BIOL 3315 1st Edition Lecture 5 Outline of Last Lecture Chapter 2 (continue)I. “C-value paradox”II. Why are some genomes so largeIII. Eukaryotic nuclear genomesIV. Eukaryotic chromosomesV. Eukaryotic chromosomesVI. Chromosomes of 2 tomatoVII. Nuclear DNAVIII. Comparative genomicIX. X chromosomes in 3 species Outline of Current Lecture Chapter 3X. OverviewXI. Phenylketonuria or PKU testXII. Malfunctional form of a gene pictureXIII. Transfer of informationXIV. RNAXV. Classes of RNAXVI. TranscriptionXVII. RNA polymerase XVIII. Transcription stepsXIX. Eukaryote RNA processingXX. Protein structure XXI. 20 common amino acidsXXII. TranslationCurrent LectureChapter 3 Gene FunctionX. OverviewA. RNA is transcribed using the rules of base pairing from the template strand of DNAThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.B. Genes code for proteins or untranslated functional RNAsC. The nucleotide sequence of the protein-coding gene determines the order of amino acids in a protein, which determines shape, size, and protein functionD. mRNA is translated into groups of three nucleotide (codon) at the ribosome through pairing of tRNA anticodon with the mRNA codonE. Change in nucleotide sequence (mutation) in a protein-coding gene may cause change in amino acid sequence, altering functionXI. Phenylketonuria or PKU testA. Phenylketonuria (PKU) is an autosomal recessive metabolic genetic disorder. The disorder occurs when the enzyme phenylalanine hydroxylase produce in the liver is absent. Phenylalanine hydroxylase metabolizes the amino acid phenylalanine into the amino acid tyrosine. If this reaction does not occur and phenylalanine is high phenylpyruvic acid is producedB. PKU test is performed routinely in newborns in the USA. Blood is collected and either mass spectrometry is used to detect high levels of phenylalanine or using abacterial assay (Guthrie test). ....XII. Malfunctional form of a geneXIII. Transfer of informationA. DNA RNA polypeptide (CENTRAL DOGMA)B. Complementary base pairing transfers information1. During transcription to form RNA2. During translation between codon and anticodonC. DNA binding proteins1. Recognize double or single stranded DNA2. Recognize specific nucleotide sequence3. Are coded by genes4. Have variety of important functionsXIV. RNAA. Transcription: coping nucleotide sequence of DNA into RNA1. Forms RNA transcript2. DNA may be transcribed multiple timesB. RNA 1. Single stranded polynucleotide2. Contain the pyrimidine uracil (U)a. Hydrogen bonds with A (in place of T in DNA)3. 5’ and 3’ end critically importantXV. Classes of RNAA. Informational RNA: protein encoding mRNA1. Primary transcript in prokaryotes2. Processed transcript in eukaryotesa. 5’ and 3’ end modificationb. Intron removal3. Translated into amino acid sequenceB. Functional RNA1. tRNA: transport amino acid to ribosome2. rRNA: structural and catalytic component of ribosomes3. snRNA: structural and catalytic component of spliceosome snRNPs4. snRNAs, siRNAs, etc.XVI. TranscriptionA. RNA polymerase1. Locally unwinds DNA 2. Uses one DNA strand as templatea. Same strand for a given gene3. Adds free nucleotides to growing RNA strand at 3’ enda. 5’ to 3’ RNA synthesisb. Template reads 3’ to 5’c. Uses rules of base pairing to synthesize complementary RNA moleculeB. Transcript is identical in sequence to non-template strand, except T’s replaced by U’sXVII. RNA polymeraseA. Prokaryotes: single RNA polymeraseB. Eukaryotes: three RNA polymerase1. RNA polymerase I transcribed rRNA genes2. RNA polymerase II transcribes protein-encoding genes and microRNAsa. Primary transcript will be processed3. RNA polymerase III transcribes tRNA genes, SS rRNA genes, snRNAs4. Both DNA strands can be codingXVIII. Transcription stepsA. Initiation 1. at 5’ end of gene 2. Binding of RNA polymerase to promoter3. Unwinding of DNAB. Elongation1. Addition of nucleotides to 3’ end 2. Rule of base pairing3. Requires Mg3+4. Energy from NTP substratesC. Termination1. At 3’ end of gene2. Terminator loop (prokaryote) or processing enzyme3. UTR- untranslated regionXIX. Eukaryote RNA processingA. *5’ end: capping1. Addition of 7- methylguanosine2. Linked by here phosphatesB. 3’ end: poly(A) tail1. Addition of up to 200 adenine nucleotides2. Downstream of AAUAAA polyadenylation signalC. Intron removal by spliceosome1. All introns have 5’ GU and 3’ AG recognition sequence (GU-AG rule)2. snRNPs of spliceosome provide catalysis3. intron excised at lariat, destroyed D. snRNPs of spliceosome provide catalysisXX. Protein structureA. Protein is a polymer of amino acids (polypeptide)1. Each amino acid as R group conferring unique properties2. Amino acids connected by peptide bond3. Each polypeptide has amino end and carboxyl endB. Structures1. Primary: amino acid sequence2. Secondary: hydrogen bonding, α-helix and β-sheet3. Tertiary: folding of secondary structure4. Quaternary: two or more tertiary structuresC. Shape and function determined by primary structure encoded by geneXXI. *20 common amino acids A. Amino acids are connected by peptide bondB. Alpha helix has hydrogen bond between coilsC. Shape and function determined by primary structure encoded by geneXXII. TranslationA. mRNA is translated by tRNAs at ribosomesB. nucleotide sequence is read three nucleotides at a time1. each triplet is called a codon2. each amino acid has one or more codona. 64 possible codons (4x4x4)= genetic code3. Used by all organisms with few exceptionsC. Genetic code specifies 20 different amino acids (sometimes