Bio. 110 - Exam 3 Learning Outcomes

Lecture 19-26 & Discussion 3

Lecture 19

  • Describe the structure of nucleic acids.

Prokaryotes and eukaryotes use DNA as their genetic material, but prokaryotes are studied better at a molecule level because they can be model organisms. Nucleic acids are polymers of nucleotide monomers that have 3 parts, a 5 C sugar (either ribose or deoxyribose), a phosphate group attached to the 5’ C. and a nitrogenous base (either purines or pyrimidines). DNA is double stranded and coiled as double helix because of complementary base-pairing and run antiparallel. While DNA is double-stranded, DNA is single.

  • Explain some of the experiments that demonstrated that DNA is the molecule of genetics.

One experiment was performed in 1928 by Griffith using 2 strains of pneumonia where dead strains of pneumonia transformed live non-pneumonia strains. This “something” would become known as DNA. During 1950, Chargaff showed the composition of nucleotides, revealing the chemical components that composes everything. Chargaff realized A & T and G & C are complementary. In 1952, Hershey and Chase uncovered a T2 phage in E. coli. Finally, Watson and Crick figured out the structure of DNA after Franklin used x-ray crystallography to show a double stranded helix and applied Chargaff’s complementary base-pairing rule to form the structure of DNA.

Lecture 20

  • Describe the process of DNA replication, including the roles of specific proteins.

Because of complementary base-pairing, when we know one strand, we know the other. The daughter molecules are semiconservative and each has a parent/template strand. Replication will begin at the origin of replication, which is a specific sequence of nucleotides where replication starts, and the replications fork will start in the middle of the strand and move in both directions, while the enzyme helicase splits the strands and single binding proteins separate the strands. Topoisomerase moves ahead of the replication fork to unwind the double strands. As the replication fork moves ahead, DNA polymerase “reads” the pare/template strand and adds in the complementary nucleotide, but can only be attach to the 3’ end. In order to attach, DA pol requires an RNA primer enzyme, primase, on the template strand. The above includes the template/parent strand, or the leading strand. Because replication works in the 5’ to 3’ end, a different process is required for the lagging strand that needs to be built in the 3’ to 5’ direction, away from the replication fork. Primase will drop in an RNA primer and DNA pol lll makes an Okazaki fragment, which detaches so more fragments can be placed. DNA pol l will then replace the RNA primer with DNA and DNA ligase bonds the two Okazaki fragments to make one, continuous strand.

  • Describe the chromosome structure in both prokaryotes and eukaryotes.

In prokaryotes, chromosomes are circular and are approximately 1 mm long with nucleotides super coiled with proteins. In eukaryotes, chromosomes are linear and about 4 cm long. Eukaryotes have histones and those bundle together to form nucleosomes that form the condensed chromatin, DNA and proteins.

Lecture 21

  • Understand what is meant by the “Central Dogma”: DNA → RNA → Protein.

All cells have the same genome, but they express differently due to proteins, which do all the work in cells thanks to enzymes. The questions remain, though, “What do genes have to do with proteins?” and “How is gene expression controlled?”

Proteins that are made for cells are determined by what genes are expressed. Essentially, your genes are instructions to make proteins. In order for information to be passed to proteins, for genes to be expressed, DNA must transcribe to RNA by mRNA and then translate into polypeptides to make amino acids to form proteins.

  • Describe the process and recognize the key components of transcription, where mRNA is produced from a DNA template.

Transcription makes mRNA that is used to make a temporary copy of a gene to be used for translation. RNA polymerase makes mRNA to DNA and separates the DNA strands because only one RNA strand is being made from the template that forms the coding strand. When the transcript is done, it is moved to the ribosomes. In eukaryotes, the primary transcript is modified before leaving the nucleus.

  • Describe the process of recognize the key components of translation, where a polypeptide is produced from the information in mRNA.

Translation is making proteins from the instructions in mRNA where 3 components are present in order to explain translation. One is codons, which are nucleotide triplets that code for amino acids from mRNA. The second is tRNA, or the adapters between codons and the actual amino acids. The third are the ribosomes, or machinery, which catalyzes the interactions between mRNA and tRNA>

Codons are a “3-letter word” for a specific amino acid. Codons are dictated by nucleotides and designate the start and stop for an amino acid chain. For each codon, there is an anticodon that is complementary which leads to the formation of a polypeptide chain to make a protein.

Lecture 22

  • Explain the processes involved in the initiation, elongation, and termination of an mRNA during transcription.

Transcription is initiated at the promoter, the are of a DNA molecule where RNA pol attaches to begin transcription, and is the starting point that is “upstream” of the transcription unit. In prokaryotes, RNA pol is able to recognize the promoter because of its size, but in eukaryotes, the promoter is much larger and flags where RNA pol needs to bind to transcription factors. One factor is the TATA Box, which dictates where nucleotide sequences can start to be read and decoded. Once transcription factors bind, RNA pol beings to form the transcription initiation complex.

Transcription elongation is accomplished by RNA pol, as complementary bases are being paired. Termination in prokaryotes requires a certain nucleotide sequence to be read, while in eukaryotes require RNA pol to transcribe a polyadenylation signal (AAUAA) to signal for RNA pol to cut the nucleotide sequence about 30 nucleotides downstream.

  • Understand how the mRNA transcript is modified in eukaryotes before translation.

Prokaryote mRNA is ready to be translated after elongation but eukaryote mRNA needs to undergo mRNA processing. Modification takes place on the prime ends. The 5’ end receives a 5’ cap, or the addition of a G and the 3’ end receives the addition of hundreds of A’s, or a poly-A tail. The goal of mRNA processing is meant to prevent RNA from being broken down to quickly by nucleases (proteins that break down RNA). Once the mRNA strands are prepped with 5’ end caps and a poly-A tails, the sequence undergoes a splice. Splicing is when the introns, non-coding/junk regions, are removed and the extrons, coding regions, are moved together by spliceosomes. A reminder is prokaryotes are ready after elongation, only eukaryotes need to undergo mRNA processing.

  • Describe the processes involved in the initiation, elongation, and termination of a polypeptide during translation.

During initiation, the small ribosomal subunits and tRNA read for a start codon, AUG or Met. Then in elongation, the ribosome has 3 sites that uses tRNA’s to add amino acids to a polypeptide. For temination, the ribosome reads for a stop codon. Stop codons can read as UAG, UAA, or UGA. Hydrolysis separates the polypeptide from tRNA and all parts of the unit fall apart and all proteins are recycled.

  • Recognize how some DNA mutations lead to changes in protein structure that affect function and other mutations have no effect.

Changes in DNA sequence can lead to changes in the protein, or a mutation. Some mutations are called a substitution, or point, mutation and there are several types. A missense mutation results in a different amino acid being coded for when a nucleotide is swapped. Silent mutations have a change in nucleotides, but the amino acid is unaffected, so the protein is unchanged. Nonsense mutations results in a nucleotide sequence that codes a stop codon prematurely. Frameshift mutations result from an instance of insertion/deletion, where the reading of an mRNA from DNA changes because a nucleotide is added or removed.

Lecture 23

  • Describe the difference between, and provide examples of, repressible and inducible operons in prokaryotes.

The “regulation of gene expression” means genes are on/off when they need to be. This require repressible and inducible operons for proteins to result in specific genes. An example of a repressible operon is for tryptophan synthesis. Feedback inhibition, the presence of a protein turns off the enzyme to express the gene. For the trp gene, it is an operon, a part of the DNA sequence of a chromosome, that controls

Lecture 24

  • Describe how transcription is regulated by control elements and activators in eukaryotes.

  • Explain how regulation of gene expression is related to the cell cycle and how mutations can result in cancerous oncogenes.

Lecture 25

  • Describe the characteristics that distinguish different types of viruses, including genome and capsid structures and modes of replication.

  • Understand how viruses have only some of the characteristics of living systems.

  • Describe some examples of viruses and how they function.

  • Describe prions and how they are different from viruses,

Lecture 26

  • Describe how PCR (polymerase chain reaction) and gene cloning can be used to make lots of copies of a specific gene.

  • Explain how gel electrophoresis can be used to visualize and separate DNA fragments.

  • Recognize how DNA polymerase and electrophoresis can be used for DNA sequencing.

Discussion 3

  • Describe the nature of epigenetic inheritance, including the evidence non-genetic inheritance of certain traits occurs and its roles in disease.