B) STARTS AT ORIGIN Initiator proteins identify specific base sequences on DNA called sites of origin Prokaryotes – single origin site E.g E.coli - oriC Eukaryotes – multiple sites of origin (replicator) E.g. yeast ARS (autonomously replicating sequences) Intiator (a) Original double hlix. - Origin of replication Rapicator Rocogntion Intlator Termination region Meting Prokaryotes Eukaryotes
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Slide 9
Please help me explain this on my report on cell dogma
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- vvnicn the following statements are correct about the repair of a DNA duplex containing the sequence below that is grown INE coli (select all that apply)? Strand A Strand B GATCTAGCCGGCATCCGAT CTAGATCGGACGTAGGCTA Methyl ✔A. MutH cleaves Strand A O B. DNA repair will result in the bold A in strand B being replaced with a C O C. DNA repair will result in the bold G in strand A being replaced with a T ✔ D. Defect will not be properly repaired in dam(-) E coli O E. The mammalian repair system would also correct the mismatch shown based on the methylation status of the DNAGive the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: bottom strand is the noncoding strand). 5'-ААCGCATGAGAAAGCCCCCCGGAAGATCACСТТСCGGGGGCТТТАТАТААТТАGC-3' 3'-тTGCGTACтстттCGGGGGGCCTTCTAGTGGAAGGCCCCCGАААТАТАТТААТтCG-5' (i) Draw the structure of hairpin loop that will be formed during transcription. (ii) Illustrate how the hairpin loop structure initiates the termination of transcription.
- Cynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X GConsider the following DNA template: 5’-AAGAGGTTCCAATGCAGGCACTCACCAACTCTTAAATAAA-3’ 3’-TTCTCCAAGGTTACGTCCGTGAGTGGTTGAGAATTTATTT-5’ If the bottom DNA strand is used as template to transcribe mRNA, predict the amino acid sequence that would result from the process of translation. Met-Ala-Leu-Thr-Gln-Glu-Gly Met-Gly-Ser-Leu-Asn-Ser-Gln Met-Thr-Asn-Ser-Leu-Ala-Gln Met-Gln-Ala-Leu-Thr-Asn-Ser Met-Glu-Ala-His-Trp-Ser-TyrNAME Tarissa shyll CENTRAL DOGMA REVIEW REPLICATION Use the DNA code provided and fill in the complementary DNA strand. Which nitrogen base CAN'T you use during replication? Uracil ATTCGATGC TACGGATCG TRANSCRIPTION Use the DNA code provided to copy an m-RNA message. Which nitrogen ase CAN'T you use during transcription? ACTGGATAC ACGGAT CGT CERDANCE TRANSLATION: USE the DECODING WHEEL to DETERMINE the AMINO ACID that corresponds to the m-RNA CODE GIVEN mRNA CODE AMINO ACID AAA GCG GAU CAA CAC UUU Which two mRNA codes correspond to histidine? GACU How many different mRNA codes correspond to Threonine? 2 CAGTGACTT TGACAGCTA Which amino acids have ONLY ONE codon? COTGOHGUCAG|UC|AGUCAGS GU AC CUGA U/G/ACU GACUCAK Alanine Valine Arginine Serine ان داد Lysine Asparagine A Threonine Serine Tyrosine Stop Cysteine Stop Tryptophan Leucine TOPPLY Histidine Proline
- DNA is made of two strands that are antiparallel. If one strand runs from 3’ to 5’ direction the other one will go from 5’ to 3’ direction. During replication or transcription, whatever the process is, it will always follow the 5’ to 3’ direction using the 3’ to 5’ directed strand as the template strand. Therefore, if following is the DNA sequence 5’-CCG ATC GCA CAA-3’ Using this sequence as template after transcription no protein can be translated. Why? Presence of start codon Absence of start codon Due to mutation If you want to start the translation, what change you need in the second codon (from 5’ to 3’ direction)? Substitution of C with G No change4 Deletion of Both I & IIIIf this is the original sequençe of a DNA strand: CCAGGTCCATGACTTAGC, how would you labeled the one below CCAGGTCCATGACTTTTTTAGC Deletion Insertion (addition) Frameshift In frame insertionDNA is made of two strands that are antiparallel. If one strand runs from 3’ to 5’ direction the other one will go from 5’ to 3’ direction. During replication or transcription, whatever the process is, it will always follow the 5’ to 3’ direction using the 3’ to 5’ directed strand as the template strand. Therefore, if following is the DNA sequence5’-CCG ATC GCA CAA-3’a) Using this sequence as template after transcription no protein can be translated. Why? I. Presence of start codonII. Absence of start codonIII. Due to mutationb) If you want to start the translation, what change you need in the second codon (from 5’ to 3’ direction)?I. Substitution of C with GII. No changeIII. Deletion of CIV. Both I & III
- Replication involves a period of time during which DNA is particularly susceptible to the introduction of mutations. If nucleotides can be incorporated into DNA at a rate of 20 nucleotides/second and the human genome contains 3 billion nucleotides, how long will replication take? How is this time reduced so that replication can take place in a few hours?Semiconservative or Conservative DNA Replication If 15N-Iabeled E. coli DNA has a density of 1.724 g/mL, 14N-labeled DNA has a density of 1.710 g/mL, and E. coli cells grown for many generations on 14NH4+as a nitrogen source are transferred to media containing 15NH4+as the sole N-source, (a) What will be the density of the DNA after one generation, assuming replication is semiconservative? (b) Suppose replication took place by a conservative mechanism in which the parental strands remained together and the two progeny strands were paired. Design an experiment that could distinguish between semiconservative and conservative modes of replication.Transcribe and translate the following DNA sequence (nontemplate strand): 5’-ATGGCCGGTTATTAAGCA-3’