Genomics Quiz 5

Card Set Information

Author:
johnpc
ID:
311272
Filename:
Genomics Quiz 5
Updated:
2015-11-10 17:52:37
Tags:
genomics
Folders:

Description:
genomics quiz
Show Answers:

Home > Flashcards > Print Preview

The flashcards below were created by user johnpc on FreezingBlue Flashcards. What would you like to do?


  1. To determine how many genes were necessary for
    survival
    single-gene knockdowns
  2. Gene knockdowns
    Any manipulation that reduces expression of a specific gene.
  3. Gene knockdowns approaches
    • -Overexpress a dominant negative
    • -Inject a morpholino
    • -Homologous recombination
    • -RNAi
  4. Overexpress a dominant negative Example
    • Tyrosine kinase receptor
    • Inject mRNA that will be translated into a mutant subunit protein
    • that lacks the site needed to initiate signaling
    • Ligand binds to dimer, but no signaling activated
  5. Morpholinos
    • •Like RNA but ribose replaced with morpholine group;
    • phosphodiester backbone modified.
    • •Resists degradation by cell.
    • •Complementary to RNA sequences.
  6. Translation blocking morpholinos
    • Binds near initiation site (overlapping portions of the 5’UTR
    • and the first exon), interfering with ribosome assembly
    • and prevents translation.
  7. Spliceblocking morpholinos
  8. Homologous recombination (knockout)
    • Relies on recombination between a vector carrying positive and negative
    • selectable markers.
    • Recombination occurs between identical sequences in vector and chromosomal DNA.
  9. Often organisms appear not to “need” many of their genes
    -Evidence?
    • -If you knockdown or knockout any single gene in a mouse, most of
    • the time the mouse will survive.
    • -Only about 30% of the genes are necessary for survival!
    • -In yeast, more than 4,700 single-gene knockouts were performed in
    • homozygous diploid lines. Only 10.7% exhibited reduced growth/
    • viability. Growth of 83.5% of the knockouts was unaffected
  10. redundancy refers to pairs of homologous genes
    • with functional overlap where one can compensate
    • for loss of the other.
  11. Sources of genetic redundancy?
    • -Gene duplication
    • -Genome duplication
    • -Convergent evolution
  12. Implications of high occurrence of redundancy in signaling components?
    • •Functional overlap in redundant genes may be beneficial in
    • maintaining ability to signal.
  13. Examples of redundant signaling proteins?
    • -Hox genes
    • -Wnt proteins
    • -Myogenic regulators (MyoD, Myf5, myogenin, Mrf4)
  14. -In study of 59 pairs of redundant genes (yeast), the
    redundant forms were
    • not expressed at the same
    • time and/or place. Expression patterns didn’t overlap.
  15. In vertebrate developmental pathways, redundant
    duplicates are
    • expressed in spatially or temporally
    • distinct areas.
  16. Knocking out one gene often results in
    • upregulation of the
    • redundant partner.
  17. Redundancy may allow
    compensation when one isoform lost
  18. somites
    • -Early in vertebrate development, segmented blocks of tissue appear on either side of the nerve cord
    • -The somites form muscle, tendons, endothelial cells, dermis, and cartilage.
  19. Portions of the somite forming
    adult skeletal muscle divided
    into two domains…
    • -Epaxial: dorso-medial region (ep)
    • -Hypaxial: ventrolateral region (hyp)
  20. Redundancy may allow adaptation to
    local conditions
  21. Redundancy can be used to increase ability of cell
    • to sense changes in
    • environment and respond.
  22. Redundancy may improve
    processing of external info
  23. Gene redundancy may provide important opportunities
    for the organism…
    • -Compensating for loss of another molecule
    • -Adapting to changing external factors
    • -Improved processing of external information
  24. -How are new gene families created?
    A) Divergent evolution

    B) Concerted evolution

    C) Birth-and-death evolution
  25. Divergent evolution
    • A group of genes is duplicated.
    • -Evolution of gene A doesn’t
    • affect evolution of gene B.
    • -Each gene gradually diverges
    • as mutations accumulate.
    • -Duplicate genes assume new
    • functions.
    • Ex: alpha and beta globins
  26. Concerted evolution
    • -Gene family members do not evolve
    • independently. Evolve at same time.
    • -In ribosomal RNA of frogs, find that
    • intergenic regions more similar in
    • different rRNAs of same species than
    • in two related Xenopus species.
    • -Why? If one gene acquires a mutation,
    • it spreads to the other rRNAs by
    • unequal crossing over or by
    • nonreciprocal recombination.
    • Ex: 5S rRNA in Xenopus
    • Primate U2 snRNA
  27. Birth-and-death evolution
    • -New genes made by gene duplication.
    • -Some remain, others become
    • pseudogenes or are deleted.
    • -Pattern in phylogenetic tree is more
    • difficult to interpret.
    • Ex: Major histocompatibility complex genes
    • T-cell receptors
    • MADS-box genes
    • Ubiquitins, etc.
  28. The domainome
    • Instead of focusing on genes, look at evolution
    • of protein domains.
    • All the domains from a dataset is the “domainome” for that
    • group of sequences.
  29. How can the g-value paradox be explained?
    • -More potential combinations between proteins
    • -Multifunctional genes
    • -Alternative splicing
    • -Transcriptional control
    • -Posttranslational modification
    • -Roles of non-coding RNA
  30. Dollo parsimony
    • assumes that once a complex trait is lost in a lineage, it cannot be
    • regained
  31. Domains acting in cell regulation
    • increased during
    • eukaryotic evolution.
  32. Domains related to metabolism
    • decreased during
    • eukaryotic evolution.
    • Suggests that roles of metabolic domains may be taken over by symbionts
  33. High number of domains in basal groups supports idea that
    • the last
    • common ancestor of eukaryotes was complex.
  34. Class I elements
    retrotransposons
  35. Class II elements
    DNA transposons
  36. Insertion of transposable elements can harm the host by
    • Insertional mutagensis, chimeric transcript production, antisense effects, and illegitimate recombination
  37. How does the genome defend itself from transposable elements?
    • RNAi
    • MircroRNA
    • Cytosine methylation
    • Defensive mutagenesis
  38. RNA interference
    (RNAi)
    • 1- In cytoplasm, Dicer binds
    • dsRNA.
    • 2- Dicer cleaves dsRNA to
    • form small interfering RNAs
    • (siRNAs, 21-23 bp long w/2 bp
    • overhang at 5’ end).
    • 3- The RNA-induced silencing
    • complex (RISC) forms when the
    • antisense strand of siRNA
    • associates with Argonaute 2
    • (AGO2) protein. May also include
    • other protein types.
    • 4- RISC complex scans RNAs
    • to find complementary
    • sequence to siRNA. siRNA
    • binds sense strand of
    • target mRNA and RISC
    • complex cleaves target.
  39. MicroRNAs
    • 1- MicroRNA is transcribed from DNA.
    • 2-These short RNA sequences form
    • haipin loops and are transported to
    • the cytoplasm by exportin5.
    • 3- In the cytoplasm, Dicer trims the
    • dsRNA (22 bp seq with overhang).
    • 4- miRNAs silence gene expression by
    • binding complementary regions in
    • the 3’ UTR of target mRNAs
    • (animals) or by binding coding
    • regions of target mRNAs (plants).
    • Base pairing is often partial and binding of the miRNA affects multiple mRNA types.
    • Effects? Inhibiting translation, causing loss of poly-A tail, interfering with methylated cap/
    • poly-A tail interactions, or causing mRNA degradation by exonucleases.
  40. Cytosine methylation
    • -Bacteria use DNA methylation to protect the genome from
    • degradation by restriction enzymes.
    • -Restriction enzymes cannot destroy methylated restriction sites
    • but can cleave unmethylated restriction sites.
    • -Endogenous methyltransferases attach methyl groups to
    • cytosines or adenines.
    • -In eukaryotes, modified 5-methyl-
    • cytosine is made by adding a methyl
    • group to the 5 position of cytosine.
    • -Causes transcriptional silencing when
    • promoter region methylated. Good for
    • long-term silencing. Doesn’t appear to
    • be reversible.
    • -Repetitive DNA in plants and
    • mammals is usually methylated.
  41. Defensive mutagenesis
    • -A method to block retroviral replication.
    • -In primates, during reverse transcription,
    • the host protein APOBEC3G is incorporated
    • in 1st strand cDNA.
    • -It deaminates cytosines in the retrovirus
    • cDNA strand, converting them to uracils.
    • -During second strand synthesis, uracil is
    • recognized as thymidine, so adenine is
    • inserted in the new DNA strand.
    • -Deactivates virus by mutating up to 25%
    • of cDNA guanine residues.
    • -Not as effective in HIV-1 retrovirus. This
    • virus makes Vif (viral interference factor)
    • which inhibits activity of the APOBEC3G protein.
  42. WHY Protein alignments are more useful if you’re comparing distantly
    related sequences.
    • -Peptide sequences are more likely to be conserved than nucleotide
    • sequences since there are multiple codons for the same amino acid.
    • - Some amino acids have similar biophysical properties. Similarities can
    • be accounted for in the protein scoring matrices.
    • -Similarities arising early in the evolutionary process may be detectable
    • using a protein alignment, but not a DNA alignment.
  43. WHY Nucleotide alignments can be more useful if you’re comparing closely
    related sequences.
    • -If sequences are very closely related, the amino acid sequences may
    • not vary much. Nucleotide sequence will vary more, allowing easier
    • detection of differences.
  44. Global Sequence Alignment
    • -Tries to align two sequences along entire length.
    • -Best for highly similar sequences of same length.
    • -As similarities decrease, misses important relationships.
  45. Local Sequence Alignment
    • -Looks for the most similar regions in sequence instead
    • of trying to align entire length.
    • -May return more than 1 result if there is more than
    • 1 subsequence in common.
    • -Good method to use if sequences differ in length or
    • share partial similarity.
  46. Dot plots
    • -A simple way to visually compare 2 sequences to
    • find local alignments, direct repeats, inverted
    • repeats, insertions, deletions or low-complexity regions
  47. Why aren’t dot plots used more?
    Do not provide a measure of statistical similarity
  48. Blosum62 matrices
    • -Score is the log of an odds ratio. Considers how often, in nature,
    • a particular residue is substituted for another versus how often
    • this substitution would occur if by random chance.
    • Si,j = log [ ( qi,j ) / ( pi
    • pj ) ]
    • Si,j is the score for a replacement of residue i with residue j.
    • qi,j represents how often the two amino acids align with each
    • other in multiple sequence alignments of protein groups.
    • pi
    • is the probability that residue i will occur among all proteins
    • pj is the probability that residue j will occur among all proteins
  49. BLAST
    • Relies on “seeding”. Looks for a short query word. Finds this and
    • related words.
    • -To score related words, uses a scoring matrix called “The neighborhood.”
    • -Threshold setting controls how many options allowed in the neighborhood.
    • -Then performs local alignment. Extends until gaps and mismatches decrease
    • score below the score threshold (S)…This info recorded by BLAST.
  50. BLAST
    Problems?
    • -If enter a sequence from a low complexity region can
    • inflate BLAST scores.
    • (Masking with DUST or SEG counter this.)
    • -The hit list contains entries that represent hypothetical
    • proteins.
    • -Hits to ESTs should be treated with caution. Sequencing
    • accuracy is lower than in “finished”
    • sequences.
  51. How does BLAST determine length
    of alignment?
    • Program measures cumulative
    • score as alignment extended
    • •If angle of drop off after a peak
    • exceeds a threshold value (X),
    • extension terminated and trims
    • alignment to preceding peak in
    • curve.
    • •HSP = high-scoring segment pair
    • and is the trimmed alignment •Then calculates E value to determine if alignment
    • is significant.

What would you like to do?

Home > Flashcards > Print Preview