelihood of being separated by recombination
  • Affects equal percentage of gamete formation
  • Affects 9:3:3:1 ratio
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3
Q

What are the types of gametes?

A

“parental types” gene combination from a certain parent
“recombinant types” gene combination because of recombination
- percentage decreases with linkage

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4
Q

What are syntenic genes?

A

genes on the same chromosomes

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5
Q

What is the “+” notation?

A

normal or wild type

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6
Q

What is a testcross parent?

A

a parent with either both recessive or both mutated alleles

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7
Q

How does gene location affect type of chromosome?

A

As genes get closer % parental types increases, % of recombinant type decreases.
- if gametes all have equal change then the genes are greater than 50 map units apart

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8
Q

What happened in 1909 concerning crossing over?

A

observed chiasmata at chromosomes during prophase of meiosis I (microscope help progression of crossing over)

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9
Q

How did Thomas Hunt Morgan influence crossing-over?

A

suggest chiasmata were sites of chromosome breakage and exchange

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10
Q

What happened in 1931 concerning crossing over?

A

H. Crighton & B. McClintock (corn) & C. Stern: direct evidence that genetic recombination depends on reciprocal exchange of chromosomes
- physical markers used identify specific chromosomse
- genetic markers used as pointes of reference for recombination

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11
Q

What did H. Sturtevant propose?

A

Recombination frequencies (RF) can be used as a measure physical distance between two linked genes
1 % RF = 1 map unit (m.u.) = 1 centiMorgan (cM)

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12
Q

What is important about recombination frequency percentage?

A

Never exceeds 50%
- RF Unlinked genes = 50% because of independent assortment
- RF linked genes cannot exceed 50% know dealing with linked
– no crossovers = parental types
– single & double crossovers –> 1:1 parental:recombinant types on average
Parentals 50-100%
Recombinants 0-50%

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13
Q

What is important about mapping genes?

A

To determine correct gene order
- Left-right orientation is arbitrary
- best: summing many small intervening distances

(start with homozygous parent because then no crossing over)

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14
Q

What are limitations of two point crosses?

A
  • Difficult to determine gene order if two genes are close together
  • Actual distances between genes do not always add up
  • Pairwise crosses are time and labor consuming
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15
Q

What is the benefit of three point crosses? What are the three things to consider in analysis?

A

Faster & more accurate mapping
- double cross between genes: should see gene from parent gene from other parent gene from first parent

Analysis:
- most common: parental configuration
- least common: double crossovers
- double crossover shows which gene in middle

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16
Q

What is the use of the Chi square test? What can it do?

A

It pinpoints the probability that ratios are evidence of linkage
- “goodness of fit” between observed and expected values
- can reject the null but not prove a hypothesis
(Cut off 0.05) > 0.05 no difference, no linkage, < 0.05 difference, linkage

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17
Q

How is the Null Hypothesis used in the Chi square test?

A
  • Null hypothesis: observed values are no different from expected values
  • linkage studies: null hypothesis = no linkage
  • then expect 1:1:1:1 in F2 progeny
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18
Q

What information is needed for the chi-square test? Then what?

A

breeding experiment:
- total # progeny
- # progeny classes
- # offspring observed in each class
–> calculate number of offspring expected in each class if no linkage

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19
Q

What is the equation for the chi-square test?

A

x^2 = (# observed - # expected)^2/# expected

  • Consider: degrees of freedom (df) = N-1
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20
Q

How is p value determined for chi-square test?

A

Chi-square value and df
- probability that deviation from expected numbers had occurred by change

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21
Q

How do we know the genetic function of DNA? And where does this lead us? What is it dependent on?

A

By knowing the molecular structure
Leads to an increased understanding of the biochemical process
Depends on: proteins to read the information as DNA itself is chemically inert

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22
Q

What are phosphodiester bonds?

A

Covalent bonds between adjacent nucleotides (in the backbone) that are fairly strong between 3’ (growing end) and 5’ (toward Phosphate)

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23
Q

What was Griffith’s contribution to DNA?

A
  • Bacterial transformation experiments showed DNA as the substance of genes
  • Streptococcus pneumoniae: S - virulent, R - nonvirulent,
  • S dead: mouse alive, S dead & R alive: mouse dead
    (1928)
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24
Q

What were Avery, MacLeod, and McCarty’s contribution to DNA?

A

Identified DNA as the transforming principle
- Protease, RNase, DNase, ultracentrifugation (fats),
- Only DNase meant no transformation of S to R cells (no DNA)

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25
Q

What were R. Franklin and M. Wilkins contribution to DNA?

A

XR diffraction
- helical structure w/ 20 A diameter
- Space between repeating units: 3.4 A
- Complete turn every 34 A (10 nucleotides)

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26
Q

What was E. Chargaff’s contribution to DNA? What is the reason for this?

A

A:T ratio is 1:1 and G:C ratio is 1:1
A:T has two H bonds
G:C has three H bonds

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27
Q

What are the two groups of nitrogenous bases?

A

Purines:
- Adenine
- Guanine
Pyrimidines:
- Cytosine
- Uracil
- Thymine

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28
Q

How do the DNA strands relate to one another? What are the kinds of spirals/backbones?

A
  • Antiparallel
  • B-form DNA: right-handed spiral & smooth backbone (MOST DNA)
  • Z-form DNA: left-handed & irregular backbone
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29
Q

What are the ways DNA are read?

A
  • most of the time from unwound DNA
  • some information from double-stranded DNA (DNA-binding proteins regulate gene expression)
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30
Q

What are the three chemical differences between RNA and DNA?

A
  • Sugar (deoxyribose/ribose)
  • Nitrogenous Base: thymine & uracil
  • Strand: DNA - Double, RNA - single
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31
Q

What are the three models of DNA replication?

A

1) Semiconservative: each strand is a template for a new strand by insertion of complementary base pair producing two identical daughter double helices (Watson-Crick)
2) Conservative: parental double helix remains intact & daughter helices newly synthesized
3) Dispersive: both strands of daughter helices contain original and newly synthesized DNA

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32
Q

What are the three requirements for DNA polymerase action?

A

1) Four dinucleotide triphosphates (for incorporation into chain and energy)
2) Single-stranded template (unwound by other proteins)
3) Primer with exposed 3’ hydroxyl

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33
Q

How are strands read verses synthesized?

A

Read 3’-5’
Synthesized 5’-3’

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34
Q

What are the four aspects of the mechanisms of DNA replication in initiation?

A

1) initiator protein: binds to origin of replication (replication will occur in two directions)
2) Helicase: unwinds the helix (regions are worked on in replication bubble- proceeds in two directions)
3) Single-strand binding proteins: keep the DNA helix open (get in way of bases so do not snap back together)
4) Primase: synthesizes RNA primer

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35
Q

What do primers do?

A

They are complementary and antiparallel to each template strand

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36
Q

What are the six aspects of the mechanism of DNA replication: Elongation?

A

Elongation: the correct nucleotide sequence is copied from template strand to newly synthesized strand of DNA
1) DNA polymerase III: catalyzes phosphodiester bond formation between adjacent nucleotide (polymerization)
2) Leading strand: continuous synthesis
3) Lagging strand: discontinuous synthesis
4) Okazaki fragment: short DNA fragments on lagging strand
5) DNA polymerase I: replaces RNA primer with DNA sequence
6) DNA ligase: covalently joins successive Okazaki fragments (phosphodiester bonds)

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37
Q

What is important about a circular bacterial chromosome and DNA replication?

A

Since replication proceeds in two directions from a single Origin of replication
- Unwinding creates supercoiled DNA

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38
Q

What is the function of DNA topoisomerase?

A

relax supercoils by cutting sugar phosphate backbone bonds strands of DNA
- unwound broken strands sealed by ligase
(synthesis continues bidirectionally until replication forks meet)

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39
Q

What are the recombinations at the DNA level?

A

(new combinations of alleles created by two types of events in meiosis)
- Independent assortment:
- Crossing over:

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40
Q

What is independent assortment?

A

each pair homologous chromosomes segregated freely from the other (new allele combinations of unliked genes)

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41
Q

What is crossing over?

A

two homologous chromosomes exchange portions of DNA
(new allele combinations for genes, ensures proper chromosome segregation during meiosis)

42
Q

What are mutations and what are the two categories and the types?

A

Changes in replicated DNA base sequence
- Forward mutation: wild-type allele to different allele
- Reverse mutation: mutant allele back to wild-type
Types: substitution, deletion, insertion

43
Q

What is substitution and what are the two types?

A

Mutation by replacement of a base by another base
- Transition: purine by purine or pyrimidine by pyrimidine
- Transversion: purine by pyrimidine (& vice versa)

44
Q

What is deletion?

A

Mutation by block of one or more base pairs lost from DNA

45
Q

What is insertion?

A

Mutation by block of one or more base pairs being added to DNA

46
Q

What are the two mutations rates?

A

1) Parents to children
- rate: 1 x 10^-8
- each child 60 mutations
- most don’t influence phenotypes
2) In sperm
- rate: 2-4 x 10^-8
- Because continual mitosis
- Older fathers means more mutations

47
Q

What did Luria-Delbruck do in 1943?

A

Fluctuation test
Purpose: origin of bacterial resistance to phage infection
Procedure: Infect wild-type bacteria with phage: use velvet to replicate bacterial colonies
Result: Majority of cells die but some grow and divide
Questions:
- Cells undergo biochemical alteration?
- Cells have heritable mutations for resistance?
- Cell mutations by chance or response to phage?
Conclusion: by chance

48
Q

What are the four natural processes that damage DNA?

A
  • Depurination: mutation 75% of time, 1000/hr in every cell, Guanine no longer function as Guanine
  • Deamination of Cytosine: changed to U lead to A-T after replication
  • X-Rays: break phosphate backbone (deletion)
  • UV light: thymine dimers (adjacent thymines form abnormal covalent bonds)
49
Q

What are trinucleotide repeats?

A
  • repeats of three nucleotide
  • Unstable trinucleotide repeats cause 20 human disease genes: Fragile X syndrome, Huntington disease
  • Children of carriers may have expanded number of repeats
    25 ok, 50 not good, 200 bad
50
Q

What are the two DNA repair mechanisms?

A

Accurate repair systems:
Error-prone repair systems

51
Q

What is proofreading?

A

What DNA polymerase does to recognize and excise mismatches in DNA (rare incorrect information of bases by DNA polymerase)

52
Q

What does DNA glycosylases do? What comes after?

A

Fixes damaged DNA post replication (usually chemical change in base ie deamination of C)
- endonuclease cleans up surrounding bases
- DNA polymerase patches the hole
- DNA ligase reattaches backbone

53
Q

How does is DNA damage from UV light repaired?

A
  • UvrA - UvrV complex: scans for double helix distortions
  • UvrB - UvrC complex: nicks damaged DNA
  • DNA polymerase fills the gap
54
Q

What directionality do polypeptides have?

A

N terminus (amino acid side) and C terminus (carboxyl side)

55
Q

What causes sickles cell disease?

A

Gly for Val at 6th amino acid affecting 3D structure of hemoglobin B chain
- Abnormal protein aggregates causes sickle shape of RBC

56
Q

How is sickle-cell anemia pleiotropic?

A

(Gene functioning in several pathways)
A) Rapid destruction of sickle cells –> anemia –> fatigue, heart damage, overactivity of bone marrow
B) Clumping of cells, interference with circulation –> local failures in blood supply –> Damage to heart, kidney, muscle/joints, brain, lung, GI
C) Accumulation of RBC in spleen –> enlargement & damage to spleen

57
Q

What are the two steps of gene expression?

A
  • Transcription: RNA polymerase transcribes DNA to produce an RNA transcript
  • Translation: Ribosomes translate the mRNA sequence to synthesize a polypeptide (follows the genetic code)
58
Q

What are the names of the strands involved in gene expression?

A
  • Template strand: of DNA is complementary to mRNA
  • RNA-like strand: of DNA has the same polarity and sequence of mRNA
59
Q

What is the DNA, mRNA, and polypeptide directionality?

A

DNA read 3’ to 5’
mRNA is transcribed 5’ to 3’
polypeptide: N to C terminus

60
Q

What is the genetic code (7)?

A

1) It has triplet codons
2) Codons are non-overlapping
3) Three nonsense codons don’t encode an amino acid; UAA, UAG, and UGA
4) It is unambiguous
5) Reading frame is established from fixed starting point - AUG (translation initiation)
6) mRNAs and polypeptides have corresponding polarities
7) Mutations: frameshift, missense, and nonsense

61
Q

What are the aspects of the accurate repair system for DNA?

A
  • Reversal of DNA base alterations (excision repair)
  • Homology-dependent repair of damaged bases or nucleotides
  • Correction of DNA replication errors
  • Double-strands break repair
62
Q

What are the aspects of the error-prone repair systems?

A
  • SOS systems
  • Microhomology-mediated end-joining (MMEJ)
63
Q

What is in vitro? Why does this work?

A

Translational systems from one organisms can use mRNA from another organism to generate protein
- works because most cells same basic genetic code
- perfect correspondence between codons & amino acids

Exceptional: ciliates, mitochondria, some prokaryotes

64
Q

What is the protein generating transcription process in prokaryotes?

A
  • Catalyzed: RNA polymerase which is directed by…
  • Promoters: DNA sequence signal start transcription
  • Function: adds nucleosides 5’-3’ forming phosphodiester bonds using ribonucleotide triphosphates (A/C/G/UTP), energy from hydrolysis of NTP bonds
  • Terminators: sequence signal stop transcription
65
Q

What is the initiation step of transcription in prokaryotes?
(Protein end)

A

RNA polymerase binds to promoter sequence
- Sigma factor binds to RNA polymerase making holoenzyme
- Region of DNA unwound to form promoter complex
- Phosphodiester bonds formed between first two nucleotides

66
Q

What is the elongation step of transcription in prokaryotes?
(Protein end)

A
  • Sigma factor separates from RNA polymerase leaving core enzyme
  • Core RNA polymerase loses affinity for promoter moves 3’-5’ on template strand
  • nucleotides added to 3’ of mRNA
67
Q

What is the termination phase of transcription in prokaryotes?
(Protein end)

A

Signal end
- Extrinsic kind: require rho factor
- Intrinsic kind: no additional factors required
- Usually form hairpin loops (intramolecular H-bonding)

68
Q

What is the difference between prokaryotic and eukaryotic transcription?

A

Eukaryotes often have enhancers
- may be thousands of base pairs away from promoter
- Required for efficient transcription

69
Q

What is the difference between the product of transcription for eukaryotic or prokaryotic organisms?
(Protein end)

A
  • Pro: primary transcript = mRNA
  • Euk: primary transcript = processed to make mRNA: 5’ methylated cap, 3’ poly-A tail, introns removed by RNA splicing
70
Q

What is a 5’ methylated cap?

A

A backward G added to the first nucleotide of a primary transcript

71
Q

What is a 3’ poly A tail?

A

Adenosines added onto the 3’ end by Poly-A polymerase

72
Q

What are Exons and Introns?

A

Exons: DNA sequence found in gene & mature mRNA (expressed)
Introns: DNA sequence found in gene but not mRNA (intervening regions)

73
Q

What is RNA splicing? Creates, catalyst, alternatives

A

Two sequential cuts to remove an intron
- creates Lariat (loop)
- catalyst: spliceosome
- alternative splicing: produces different mRNAs from same primary transcript (ie intron in membrane-bound, exon in secreted Ig)

74
Q

What are the three levels of tRNA structure?

A

Primary: nucleotide sequence
Secondary: (cloverleaf) b/c short complementary sequences
Tertiary: (L shape)b/c 3D folding

75
Q

What connects tRNA and amino acids?

A

Aminoacyl-tRNA synthetase - recognizing specific amino acid and structural features of corresponding tRNA

76
Q

What is the tRNA Wobble effect?

A

Some tRNAs recognize more than one codon
(wobble position: last position on the right, not as strong: more flexible)

77
Q

What are the composition of Prokaryotic and Eukaryotic ribosomes?

A
  • Pro: 70S, 50S & 30S (subunits)
  • Euk: 80S, 60S & 40S (subunits)
    Subunits are composed of RNA & Protein
78
Q

What are the different functions of the ribosome subunits?

A
  • Small subunit: binds to mRNA
  • Large subunit: has peptidyl transferase activity
    – Three distinct tRNA binding areas: Aminoacyl, Peptidyl, and Exit sites
79
Q

What is the mechanism of translation?
(Protein end)

A
  • Initiation stage: start codon, AUG 5’ end (bacteria: initiator tRNA has formylated Met: fMet)
  • Elongation stage: aa added to growing polypeptide, 5’-3’ direction along mRNA
  • Termination stage: polypeptide synthesis stops at 3’ end (recognize nonsense codons, synthesis halted by release factors, release of ribosomes, polypeptides, and mRNA)
80
Q

What is the initiation stage of translation in prokaryotes? (protein end)

A

Ribosome binds to Shine-Dalgarno box (AGGAGG) and in AUG in 3 subsequential steps:
1) Small ribosomal subunit binds
2) fMet-tRNA positioned in P site
3) Large subunit binds

81
Q

What is the initiation stage of translation in eukaryotes? (protein end)

A

Small ribosomal subunit binds to 5’ methylated cap then migrates to first AUG codon, initiator tRNA carries Met

82
Q

What is the elongation phase of translation? (protein end)

A
  • add amino acids to C-terminus of polypeptide
  • Charged tRNAs into A site by elongation factors
83
Q

What is the termination phase of translation? (protein end)

A
  • stop codons do not have normal tRNAs anticodons
  • Release factors bind to stop codons: release ribosomal subunit, mRNA, & polypeptide
84
Q

What do polyribosomes do?

A

Several ribosomes translating same mRNA: simultaneous synthesis of many copies of polypeptide from single mRNA
(typical action)

85
Q

What is enzymatic cleavage? What are the three types?

A

A form of post translational processing
1) N-terminal Met removal: fMet
2) Polyprotein processes (polyprotein cleaved into multiple smaller polypeptides) producing more protein by creating a long polypeptide (more efficient than starting and stopping) then cut into smaller polypeptides
3) Zymogen activation - inactive form of dangerous (digestive) protein has prosegment, cleaved to activate protein

86
Q

Prokaryotic vs eukaryotic cells what is the big difference in replication because of the cell structure?

A
  • Prokaryotes: do not have a nucleus so transcription and translation can happen simultaneously (leads to greater growth rate)
  • Eukaryotes: Transcription in nucleus, translation in cytoplasm sequentially (takes time)
87
Q

What are additional ways to have mutations? (not in coding sequence)

A
  • Changes in enhancer region or promotor regions not part of DNA code (exons)
  • splicing not only/all intron
88
Q

What was the Human Genome Project? What has happened since? (Ch 9)

A

Completed in 2003 - an accurate sequence of the human genome
- now faster
- Thousands of species have been sequenced

89
Q

What are the general ideas behind genome sequencing? (Ch 9)

A
  • Fragmenting the genome (need some overlap)
  • Cloning DNA fragments
  • Sequencing DNA fragments
  • Reconstructing the genome sequence from fragments
  • Analyzing the information found in genomes
    (info to grow, maintain, reproduce human)
    (working on technology to read DNA strand in entirety)
90
Q

Why were we able to fragment the DNA?
Where does it come from and what is the purpose? (Ch 9)

A

Discovery of restriction enzymes “molecular scissors” that recognize specific sequence of bases
- cuts both strands of sugar-phosphate backbones
- generates restriction fragments (by digestion of DNA)
- hundreds now available
(straight or sticky ends)
- defense mechanism in bacteria against virus (chop up invading DNA)

91
Q

What are the characteristics of recognition sites for restriction enzymes? (Ch 9)

A

Usually 4-8 bp
- often palindromic: base sequences of each strand are identical when read 5’ - 3’
- Each enzyme cuts at same place relative to its specific recognition sequence

92
Q

What are the type of ends restriction enzymes produce? (Ch 9)

A
  • Blunt ends: cuts are straight through both DNA strands at the line of symmetry
  • Sticky ends: cuts are displaced equally on either side of the line of symmetry (5’ or 3’ overhangs)
  • Will recombine after cut unless something gets in way
93
Q

How do you calculate frequency of cutting? (Ch 9)
*

A
  • Ave fragment length is 4^n
  • n = number of bases in the recognition site
  • 4-base recognition site occurs every 4^4 bp
    – ave restriction fragment size = 256 bp
    – 3 billion bp grnome/256 = 12 million fragments
  • 6-base recognition site occurs every 4^6 bp
    – ave restriction fragment size is 4100 bp
    – 3 billion bp genome/4100 = 700,000 fragments
  • 8-base recognition site occurs every 4^8
    – ave restriction fragment size is 65,536 bp
    – 3 billion bp genome/65,536 = 45,776 fragments
94
Q

What is the purpose of gel electrophoresis? (Ch 9)

A
  • separate DNA by size
  • linear DNA: distance through gel depends on size
  • determine size of fragments by comparison with markers of known size (ladder)
95
Q

What is the purpose of molecular cloning? (Ch 9)

A
  • allow for focusing on certain information (get around information overload)
  • Definition: means to purify a specific DNA fragment away from all other fragments and make many identical copies of the fragment
96
Q

What are the two steps of molecular cloning? (Ch 9)

A
  • Insert DNA fragments into cloning vectors to make a recombinant DNA molecule
  • Transport recombinant DNA into living cell to be copied
97
Q

What are the main features of plasmid vectors (molecular cloning step 1)? (Ch 9)

A
  • origin of replication (certain gene will be copied)
  • known restriction site for cloning insert DNA
  • selectable marker (ie antibiotic resistance) to recover plasmids
98
Q

How are recombinant DNA molecules made with plasmid vectors? (Ch 9)

A
  • Cut (digestion) plasmid and open up using restriction enzyme (results in complementary sticky ends
  • insert gene of interest (plasmid may or may not take up gene)
  • ligase seals phosphodiester backbones between vector and insert
99
Q

What is step 2 of molecular cloning? (Ch 9)

A

Host cells take up and amplify recombinant DNA
- Transformation: the process by which a cell or organism takes up foreign DNA (via shock- daze w/o death)
- success ~ 0.1%
- Selection: only cells with plasmid grow in media with ampicillin
- cell multiplies with recombinant plasmid

100
Q

What is the challenge of reading DNA without knowing where the gene is? (Ch 9)

A
  • there are six potential reading frames due to stop codons and either strand can be template strand. LOOKING FOR ORF
  • open reading-frame (ORF) = reading-frame uninterrupted by stop codons: indicates this is the reading frame to focus on
101
Q

What is the arrangement of genes in the genome?
What is the exome and what is the percentage of the DNA is considered the exome? (Ch 9)

A
  • 25,000 genes (humans)
  • part corresponding to exons exome: DNA actually codes for proteins
  • Most genome is non-coding
    – Exome = ~ 2%
    – Introns
    – Centromeres, telomers, transposable elements
    – Simple repeating sequences
102
Q

What are the regions of the DNA that contain genes? (2 types) (Ch 9)

A
  • Gene-rich regions: chromosomal regions that have many more genes than expected from average gene density over entire genome
  • ie: class II region of major histocompatibility complex
  • Gene deserts: Regions of >1 Mb that have no identifiable genes
  • 3% human genome comprised of gene deserts
  • Exist because hard to identify?

Biological significance of regions is unknown

CU: Genetics (7 decks)

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