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BIOL 2500 > Definitions > Flashcards

Definitions Flashcards

1
Q

Def: Point mutations

A
  • Change in a nucleotide(s)

- New genetic variation

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2
Q

Def: Chromosomal mutations

A
  • Important source of new genetic variation
  • Can cause developmental defects and disease in humans
  • Are structural changes in a chromosome OR duplications/deletions of whole chromosomes
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3
Q

Name the four types of chromosomal mutations?

A

1) deletion
2) duplication
3) inversion
4) translocation (chromosomes exchange segments)

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4
Q

What are the causes of chromosomal mutations?

A
  • chromosome breaks during DNA synthesis and crossing over
  • radiation, chemicals
  • transposable elements of DNA
  • recombination errors during cross-over
  • broken ends are sticky and may rejoin in a new configuration
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5
Q

What may be a significant factor in maintaining reproductive isolation between species?

A

Non-lethal chromosomes

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6
Q

What are the effects of non-lethal mutations?

A
  • alter meiotic pairing
  • alter effects of recombination (crossing over) and the genotypes of gametes
  • can be studied genetically
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7
Q

Pericentric vs Paracentric inversions

A

Pericentric: cross-over products inviable, centromere involved
Paracentric: cross-over products inviable, centromere not involved

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8
Q

How do chromosomal mutations play a role in Downs syndrome, leukemia, lymphoma, and cancer cells?

A

Downs syndrome: a reciprocal translocation causes one form of this
Leukemia and Lymphoma: translocation interrupts a gene involved in cell cycle regulation
Cancer cells: have “genomic instability” due to mutations in DNA repair genes

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9
Q

Are there chromosomal mutations on the human Y chromosome?

A
  • none fully proven to be truly Y-linked

- Chinese study of hearing impairment?

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10
Q

How do deletions affect a homozygote where the centromere is retained?

A

-the gametes are inviable due to missing genes

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11
Q

If a chromosome loses it centromere due to deletion, what happens?

A
  • the whole chromosome is lost

- usually lethal

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12
Q

How do deletions affect a heterozygote where the centromere is retained?

A
  • if wild type alleles are on the complete homologue, phenotype is normal/wild type
  • if recessive mutation on the complete homologue, recessive phenotype expressed = pseudodominance (unexpected expression of recessive allele due to absence of dominant allele)
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13
Q

Def: Pseudodominance

A

unexpected expression of recessive allele due to absence of dominant allele

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14
Q

Name the three types of duplications.

A

1) tandem repeat
2) reverse tandem repeat
3) terminal tandem repeat

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15
Q

What are tandem repeats used for today?

A

-DNA fingerprinting

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16
Q

What is an example of duplication where in evolutionary time, the extra gene copies took on new functions?

A
  • globin genes
  • code for protein subunits of hemoglobin
  • multiple globin genes in humans
  • genes with greater 02 affinity expressed in the fetus
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17
Q

What are some genetic consequences of inversions?

A
  • homozygous mutant still has all the genes and is normal unless there is an inversion break or the position affects gene expression
  • heterozygous mutant often have decreased fertility due to cross-over events within the inversion
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18
Q

Are the gametes viable or inviable after a:
A) paracentric inversion
B) pericentric inversion

A

A) 50% viable and 50% inviable, inversion heterozygotes may have reduced fertility
B) 50% viable and 50% inviable

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19
Q

Def: Reciprocal translocation

A

2 non-homologous chromosomes exchange segments

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20
Q

During translocations, what are the two possibilities of what happens at anaphase I?

A

1) alternate segregation
- alternate centromeres segregate to same pole (50%)
- (N1+N2) vs (T1+T2)
- all gametes are viable (receive all genes)
2) adjacent segregation
- 50% of the time
- (N1+T2) vs (T1+N2)
- both not viable since some of the genes are missing, deletions and duplications
- fertility is reduced by about 50% in translocation heterozygotes

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21
Q

Def: Aneuploidy

A
  • changes in 1 or a few chromosomes
  • due to non-disjunction during meiosis
  • often decreases fertility because meiosis is abnormal
  • can change phenotypic ratios in offspring
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22
Q

Def: Euploid

A
  • complete set(s) of chromosomes

- euploid number varies among species

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23
Q

In wild species, what is the frequency of polyploidy?

A

-unreduced gametes occur at about the same rate as point mutations

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24
Q

Rank the frequency of polyploidy in the following species: invertebrates, vertabrates and plants

A

highest to lowest frequency:

1) plants (40-70%) = high
2) invertebrates (42%) =variable
3) vertebrates (0%) =low

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25
Q

Why is there a higher frequency of polyploidy among plants?

A
  • plants may have a higher tolerance of extra genes
  • fewer cell and tissue types
  • easier to accommodate larger cells, slower cell division
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26
Q

Name the mechanisms by which polyploids can arise.

A

A) somatic doubling
B) unreduced gametes (most common)
C) triploid bridge
D) multiple fertilizations

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27
Q

What are the three types of polyploidy and how do they differentiate?

A

A) autopolyploids

  • chromosome sets from the same species
  • spontaneous doubling of chromosomes
    ex) alfalfa (4N), fireweed, bananas (3N) have low fertility or sterile due to uneven number of chromosomes during division.

B) allopolyploids

  • gametes from different species are combined
  • may be natural or synthesized
  • crossing
  • ex) canola, wheat

C) combined ploidy levels

  • ex) social insects
  • males are monoploid (N) and females are diploid (2N)
  • develop from unfertilized eggs
  • produce gametes through mitosis
  • colony can control sex ratio through fertilization rates
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28
Q

What is the problem with the uneven number of polyploids like bananas who are 3N?

A

-low fertility or sterile due to uneven number of chromosomes during division

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29
Q

Def: Population genetics

A

-The study of genetic variation within populations and the genetic basis of evolution

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30
Q

Empirical vs theoretical component of population genetics

A

Empirical: measuring genetic variation
Theoretical: using mathematical models to explain patterns

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31
Q

What are two important formulas when calculating allele frequencies in a population that is in Hardy-Weinburg equilibrium?

A

1=p+q

1=p^2 +2pq + q^2

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32
Q

Def: Population

A
  • group of interbreeding individuals belonging to a single species
  • entire group of interest
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33
Q

Def: Gene pool

A

all alleles in a population and their distribution into genotypes

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34
Q

Def: Genotype frequencies

A

the proportion of each genotype in a population

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35
Q

Def: Allele frequencies

A

the proportion of each allele in a population

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36
Q

What are the assumptions in the Hardy-Weinburg Law?

A
  • infinite population size
  • no mutation
  • random mating
  • no migration
  • no selection
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37
Q

What does the Hardy-Weinburg Law allow us to do?

A

-relates Mendelian Segregation to allele and genotype frequency in an “ideal” situation

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38
Q

In natural populations, the rare allele occurs in which individuals?

A
  • mostly in heterozygotes

- most likely to have the disease allele

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39
Q

When can you apply the Hardy-Weinburg Law with the recessive human allele (ex-albinism, cystic fibrosis)

A

-only works if the estimate is made BEFORE selection (ex-newborns)

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40
Q

Does a dominant phenotype mean a dominant allele?

A

NO! Just because a phenotype appears dominant in a population, doesn’t mean that genetically it is the dominant allele!

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41
Q

What are the results from the Hardy-Weinburg Law?

A
  • allele frequencies do not change over time

- genotype frequencies can be predicted from allele frequencies and vice-versa

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42
Q

Def: Evolution

A

changes in allele/traits and genotype frequency over generations or over time

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43
Q

What happens if there are violations of the Hardy-Weinburg conditions?

A

causes evolution

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44
Q

Def: Molecular clock

A
  • mutations alone are slow and many new ones are lost to drift (chance)
  • neutral mutations accumulate over time
  • can estimate time since a common ancestor for two groups
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45
Q

What is the formula for calculating relative fitness of genotype ij?

A

Wij= (survivalreproduction of genotype ij) / (survivalreproduction of most fit genotype in population)

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46
Q

ON EXAM

What does the most fit genotype depend on?

A

Fitness depends on phenotype if A is dominant then AA, Aa have same phenotype and therefore same fitness (wA_)

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47
Q

Def: Population mean fitness (w bar)

A

Frequency of each genotype in population weighted by its fitness and summed over all genotypes

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48
Q

How is the Hardy-Weinburg Law regenerated with each generation?

A

random mating

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49
Q

What are the two components explaining population genetic structure?

A

A) high genetic variation within populations (many alleles, even allele frequency, all possible genotypes present)
B) Differences among populations (divergent/differentiated populations have different allele frequencies)

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50
Q 
How do the following mutations have significant evolutionary impact?
A) point mutation
B) chromosomal inversion
C) gene duplication
D) genome duplication
A

A) creates new alleles
B) alleles inside inversion are transmitted together as a unit
C) redundant genes may acquire new functions through accumulation of additional mutations
D) may create new species, massive gene duplication

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51
Q

What are the conditions for evolution by natural selection as proposed by Charles Darwin in “Origin of Species” (how populations evolve)?

A
  • phenotypes vary within populations
  • some variation is heritable (geneticc)
  • more offspring are born than can survive to reproductive maturity =struggle for existence
  • some genetic variants produce more offspring than others
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52
Q

Def: Directional Selection

A
  • 1 allele favoured over other(s) so there is a decrease in genetic variation at selected locus
  • Change due to selection is much faster than changes due to mutation
  • Favoured allele may depend on environment
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53
Q

Def: Non-Directional Selection

A
  • Maintain more than 1 allele in a population
    1) Heterozygote advantage
    • Heterozygote more fit than either of the two homozygotes
    • Ex) B hemoglobin and sickle cell anemia (Sickle cell allele is maintained in areas with high levels of malaria)
      2) Negative frequency dependence
    • Rare genotypes most fit
    • As their frequency increases, fitness decreases
    • Ex) predator-prey interactions
    • Phenotypes and genotypes maintained at intermediate frequencies
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54
Q

Def: Inbreeding

A
  • Mating among relatives
  • ex) Self-fertilization (extreme)
  • Milder inbreeding (cousins) also get less heterozygosity but more slowly
  • affects all genes (entire genome)
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55
Q

Def: Assortative Mating

A

Affects genes associated with mating patterns
A) Positive
-Similar phenotypes tend to mate (like with like)
-Tendency for talltall and short*short
-Increase homozygosity, decrease heterozygosity and decrease genetic variation
B) Negative
-Opposites attract (like with unlike)
-MHC alleles in humans, attracted to other genotypes
-Decrease homozygosity, increase genetic variation

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56
Q

What are the effects of a finite population size?

A
  • Causes genetic drift: changes in allele frequency due to sampling errors that occur when gametes are sampled to form zygotes
  • Allele frequency in the gametes matches the parental generation because there are 1000s to millions of gametes
  • Offspring (zygotes) are a small sample of the gamete pool and may not be representative of the parental generation
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57
Q

What are the outcomes of genetic drift?

A
  1. Loss and fixation of alleles within populations
  2. Loss of genetic variation within populations
  3. Divergence (or differences) among populations (Some all AA or some all aa)
58
Q

Def: Migration

A
  • Causes gene flow, movement of alleles between populations
  • 2 components:
    a. Gene movement: travel by gametes or adults
    b. Gene establishment: depends on fitness of migrants and residents
  • increase genetic variation
  • decrease divergence due to drift
59
Q

How can gene flow (migration) be determined to give the overall divergence among populations?

A

N=population size
M=migration
If high NM=low divergence (NM>2)
If low NM=high divergence (NM<1)

60
Q

Name the features of genetic drift.

A
  • drift occurs in all populations, but is most dramatic in smaller populations
  • affects all genes
  • no allele or genotype is “favoured”because sampling error is random
  • allele frequency in one generation influences gametes for the next generation
  • as a result, sampling errors accumulate and populations tend to diverge over time
61
Q

Which alleles are rapidly lost and rapidly fixed?

A
  • rare alleles are usually rapidly lost

- common alleles are usually rapidly fixed

62
Q 
Summarize the evolutionary forces (violations of Hardy-Weinburg) - Action and Effect on Populations:
A) Mutation
B) Selection
C) Non-random mating
D) Drift
E) Migration (gene flow)
A 
.Action
A) generates new alleles
B) increases frequency of favoured allele/genotype
C) inbreeding, assortative mating
D) sampling errors during random mating
E) movement of alleles among populations

Effect on Populations
A) increase genetic variation
B) decrease or increase genetic variation and population divergence
C) increase homozygotes, increase or decrease homzygotes
D) decrease genetic variation, increase population divergence
E) increase genetic variation, decrease population divergence

63
Q

Contributing vs Non-contributing alleles

A

Contributing alleles: influence trait value (amount of pigment)
Non-contributing alleles: don’t add to trait value

64
Q

What are the formulas used to estimate allele number if:
A) you only have phenotypic ratios
B) you are only given the extremes

A

A) # of alleles (N) = #classes - 1

B) (1/2)^N

65
Q

Name the two categories of traits.

A

1) Categorical/discontinuous: a few discrete phenotypes (red versus white flowers, blood types)
2) Quantitative/continuous: a range or continuous distribution of phenotypes

66
Q

Def: Polygenic

A

controlled by many genes

67
Q

Def: Multifactored

A
  • continuous, quantitative traits

- reflect genetic and environmental influences

68
Q

Def: Quantitative genetics

A
  • study of inheritance of continuous traits

- statistical analysis of resemblance between relatives

69
Q

Def: Sample

A

subset of population

70
Q

Def: Frequency distribution

A
  • graph showing the frequency of phenotypes within a sample
  • represents variation within the sample
  • estimates variation within the population
71
Q

Def: Familial traits

A

shared by family members for any reason

72
Q

Def: Heritable traits

A

similar in family members due to shared genes

73
Q

Def: Heritability

A

-proportion of phenotypic variation (Vp) that is due to genetic variation

74
Q

Broad-sense (H^2) vs Narrow-sense (h^2) heritability

A

Broad-sense: all genetic variation, how much genetic variance (Va,Vd,Vi) contribute to phenotypic variance
Narrow-sense: how much additive genetic factors contribute to phenotypic variance, contributes directly to resemblance between parents and offspring

75
Q

What is the “Homo floresiensis”, the dwarf hominid?

A
  • fossil hominid discovered in 2003 on Flores Island, East Indonesia
  • 17000 to 95000 bp
  • adult female approximately 1 m tall
  • brain the size of a modern newborn
  • stone tools and charred animal remains were also recovered
76
Q

What is the importance of genetic correlation?

A
  • influence responses to selection
  • important in breeding
  • important in evolution
77
Q

How is total phenotypic variation (Vp) calculated?

A

Vp=Va+Vd+Vi+Ve+Vg*e
Vp=Vg+Ve

Va=alleles of additive effect (# of contributing alleles)
Vd=dominance within loci
Vi=interactions between loci
Ve=effects of environment (UV exposure, liver function, nutrition)
Vg*e=genotype and environment interactions (freckles vs tan)

78
Q

How is broad-sense heritability calculated?

A

H^2 =Vg/Vp

79
Q

How is narrow sense heritability calculated?

A

h^2=Va/Vp
=Va/(Vg+Ve)
=selection response/selection differential

80
Q

Nucleoside vs nucleotide

A

Nucleoside=sugar+base

Nucleotide=nucleoside+phosphate

81
Q

What is the breeder’s equation?

A

R=h^2 * S

-can accurately predict 5-10 generations and “decent” predictions in the long term

82
Q

What is the structure of DNA?

A
  • genetic material is composed of nucleic acids

- basic unit of nucleic acids is a nucleotide

83
Q

What are the three components of a nucleotide?

A
  • 5-carbon sugar
  • base
  • phosphate group
84
Q

How is the sugar arranged in DNA and RNA?

A

-5’ carbon ring
-numbering gives directionality
A) Deoxyribose (in DNA)
-2’-OH missing
B) Ribose sugar (in RNA)
-1’ attaches to the base
-2’ has no -OH in DNA and -OH in RNA
-3’ -OH attaches to the next nucleotide in macromolecule
-4’ is just there
-5’ site of phosphate (PO4) attachment

85
Q

How is the base arranged in DNA and RNA?

A 
A) Purine (2 rings)
-Adenine (A), guanine (G)
-PUAG: short name, large molecules
B) Pyrimidine (1 ring)
-Cytosine (C), thymine (T) in DNA
-Uracil (U) in RNA
-PYCTU: longer name, smaller molecule
86
Q

How is the phosphate (PO4) group arranged in DNA and RNA?

A

-Polynucleotide chain: nucleotides joined by phospho-diester bonds between 3’OH+5’PO4

87
Q

Where does DNA replication occur?

A
  • S phase of cell cycle
  • Semi-conservative
  • One strand forms the complete template for each new strand
88
Q

Name the requirements for DNA synthesis.

A
  1. Template DNA
  2. dNTP’s
  3. Primer (short segment of nucleic acid provides free 3’ OH group)
  4. DNA polymerase (catalyses formation of phosphodiester bonds between existing 3’ OH and incoming phosphate
89
Q

***Which are the directions of the new strand and the template strand?

A

New strand direction=5’ to 3’

Template strand direction=5’ to 3’

90
Q

What are the various functions of DNA polymerase I and III in E.coli?

A

Functions:

  • 5’ to 3’ synthesis: DNA pol I and III
  • 3’ to 5’ exonuclease activity (repair of mistakes, remove nucleotides that just came in): DNA pol I and III
  • 5’ to 3’ exonuclease: DNA pol I only
  • Processivity (how long they would stay on a DNA molecule): DNA pol I is low, DNA pol III is high so stays on longer
91
Q

What is the MAIN function of DNA polymerase I and III?

A

DNA pol I: replication and repair

DNA pol III: main enzyme for replication

92
Q

What is the role of helicase in the DNA replication process?

A

-unwinds the DNA

93
Q

What is the role of a replisome in the DNA replication process?

A

-a replication protein that are associated physically

94
Q

What explains the shortening of telomeres?

A
  • When RNA primer is removed, no 3’OH group for synthesis

- Explains the shortening of chromosomes gradually

95
Q

During DNA replication, in which direction does the RNA primer work?

A

5’ to 3’ direction

96
Q

What is the whole process from DNA to protein?

A

DNA-Transcription-RNA-Translation-Protein

97
Q

What is the role of RNA polymerase in the transcription process?

A

enzyme that catalyzes polymerization of RNA

98
Q

What are key points in the transcription process?

A

a. 5’ to 3’ direction
b. No primer required (difference between RNA and DNA)
c. DNA template required (Only one of the two DNA strands is copied)
d. Nucleotides added to 3’ end are NTP’s (not dNTP’s) so release of 2 PO4 gives energy for phosphodiester bond formation

99
Q

What is the directionality of the DNA coding strand (non-template), DNA non-coding strand (template) and the RNA?

A

Coding strand (non-template): 5’ to 3’
Non-coding strand (template): 3’ to 5’
RNA: 5’ to 3’

100
Q

What are the complementary pairs?

A

A=T and C=G

101
Q

Why is DNA a helix?

A
  • helix has a right hand twist of 10bp/turn
  • twist creates major and minor groove
  • creates two surfaces for protein interactions
102
Q

Explain the process of DNA replication.

A

1) Initiator proteins bind to “replicator” to form replication bubble.
2) Helicase breaks hydrogen bonds to unwind DNA strands
3) SSB coats single-stranded DNA
4) Primase synthesizes RNA primers on leading and lagging strands
5) DNA polymerase III synthesizes DNA 5’ to 3’ continuously along leading strand and discontinuously along lagging strand
6) DNA polymerase I removes nucleotides of RNA primers, replacing them with DNA nucleotides
7) DNA ligase catalyzes phosphodiester bonds to join DNA segments

103
Q

What are the three stages in the transcription process in bacteria (prokaryotes)?

A

1) initiation (in promoter region of the gene)
2) elongation (in the coding region of the gene)
3) termination (in termination region of the gene)

104
Q

What are the differences between the prokaryotic and eukaryotic transcription process?

A

E. coli:

  • RNA polymerase (RNA pol) makes all the RNA and is a protein complex with a number of subunits (2 alpha subunits, 2 beta subunits, Sigma subunit at initiation =“sigma factor”)
  • Both initiation and termination are triggered by sequence cues in DNA
  • Transcribing=adding nucleotides
  • just know that there are consensus regions in the promoter region, don’t need to know details

Eukaryotes:

  • More complex
  • Pre-mRNA processed
  • mRNA transported from nucleus
  • More enzymes
  • RNA pol I: makes rRNA
  • RNA pol II*: makes mRNA and snRNA
  • RNA pol III: makes tRNA, rRNA, snRNA
105
Q

How does the initiation process work in the transcription process in eukaryotes?

A

-Involves multiple promoter elements, variable among genes
Ex) TATA box -30 or -25, CAAT box -80, GC box -90 (further downstream)
-RNA polymerase II binds to promoter elements in association with other proteins called “transcription factors” (TF)
-Initiation complex: know that it is the promoters and TF together

106
Q

How does the elongation process work in the transcription process in eukaryotes?

A
  • RNA poly II released from TF to begin transcription
  • Transcribes pre-mRNA (mRNA before processing)
  • Low base transcription rate that can be altered
  • Binding proteins are cell and tissue specific, their presence is regulated by the hormonal and developmental stage
107
Q

Which DNA elements are involved in the elongation process of transcription in eukaryotes?

A

DNA elements Binding proteins Result
Enhancers Activators Increase transcription rates
Silencers Repressors Decrease transcription rate

108
Q

How does mRNA processing (includes termination) work in the transcription process in bacteria?

A
  • No nucleus
  • No mRNA processing
  • Transcription and translation are directly coupled
109
Q

How does mRNA processing (includes termination) work in the transcription process in eukaryotes?

A
  • Transcription in the nucleus
  • Stage 1 - pre mRNA synthesis
  • Stage 2 - RNA processing
  • Mature mRNA is transported to the cytoplasm and translation can begin

-3 steps to RNA processing
A. 5’ cap
B. 3’ Poly A tail is added (involved in termination)
C. Splicing of introns (removal)

110
Q

Name some features of RNA.

A

1) OH on 2’ carbon of sugar
2) uracil instead of thymine
3) usually single stranded
4) can form short double stranded hairpin regions leading to the potential for complex tertiary structures

111
Q

What is a common feature between the DNA coding strand and the mRNA transcript?

A

have same polarity (go from 5’ to 3’ direction) and sequence except for substituting U in mRNA for T in DNA

112
Q

What happens to DNA during the elongation process in bacteria of the transcription process?

A

DNA untwists and re-twists as RNA polymerase proceeds

113
Q

What is the general sequence of steps in the formation of eukaryotic mRNA?

A

1) DNA transcribed
2) pre-mRNA
3) mRNA translated
4) polypeptide

114
Q

What is the role of the 5’ cap during mRNA processing in eukaryotes?

A
  • protects mRNA from degradation
  • binds ribosome to initiate translation
  • added early in transcription
  • methylation of first two nucleotides of transcript
115
Q

What is the role of the Poly A tail during mRNA processing in eukaryotes?

A
  • also protects mRNA and involved in initiation of translation
  • plays a role in termination (no clear terminators in eukaryotes)
  • at 3’ end, added by polyA polymerase enzyme
116
Q

What is splicing during the mRNA processing in eukaryotes?

A
  • introns (intervening regions) are removed from pre-mRNA

- exons (expressed regions) remain in message and are translated

117
Q

What is the role of mRNA in the translation process?

A

provides a template

118
Q

How many amino acids exist and what are their common structure?

A
  • 20 amino acids

- an amino group (N-terminus), a side chain (unique to each), and a carboxyl group (C-terminus)

119
Q

***MEMORIZE

What are the 3 different types of R groups?

A

A) Acidic

  • Aspartic acid (Asp)
  • Glutamic acid (Glu)

B) Basic

  • Lysine (Lys)
  • Arginine (Arg)
  • Histidine (His)

C) Neutral (all others)

120
Q

What is a polypeptide?

A
  • Linear chains of amino acids
  • Macromolecular subunit of proteins
  • Amino acids joined by peptide bonds
121
Q

Def: Peptide bond

A

covalent bond between the carbonyl group and amino group of another

122
Q

During translation, what are the directions?

A
  • mRNA processed from 5’ to 3’

- polypeptide generated from N-terminus to C-terminus

123
Q

Why is the genetic code used?

A
  1. Information coding for complex biology (evolutionary reason)
  2. ID mutations that cause disease (Compare DNA sequences and proteins in individuals with and without the disease - Ex) Cystic fibrosis, anemias)
  3. Start point for genetic engineering (Know gene and code, alter to adjust product)
124
Q

What are the two types of point mutations?

A

1) Transition=purine to purine or pyrimidine to pyrimidine

2) Transversions=purine (PUAG) to pyrimidine (PYCTU) or pyrimidine to purine

125
Q

Name the consequences for gene product (in order of increasing severity)

A

1) silent mutations (same amino acid)
2) neutral mutation (amino acid chemically similar; protein function is normal)
3) missense mutation (amino acids are chemically distinct)
4) nonsense mutation (Introduces premature stop codon;
Translation interrupted; Causes chain termination and truncated polypeptide (lethal if an essential protein)

126
Q

Name the four levels of protein structure.

A

1) Primary: amino acid sequence
2) Secondary: pattern of folding and twisting
3) Tertiary: conforming, 3D shape
4) Quaternary: complex of 2+ polypeptide subunits

127
Q

Name the characteristics of the genetic code.

A

1) triplet (1 codon=1 amino acid, each tRNA brings 1 amino acid per codon)
2) continuous (no nucleotides skipped)
3) non-overlapping (successive groups of 3)
4) virtually universal
5) degenerate/redundant (more than 1 codon codes for each amino acid = wobble)
6) start and stop (non-sense) signals (AUG=start and UAG, UAA, UGA=stop codons)
7) non-ambiguous (more than one codon per amino acid but only 1 amino acid for any codon

128
Q

What is the general significance of mutations?

A

1) mutations are changes in DNA structure (nucleotide or chromosomal level)
2) nucleotide changes sometimes affect phenotypes
3) mutations have random effects (good, bad or neutral)
4) mutation rate=probability of mutation in a given period of time, reflects rates of mistake in DNA process and repair of mistakes by enzymes
5) mutagens are chemicals, radiation, heat that increase mistake rate and interfere with repair mechanism

129
Q

What happens if you have insertions or deletions in the genetic code?

A

=frameshift mutations

  • affects all amino acids following the mutation
  • can lead to a shorter or longer polypeptide
130
Q

In animals, which mutations matter?

A
  • somatic mutations affect individuals but not gametes (skin cancer, asbestosis), not heritable
  • germ line mutations affect gametes, are heritable and are directly important in evolution
131
Q

In plants, which mutations matter?

A
  • have no germ line
  • meristematic tissue throughout the plant body
  • plant body has vegetative growth and gamete production
  • can acquire heritable mutations through their lifetime
132
Q

Which machinery is involved in the translation process?

A
  • mRNA provides template
  • Ribosome
  • Transfer RNAs (tRNA)
  • Amino acids
133
Q

What is the functions of tRNA in the translation process?

A
  • tRNAs function to recognize codons and bring the appropriate amino acids
  • 61 sense codons
  • 30-50 tRNAs in most species
  • amino acid is on the 3’ end of tRNA
  • anticodon reads codon in mRNA from 3’ to 5’
  • 3rd position on the mRNA is the “wobble position” where the codon is variable = degenerate code
  • Same tRNA (and amino acid) recognizes >1 codon
134
Q

The wobble position (=degenerate code) can be found where?

A

in mRNA codon, not in tRNA anticodon since tRNA recognizes more than 1 codon

135
Q

What is the difference between a charged and uncharged tRNA?

A

Charged tRNA = aminoacyl tRNA: tRNA with the amino acid attached (what we drew)
Uncharged tRNA: amino acid has been removed

136
Q

What is the function of Aminoacyl tRNA synthetases?

A
  • Specific to tRNAs
  • Add amino acid to anticodon
  • Recognize anticodon in tRNA
137
Q

What is the structure of ribosomes of bacteria?

A

-rRNA and ribosomal proteins
-Large subunit (50S) has 3 sites:
§ E=exit
§ P=peptidyl (peptide bond formation)
§ A=aminoacyl (takes charged tRNA)
-Small subunit (30S)

138
Q

In which direction does the ribosome move during translation?

A

moves from left to right (5’ to 3’ direction)

139
Q

In which direction does the charged tRNA move during translation?

A

From A site, to P site (where peptide bond forms), to E site and comes out an uncharged tRNA
*3’ to 5’ of mRNA

140
Q

Describe the process of initiation for translation.

A

-small subunit and mRNA assemble
-guided by ribosome binding sequence (upstream) and start codon (AUG)
-initiator tRNA (fmet) binds to start codon
-large subunit binds tRNA in P site
=initiation complex

141
Q

Describe the process of elongation for translation.

A
  • charged tRNA binds next codon in A site
  • peptide bond forms between fmet and next amino acid (initial tRNA is now uncharged)
  • translocation happens where the ribosome shifts forward, the shift moves the intitial tRNA to exit site, Peptidyl-tRNA now in P-site (has growing polypeptide) and A site is now empty
  • uncharged initial tRNA released from E site
  • process repeats
142
Q

Describe the process of termination for translation.

A
  • ribosome encounters stop codon (UAA, UAG, UGA0
  • release factors (RF) bind to stop codon
  • additional release factors stimulate termination where polypeptide chain is released, tRNA released, and ribosome dissociates
  • usually several ribosome translate simultaneously (polyribosome or polysome)

BIOL 2500 (3 decks)

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