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1) Transcription or Copying mRNA molecules are synthesized as complementary copies of DNA template

Expression of genetic information (Part II). Definition: a complex process in which the genetic information in DNA is decoded and used to specify the manufacture of specific protein in the cell. Includes 2 steps:. 1) Transcription or Copying

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1) Transcription or Copying mRNA molecules are synthesized as complementary copies of DNA template

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  1. Expression of genetic information (Part II) Definition: a complex process in which the genetic information in DNA is decoded and used to specify the manufacture of specific protein in the cell. Includes 2 steps: 1) Transcription or Copying mRNA molecules are synthesized as complementary copies of DNA template 2) Translation Conversion of genetic language in mRNA molecule (codons) into amino acid language of protein

  2. Translation = Protein synthesis: Copy of genetic information in mRNA (codons) Conversion into by Polypeptide chain (protein) Requirements mRNA Active tRNA Ribosomes

  3. What are the kinds of processes that take place in nucleus and cytoplasm during gene expression ? tRNA (RNA polymerase III) mRNA (RNA polymerase II) Nuclear envelope protein large subunit tRNA mRNA Ribosome: 2 subunits DNA cisterone nucleolus Nuclear pore Small subunit DNA d.h.

  4. Function: 1- translate the information in mRNA into protein. 2- hold mRNA, amino-acyl tRNA & polypeptide chain in a correct orientation during translation. 3- form the peptide bonds between amino acids • Ribosome= machines of protein synthesis • Properties: • Large subunit: • contains 2 binding site. • P= For peptidyl-tRNA (hold • peptide chain). • A= For amino-acyl-tRNA, that • delivers the next amino acid. • Small subunit • contains mRNA binding site. • - Separate entities are inactive. • - become active when the 2 subunits are linked to mRNA during translation. • Polyribosomes (polysomes) are mRNA -linked clusters of ribosomal molecules . P- site A - site L. Subunit S. subunit mRNA mRNA –binding site

  5. Ribosomestructure

  6. tRNA= transfer RNA 3’ Amino acid acceptor end Anticodon Amino acid 5’ • Small molecules that are transcribed from tRNA gene. • Properties: • Recognized by amino-acyl-tRNA synthetase & ribosomes. • Have amino acid acceptor 3’ end. • Have anticodon that recognizes and links the complementary mRNA –codon. • Types: • 1)initiator tRNA (carries methionine recognizes & links the start codon; AUG). • 2) regular tRNA for other amino acids. • -there are (40 tRNA, 20 A.A., 61 codons). + + ATP Amino-acyl – tRNA synthetase ADP + P Active tRNA

  7. Structure of tRNA

  8. Steps ofProtein Synthesis Anticodon 5’ 3’ 5’ 3’ Methionine Initiator-tRNA Start codon 1) Initiation 5’ 3’ mRNA Small subunit - Begins when initiator complex (initiator tRNA & small ribosomal subunit) binds the start codon of mRNA. - That followed by formation of functioning ribosome by binding of large ribosomal subunit to initiator complex A -site P -site 5’ mRNA Small subunit

  9. 2) Elongation 1st step • Begins byoccupying A- site by 2nd tRNA with its amino acid, that carries anticodon complementary to the next mRNA –codon. • 2) Peptide bond formation; Amino acid is detached from tRNA in P-site & joined by peptide bond to amino acid linked to tRNA in A-site. • 3) Translocation, • Ribosome moves one codon in direction 5’ – 3’, that leads to: • - release tRNA from P-site to the • cytoplasm. • - transfer tRNA & the peptide chain • transfer from A- site to P-site. • - A- site become open to receive • another suitable tRNA. • by repeating steps 1, 2, 3 (elongation • cycle), the peptide chain elongates till • the ribosome reaches the stop codon. 5’ Small subunit 2nd step 5’ Small subunit 3rd step GTP GDP+P+E 5’ mRNA

  10. 3) Termination Releasing factor Polypeptide chain 5’ dissociation • At stop codon: • Release factor recognizes and binds the mRNA - stop codon. It terminates the protein synthesis by releasing: • Large & small ribosomal subunits • Polypeptide chain • tRNA molecule • Releasing factor • mRNA

  11. In Eukaryotic cells: Transcript mRNA is immature (called pre-mRNA), why? a) - contains non-coding regions (introns) between protein coding regions (exons). b) - needs modifications before it becomes competent for transport & translation. • Posttranscriptional modification & processing of mRNA in Eukaryotic Cell • In prokaryotic cell (bacteria) • - translation is coupled with transcription, • Protein synthesis begins while the mRNA is being completed, as multiple ribosomes attach to the mRNA to form a polyribosomes • - No modification or processing, why ?

  12. 1) Capping: - Begins when mRNA is about 20-30 nucleotides long.- (7 –methyl guanosine capis added to 5’ end of mRNA). Importance: 1) Non-capped mRNA can not be recognized by ribosomes. 2) Protection against degradation. 3) Increase the stability of mRNA. Half life: - about 10 hs. In Eukaryotic cell. - about 2 mins. In prokaryotic cell. 2) Polyadenylated tail ( Poly-A tail) addition of 100 – 250 adenine nucleotides to the 3’ end of completed mRNA. Importance: 1) Help in passage of mRNA from nucleus into the cytoplasm. 2) Stabilizes the mRNA against degradation (increase life span of mRNA in cytoplasm) • Types of modifications

  13. Intron II Intron I • 3) Splicing: • = Means removing the introns(non-coding regions) and splicing the exons(coding regions) together. • Removal of introns is performed by small nuclear ribonucleoprotein complex enzyme (snRNPC). • Function: • formation a continuous protein coding message. Immature mRNA 3’ 5’ -OH P-P-P- 1) capping by addition 7- methyl Guanosine to 5” end. -OH G-P-P-P- 2) Addition polyadenylate tailto 3’end. -AAAAAA… G-P-P-P- 3) Splicing the exons to form continuous coding message Remove introns I & II Mature mRNA --AAAAAA… G-P-P-P-- Nuclear envelope Mature mRNA becomes competent for transport & translation in cytoplasm --AAAAAA… G-P-P-P--

  14. 1) Transcription & translation are not coupled. 2) Average half life 10 hours (translation continuous about 10 hours). 3) After transcription, mRNA is modified & processed. 1) Transcription & translation are coupled. 2) Average half life 2 minutes 3) mRNA is ready for translation as soon as it transcribed (no modification). Gene expression inEukaryotic cell prokaryotic cell Half life = the time that required for deterioration half of the molecule forming mRNA.

  15. Characters of genetic code 1) Triplet code (three bases) 2) Universal for all organisms ex. : UUU = phenylalanine in all organisms. 3) Redundant, redundancy of genetic code: more than one code specify one amino acid. (61 codon, 40 tRNA, 20 amino acid). 4) Reads as a series of nucleotide. - No comma between the codon - The start codon determines the reading frame. 5) The codon could undergo mutation

  16. The Genetic Code

  17. Mutations • Definition: Any change in the nucleotide sequence of DNA. • Two types: 1) Unstable mutation: Revert back to original sequence. 2) Stable mutation: Change the characteristics (phenotype) of organism.

  18. Definition :An organism showing deviation in some characters (phenotype), whose progeny maintain these deviances.Cause:molecular change in hereditary material.Types:1- Base substitution (Missense & nonsense mutants).2- Frame –shift mutants. • Mutants

  19. 2) Missense & Nonsense mutants: • a) Missense mutation Cause: Base substitution mutation in DNA. Results: one amino acid is replaced with another in a protein. The effect: depends upon the position of the replaced amino acid. Missense mutation (Base substitution mutation) The amino acid is not a part of the active site The amino acid is located at or near the active site Reduce or absence the enzyme activity ( in Sickle cell disease; glutamic acid is replaced with valine in Hemoglobin) No change in enzyme activity

  20. 5’____ ATG ……………..T A C……………TAG____3’ 3’____ TAC …………….. AT G……………ATC____5’ b) Non-sense mutation: Cause: mutation that creates an internal stop codon in a gene, will prematurely terminates the encoded protein. DNA d.h. mutation 5’____ ATG ……………..T A G……………TAG____3’ 3’____ TAC …………….. AT C……………ATC____5’ transcription mRNA 5’____ AUG …………….. UAG……………UAG____3’ Start codon internal stop normal stop codon codon translation Meth. ________________ Protein terminates prematurely shorter or nonfunctional protein Base –pair substitution leads to formation internal stop codon (UAG, UGA, UAA), that replaces the amino acid codon.

  21. 3) Frame –shift mutants • Cause: - Insertion or deletion base – pair. Results: - Misreading of all codons from the point of deletion or insertion to the end of the message. - Change the amino acid sequence of encoded polypeptide chain. AUG ACA CAA UGG ACU GAC ………… mRNA Meth. ………………………………………… protein (normal) Deletion of a single base A AUG ACA CAU GGA CUG AC ………… mRNA Meth. ………………………………………… protein (unuseful) • Results:- formation of unuseful new polypeptide chain.

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