FSU BSC 2010 - Chapter 5: The Structure and Function of Large Biological Molecules

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Chapter 5 The Structure and Function of Large Biological Molecules Place the nitrogenous bases A T G C and U into groups Purines Adenine and Guanine Pyrimidines Thymine Cytosine and Uracil G C form hydrogen bonds A T form hydrogen bonds A U form Be able to compare and contrast the structures of DNA and RNA RNA Sugar Ribose Nitrogenous bases C G A U Single Stranded Basically you should be able to draw a DNA molecule and identify the hydrogen bonds DNA Sugar deoxyribose Nitrogenous bases C G A T Double Stranded following parts o 5 and 3 ends o Phosphate groups o Ribose or deoxyribose sugar o Nitrogenous bases Phosphate group Sugar deoxyribose Study Guide for Exam III November 6th Nitrogen bases Fall 2014 Chapter 16 The Molecular Basis of Inheritance You should be able to describe label draw a replication bubble with respect to the following aspects Location and function of helicase single strand binding proteins topoisomerase primase DNA polymerase III DNA polymerase I ligase Study Guide for Exam III November 6th The difference between leading and lagging strands is o the leading strand is synthesized in the same direction as the movement of the replication fork o the leading strand is synthesized at twice the rate of the lagging strand Study Guide for Exam III November 6th strand is synthesized in short fragments that are stitched together o the leading strand is synthesized by adding nucleotides to the 3 Fall 2014 o the leading strand is synthesized continuously whereas the lagging end where the lagging strand is synthesized by adding nucleotides to the 5 end An RNA primer allows the DNA polymerase to start adding new nucleotides and it ultimately gets degraded Okazaki fragments are segments of DNA that is synthesized by DNA polymerase III and they are eventually joined together by DNA polymerase I which removes RNA nucleotides and fills in the gap with DNA nucleotides Understand the differences between the replication of circular and linear chromosomes Bacteria Prokaryotic Circular Genome Single DNA molecule 1 origin of replication Humans Eukaryotic Linear genome multiple linear chromosomes multiple origins of replication Study Guide for Exam III November 6th Telomeres are special repetitive non coding nucleotide sequences at the end of eukaryotic chromosomal DNA molecules They postpone the erosion of genes near the ends of DNA molecules You should be able to describe chromatin Certain degrees of DNA packing make genes available or unavailable Dense packing makes it difficult for the cell to express genetic info Histone proteins in a nucleosome can change shape so that the chromatin is easier to transcribe o In histone acetylation acetyl groups are attached to positively charged lysines in histone tail which loosens chromatin structure promoting transcription o Methyl group can condense chromatin Heterochromatin transcriptionally inactive genes centromere telomere tightly packed Euchromatin transcriptionally active loosely packed Chapter 17 From Gene to Protein Know the process of transcription Prokaryotes Study Guide for Exam III November 6th Translation and transcription can happen simultaneously No processing of RNA before translation No nuclear envelope to separate DNA Translation occurs at ribosomes plasma membrane or free in cytoplasm Eukaryotes Translation and transcription are completely separate RNA is processed in the nucleus Fall 2014 RNA translocates from nucleus to the cytoplasm Translation occurs at ribosomes free nuclear envelope or rough ER Promoters signal the transcriptional start point and usually extend several dozen nucleotide pairs upstream of the start point RNA polymerase catalyzes RNA synthesis by prying the DNA strands apart hooking together the RNA nucleotides Study Guide for Exam III November 6th Study Guide for Exam III November 6th Fall 2014 Understand the processes of elongation and termination including the proteins responsible for elongation and the DNA sequences responsible for termination ELONGATION Elongated by the protein RNA polymerase which breaks two strands of DNA apart and adds RNA nucleotides to the template strand of DNA TERMINATION DNA sequence to stop transcription AAUAAA which releases the RNA transcript is released 10 35 nucleotides past sequence How are Eukaryotic RNA transcripts modified and for what purpose Enzymes modify pre mRNA to mRNA to facilitate export of mRNA to the cytoplasm protect mRNA from hydrolytic enzymes and help ribosomes attach to the 5 end 5 Cap modifided guanine molecule G P P P 3 poly A tail AAAA AAAA Most RNA transcripts have long noncoding stretches of nucleotides Introns which RNA splicing removes joining the coding stretches together exons Spliceosomes carry out RNA splicing The consist of proteins and snRNA small nuclear ribonucleoacids Know the translation initiation complex Study Guide for Exam III November 6th The first codon that begins translation is AUG methionine tRNA is non coding RNA that is not translated into a protein It bridges the gap between mRNA codons and amino acids with its two binding sites one is an anti codon which base pairs with a codon forming hydrogen bonds and the other is a 3 end which binds to a specific amino acid o Translation requires a match between a tRNA and an amino acid done by the enzyme aminoacyl tRNA synthetase and a match between the tRNA anticodon and an mRNA codon Translation ends when the release factor causes the addition of a water Study Guide for Exam III November 6th molecule instead of an amino acid which releases the polypeptide causing translation assembly to come apart Fall 2014 Know the different types of DNA mutations that we discussed Silent Mutations Have no effect on the amino acid sequence of the protein produced Missense Mutations Change a Codon so that a different amino acid is inserted at that position in the protein Nonsense Mutations Change a codon to a stop codon so that the growing polypeptide is truncated Insertions and deletions are additions or losses of nucleotide pairs in a gene usually most disastrous effect on protein function Be able to figure out the change in amino acid sequence caused by a specific mutation in a DNA sequence Study Guide for Exam III November 6th Chapter 18 Regulation of Gene Expression Understand the reasons why a cell needs to regulate the expression of the genes contained in its genome It regulates development and is responsible for differences in cell types Abnormalities in gene expression can lead to


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FSU BSC 2010 - Chapter 5: The Structure and Function of Large Biological Molecules

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