Central Dogma and Gene Expression and Control ECA

Central Dogma

  1. What is meant by gene expression? What is a biochemical (metabolic) pathway? What is the consequence of any genetic changes to a member (protein) of a pathway?
  2. Describe the general method of the flow of information from DNA to RNA to protein. Who coined the term “central dogma”? What is the exception to central dogma (see page 328 in Campbell 8). Compare and contrast informational flow in prokaryotes with eukaryotes. Discuss mRNA processing of the primary transcipt.
  3. What is the relationship of complementary nucleotides to information flow in DNA and RNA? Why did evolution settle on codons as triplet nucleotides? What does the “dictionary” of genetic code (figure 17.5) reveal about the relationship between triplet codons and amino acids? What are some of the exceptions to the universality of the genetic code in terms of amino acids? (page 331).
  4. In general, describe the process of initiation, elongation, and termination in transcription. How is the product of eukaryotic transcription processed? (hint: 5' cap, poly-A tail, RNA splicing, exons, introns, snRNPs, splicesomes).
  5. Is there any use for excised introns (alternative RNA splicing and exon shuffling)?
  6. In general, describe the process of translation. (include tRNA, rRNA, anti-codon, AUG (methionine), codon recognition, peptide formation, translocation, termination).
  7. Describe the features of post-transcriptional modifications (what happens to the protein after translation?). (targeting polypeptides for specific locations).
  8. What are the various ways (and types) mutations can affect protein structure and function? (types of point mutations-single nucleotide polymorphisms (SNP)... base pair substitution, silent mutation, misssense mutation, nonsense). Base pair insertion or deletion...frameshift.
  9. Compare gene expression in bacteria to archaea and eukarya (section 17.5).
  10. How has the definition of a gene been modified since genomic sequencing? (17.5)

Gene Expression in Prokaryotes and Eukaryotes

  1. Explain the operon in prokaryotes. Contrast inducible with repressible operons.
  2. What is meant by differential gene expression?
  3. Discuss the different ways (levels) that eukaryotic genes are regulated. Include:

a) regulation of chromatin (histone modifications, high levels acetylation (-COCH3) of lysine or deacetylation and methylation , heterochromatin, euchromatin),

b) DNA methylation (-CH3) of cytosine and genomic imprinting of maternal/paternal genes;

c) Epigenetic inheritance outside the nucleotide sequence (reversible changes) mystery yet to be solved

  1. How do activators and enhancers help with the regulation of transcription initiation?
  2. What is the role of  RNA processing and mRNA degradation in “fine tuning” transcription? What can be done to control translation? What is the role of protein processing and degradation in “post” translational?
  3. Describe the role of non-coding RNA's (not expressed as proteins) in controlling gene expression.
  4. Discuss the importance of cell differentiation and morphogenesis shape organismal shape. Include the role of cytoplasmic determinants, induction, pattern formation,  HOX (homeobox or homeotic) genes, maternal effect genes, egg-polarity, bicoid, and morphogens).
  5. How do cells become cancerous?

Viruses:

1.       What is the probable origin of viruses? Describe the basic structure of a virus.

2.       Compare and contrast virus with viroid and prions.

3.       Compare and contrast lytic with lysogenic virus life cycles.

4.       What role might viruses play in the evolution of species?

5.      Compare a retrovirus like HIV with retrotransposons and other elements.

 

ECA: DNA and Central Dogma Learning Objectives

Organisms share many conserved core processes and features that evolved and are widely distributed among organisms today. a. Structural and functional evidence supports the relatedness of all domains. Evidence of student learning is a demonstrated understanding of the following concepts:

 

1. DNA and RNA are carriers of genetic information.

 

2. Major features of the genetic code are shared by all contemporary living systems.

 

3. Metabolic pathways are conserved across all three domains.

Non-eukaryotes can transfer genetic information laterally through the mechanisms of transformation, transduction and conjugation; most eukaryotes do not transfer information laterally.

 

a. Horizontal transfer of genetic material is different from vertical transfer of genetic information. Evidence of student learning is a demonstrated understanding of these concepts:

 

1. Transformation is the uptake of genetic material from the environment.

 

2. Transduction is the transfer of genetic material by a virus.

 

3. Conjugation involves cell-to-cell contact and transfer of genetic material following fusion of cell membranes

Populations of organisms continue to evolve. a. Evidence supports the idea that evolution has occurred in all organisms. b. Evidence supports the idea that evolution continues to occur in all organisms. Students should be able to demonstrate understanding of the above concept by using an illustrative example such as:

 

Chemical resistance (in that mutations for resistance to antibiotics occur in bacteria regardless of the presence of antibiotics; pesticides, herbicides; cancer chemotherapy)

 

Emergent diseases

 

Timing and coordination of several events are necessary for the normal development of an organism, and these events require regulation by multiple mechanisms. a. Observable cell differentiation results from the expression of genes for tissue- specific proteins. b. Induction of transcription factors during development result in sequential gene expression. Evidence of student learning is a demonstrated understanding of the following concepts:

 

1. Homeotic genes are involved in developmental patterns and sequences.

 

2. Embryonic induction in development results in the correct timing of events. .

 

4. Genetic mutations can result in abnormal development.

 

5. Genetic transplantation experiments support the link between gene expression and normal development.

 

DNA, and in some cases RNA, is the primary source of heritable information. a. Genetic information is transmitted from one generation to the next through DNA or RNA.

 

1. Genetic information is stored in DNA molecules through the linear sequence of the nitrogenous bases.

 

2. The structural properties of DNA and RNA allow for transmission of genetic information.

 

3. The base pairing between the DNA strands and DNA to RNA is complementary so that one strand of contains the information necessary for the synthesis of its complement.

 

4. The organization of DNA into chromosomes differs between prokaryotic and eukaryotic cells, but the structure of DNA and the genetic code are the same.

 

5. Prokaryotic cells have circular chromosomes while eukaryotic cells have linear chromosomes with special structures at the end called telomeres.

 

6. Both prokaryotes and eukaryotes can contain extra small circular DNA molecules called plasmids.

 

b. DNA and RNA have structural similarities and differences.

1. The structure of DNA and RNA contributes to their functions.

2. The chemical structure of sugar (deoxyribose or ribose) determines whether the molecule is a DNA or RNA. In addition, RNA substitutes the nitrogenous base uracil for thymine found in DNA. DNA is usually double-stranded, RNA is single-stranded.

3. The phosphate groups give DNA and RNA acidic properties and connect the sugar molecules at the 3’ and 5’ carbon positions resulting in a molecule that is helical and orientated along a 3’ to 5’ direction with the nitrogenous bases perpendicular to the backbone.

4. The three components-- sugar, phosphate, and nitrogenous base--are connected by covalent bonds.

5. The chemical structure of the bases allows the formation of specific hydrogen bonding such that adenine pairs with thymine or uracil (A-T or A-U) and cytosine pairs with guanine (C-G).

6. This specific pairing of nitrogenous bases allows DNA to form a stable double helix with the nitrogenous bases forming rungs within a helical ladder composed of two strands of dexoyribose and phosphate groups.

7. The presence of the extra OH group in the ribose sugar in RNA prevents the formation of a stable, double ladder-type helix.

8. The two DNA strands have opposite orientations, i.e. are antiparallel.

9. Cells contain several different types of RNA molecules; messenger RNA (mRNA), transfer RNA (t-RNA), and ribosomal RNA (rRNA), each with a different cellular function but with the same basic single-stranded chemical structure.

10. The sequence of the RNA bases determines RNA function. The three different RNA molecules have three different functions:

 

mRNA carries information from the DNA to the ribosome for translation.

 

tRNAs bind specific amino acids and allow information in the mRNA to be translated to a linear peptide sequence. Each tRNA has a unique cloverleaf structure that allows it to interact with the ribosome and enzymes that attach specific amino acids to it.

 

rRNA molecules are functional building blocks of ribosomes

 

The flow of genetic information is from a sequence of nucleic acids in a gene to a sequence of amino acids in a protein. Evidence of student learning is a demonstrated understanding of the following concepts:

 

1. The enzyme RNA-polymerase reads the DNA template from 3’ to 5’ and builds a messenger RNA molecule based on the complementary (base-pairing) nature of the nucleotides. This mRNA transcript determines the order of amino acids in the polypeptide.

 

 

2. In eukaryotic cells the mRNA transcript undergoes a series of enzyme regulated modifications that include a poly-A tail, a GTP cap, and intron excisions before it is transported from the nucleus and translated in the cytoplasm.

 

3. The mRNA transcript is transported from the nucleus to the site of translation.

f. Prokaryotic cells lack a nucleus which allows transcription to be coupled with the translation of the message.

 

1. Translation occurs on the ribosome and involves energy and many steps including initiation, elongation and termination. The details and names of the enzymes and factors involved in each of these steps is beyond the scope of the course and exam.

 

2. The mRNA interacts with the rRNA of the ribosome to initiate translation.

 

3. The t sequence of nucleotides on the mRNA is read in triplets called codons.

 

4. Each codon codes for a specific amino acid, which can be determined by using a Genetic Code chart. Many amino acids have more than one codon. Memorizing the code is beyond the scope of the course and exam.

 

5. tRNA brings the correct amino acid to the place on the mRNA.

 

6. The amino acid is transferred to the growing peptide chain.

 

7. The process is repeated until a ―stop codon is reached.

 

8. The process terminates by release of the newly synthesized peptide.

 

9. Proteins (enzymes) determine phenotypes through regulation of chemical reactions responsible for cell development, growth, and maintenance.

 

10. Genetic information in retroviruses have an alternate flow from RNA to DNA

made possible by reverse transcriptase, an enzyme that copies the viral RNA genome into DNA. This DNA is then used to direct the assembly of new viral particles.

g. Genetic engineering techniques can manipulate the heritable information of DNA or RNA. Students should be able to demonstrate understanding of the above concept by using an illustrative example such as:

 

Genetically modified foods

 

Transgenic animals

 

Cloned animals

 

Pharmaceuticals such as human insulin or factor X

 

Learning Objective: The student is able to construct scientific explanations that DNA, and in some cases RNA, is the primary source of heritable information.

Learning Objective: The student is able to use data to justify why DNA, and in some cases RNA, is the primary source of heritable information.

Learning Objective: The student is able to describe representations and models of how the cellular processes of how genetic information is passes from generation to generation.

Learning Objective: The student is able to describe representations and models by which genetic information is translated to the production of polypeptides. Learning Objective: The student is able to describe how humans can modify heritable information.

 

Cells can be activated, produce new products, and retain their activated state through gene regulation. a. Gene regulation plays a role in determining structure and function of cells. b. Both DNA regulatory sequences and regulatory genes are involved in gene expression.

 

1. Regulatory sequences are stretches of DNA that interact with regulatory proteins to control transcription. Students should be able to demonstrate understanding of the above concept by using an illustrative example such as:

 

Promoters

 

Terminators

 

Enhancers

 

2. A regulatory gene is a sequence of DNA that encodes information for regulatory proteins or RNAs.

 

c. Both positive and negative control mechanisms regulate gene expression regulation in bacteria and viruses.

 

1. Specific genes can be turned on by the presence of an inducer

 

2. Specific genes can be inhibited by the presence of a repressor

 

3. Inducers and repressors are small molecules that interact with regulatory proteins and regulatory sequences.

 

4. Regulatory proteins inhibit gene expression by binding to DNA and blocking transcription (negative control)

 

5. Regulatory proteins stimulate gene expression by binding to DNA and stimulating transcription (positive control).

 

6. Certain genes are continuously expressed; that is, they are always turned "on" or constitutively expressed, e.g. the ribosomal genes.

 

d. In eukaryotes, gene expression is more complex and controlled by regulatory genes, regulatory elements, and transcription factors which act in concert to modulate expression.

1. Transcription factors bind to specific DNA elements (sequences) and or other regulatory proteins. 2. Some of these transcription factors are activators (increase expression), while others are repressors (decrease expression). 3. The combination of transcription factors binding to the regulatory regions at any one time determines how much, if any, of the gene product will be produced.

 

e. Gene regulation explains the differences between organisms with similar genes.

 

Learning Objective: The student is able to articulate the connection between regulation of gene expression and observed differences between organisms.

Learning Objective: The student is able to explain how the regulation of gene expression is essential for efficient cell function.

 

Changes in genotype can result in changes in phenotype. a. Alterations in a DNA sequence may lead to changes in the polypeptide produced and the consequent phenotype. b. Normal errors in DNA replication and repair and external factors, including radiation and reactive chemicals can cause random changes, mutations in the DNA.

 

 

1. Mutations are generally detrimental however some can be beneficial and confer increased survival allowing for increased reproduction. Students should be able to demonstrate understanding of the above concept by using an illustrative example such as:

 

Antibiotic resistant mutations

 

Pesticide resistance

 

Sickle cell

Learning Objective: The student is able to analyze through visual representation how changes in a DNA nucleotide sequence can result in a change in the polypeptide produced. Learning Objective: The student is able to predict that changes in a specific DNA or RNA sequence can result in changes in gene expression (phenotype).

 

Viruses reproduce and can introduce genetic variation into their hosts. a. Viral reproduction differs from other reproductive strategies but allows for variation via different mechanisms.

 

1. Viruses replicate via a component assembly model allowing one virus to produce many offspring simultaneously via the lytic cycle.

 

2. Virus replication allows for mutations to occur through normal host pathways.

 

3. RNA viruses replicate via a DNA intermediate, but lack a replication error checking mechanism, and thus have a higher rate of mutation.

 

4. Different viruses can combine/recombine information if they infect the same host cell.

 

5. Some viruses are able to integrate into the host DNA and establish a lysogenic infection. These latent viral genomes can result in new properties. Evidence of student learning is knowledge of increased pathogenicity in bacteria or ONE other example.

 

b. The basic structure of viruses facilitates transfer of genetic information

 

1. Viruses transmit DNA or RNA when they infect a host cell. Students should be able to demonstrate understanding of the above concept by using an illustrative example such as:

 

Transduction in bacteria

 

Transposons present in incoming DNA

 

2. Viruses have very large reproductive capabilities that allow for rapid evolution and acquisition of new phenotypes.

 

3. HIV is a model system for the rapid evolution of a virus within the host that leads to that pathogenic nature of the virus to the host.

 

Learning Objective: The student is able to construct an explanation of how viruses introduce genetic variation in host organisms.

Learning Objective: The student is able to describe using representations and appropriate models how viruses reproduce in host organisms.

Learning Objective: The student is able to describe using representations and appropriate models how viruses introduce genetic variation into their host organisms