BIO 011 Lecture Notes - Lecture 33: Omics, Notochord, Gastrulation

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12 Jun 2018

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End of Bio 11 -omics

Jadzia Wray

Genome sequencing: determine the nucleotide base sequence of an entire genome.

- Determine evolutionary relationships

- Identify mutations that affect phenotypes

- Identify genes for particular traits, such as genes associated with diseases

Sanger (dideoxy) sequencing

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Automated DNA Sequencing

reaction products are

separated by size and

identified by color

each dNTP is

attached to

a fluorescent

molecule

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Comparative Genomics

- There are major differences between eukaryotic and prokaryotic genomes:

o Eukaryotic genomes are larger and have more protein-coding genes.

o Eukaryotic genomes have more regulatory sequences. Greater complexity requires more

regulation.

- Much of eukaryotic DNA is noncoding, including introns, gene control sequences, and repeated

sequences.

- Eukaryotes have multiple chromosomes; each must have an origin of replication, a centromere,

and a telomeric sequence at each end.

Proteomics

- Proteome: sum total of proteins produced by an

organism; it is more complex than the genome.

- Proteomics seeks to identify and characterize all

the expressed proteins in an organism.

Metabolomics

- Metabolome: quantitative description of all of the small molecules in a cell or organism.

- Primary metabolites: involved in normal processes such as pathways like glycolysis (ex.

hormones and signaling molecules).

- Secondary metabolites: often unique to particular organisms or groups (ex. antibiotics).

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Document Summary

Genome sequencing: determine the nucleotide base sequence of an entire genome. Identify genes for particular traits, such as genes associated with diseases. Automated dna sequencing each dntp is attached to a fluorescent molecule reaction products are separated by size and identified by color. There are major differences between eukaryotic and prokaryotic genomes: eukaryotic genomes are larger and have more protein-coding genes, eukaryotic genomes have more regulatory sequences. Much of eukaryotic dna is noncoding, including introns, gene control sequences, and repeated sequences. Eukaryotes have multiple chromosomes; each must have an origin of replication, a centromere, and a telomeric sequence at each end. Proteome: sum total of proteins produced by an organism; it is more complex than the genome. Proteomics seeks to identify and characterize all the expressed proteins in an organism. Metabolome: quantitative description of all of the small molecules in a cell or organism. Primary metabolites: involved in normal processes such as pathways like glycolysis (ex. hormones and signaling molecules).

Related Questions

1. Genomic DNA is digested with a restriction enzyme that has a 4 bp recognition (restriction) site. What would be the expected average size of the fragments that would be produced by digesting DNA of random sequence with this enzyme? A) 8 bp B) 16 bp C) 125 bp D) 256 bp E) 512 bp

2.insertion or removal of one or more nucleotide base pairs in DNA within a gene often results in a ____________ mutation. A) transition B) frameshift C) reversion D) transversion E) suppressor

3. Which ingredient is NOT required in Sanger DNA sequencing? A) Primers for DNA polymerase. B) Reverse transcriptase. C) Dideoxy nucleotides. D) Deoxy nucleotides. E) DNA template.

4. What is the purpose of Taq polymerase in a PCR reaction? a. DNA denaturation b. Primer annealing c. DNA synthesis d. Heating of the reaction e. All of the above are associated with Taq polymerase.

5.You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny: Blue shell, long antenna 92 Green shell, short antenna 88 Blue shell, short antenna 31 Green shell, long antenna 29 Total 240 Assuming that the genes are linked, what is the map distance between them in cM? A) 75.0 cM B) 25.0 cM C) 12.1 cM D) 12.9 cM E) The genes are actually assorting independently. Genes A and B are linked with 10 map units (cM) between them. A heterozygous individual AB/ab (note that the dominants are in coupling) would be expected to produce gametes in which of the following frequencies?

A) 45% AA; 10% Ab; 45% bb

B) 5% AB; 45% Ab; 45% aB; 5% ab

C) 10% AB; 40% Ab; 40% aB; 10% ab

D) 45% AB; 10% Ab; 10% aB; 45% ab

E) 45% AB; 5% Ab; 5% aB; 45% ab

True or False????? HELP

1. In the field of epidemiology, increased prevalence of a disease in a population is the universal signal that disease control and treatment are not effective.

2. Cytoplasmic membranes of prokaryotes typically contain more proteins than cytoplasmic membranes of eukaryotes microorganisms.
3. Sanger (or dideoxy) DNA sequencing produces relatively short DNA sequence data, but is by far the least expensive DNA sequencing method currently available.
4. The (italicizes) abbreviation lac Z refers explicitly to the beta galactosidase protein/ enzyme.
5. Archaeal pathogens have been linked to some of the most virulent infectious diseases. 6. A bacterial mutant that is unable to make an amino acid required for growth (relative to its wild-type strain is termed an auxotroph.
7. When counting plaques to enumerate phage particles (as in the BIOL2026 Laboratory), the plaque represents lacK, not presence, of bacterial growth.
8. Because of competition with plant roots for nutrients, microbial numbers in the rhizosphere are typically much lower than in surrounding bulk soil.
9. Chemolithoautotrophy is the strategy that facilitates the existence of biological communities deep beneath the earthâs surface and near hydrothermal vents.

10. In a primary treatment (1 Â°) system of a municipal sewage/ wastewater treatment plant, the main goal is to remove ammonium (NH4+) using specific bacterial enrichment cultures.
11. The small intestine in human is the main site of microbial breakdown of partially digested food.
12. Root nodules of legumes contain nitrogen fixing bacteria.
13. Anoxygenic photosynthesis involves using/ removing electrons from H2O.
14. Algae and Cyanobacteria utilize generally the same two-photosystem biochemical mechanisms in the light- reaction of photosynthesis.
15. In bacteria, the only known reason proton motive force is generated is to power ATP synthase in cellular respiration/ oxidative phosphorylation.
16. Anaerobic respiration is only carried out by prokaryotes, however fermentation can be carried out by both prokaryote and eukaryal microorganism.

17. The general structure and biochemical âmotorsâ of flagella are similar between Bacteria and eukaryotic cells, but fundamentally different in the Archaea.
18. Antibiotic resistance genes appeared in microbes (for the first time ever) only after human .. producing and releasing massive amounts of antibiotics to the environment.
19. Experiment by Esther Lederberg, and then later by Luria and Delbruck (that won them a Nobel prize), illustrated that genetic mutations in bacteria arose in the absence of selective pressures, rather than in response to selection.
20. Quaternary sensing is a mechanism used by bacteria to sense other bacteria and regulate gene expression accordingly (e.g. In the case of expressing the liac operon or initiating EPS production in biofilms).

Cystic fibrosis is caused by mutations in the CFTR gene, whichencodes for a chloride channel. You are studying cystic fibrosispatients to determine what mutation they possess in the CFTR gene.The difference between the mutant and wild type CFTR genes can beuncovered by examining the CFTR:

(1 point - partial possible)

DNA

mRNA

protein

You decide to identify the CFTR mutation by analyzing thegenomic DNA of your patients compared to genomic DNA of healthyindividuals. You specifically are looking to see whether a specificgene truncation has occurred in the patients. You will determinethis using hybridization techniques. Which of the following willallow you to accomplish this?

(1 point - partial possible)

Using an RNA probe complementary to the region not removed bythe truncation.

Using an RNA probe complementary to the region removed by thetruncation.

Using an DNA probe complementary to the region not removed bythe truncation.

Using an DNA probe complementary to the region removed by thetruncation.

You would like to ensure that this experiment (to determinewhether patients have a specific CFTR gen truncation usinghybridization) is properly controlled. Which of the followingsamples must you include?

(1 point - partial possible)

The genome of a healthy individual who does not have CFTR.

The genome of a CFTR patient known to have the specifictruncation you are trying to identify.

The genome of a CFTR patient with a point mutation in a separateregion of the gene but without the disease truncation.

The genome of a healthy individual married to a CFTR patientwith the specific truncation you are trying to identify.

The genome of a patient with muscular dystrophy, which can bedue to a trucation in the dystrophin gene.

To conduct the hybridization experiment, you are trying todecide between using a DNA or RNA probe. Which would be a betterchoice and why?

(1 point)

An RNA probe because RNA has uracil bases.

An RNA probe because it could also be used in a translationexperiment.

A DNA probe because it is more stable than RNA.

A DNA probe because RNA cannot bind to DNA.

Which of the following will lower the Tm of a given DNAstrand?

(1 point - partial possible)

Increasing the percentage of GC base pairs.

Raising the pH of the solution from extremely acidic toneutral.

Decreasing the salt concentration in the DNA solution from 50mMNaCl to 5mM NaCl.

None of the above.

Imagine Hershey/Chase had used an RNA virus (genome composed ofRNA) instead of a DNA virus in their experiment. Wouldradioactivity still have been found in the pellet?

(1 point)

A.No, because only DNA can be labeled with radioactivity.

B.No, because the RNA genome would not enter the bacteria uponinfection.

C. No, because while DNA and RNA nucleotides are similar, theyare not identical.

D. Yes, because DNA and RNA nucleotides are similar.

Griffith and Avery's transformation experiments allowed us toidentify that DNA is our genetic information. Which of thefollowing scenarios would result in bacterial cells that arecapable of killing mice upon injection?

(1 point - partial possible)

A. Heat killed non-virulent bacteria is added to a live virulentbacteria strain.

B. Heat killed virulent bacteria is added to a heat killednon-virulent bacteria strain.

C. A heat killed virulent bacteria that is treated with anuclease, is then added to a non-virulent bacteria strain.

D.Heat killed mouse cells are added to a non-virulentbacteria.

The human genome consists mostly of non-coding DNA. Which of thefollowing are benefits of this?

(1 point - partial possible)

A. Random DNA mutations generally won't affect RNA and proteinfunction.

B. It is faster to duplicate the genome when these arepresent.

C. The existence of introns can lead to multiple variations ofproteins encoded by a single gene.

D. Abundant non-coding genome is a protective feature againstthe consequences of transposition of mobile elements.

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BIO 011 Lecture Notes - Lecture 33: Omics, Notochord, Gastrulation

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