Review #2 Flashcards
Explain the process of DNA replication (Hint: know that it’s basically the same fundamental process, whether IB says to describe just the process or the process in prokaryotes or eukaryotes).
- HELICASE “unzips” the parental DNA molecule (breaking the hydrogen bonds between bases).
- In eukaryotes, gyrase and single-strand binding proteins stabilizes unzipped DNA molecules at many sites.
- PRIMASE adds a sequence of RNA bases (a primer) to each parental DNA molecule at the replication origin (each parental molecule serves as a template).
- DNA POLYMERASE III adds new nucleotides (deoxynucleoside triphosphates - two phosphates lost to provide energy for binding) to the RNA primer (at the 3’ end) to create a new complementary strand.
- Continuous in the leading strand, as the Okazaki fragments in the lagging strand (moves in a 5’ to 3’ direction - adding new nucleotides to the 3’ end only).
- In the lagging strands, DNA LIGASE fills the gap between fragments (5’ to 3’).
- DNA POLYMERASE I removes the RNA primers and replaces them with DNA nucleotides (5’ to 3’ direction - DNA bases left unpaired at the tip of the 5’ end after primers are removed)
Explain the process of transcription (including post-transcriptional processing in eukaryotic cells).
INITIATION:
- RNA polymerase unwinds DNA strands and binds to promoter on DNA (antisense - serves as a template for mRNA to be built off of)
ELONGATION:
- RNA polymerase adds RNA nucleotides (nucleoside triphosphates - two phosphates lost to provide energy for binding) to 3’ end of growing mRNA based on code in antisense strand of DNA.
- Works in a 5’ to 3’ direction.
- Bases added using complementary base pairing rules (A+U and C+G).
TERMINATION:
- RNA polymerase continues until terminator sequence reached (passes in eukaryotes).
- mRNA molecule detaches from DNA.
- RNA polymerase detaches from DNA.
Post-transcription:
- In EUKARYOTES: Introns removed to form mature mRNA (only exons remain to be translated into amino acids/ protein) – one gene = many polypeptides (many mRNA from one DNA sequence - alternative RNA splicing by spliceosomes and snRNA’s)
Explain the process of translation/ polypeptide synthesis/ polypeptide formation.
INITIATION:
- Small subunit of ribosome binds to mRNA at start codon (AUG)
- tRNA (with complementary anticodon UAC) binds to mRNA (complementary base pairing).
- Large subunit binds (with 1st tRNA in P site) - GTP
ELONGATION (and translocation):
- 2nd tRNA comes into A site (complementary base pairing with mRNA codon).
- Peptide bond forms between amino acids of two tRNA molecules.
- 1st tRNA translocates into E (exit) site and leaves ribosome.
- 2nd tRNA translocates into P (polypeptide) site.
- Ribosome moves along mRNA in 5’ to 3’ direction.
- 3rd tRNA comes into A site.
- Peptide bond forms between amino acids of two tRNA molecules (one in P site and one in A site) and so on… until….
TERMINATION: - Stop codon (on mRNA) reached. - Polypeptide chain released from tRNA in P site. - Ribosome disassembles. TERMINATION:
Explain factors that affect enzymes/ enzyme-controlled reactions.
Temperature:
- Increases enzyme activity (more kinetic energy = more collisions).
- Enzyme activity peaks at an optimal temperature.
- Higher temperatures decrease activity (causes denaturation).
pH:
- Enzyme activity is highest at an optimal pH range.
- Activity decreases outside of this range (due to denaturation).
Substrate Concentration:
- Increases enzyme activity (more particles = more collision).
- At a certain point, activity plateaus (saturation of active site).
Outline the structure of DNA (and be able to identify structures in DNA in diagrams and diagram/ label double stranded AND single stranded DNA).
DNA (deoxyribonucleic acid) is a nucleic acid (made up of four nucleotides; 4 of them in DNA, each with a different base). Each nucleotide is made of a phosphate group covalently bonded to a pentose sugar (deoxyribose bonded to a nitrogenous base). DNA nucleotides are linked together by phosphodiester bonds in a 5’ to 3’ direction to form a single strand. Double-stranded DNA is created when hydrogen bonds form between bases (A+T and C+G) between single strands. Strands in double stranded DNA are ANTI-PARALLEL.
Outline the structure of the nucleosome and its role in eukaryotic cells.
Core = 8 histone proteins (+ charged) with DNA molecule (- charged) wrapped TWICE around (like beads on a string). Nucleosomes are a fundamental unit of DNA packaging, as it allows supercoiling of DNA into chromosomes. Supercoiling prevents certain genes from being accessed by transcription factors/enzymes, as it regulates the process.
Outline the structure of an amino acid and formation of a dipeptide (be able to diagram/ label this and identify an amino acid/ dipeptide/ peptide bond in a diagram too).
Amino acids are put together using mRNA bases in sets of 3 called codons. They are put together using the genetic code, which is universal for all life. Amino acids are covalently joined by peptide bonds to form polypeptide chains. This requires condensation reactions, in which water molecules are removed. Dipeptide bonds, in specifics, are formed when TWO amino acids are joined together by a hydrogen bond.
Compare DNA and RNA structure.
DNA:
- Double-stranded
- Deoxyribose (sugar)
- Guanine, cytosine, adenine, THYMINE
RNA:
- Single-stranded
- Ribose (sugar)
- Guanine, cytosine, adenine, URACIL
Outline ribosome structure.
Ribosomes are composed of ribosomal rRNA (nucleolus) and protein. They consist of two subunits: a small subunit that contains an mRNA binding site and a large subunit that contains three tRNA binding sites (E, P, and A). In prokaryotes, ribosomes are 70S while in eukaryotes, they are 80S.
Outline tRNA structure.
Transfer RNA carries amino acids to the ribosome, as amino acids are attached by tRNA activating enzymes. The tRNA-activating enzyme works in two steps will lead to:
- The enzyme joining ATP to an amino acid.
- “Charged amino acid is linked to tRNA (AMP is released).
The purpose of the “charging” of the amino acid is to create a high energy bond that can be used during translation. Ribosomes use this energy to synthesize peptic bonds. Each tRNA-activating enzyme is specific to a particular amino acid, but may bind to multiple tRNA (due to degeneracy).
Outline types of DNA sequences and their functions.
- Unique (single-copy) sequences = genes (code for proteins)
-2% of genome - Highly repetitive sequences = found
between genes (form barriers of non-
coding regions between genes)
-5 to 45% of genome
-Short-tandem repeats (STR’s): form
polymorphisms (significant variation
between individuals – used to create
DNA profiles)
-Transposable (moveable = shuffle
genes) - Structural Sequences = pseudogenes (highly coiled at centromeres and telomeres)
-20% of genome
Outline the factors that comprise the epigenome and their effect(s) on gene expression.
The gene expression of a cell is the epigenome, which is a collection of all of the factors that modify/impact the expression of genes without altering their DNA sequences. These factors include:
Nucleosomes: more nucleosomes = more DNA packaged tightly together, making genes less accessible to RNA polymerase (less transcription, leading to less mRNA, leading to less protein synthesis from those genes).
Methylation: Methyl groups (CH3) bind to DNA, causing it to wrap more tightly around histones. More methylation = less transcription/ less protein from those genes (if any). Highly methylated genes are usually not expressed at all, and methylation of DNA is maintained through cell division and even from parent to offspring!
Proteins:
Transcription factors – aid in RNA polymerase binding to DNA. Transcription activators/ transcription repressors.
ENVIRONMENT: Can change methylation patterns and/ or affect proteins involved in regulating gene expression/ splicing (wrong genes on or off etc.). Chemicals (cigarette smoke, preservatives, pollutants, topical medications/ creams etc.) Infectious agents (bacteria, viruses, prions).
Outline the four levels of protein structure and their significance to proteins.
Primary Structure: Order of amino acid sequence that is formed by covalent peptide bonds (which are formed during translation).
Secondary Structure: Folding polypeptides into repeat patterns (alpha helix (keratin in hair) or beta-pleated sheets (spider silk)). These are held together by hydrogen bonds that stabilizes structure of fibrous proteins. They are formed by interactions between amino acids and carboxyl groups.
Tertiary Structure: Folding pattern on polypeptide into 3D shape (for forming an active site for an enzyme), which are known as “globular proteins.” They are stabilized by disulfide bridges. ionic bonds, hydrogen bonds, and hydrophobic (Van der Waals) interactions, as well as interactions between R groups.
Quaternary Structure: These are not in all proteins. They link several polypeptide chains together, using the same bonding as tertiary structures. They are formed when prosthetic groups are linked to pokypeptides. An example would include the haem in haemoglobin.
Outline the process of end-product inhibition (allosteric inhibition) in the control of metabolic pathways.
End-product inhibition is a negative feedback mechanism. Metabolic pathways are regulated by end-product inhibition. Every metabolic pathways start with a specific molecule and end with a specific product.There are many steps in a metabolic pathway, each of which is catalyzed by its own specific enzyme (because the substrate changes shape after each reaction!). When the end-product is present in sufficient amounts it is able to bind to an allosteric site on the first enzyme in the metabolic pathway, causing its active site to change shape, making it non-functional and shutting down the entire pathway.
