BIOLOGY 111 1st Edition Lecture 20 Outline of Last Lecture I Molecular Basis of Inheritance II Chargaff s Rule III DNA IV Base Pairing V Replication of E coli VI Replication Problems VII DNA packaging Outline of Current Lecture I Transcription II Translation III Structural Differences between DNA and RNA IV DNA Template Strands V Codons Current Lecture Gene Expression involves instructions from DNA that gets translated into some product One gene signals for a polypeptide sequence used to be thought was that one gene signaled for one enzyme instead it codes for many enzymes Transcription DNA makes RNA o Structural genes produce messenger RNA mRNA that specify the amino acid sequence of a polypeptide o For eukaryotes transcription occurs in the nucleus and have an additional step called RNA processing where pre mRNA is processed into functionally active mRNA o For prokaryotes transcription occurs in the cytoplasm Translation RNA makes protein o Synthesis of a specific polypeptide on a ribosome for both prokaryotes and eukaryotes Structural Differences in DNA and RNA o DNA Double stranded Base pairs are Adenine A Thymine T Guanine G and Cytosine C Deoxy in deoxyribose nucleic acid means it contains one less oxygen o RNA Single stranded These 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 Base Pairs Thymine T is replaced with Uracil U Ribose means it contains the extra oxygen that DNA is missing Template Strands o There are two strands of DNA because it is double stranded each of which contain several different genes and are available to be used as a template for RNA transcription o For each different gene either one of the two DNA strands can be used but for a single gene only one DNA strand can be used for transcription i e gene 1 may use a different DNA template strand than gene 3 BUT gene 1 will always that same strand of DNA as a template for transcription and translation Codons o Ribosomes read nucleotides of mRNA in groups of three or codons o There are 20 amino acids and each codon specifies one amino acid refer to charts some codons signal to stop UAA UAG and UGA and one codon AUG codes for methionine which acts as a start codon o There need to be enough arrangements of codons to call for all 20 amino acids Per 1 codon signals for four amino acids one for each codon Per 2 codon signals for 15 amino acids Per 3 codon signals for 64 amino acids this is enough to meet all 20
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